Theorem. Let $n$ be a positive integer and let $A$ be an $n\times n$ matrix. Then, $A$ is symmetric if and only if it is orthogonally diagonalizable.
-
Today I want to present a consequence of the Singular Value Decomposition of a matrix that is not presented in the textbook. But, before that I want to recall a matrix factorization presented in Section 2.5 of the textbook in Theorem 10: Column-Row Expansion of a Matrix Product.
This proof is concise. To make it easier for you to internalize it, I will elaborate on it below.
- Let us first introduce the concept of the outer product of two vectors. Let \(m, n \in \mathbb{N}\), and let \(\mathbf{x} \in \mathbb{R}^m\) and \(\mathbf{y} \in \mathbb{R}^n\). That is, \[ \mathbf{x} = \begin{bmatrix} x_1 \\ \vdots \\ x_m \end{bmatrix}, \quad \mathbf{y} = \begin{bmatrix} y_1 \\ \vdots \\ y_n \end{bmatrix}, \quad \text{where} \quad x_1, \ldots, x_m, y_1, \ldots, y_n \in \mathbb{R}. \] Recall that \(\mathbf{x}\) is an \(m \times 1\) matrix, while \(\mathbf{y}^\top\) is a \(1 \times n\) matrix, making the product \(\mathbf{x} \mathbf{y}^\top\) a well-defined \(m \times n\) matrix. The \(m \times n\) matrix \(\mathbf{x} \mathbf{y}^\top\) is called the outer product of the vectors \(\mathbf{x}\) and \(\mathbf{y}\). Specifically, \[ \mathbf{x} \mathbf{y}^\top = \mathbf{x} \bigl[ y_1 \ \cdots \ y_n \bigr] = \bigl[ y_1 \mathbf{x} \ \cdots \ y_n \mathbf{x} \bigr] = \begin{bmatrix} x_1 y_1 & \cdots & x_1 y_n \\ \vdots & \ddots & \vdots \\ x_m y_1 & \cdots & x_m y_n \end{bmatrix} = \begin{bmatrix} x_1 \mathbf{y}^\top \\ \vdots \\ x_m \mathbf{y}^\top \end{bmatrix}. \]
- Assume that the vectors \(\mathbf{x}\) and \(\mathbf{y}\) are nonzero vectors. Then we have \[ \operatorname{Col}\bigl(\mathbf{x} \mathbf{y}^\top\bigr) = \operatorname{Span}\{\mathbf{x}\} \quad \text{and} \quad \operatorname{Row}\bigl(\mathbf{x} \mathbf{y}^\top\bigr) = \operatorname{Span}\{\mathbf{y}\}. \] Consequently, the rank of \(\mathbf{x} \mathbf{y}^\top\) is \(1\). Conversely, every matrix of rank \(1\) is an outer product.
- In this item I will give a slightly different proof of the Column-Row Expansion of a Matrix Product Theorem. Let \(l,m,n \in \mathbb{N}\), let \(A\) be an \(m\times l\) matrix and let \(B\) be an \(l\times n\) matrix. Let \(i\in \{1,\ldots,m\}\), \(k\in\{1,\ldots,l\}\) and \(j\in\{1,\ldots,n\}\). Denote by \(a_{ik}\) the entry in \(i\)-th row and \(k\)-th column of \(A\) and denote by \(b_{kj}\) the entry in \(k\)-th row and \(j\)-th column of \(B\). Set \[ \operatorname{col}_k(A) = \mathbf{a}_k = \begin{bmatrix} a_{1k} \\ \vdots \\ a_{mk} \end{bmatrix}, \quad \operatorname{col}_j(B) = \begin{bmatrix} b_{1j} \\ \vdots \\ b_{lj} \end{bmatrix}, \quad \operatorname{row}_k(B) = \Bigl[ b_{k1} \ \cdots \ b_{kn} \Bigr]. \] Then \begin{align*} AB & = A \Bigl[ \operatorname{col}_1(B) \ \cdots \ \operatorname{col}_n(B) \Bigr] \\[7pt] & = \Bigl[ A \operatorname{col}_1(B) \ \cdots \ A \operatorname{col}_n(B) \Bigr] \\[7pt] & = \left[ A \begin{bmatrix} b_{11} \\ \vdots \\ b_{l1} \end{bmatrix} \ \phantom{\begin{array}{c} \vdots \\ \vdots \\ \vdots \end{array}}\mkern-10mu \cdots \ \ A \begin{bmatrix} b_{1n} \\ \vdots \\ b_{ln} \end{bmatrix} \right] \\[7pt] & = \Biggl[ \sum_{k=1}^l b_{k1} \mathbf{a}_k \ \ \cdots \ \ \sum_{k=1}^l b_{kn} \mathbf{a}_k \Biggr] \\[7pt] & =\sum_{k=1}^l \biggl[ b_{k1} \mathbf{a}_k \ \ \cdots \ \ b_{kn} \mathbf{a}_k \biggr] \\[7pt] & =\sum_{k=1}^l \mathbf{a}_k \Bigl[ b_{k1} \ \ \cdots \ \ b_{kn} \Bigr] \\[7pt] & =\sum_{k=1}^l \operatorname{col}_k(A) \operatorname{row}_k(B) \\[7pt] & = \operatorname{col}_1(A) \operatorname{row}_1(B) + \cdots + \operatorname{col}_l(A) \operatorname{row}_l(B). \end{align*}
- In the sum \[ \operatorname{col}_1(A) \operatorname{row}_1(B) + \cdots + \operatorname{col}_l(A) \operatorname{row}_l(B) \] we can drop the terms which equal to the zero matrix. In this way we have written the product \(AB\) as a sum of rank-one matrices.
- Next we implement the column-row expansion of a product of two matrices to a reduced singular value decomposition of a nonzero \(m\times n\) matrix \(A\) of rank \(r \in \bigl\{1,\ldots,\min\{m,n\}\bigr\}\) which we studied yesterday: \begin{align*} A & = U_r D_r V_r^\top \\ & = \Bigl[ \mathbf{u}_1 \ \cdots \ \mathbf{u}_r \Bigr] \begin{bmatrix} \sigma_1 & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & \sigma_r \end{bmatrix} \begin{bmatrix} \mathbf{v}_1^\top \\ \vdots \\ \mathbf{v}_r^\top \end{bmatrix} \\ & = \Bigl[ \sigma_1 \mathbf{u}_1 \ \cdots \ \sigma_1 \mathbf{u}_r \Bigr] \begin{bmatrix} \mathbf{v}_1^\top \\ \vdots \\ \mathbf{v}_r^\top \end{bmatrix} \\ & = \sigma_1 \mathbf{u}_1 \mathbf{v}_1^\top + \cdots + \sigma_r \mathbf{u}_r \mathbf{v}_r^\top. \end{align*} Here \(\{\mathbf{v}_1 \ \cdots \ \mathbf{v}_r\}\) is an orthonormal basis of \(\operatorname{Row}(A)\) and \(\{\mathbf{u}_1 \ \cdots \ \mathbf{u}_r\}\) is an orthonormal basis of \(\operatorname{Col}(A)\) and \(\sigma_1 \geq \) In this way we have expressed the matrix \(A\) as a linear combination of rank-one matrices (which are outer products of left and right singular vectors).
- Today we discussed Example 7 in Section 7.4 about reduced SVD and the Pseudoinverse (Moore-Penrose inverse).
-
Example 2. Here is a calculation of a singular value decomposition of the matrix
\[
A = \left[\!\begin{array}{rrr}
3 & -1 & 1 \\
-1 & 3 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1 \end{array}\right].
\]
- (I) To find the singular values and right singular vectors we calculate the matrix \[ A^\top \!A = \left[\!\begin{array}{rrrr} 3 & -1 & 1 & 1 \\ -1 & 3 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{array}\right] \left[\!\begin{array}{rrr} 3 & -1 & 1 \\ -1 & 3 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right] = \left[\!\begin{array}{rrr} 12 & -4 & 4 \\ -4 & 12 & 4 \\ 4 & 4 & 4 \end{array}\right] = 4 \left[\!\begin{array}{rrr} 3 & -1 & 1 \\ -1 & 3 & 1 \\ 1 & 1 & 1 \end{array}\right]. \] Observe that adding the first two columns and subtracting twice the third column gives the zero vector. Hence $\lambda_3 = 0$ is an eigenvalue of $A^\top\!A$ and a corresponding eigenvector is $\bigl[ -1 \ -1 \ \ 2 \bigr]^\top$. Since each row of $A^\top\!A$ sums to $12$, $\lambda_2 = 12$ is an eigenvalue of $A^\top\!A$ and a corresponding eigenvector is $\bigl[ 1 \ \ 1 \ \ 1 \bigr]^\top$. Since the vector $\bigl[ 1 \ -1 \ \ 0 \bigr]^\top$ is orthogonal to both earlier found eigenvectors it also must be an eigenvector of $A^\top\!A$. The corresponding eigenvalue is $\lambda_1 = 16$. Thus the singular values of $A$ are $\sigma_1 = 4$ and $\sigma_2 = 2\sqrt{3}$, and the matrices $\Sigma$ and $V$ are as follows \[ \Sigma = \left[\!\begin{array}{rrr} 4 & 0 & 0 \\ 0 & 2\sqrt{3} & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \qquad V = \left[\!\begin{array}{rrr} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{6}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{6}} \\ 0 & \frac{1}{\sqrt{3}} & \frac{2}{\sqrt{6}} \end{array}\right] = \bigl[ \mathbf{v}_1 \ \mathbf{v}_2 \ \mathbf{v}_3 \bigr]. \]
- (II) To find a $4\!\times\!4$ orthogonal matrix $U$ we first normalize vectors \[ A \left[\!\begin{array}{r} 1 \\ -1 \\ 0 \end{array}\right] = \left[\!\begin{array}{rrr} 3 & -1 & 1 \\ -1 & 3 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right] \left[\!\begin{array}{r} 1 \\ -1 \\ 0 \end{array}\right] = \left[\!\begin{array}{r} 4 \\ -4 \\ 0 \\ 0 \end{array}\right] = 4 \left[\!\begin{array}{r} 1 \\ -1 \\ 0 \\ 0 \end{array}\right], \quad \text{hence} \quad \mathbf{u}_1 = \left[\!\begin{array}{r} \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} \\ 0 \\ 0 \end{array}\right], \] and \[ A \left[\!\begin{array}{r} 1 \\ 1 \\ 1 \end{array}\right] = \left[\!\begin{array}{rrr} 3 & -1 & 1 \\ -1 & 3 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right] \left[\!\begin{array}{r} 1 \\ 1 \\ 1 \end{array}\right] = \left[\!\begin{array}{r} 3 \\ 3 \\ 3 \\ 3 \end{array}\right] = 3 \left[\!\begin{array}{r} 1 \\ 1 \\ 1 \\ 1 \end{array}\right], \quad \text{hence} \quad \mathbf{u}_2 = \left[\!\begin{array}{r} \frac{1}{2} \\ \frac{1}{2} \\ \frac{1}{2} \\ \frac{1}{2} \end{array}\right]. \] From the general considerations about the singular value decomposition we know that the singular values and left and right singular vectors must satisfy: $A\mathbf{v}_1 = \sigma_1 \mathbf{u}_1$ and $A\mathbf{v}_2 = \sigma_2 \mathbf{u}_2$. Next we verify these equalities: \[ \left[\!\begin{array}{rrr} 3 & -1 & 1 \\ -1 & 3 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right] \left[\!\begin{array}{r} \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} \\ 0 \end{array}\right] = 4 \left[\!\begin{array}{r} \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} \\ 0 \\ 0 \end{array}\right] \quad \text{and} \quad \left[\!\begin{array}{rrr} 3 & -1 & 1 \\ -1 & 3 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right] \left[\!\begin{array}{r} \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \end{array}\right] = 2\sqrt{3} \left[\!\begin{array}{r} \frac{1}{2} \\ \frac{1}{2} \\ \frac{1}{2} \\ \frac{1}{2} \end{array}\right] \] It has been established in class that $\mathbf{u}_1$ and $\mathbf{u}_2$ form an orthonormal basis for $\operatorname{Col}A$.
- (III) To complete the matrix $U$ we need an orthonormal basis for $\mathbb{R}^4$. Since the space $\operatorname{Nul}\bigl(A^\top\bigr)$ is the orthogonal complement of $\operatorname{Col}A$, we can simply find the nullspace of $A^\top$, and then find two orhonormal vectors in $\operatorname{Nul}\bigl(A^\top\bigr).$ Here we go: \[ \textstyle \left[\!\begin{array}{rrrr} 3 & -1 & 1 & 1 \\ -1 & 3 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{array}\right] \sim \left[\!\begin{array}{rrrr} 1 & 1 & 1 & 1 \\ 0 & 4 & 2 & 2 \\ 0 & -4 & -2 & -2 \end{array}\right] \sim \left[\!\begin{array}{rrrr} 1 & 1 & 1 & 1 \\ 0 & 1 & 1/2 & 1/2 \\ 0 & 0 & 0 & 0 \end{array}\right] \sim \left[\!\begin{array}{rrrr} 1 & 0 & 1/2 & 1/2 \\ 0 & 1 & 1/2 & 1/2 \\ 0 & 0 & 0 & 0 \end{array}\right] \] Thus, \[ \operatorname{Nul}\bigl(A^\top\bigr) = \left\{ s \left[\!\begin{array}{r} -1 \\ -1 \\ 0 \\ 2 \end{array}\right] + t \left[\!\begin{array}{r} -1 \\ -1 \\ 2 \\ 0 \end{array}\right] \ : \ s, t \in \mathbb{R} \right\}. \] All the vectors in $\operatorname{Nul}\bigl(A^\top\bigr)$ are orthogonal to $\mathbf{u}_1$ and $\mathbf{u}_2$ (verify this). There are many pairs of orthonormal vectors in $\operatorname{Nul}\bigl(A^\top\bigr).$ One pair that cough my attention is obtained with $s=1/2$, $t=1/2$ and $s=1/2$, $t=-1/2$ and then normalized. That is the pair \[ \mathbf{u}_3 = \left[\!\begin{array}{r} -\frac{1}{2} \\ - \frac{1}{2} \\ \frac{1}{2} \\ \frac{1}{2} \end{array}\right] \quad \text{and} \quad \mathbf{u}_4 = \left[\!\begin{array}{c} 0 \\ 0 \\ -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{array}\right] \] Finally, \[ U = \left[\!\begin{array}{rrrr} \frac{1}{\sqrt{2}} & \frac{1}{2} & -\frac{1}{2} & 0 \\ -\frac{1}{\sqrt{2}} & \frac{1}{2} & -\frac{1}{2} & 0 \\ 0 & \frac{1}{2} & \frac{1}{2} & -\frac{1}{\sqrt{2}} \\ 0 & \frac{1}{2} & \frac{1}{2} & \frac{1}{\sqrt{2}} \end{array}\right]. \]
- Remark To find vectors $\mathbf{u}_3$ and $ \mathbf{u}_4$ it might be slightly more efficient to proceed in the following way. Since we know that $\mathbf{u}_1$ and $ \mathbf{u}_2$ form a basis for $\operatorname{Col} A$ we can find a basis for $(\operatorname{Col} A)^{\perp}$ by solving the system \[ \left[\!\begin{array}{rrrr} 1 & -1 & 0 & 0 \\ 1 & 1 & 1 & 1 \end{array}\right] \left[\!\begin{array}{c} x_1 \\ x_2 \\ x_3 \\ x_4 \end{array}\right] = \left[\!\begin{array}{c} 0 \\ 0 \end{array}\right] \] The row reduction of the matrix \[ \left[\!\begin{array}{rrrr} 1 & -1 & 0 & 0 \\ 1 & 1 & 1 & 1 \end{array}\right] \sim \cdots \sim \left[\!\begin{array}{rrrr} 1 & 0 & 1/2 & 1/2 \\ 0 & 1 & 1/2 & 1/2 \end{array}\right] \] might be simpler than the row reduction that we did in (III).
- To celebrate our work we verify \[ \left[\!\begin{array}{rrr} 3 & -1 & 1 \\ -1 & 3 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right] = \left[\!\begin{array}{rrrr} \frac{1}{\sqrt{2}} & \frac{1}{2} & -\frac{1}{2} & 0 \\ -\frac{1}{\sqrt{2}} & \frac{1}{2} & -\frac{1}{2} & 0 \\ 0 & \frac{1}{2} & \frac{1}{2} & -\frac{1}{\sqrt{2}} \\ 0 & \frac{1}{2} & \frac{1}{2} & \frac{1}{\sqrt{2}} \end{array}\right] \left[\!\begin{array}{rrr} 4 & 0 & 0 \\ 0 & 2\sqrt{3} & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \left[\!\begin{array}{rrr} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\ -\frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{6}} & \frac{2}{\sqrt{6}} \end{array}\right] . \]
- The reduced Singular Value Decomposition of $A$ is (we drop the vectors that belong to the nullspaces of $A$ and $A^\top$ and the zero rows and columns of $\Sigma$) \[ \left[\!\begin{array}{rrr} 3 & -1 & 1 \\ -1 & 3 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right] = \left[\!\begin{array}{rr} \frac{1}{\sqrt{2}} & \frac{1}{2} \\ -\frac{1}{\sqrt{2}} & \frac{1}{2} \\ 0 & \frac{1}{2} \\ 0 & \frac{1}{2} \end{array}\right] \left[\!\begin{array}{rr} 4 & 0 \\ 0 & 2\sqrt{3} \end{array}\right] \left[\!\begin{array}{rrr} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \end{array}\right] . \]
- The pseudoinverse of $A$ is \begin{align*} A^+ & = \left[\!\begin{array}{rr} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} \\ 0 & \frac{1}{\sqrt{3}} \end{array}\right] \left[\!\begin{array}{cc} \frac{1}{4} & 0 \\ 0 & \frac{1}{2\sqrt{3}} \end{array}\right] \left[\!\begin{array}{rrrr} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 & 0 \\ \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \end{array}\right] \\ & = \left[\!\begin{array}{rrrr} \frac{5}{24} & -\frac{1}{24} & \frac{1}{12} & \frac{1}{12} \\ -\frac{1}{24} & \frac{5}{24} & \frac{1}{12} & \frac{1}{12} \\ \frac{1}{12} & \frac{1}{12} & \frac{1}{12} & \frac{1}{12} \end{array}\right] \\ & = \frac{1}{24} \left[\!\begin{array}{rrrr} 5 & -1 & 2 & 2 \\ -1 & 5 & 2 & 2 \\ 2 & 2 & 2 & 2 \end{array}\right] \end{align*}
-
The following four properties make the pseudoinverse unique and special:
- The matrix $A^+A$ is symmetric.
- The matrix $AA^+$ is symmetric.
- $AA^+A = A$
- $A^+AA^+ = A^+$
- Suggested problems for Section 7.4: 3, 7, 11, 13, 14, 15, 17, 18, 19 (Notice a mistake in Practice Problem 2.)
- I post my notes on the Singular Value Decomposition that I wrote in GoodNotes on my iPad, trying to replicate what I did on the blackboard.
-
I believe that colors and pictures will help you internalize the process of the construction of the Singular Value Decomposition.
-
Example 1. As the first example consider $3\!\times\!2$ matrix
\[
A = \left[\begin{array}{rr}
1 & 2 \\
2 & 1 \\
1 & -1
\end{array}\right].
\]
- To find a Singular Value Decomposition of the $3\!\times 2$ matrix $A$ given above, we are looking for a $3\!\times\!3$ orthogonal matrix $U$, the $3\!\times\!2$ matrix $\Sigma$ with the singular values of $A$ on the "diagonal", and a $2\!\times\!2$ orthogonal matrix $V$, such that $A = U\Sigma V^\top$.
- (I) To find the singular values and right singular vectors of $A$ we calculate the symmetric matrix \[ A^\top A = \left[\begin{array}{rrr} 1 & 2 & 1 \\ 2 & 1 & -1 \end{array}\right] \left[\begin{array}{rr} 1 & 2 \\ 2 & 1 \\ 1 & -1 \end{array}\right] = \left[\begin{array}{rr}6 & 3 \\ 3 & 6 \end{array}\right]. \] The eigenvalues of this matrix $A^\top A$ in nonincreasing order are $9$ and $3.$ The orthogonal diagonalization of the matrix $A^\top A$ is \[ A^\top A = \left[\begin{array}{rr}6 & 3 \\ 3 & 6 \end{array}\right] = \left[\begin{array}{rr} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} &\frac{1}{\sqrt{2}} \end{array}\right] \left[\!\begin{array}{cc} 9 & 0\\ 0 & 3 \end{array}\right] \left[\begin{array}{rr} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ - \frac{1}{\sqrt{2}} &\frac{1}{\sqrt{2}} \end{array}\right]. \] Here we found the matrix $V$ in the singular value decomposition \[ V = \left[\begin{array}{rr} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} &\frac{1}{\sqrt{2}} \end{array}\right]. \] The singular values of $A$ are $3$ and $\sqrt{3}.$ The matrix $\Sigma$ is \[ \Sigma = \left[\!\begin{array}{cc} 3 & 0\\ 0 & \sqrt{3} \\ 0 & 0 \end{array}\right] \]
- (II) To find a $3\!\times\!3$ orthogonal matrix $U$, we rewrite the equality $A = U \Sigma^\top V^\top$ as \[ A V = U \Sigma. \] It is easier to see what is going on in the above equality if we write \[ V = \bigl[ \mathbf{v}_1 \ \mathbf{v}_2 \bigr], \quad U = \bigl[ \mathbf{u}_1 \ \mathbf{u}_2 \ \mathbf{u}_3 \bigr], \quad \Sigma = \left[\!\begin{array}{cc} \sigma_1 & 0\\ 0 & \sigma_2 \\ 0 & 0 \end{array}\right] \] Thus, \[ A V = \bigl[ A \mathbf{v}_1 \ A \mathbf{v}_2 \bigr] \] and \[ U \Sigma = \bigl[ \mathbf{u}_1 \ \mathbf{u}_2 \ \mathbf{u}_3 \bigr] \left[\!\begin{array}{cc} \sigma_1 & 0\\ 0 & \sigma_2 \\ 0 & 0 \end{array}\right] = \bigl[ \sigma_1 \mathbf{u}_1 \ \sigma_2 \mathbf{u}_2 \bigr]. \] Hence \[ A \mathbf{v}_1 = \sigma_1 \mathbf{u}_1, \qquad A \mathbf{v}_2 = \sigma_2 \mathbf{u}_2. \] Therefore, \[ \mathbf{u}_1 = \frac{1}{\sigma_1} A \mathbf{v}_1 = \frac{1}{3} \left[\begin{array}{rr} 1 & 2 \\ 2 & 1 \\ 1 & -1 \end{array}\right] \left[\begin{array}{c} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{array}\right] = \left[\begin{array}{c} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \\ 0 \end{array}\right] \] and \[ \mathbf{u}_2 = \frac{1}{\sigma_2} A \mathbf{v}_2 = \frac{1}{\sqrt{3}} \left[\begin{array}{rr} 1 & 2 \\ 2 & 1 \\ 1 & -1 \end{array}\right] \left[\begin{array}{c} -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{array}\right] = \left[\begin{array}{c} \frac{1}{\sqrt{6}} \\ -\frac{1}{\sqrt{6}} \\ -\frac{2}{\sqrt{6}} \end{array}\right]. \] Thus, we have the first two columns of $U$. Those two columns form a basis for the column space of $A$.
- (III) The next step is to find the remaining column of $U.$ Since the first two columns of $U$ form an orthonormal basis for $\operatorname{Col}(A)$, the remaining column of $U$ will be a unit vector in $\operatorname{Nul}(A^\top).$ That vector is \[ \mathbf{u}_3 = \left[\!\begin{array}{r} \frac{1}{\sqrt{3}} \\ -\frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \end{array}\right]. \] Finally we have the complete $3\!\times\!3$ matrix $U$ \[ U = \left[\!\begin{array}{ccc} \frac{1}{\sqrt{2}} &\frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{2}} & - \frac{1}{\sqrt{6}} & - \frac{1}{\sqrt{3}} \\ 0 & -\frac{2}{\sqrt{6}} & \frac{1}{\sqrt{3}} \end{array}\right]. \]
- To celebrate our work we verify \[ A = \left[\begin{array}{rr} 1 & 2 \\ 2 & 1 \\ 1 & -1 \end{array}\right] = \left[\!\begin{array}{ccc} \frac{1}{\sqrt{2}} &\frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{2}} & - \frac{1}{\sqrt{6}} & - \frac{1}{\sqrt{3}} \\ 0 & -\frac{2}{\sqrt{6}} & \frac{1}{\sqrt{3}} \end{array}\right] \left[\!\begin{array}{cc} 3 & 0\\ 0 & \sqrt{3} \\ 0 & 0 \end{array}\right] \left[\begin{array}{rr} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} &\frac{1}{\sqrt{2}} \end{array}\right]. \] Or, equivalently, what is easier $AV = U\Sigma$: \[ \left[\begin{array}{rr} 1 & 2 \\ 2 & 1 \\ 1 & -1 \end{array}\right] \left[\begin{array}{rr} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} &\frac{1}{\sqrt{2}} \end{array}\right] = \left[\!\begin{array}{ccc} \frac{1}{\sqrt{2}} &\frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{2}} & - \frac{1}{\sqrt{6}} & - \frac{1}{\sqrt{3}} \\ 0 & -\frac{2}{\sqrt{6}} & \frac{1}{\sqrt{3}} \end{array}\right] \left[\!\begin{array}{cc} 3 & 0\\ 0 & \sqrt{3} \\ 0 & 0 \end{array}\right]. \]
-
In this case, since the $A$ maps vectors in $\mathbb{R}^2$ to vectors in $\mathbb{R}^3$, we can represent the geometric significance of the singular values and the singular vectors in a picture.
The narrative is missing for the above picture. Also, I want to make the picture of the action of $A^\top$. I already made the left-hand side picture:
The right-hand side picture will follow soon.
- Today I post explorations of three quadratic forms on \(\mathbb{R}^3\).
-
Example 3.
In this item we consider the quadratic form
\[
Q(x_1, x_2, x_3) = x_1^2 - 4 x_1 x_2 +4 x_2 x_3 - x_3^2 \quad \text{where} \quad x_1, x_2 \in \mathbb{R}.
\]
- T1 We have \[ Q(x_1, x_2, x_3) = \bigl[ x_1 \ \ x_2 \ \ x_3 \bigr] \left[\! \begin{array}{ccc} 1 & -2 & 0 \\ -2 & 0 & 2 \\ 0 & 2 & -1 \end{array} \!\right] \left[\! \begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array} \!\right] = x_1^2 - 4 x_1 x_2 +4 x_2 x_3 - x_3^2 \quad \text{where} \quad x_1, x_2, x_3 \in \mathbb{R}. \]
- T2
Clearly the quadratic form $Q$ is not a zero form. To classify $Q$ as positive semidefinite, negative semidefinite, indefinite we orthogonally diagonalize the matrix of this quadratic form:
\[
\left[\!
\begin{array}{ccc}
1 & -2 & 0 \\
-2 & 0 & 2 \\
0 & 2 & -1
\end{array}
\!\right] = \left[\!\begin{array}{ccc}
-\frac{1}{3} & -\frac{2}{3} & \frac{2}{3} \\
-\frac{2}{3} & \frac{2}{3} & \frac{1}{3} \\
\frac{2}{3} & \frac{1}{3} & \frac{2}{3} \\
\end{array}
\!\right] \left[\!
\begin{array}{ccc}
-3 & 0 & 0 \\
0 & 3 & 0 \\
0 & 0 & 0
\end{array}
\!\right] \left[\!\begin{array}{ccc}
-\frac{1}{3} & -\frac{2}{3} & \frac{2}{3} \\
-\frac{2}{3} & \frac{2}{3} & \frac{1}{3} \\
\frac{2}{3} & \frac{1}{3} & \frac{2}{3} \\
\end{array}
\!\right]^\top
\]
Let us introduce two bases
\[
\mathcal{B} = \left\{ \left[\!
\begin{array}{c}
-\frac{1}{3} \\ -\frac{2}{3} \\ \frac{2}{3}
\end{array}
\!\right], \left[\!
\begin{array}{c}
-\frac{2}{3} \\ \frac{2}{3} \\ \frac{1}{3}
\end{array}
\!\right], \left[\!
\begin{array}{c}
\frac{2}{3} \\ \frac{1}{3} \\ \frac{2}{3}
\end{array}
\!\right] \right\} \qquad \text{and} \qquad
\mathcal{E} = \left\{ \left[\!
\begin{array}{c} 1 \\ 0 \\ 0 \end{array}
\!\right], \left[\!
\begin{array}{c}
0 \\ 1 \\ 0
\end{array}
\!\right], \left[\!
\begin{array}{c}
0 \\ 0 \\ 1
\end{array}
\!\right] \right\}.
\]
The above orthogonal diagonalization suggests a very useful change of coordinates:
\[
\mathbf{y} = \left[\!\begin{array}{ccc}
-\frac{1}{3} & -\frac{2}{3} & \frac{2}{3} \\
-\frac{2}{3} & \frac{2}{3} & \frac{1}{3} \\
\frac{2}{3} & \frac{1}{3} & \frac{2}{3} \\
\end{array}
\!\right]^\top \mathbf{x} = \underset{\mathcal{B}\leftarrow\mathcal{E}}{P} \mathbf{x}, \qquad
\mathbf{x} = \left[\!\begin{array}{ccc}
-\frac{1}{3} & -\frac{2}{3} & \frac{2}{3} \\
-\frac{2}{3} & \frac{2}{3} & \frac{1}{3} \\
\frac{2}{3} & \frac{1}{3} & \frac{2}{3} \\
\end{array}
\!\right] \mathbf{y} = \underset{\mathcal{E}\leftarrow\mathcal{B}}{P} \mathbf{y}.
\]
The coordinates $\mathbf{y}$ are the coordinates relative to the basis $\mathcal{B}$.
With the change of coordinates \[ \mathbf{y} = \underset{\mathcal{B}\leftarrow\mathcal{E}}{P} \mathbf{x}, \qquad \mathbf{x} = \underset{\mathcal{E}\leftarrow\mathcal{B}}{P} \mathbf{y}, \] the quadratic form $Q$ simplifies as follows \[ x_1^2 - 4 x_1 x_2 +4 x_2 x_3 - x_3^2 = - 3 y_1^2 + 3 y_2^2. \] Clearly the form $- 3 y_1^2 + 3 y_2^2$ is an indefinite form taking the value $-3$ at $(y_1,y_2,y_3) = (1,0,0)$ and the value $3$ at $(y_1,y_2,y_3) = (0,1,0).$ - T4
The above introduced change of coordinates yields
\[
\bigl\{ \mathbf{x} \in \mathbb{R}^3 \, : \, Q(\mathbf{x}) = -1 \bigr\} = \bigl\{ \mathbf{y} \in \mathbb{R}^3 \, : \, - 3 y_1^2 + 3 y_2^2 = -1 \bigr\},
\]
\[
\bigl\{ \mathbf{x} \in \mathbb{R}^3 \, : \, Q(\mathbf{x}) = 0 \bigr\} = \bigl\{ \mathbf{y} \in \mathbb{R}^3 \, : \, - 3 y_1^2 + 3 y_2^2 = 0 \bigr\},
\]
and
\[
\bigl\{ \mathbf{x} \in \mathbb{R}^3 \, : \, Q(\mathbf{x}) = 1 \bigr\} = \bigl\{ \mathbf{y} \in \mathbb{R}^3 \, : \, - 3 y_1^2 + 3 y_2^2 = 1 \bigr\}.
\]
The set
\[
\bigl\{ \mathbf{x} \in \mathbb{R}^3 \, : \, Q(\mathbf{x}) = 0 \bigr\} = \bigl\{ \mathbf{y} \in \mathbb{R}^3 \, : \, - 3 y_1^2 + 3 y_2^2 = 0 \bigr\}
\]
is a union of two planes. These planes are represented by the following two spans:
\[
\operatorname{Span} \left\{
\left[\!
\begin{array}{c}
1 \\ 1 \\ 0
\end{array}
\!\right],
\left[\!
\begin{array}{c}
0 \\ 0 \\ 1
\end{array}
\!\right]
\right\} \qquad \text{and} \qquad \operatorname{Span} \left\{
\left[\!
\begin{array}{c}
1 \\ -1 \\ 0
\end{array}
\!\right],
\left[\!
\begin{array}{c}
0 \\ 0 \\ 1
\end{array}
\!\right]
\right\}
\]
in the coordinates relative to the basis $\mathcal{B}$. To get the spans of the vertices in the original coordinate system relative to the basis $\mathcal{E}$ we apply the change of coordinates matrix $\displaystyle \underset{\mathcal{E}\leftarrow\mathcal{B}}{P}$:
\[
\operatorname{Span} \left\{
\left[\!
\begin{array}{c}
-1 \\ 0 \\ 1
\end{array}
\!\right],
\left[\!
\begin{array}{c}
2 \\ 1 \\ 2
\end{array}
\!\right]
\right\} \qquad \text{and} \qquad \operatorname{Span} \left\{
\left[\!
\begin{array}{c}
1 \\ -4 \\ 1
\end{array}
\!\right],
\left[\!
\begin{array}{c}
2 \\ 1 \\ 2
\end{array}
\!\right]
\right\}
\]
The set
\[
\bigl\{ \mathbf{x} \in \mathbb{R}^3 \, : \, Q(\mathbf{x}) = 1 \bigr\} = \bigl\{ \mathbf{y} \in \mathbb{R}^3 \, : \, - 3 y_1^2 + 3 y_2^2 = 1 \bigr\}
\]
is a hyperbolic cylinder. The equation $- 3 y_1^2 + 3 y_2^2 = 1$ represents a hyperbola in the plane spanned by the vectors
\[
\left[\!
\begin{array}{c}
-\frac{1}{3} \\ -\frac{2}{3} \\ \frac{2}{3}
\end{array}
\!\right] \quad \text{and} \quad \left[\!
\begin{array}{c}
-\frac{2}{3} \\ \frac{2}{3} \\ \frac{1}{3}
\end{array}
\!\right] \quad \text{coordinates relative to} \quad \mathcal{E}.
\]
The cylinder is formed by the parallel lines that go through the points on the hyperbola and are orthogonal to the plane spanned by the above two vectors.
The set
\[
\bigl\{ \mathbf{x} \in \mathbb{R}^3 \, : \, Q(\mathbf{x}) = -1 \bigr\} = \bigl\{ \mathbf{y} \in \mathbb{R}^3 \, : \, - 3 y_1^2 + 3 y_2^2 = -1 \bigr\}
\]
is a hyperbolic cylinder. The equation $- 3 y_1^2 + 3 y_2^2 = -11$ represents a hyperbola in the plane spanned by the vectors
\[
\left[\!
\begin{array}{c}
-\frac{1}{3} \\ -\frac{2}{3} \\ \frac{2}{3}
\end{array}
\!\right] \quad \text{and} \quad \left[\!
\begin{array}{c}
-\frac{2}{3} \\ \frac{2}{3} \\ \frac{1}{3}
\end{array}
\!\right] \quad \text{coordinates relative to} \quad \mathcal{E}.
\]
The cylinder is formed by the parallel lines that go through the points on the hyperbola and are orthogonal to the plane spanned by the above two vectors.
- T3 Since the change of coordinate matrices $\displaystyle \underset{\mathcal{E}\leftarrow\mathcal{B}}{P}$ and $\displaystyle \underset{\mathcal{B}\leftarrow\mathcal{E}}{P}$ are orthogonal we have \[ \| \mathbf{y} \|^2 = \mathbf{y}^\top \mathbf{y} = \left(\underset{\mathcal{B}\leftarrow\mathcal{E}}{P} \mathbf{x}\right)^\top \left(\underset{\mathcal{B}\leftarrow\mathcal{E}}{P} \mathbf{x}\right) = \mathbf{x}^\top \left(\underset{\mathcal{B}\leftarrow\mathcal{E}}{P}\right)^\top \underset{\mathcal{B}\leftarrow\mathcal{E}}{P} \mathbf{x} = \mathbf{x}^\top \mathbf{x} = \|\mathbf{x} \|^2. \] Therefore \begin{align*} S & = \Bigl\{ Q(\mathbf{x}) \, : \, \mathbf{x} \in \mathbb{R}^3, \ \| \mathbf{x} \| = 1 \Bigr\} \\ & = \Bigl\{ -3 y_1^2 + 3 y_2^2 \, : \, y_1^2 + y_2^2 + y_3^2 = 1, \ y_1, y_2, y_3 \in\mathbb{R} \Bigr\}. \end{align*} Assume whenever $y_1^2 + y_2^2 + y_3^2 = 1$. Then \[ -3 = -3 y_1^2 - 3 y_2^2 - 3 y_3^2 \leq -3 y_1^2 + 3 y_2^2 \leq 3 y_1^2 + 3 y_2^2 + 3 y_3^2 = 3. \] Therefore, we have that $\min S = -3$ and $\max S = 3$. The form $-3 y_1^2 + 3 y_2^2$ takes the value $-3$ when $y_1 = 1, y_2 = 0, y_3 = 0$, or $y_1 = -1, y_2 =0, y_3 = 0$, and the value $3$ when $y_1 = 0, y_2 = 1, y_3 = 0$ or $y_1 = 0, y_2 =-1, y_3 = 0$. Using the change of coordinates matrix $\displaystyle \underset{\mathcal{E}\leftarrow\mathcal{B}}{P}$ we conclude that \[ \bigl\{ \mathbf{x} \in \mathbb{R}^3 \, : \, Q(\mathbf{x}) = -3 \bigr\} = \left\{ \left[\! \begin{array}{c} -\frac{1}{3} \\ -\frac{2}{3} \\ \frac{2}{3} \end{array} \!\right], -\left[\! \begin{array}{c} -\frac{1}{3} \\ -\frac{2}{3} \\ \frac{2}{3} \end{array} \!\right] \right\}. \] and \[ \bigl\{ \mathbf{x} \in \mathbb{R}^3 \, : \, Q(\mathbf{x}) = 3 \bigr\} = \left\{ \left[\! \begin{array}{c} -\frac{2}{3} \\ \frac{2}{3} \\ \frac{1}{3} \end{array} \!\right], - \left[\! \begin{array}{c} -\frac{2}{3} \\ \frac{2}{3} \\ \frac{1}{3} \end{array} \!\right] \right\}. \]
-
Example 4.
In this item we consider the quadratic form
\[
Q(x_1, x_2, x_3) = 4 x_1 x_2 +2 x_1 x_3 + 3 x_2^2+4 x_2 x_3 \quad \text{where} \quad x_1, x_2 \in \mathbb{R}.
\]
- T1 We have \begin{align*} Q(x_1, x_2, x_3) = \bigl[ x_1 \ \ x_2 \ \ x_3 \bigr] \left[\! \begin{array}{ccc} 0 & 2 & 1 \\ 2 & 3 & 2 \\ 1 & 2 & 0 \\ \end{array} \!\right] \left[\! \begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array} \!\right] & = 4 x_1 x_2 +2 x_1 x_3 + 3 x_2^2+4 x_2 x_3 \\ & \qquad \text{where} \quad x_1, x_2, x_3 \in \mathbb{R}. \end{align*} The symmetric matrix $\displaystyle A = \left[\! \begin{array}{ccc} 0 & 2 & 1 \\ 2 & 3 & 2 \\ 1 & 2 & 0 \\ \end{array}\!\right]$ is said to be associated with the quadratic form $Q$. In the next item we will orthogonally diagonalize the matrix $A$. Finding an orthonormal basis $\{\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3 \}$ of \(\mathbb{R}^3\) that consists of unit eigenvectors of \(A\) and the corresponding eigenvalues \(\lambda_1, \lambda_2, \lambda_3\) will be a significant help in understanding the quadratic form \(Q\).
- T2 Clearly the quadratic form $Q$ is not a zero form. To classify $Q$ as positive semidefinite, negative semidefinite, indefinite we orthogonally diagonalize the matrix $A$ that is associated with this quadratic form: \[ \left[\! \begin{array}{ccc} 0 & 2 & 1 \\ 2 & 3 & 2 \\ 1 & 2 & 0 \\ \end{array} \!\right] = \left[\!\begin{array}{ccc} -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} \\ 0 & -\frac{1}{\sqrt{3}} & \frac{2}{\sqrt{6}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} \\ \end{array} \!\right] \left[\! \begin{array}{ccc} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 5 \end{array} \!\right] \left[\!\begin{array}{ccc} -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} \\ 0 & -\frac{1}{\sqrt{3}} & \frac{2}{\sqrt{6}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} \\ \end{array} \!\right]^\top. \] Here \[ \mathbf{u}_1 = \left[\!\begin{array}{c} -\frac{1}{\sqrt{2}} \\ 0 \\ \frac{1}{\sqrt{2}}\end{array}\!\right], \quad \mathbf{u}_2 = \left[\!\begin{array}{c}\frac{1}{\sqrt{3}} \\ -\frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}}\end{array}\!\right], \quad \mathbf{u}_3 = \left[\! \begin{array}{c} \frac{1}{\sqrt{6}} \\ \frac{2}{\sqrt{6}} \\ \frac{1}{\sqrt{6}} \end{array}\!\right], \] are normalized eigenvectors of\(A\). The corresponding eigenvalues, respectively, are $\lambda_1 = -1$, \(\lambda_2 = -1\), and \(\lambda_3 = 5\). Let us introduce two bases of \(\mathbb{R}^3\), the basis \(\mathcal{B}= \{\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3 \}\) which consists of the normalized eigenvectors of \(A\) and the standard basis \(\mathcal{E} = \{\mathbf{i}, \mathbf{j}, \mathbf{k}\}\). (Here I use the notation \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) for the standard coordinate vectors in \(\mathbb{R}^3\). This notation is common in multivariable calculus courses.) That is, \[ \mathcal{B} = \left\{ \left[\! \begin{array}{c} -\frac{1}{\sqrt{2}} \\ 0 \\ \frac{1}{\sqrt{2}} \end{array} \!\right], \left[\! \begin{array}{c} \frac{1}{\sqrt{3}} \\ -\frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \end{array} \!\right], \left[\! \begin{array}{c} \frac{1}{\sqrt{6}} \\ \frac{2}{\sqrt{6}} \\ \frac{1}{\sqrt{6}} \end{array} \!\right] \right\} \quad \text{and} \quad \mathcal{E} = \left\{ \left[\! \begin{array}{c} 1 \\ 0 \\ 0 \end{array} \!\right], \left[\! \begin{array}{c} 0 \\ 1 \\ 0 \end{array} \!\right], \left[\! \begin{array}{c} 0 \\ 0 \\ 1 \end{array} \!\right] \right\}. \] The above orthogonal diagonalization suggests the following very useful change of coordinates: \begin{align*} \mathbf{y} & = \underset{\mathcal{B}\leftarrow\mathcal{E}}{P} \mathbf{x} = \left[\!\begin{array}{ccc} -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} \\ 0 & -\frac{1}{\sqrt{3}} & \frac{2}{\sqrt{6}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} \end{array} \!\right]^\top \mathbf{x} = \left[\!\begin{array}{ccc} -\frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{6}} & \frac{2}{\sqrt{6}} & \frac{1}{\sqrt{6}} \\ \end{array} \!\right] \mathbf{x}, \\ \mathbf{x} & = \underset{\mathcal{E}\leftarrow\mathcal{B}}{P} \mathbf{y} = \left[\!\begin{array}{ccc} -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} \\ 0 & -\frac{1}{\sqrt{3}} & \frac{2}{\sqrt{6}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} \\ \end{array} \!\right] \mathbf{y} . \end{align*} The coordinates $\mathbf{y}$ are the coordinates of $\mathbf{x}$ relative to the basis $\mathcal{B}$, that is $\mathbf{y} = \bigl[\mathbf{x}\bigr]_{\mathcal{B}}.$ With the change of coordinates \[ \mathbf{y} = \underset{\mathcal{B}\leftarrow\mathcal{E}}{P} \mathbf{x}, \qquad \mathbf{x} = \underset{\mathcal{E}\leftarrow\mathcal{B}}{P} \mathbf{y}, \] the quadratic form $Q$ simplifies as follows \begin{align*} 4 x_1 x_2 +2 x_1 x_3 + 3 x_2^2+4 x_2 x_3 & = \bigl[ x_1 \ \ x_2 \ \ x_3 \bigr] \left[\! \begin{array}{ccc} 0 & 2 & 1 \\ 2 & 3 & 2 \\ 1 & 2 & 0 \\ \end{array} \!\right] \left[\! \begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array} \!\right]\\[5pt] \boxed{\text{orthogonal diagonalization}} & = \mathbf{x}^\top \left[\!\begin{array}{ccc} -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} \\ 0 & -\frac{1}{\sqrt{3}} & \frac{2}{\sqrt{6}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} \\ \end{array} \!\right] \left[\! \begin{array}{ccc} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 5 \end{array} \!\right] \left[\!\begin{array}{ccc} -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} \\ 0 & -\frac{1}{\sqrt{3}} & \frac{2}{\sqrt{6}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} \\ \end{array} \!\right]^\top \mathbf{x} \\[5pt] \boxed{\text{definitions of $\displaystyle \underset{\mathcal{B}\leftarrow\mathcal{E}}{P}$ and $\displaystyle \underset{\mathcal{E}\leftarrow\mathcal{B}}{P}$}} & = \mathbf{x}^\top \underset{\mathcal{E}\leftarrow\mathcal{B}}{P} \left[\! \begin{array}{ccc} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 5 \end{array} \!\right] \underset{\mathcal{B}\leftarrow\mathcal{E}}{P} \mathbf{x} \\[5pt] \boxed{\text{use the fact $\displaystyle \Bigl( \displaystyle \underset{\mathcal{B}\leftarrow\mathcal{E}}{P}\Bigr)^\top = \underset{\mathcal{E}\leftarrow\mathcal{B}}{P}$}} & = \mathbf{x}^\top \Bigl( \underset{\mathcal{B}\leftarrow\mathcal{E}}{P}\Bigr)^\top \left[\! \begin{array}{ccc} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 5 \end{array} \!\right] \underset{\mathcal{B}\leftarrow\mathcal{E}}{P} \mathbf{x} \\[5pt] \boxed{\text{properties of transpose}} & = \Bigl( \underset{\mathcal{B}\leftarrow\mathcal{E}}{P} \mathbf{x}\Bigr)^\top \left[\! \begin{array}{ccc} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 5 \end{array} \!\right] \underset{\mathcal{B}\leftarrow\mathcal{E}}{P} \mathbf{x} \\[5pt] \boxed{\text{change of coordinates}} & = \mathbf{y}^\top \left[\! \begin{array}{ccc} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 5 \end{array} \!\right] \mathbf{y} \\[5pt] & = \bigl[ y_1 \ \ y_2 \ \ y_3 \bigr] \left[\! \begin{array}{ccc} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 5 \end{array} \!\right] \left[\! \begin{array}{c} y_1 \\ y_2 \\ y_3 \end{array} \!\right]\\[5pt] & = - y_1^2 - y_2^2 + 5 y_3^2. \end{align*} Clearly the quadratic form $- y_1^2 - y_2^2 + 5 y_3^2$ is an indefinite form taking the value $-1$ at $(y_1,y_2,y_3) = (1,0,0)$ and the value $5$ at $(y_1,y_2,y_3) = (0,0,1).$ Or, using the change of coordinates matrices, the quadratic form given in the Example takes the negative value $-2$ at $(x_1,x_2,x_3) = (-1,0,1)$ and the positive value $30$ at $(x_1,x_2,x_3) = (1,2,1).$
- T4(I)
The above introduced change of coordinates yields the following set equality
\[
\bigl\{ \mathbf{x} \in \mathbb{R}^3 \, : \, Q(\mathbf{x}) = c \bigr\} = \left\{ \mathbf{x} \in \mathbb{R}^3 \, : \bigl[ \mathbf{x}\bigr]_{\mathcal{B}} = \left[\!
\begin{array}{c} y_1 \\ y_2 \\ y_3\end{array}
\!\right] \quad \text{and} \quad \, - y_1^2 - y_2^2 + 5 y_3^2 = c \right\}
\]
which holds for each $c \in \mathbb{R}.$
Since the set \[ \bigl\{ \mathbf{y} \in \mathbb{R}^3 \, : \, - y_1^2 - y_2^2 + 5 y_3^2 = 0 \bigr\} \] is a rotated cone, the set equality stated at the beginning of this item with $c=0$ implies that the set \[ \bigl\{ \mathbf{x} \in \mathbb{R}^3 \, : \, Q(\mathbf{x}) = 0 \bigr\} \] is also a rotated cone. The intersection of this cone and the plane spanned by the vectors $\mathbf{u}_1$ and $\mathbf{u}_3$ is a pair of lines given by the equations \[ y_1 = \sqrt{5} y_3 \quad \text{and} \quad y_1 = -\sqrt{5} y_3. \] These lines are spanned, respectively, by the following vectors \[ \sqrt{5} \mathbf{u}_1 + \mathbf{u}_3 = \sqrt{5} \left[\! \begin{array}{c} -\frac{1}{\sqrt{2}} \\ 0 \\ \frac{1}{\sqrt{2}} \end{array} \!\right] + \left[\! \begin{array}{c} \frac{1}{\sqrt{6}} \\ \frac{2}{\sqrt{6}} \\ \frac{1}{\sqrt{6}} \end{array} \!\right] \] and \[ -\sqrt{5} \mathbf{u}_1 + \mathbf{u}_3 = -\sqrt{5} \left[\! \begin{array}{c} -\frac{1}{\sqrt{2}} \\ 0 \\ \frac{1}{\sqrt{2}} \end{array} \!\right] + \left[\! \begin{array}{c} \frac{1}{\sqrt{6}} \\ \frac{2}{\sqrt{6}} \\ \frac{1}{\sqrt{6}} \end{array} \!\right] \] Therefore, the cone is obtained by the rotation of the line spanned by the vector \[ \sqrt{5} \mathbf{u}_1 + \mathbf{u}_3 = \sqrt{5} \left[\! \begin{array}{c} -\frac{1}{\sqrt{2}} \\ 0 \\ \frac{1}{\sqrt{2}} \end{array} \!\right] + \left[\! \begin{array}{c} \frac{1}{\sqrt{6}} \\ \frac{2}{\sqrt{6}} \\ \frac{1}{\sqrt{6}} \end{array} \!\right] \] about the line spanned by the vector \[ \mathbf{u}_3 = \left[\! \begin{array}{c} \frac{1}{\sqrt{6}} \\ \frac{2}{\sqrt{6}} \\ \frac{1}{\sqrt{6}} \end{array} \!\right]. \]The reasoning here is the same as in a multivariable calculus course when we figure out the shape of the surface given by the equation \(-x^2-y^2+5z^2 = 0\), or \[ |z| = \frac{1}{\sqrt{5}} \sqrt{x^2+y^2}. \] This cone is obtained by rotating the line \(z = \frac{1}{\sqrt{5}}x, y = 0\) about the \(z\)-axes. Or, in the language of vectors, by rotating the line spanned by the vector \(\sqrt{5} \mathbf{i}+\mathbf{k}\) about the vector \(\mathbf{k}\).
Below is a graph of the cone \[ \bigl\{ \mathbf{x} \in \mathbb{R}^3 \, : \, Q(\mathbf{x}) = 0 \bigr\} \] with the unit vectors \(\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3\) in blue and the line spanned by the vector \[ \sqrt{5} \mathbf{u}_1 + \mathbf{u}_3 = \sqrt{5} \left[\! \begin{array}{c} -\frac{1}{\sqrt{2}} \\ 0 \\ \frac{1}{\sqrt{2}} \end{array} \!\right] + \left[\! \begin{array}{c} \frac{1}{\sqrt{6}} \\ \frac{2}{\sqrt{6}} \\ \frac{1}{\sqrt{6}} \end{array} \!\right] \] in red. It is interesting to identify the blue vector in the picture below which corresponds to \(\mathbf{u}_3\), since the rotation of the red line about this vector creates the cone. Notice that \(\mathbf{u}_3\) is the only blue vector with all three coordinates positive. Also notice that the positive directions of \(x_1\)-axis and \(x_2\)-axis point away from the observer.
- T4(II)
The set
\[
\bigl\{ \mathbf{y} \in \mathbb{R}^3 \, : \, - y_1^2 - y_2^2 + 5 y_3^2 = 1 \bigr\}
\]
is a rotated two sheet hyperboloid. This two sheet hyperboloid is obtained as the hyperbola $-y_1^2 + 5 y_3^2 = 1, y_2 = 0$ rotates about the $y_3$-axis. Consequently, the set equality stated at the beginning of this item with $c=1$ implies that the set
\[
\bigl\{ \mathbf{x} \in \mathbb{R}^3 \, : \, Q(\mathbf{x}) = 1 \bigr\}
\]
is also a rotated two sheet hyperboloid. This hyperboloid is obtained as a hyperbola in the plane spanned by the vectors
\[
\mathbf{u}_1 = \left[\!
\begin{array}{c}
-\frac{1}{\sqrt{2}} \\ 0 \\ \frac{1}{\sqrt{2}}
\end{array}
\!\right], \quad \text{and} \quad
\mathbf{u}_3 = \left[\!
\begin{array}{c}
\frac{1}{\sqrt{6}} \\ \frac{2}{\sqrt{6}} \\ \frac{1}{\sqrt{6}}
\end{array}
\!\right]
\]
rotates about the vector
\[
\mathbf{u}_3 = \left[\!
\begin{array}{c}
\frac{1}{\sqrt{6}} \\ \frac{2}{\sqrt{6}} \\ \frac{1}{\sqrt{6}}
\end{array}
\!\right].
\]
In the picture below this hyperbola is colored red.
- T4(III)
The set
\[
\bigl\{ \mathbf{y} \in \mathbb{R}^3 \, : \, - y_1^2 - y_2^2 + 5 y_3^2 = -1 \bigr\}
\]
is a rotated one sheet hyperboloid. This hyperboloid is obtained as the hyperbola $-y_2^2 + 5 y_3^2 = -1, y_1 = 0$ rotates about $y_3$-axis. The set equality stated at the beginning of this item with $c=-1$ implies that the set
\[
\bigl\{ \mathbf{x} \in \mathbb{R}^3 \, : \, Q(\mathbf{x}) = -1 \bigr\}
\]
is also a rotated one sheet hyperboloid. This hyperboloid is obtained as a hyperbola in the plane spanned by the vectors
\[
\mathbf{u}_2 =
\left[\!
\begin{array}{c}
\frac{1}{\sqrt{3}} \\ -\frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}}
\end{array}
\!\right], \quad \text{and} \quad
\mathbf{u}_3 = \left[\!
\begin{array}{c}
\frac{1}{\sqrt{6}} \\ \frac{2}{\sqrt{6}} \\ \frac{1}{\sqrt{6}}
\end{array}
\!\right]
\]
rotates about the vector
\[
\mathbf{u}_3 =
\left[\!
\begin{array}{c}
\frac{1}{\sqrt{6}} \\ \frac{2}{\sqrt{6}} \\ \frac{1}{\sqrt{6}}
\end{array}
\!\right].
\]
- T3 Since the change of coordinate matrices $\displaystyle \underset{\mathcal{E}\leftarrow\mathcal{B}}{P}$ and $\displaystyle \underset{\mathcal{B}\leftarrow\mathcal{E}}{P}$ are orthogonal we have \[ \| \mathbf{y} \|^2 = \mathbf{y}^\top \mathbf{y} = \left(\underset{\mathcal{B}\leftarrow\mathcal{E}}{P} \mathbf{x}\right)^\top \left(\underset{\mathcal{B}\leftarrow\mathcal{E}}{P} \mathbf{x}\right) = \mathbf{x}^\top \left(\underset{\mathcal{B}\leftarrow\mathcal{E}}{P}\right)^\top \underset{\mathcal{B}\leftarrow\mathcal{E}}{P} \mathbf{x} = \mathbf{x}^\top \mathbf{x} = \|\mathbf{x} \|^2. \] Therefore \[ S = \bigl\{ Q(\mathbf{x}) \, : \, \mathbf{x} \in \mathbb{R}^3, \ \| \mathbf{x} \| = 1 \bigr\} = \bigl\{ - y_1^2 - y_2^2 + 5 y_3^2 \, : \, \mathbf{y} = 1, \ \mathbf{y} \in\mathbb{R}^3 \bigr\}. \] Since whenever $y_1^2 + y_2^2 + y_3^2 = 1$ we have \[ -1 = -1 y_1^2 - 1 y_2^2 - 1 y_3^2 \leq - y_1^2 - y_2^2 + 5 y_3^2 \leq 5 y_1^2 + 5 y_2^2 + 5 y_3^2 = 5 \] wededuce that $\min S = -1$ and $\max S = 5$. The form $- y_1^2 - y_2^2 + 5 y_3^2$ takes the value $-1$ when $(y_1,y_2,y_3) = (\cos \theta, \sin \theta, 0)$ and the value $5$ when $(y_1,y_2,y_3) = (0,0,1)$ or $(y_1,y_2,y_3) = (0,0,-1)$. Using the change of coordinates matrix $\displaystyle \underset{\mathcal{E}\leftarrow\mathcal{B}}{P}$ we conclude that: The minimum value $-1$ is taken at the circle on the unit sphere in $\mathbb{R}^3$. That is \begin{align*} \bigl\{ \mathbf{x} \in \mathbb{R}^3 \, : \, Q(\mathbf{x}) = -1 \bigr\} & = \bigl\{(\cos \theta) \mathbf{u}_1 + (\sin \theta) \mathbf{u}_2 \ : \ \theta \in [0, 2\pi) \bigr\} \\ & = \left\{ (\cos \theta) \left[\! \begin{array}{c} -\frac{1}{\sqrt{2}} \\ 0 \\ \frac{1}{\sqrt{2}} \end{array} \!\right] + (\sin \theta) \left[\! \begin{array}{c} \frac{1}{\sqrt{3}} \\ -\frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \end{array} \!\right] \ : \ \theta \in [0, 2\pi) \right\}. \end{align*} The situation with the maximum value $5$ is simpler, this value is taken at two vectors: \[ \bigl\{ \mathbf{x} \in \mathbb{R}^3 \, : \, Q(\mathbf{x}) = 5 \bigr\} = \{ \mathbf{u}_3, - \mathbf{u}_3 \} = \left\{\left[\! \begin{array}{c} \frac{1}{\sqrt{6}} \\ \frac{2}{\sqrt{6}} \\ \frac{1}{\sqrt{6}} \end{array} \!\right], - \left[\! \begin{array}{c} \frac{1}{\sqrt{6}} \\ \frac{2}{\sqrt{6}} \\ \frac{1}{\sqrt{6}} \end{array} \!\right] \right\}. \]
-
Example 5.
In this item we consider the quadratic form
\begin{align*}
Q(x_1, x_2, x_3) & = 2 x_1^2+2 x_1 x_2 +2 x_1 x_3 +2 x_2^2 + 2 x_2 x_3 + 2 x_3^2 \\
& \qquad \qquad \text{where} \quad x_1, x_2, x_3 \in \mathbb{R}.
\end{align*}
- T1 We have \begin{align*} Q(x_1, x_2, x_3) & = \bigl[ x_1 \ \ x_2 \ \ x_3 \bigr] \left[\! \begin{array}{ccc} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{array} \!\right] \left[\! \begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array} \!\right] \\ & = 2 x_1^2+2 x_1 x_2 +2 x_1 x_3 +2 x_2^2 + 2 x_2 x_3 + 2 x_3^2 \\ & \qquad \text{where} \quad x_1, x_2, x_3 \in \mathbb{R}. \end{align*}
- T2 Clearly the quadratic form $Q$ is not a zero form. To classify $Q$ as positive semidefinite, negative semidefinite, indefinite we orthogonally diagonalize the matrix of this quadratic form: \[ \left[\! \begin{array}{ccc} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{array} \!\right] = \left[\!\begin{array}{ccc} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\ 0 & -\frac{2}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\ \end{array} \!\right] \left[\! \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 4 \end{array} \!\right] \left[\!\begin{array}{ccc} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\ 0 & -\frac{2}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\ \end{array} \!\right]^\top \] Let us introduce two bases \[ \mathcal{B} = \left\{ \left[\! \begin{array}{c} \frac{1}{\sqrt{2}} \\ 0 \\ -\frac{1}{\sqrt{2}} \end{array} \!\right], \left[\! \begin{array}{c} \frac{1}{\sqrt{6}} \\ -\frac{2}{\sqrt{6}} \\ \frac{1}{\sqrt{6}} \end{array} \!\right], \left[\! \begin{array}{c} \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \end{array} \!\right] \right\} \qquad \text{and} \qquad \mathcal{E} = \left\{ \left[\! \begin{array}{c} 1 \\ 0 \\ 0 \end{array} \!\right], \left[\! \begin{array}{c} 0 \\ 1 \\ 0 \end{array} \!\right], \left[\! \begin{array}{c} 0 \\ 0 \\ 1 \end{array} \!\right] \right\}. \] The above orthogonal diagonalization suggests a very useful change of coordinates: \[ \mathbf{y} = \left[\!\begin{array}{ccc} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\ 0 & -\frac{2}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\ \end{array} \!\right]^\top \mathbf{x} = \underset{\mathcal{B}\leftarrow\mathcal{E}}{P} \mathbf{x}, \qquad \mathbf{x} = \left[\!\begin{array}{ccc} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\ 0 & -\frac{2}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\ \end{array} \!\right] \mathbf{y} = \underset{\mathcal{E}\leftarrow\mathcal{B}}{P} \mathbf{y}. \] The vector $\mathbf{y}$ is the coordinate vector of $\mathbf{x}$ relative to the basis $\mathcal{B}$, that is $\mathbf{y} = \bigl[\mathbf{x}\bigr]_{\mathcal{B}}.$ With the change of coordinates \[ \mathbf{y} = \underset{\mathcal{B}\leftarrow\mathcal{E}}{P} \mathbf{x}, \qquad \mathbf{x} = \underset{\mathcal{E}\leftarrow\mathcal{B}}{P} \mathbf{y}, \] the quadratic form $Q$ simplifies as follows \[ 2 x_1^2+2 x_1 x_2 +2 x_1 x_3 +2 x_2^2 + 2 x_2 x_3 + 2 x_3^2 = y_1^2 + y_2^2 + 4 y_3^2. \] The quadratic form $y_1^2 + y_2^2 + 4 y_3^2$ is positive definite since $y_1^2 + y_2^2 + 4 y_3^2 \geq 0$ for all $y_1, y_2, y_3 \in \mathbb{R}$ and $y_1^2 + y_2^2 + 4 y_3^2 = 0$ implies $(y_1,y_2,y_3) = (0,0,0).$ Therefore the given quadratic form $Q(\mathbf{x})$ is also positive definite.
- T4(I) The above introduced change of coordinates yields the following set equality \[ \bigl\{ \mathbf{x} \in \mathbb{R}^3 \, : \, Q(\mathbf{x}) = c \bigr\} = \left\{ \mathbf{x} \in \mathbb{R}^3 \, : \bigl[ \mathbf{x}\bigr]_{\mathcal{B}} = \left[\! \begin{array}{c} y_1 \\ y_2 \\ y_3\end{array} \!\right] \quad \text{and} \quad \, y_1^2 + y_2^2 + 4 y_3^2 = c \right\} \] which holds for each $c \in \mathbb{R}.$ Since \[ \bigl\{ \mathbf{y} \in \mathbb{R}^3 \, : \, y_1^2 + y_2^2 + 4 y_3^2 = -1 \bigr\} \] is the empty set, the stated set equality with $c = -1$ yields that \[ \bigl\{ \mathbf{x} \in \mathbb{R}^3 \, : \, Q(\mathbf{x}) = -1 \bigr\} \] is the empty set.
- T4(II) Since the set \[ \bigl\{ \mathbf{y} \in \mathbb{R}^3 \, : \, y_1^2 + y_2^2 + 4 y_3^2 = 0 \bigr\} \] is a singleton set consisting of the zero vector, the stated set equality with $c = 0$ yields that \[ \bigl\{ \mathbf{x} \in \mathbb{R}^3 \, : \, Q(\mathbf{x}) = 0 \bigr\} \] is a singleton set consisting of the zero vector.
- T4(III)
The set
\[
\bigl\{ \mathbf{y} \in \mathbb{R}^3 \, : \, y_1^2 + y_2^2 + 4 y_3^2 = 1 \bigr\}
\]
is a rotated ellipsoid. This ellipsoid is obtained as the ellipse
\[
\bigl\{ \mathbf{y} \in \mathbb{R}^3 \, : \, y_1^2 + 4 y_3^2 = 1, \ y_2 = 0 \bigr\}
= \left\{ \left[\!
\begin{array}{c} \cos\theta \\ 0 \\ \frac{1}{2} \sin\theta \end{array}
\!\right] \ : \ \theta \in [0, 2 \pi) \right\},
\]
which is in the $y_1y_3$-plane, rotates about the $y_3$-axis. The set equality stated at the beginning of this item with $c = 1$ yields that the set
\[
\bigl\{ \mathbf{x} \in \mathbb{R}^3 \ : \ Q(\mathbf{x}) = 1 \bigr\}
\]
is also a rotated ellipsoid obtained as the ellipse
\[
\left\{
(\cos\theta)\left[\!
\begin{array}{c}
\frac{1}{\sqrt{2}} \\ 0 \\ -\frac{1}{\sqrt{2}}
\end{array}
\!\right] + \frac{1}{2} (\sin\theta )
\left[\!
\begin{array}{c}
\frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}}
\end{array}
\!\right]
\ : \ \theta \in [0, 2 \pi) \right\},
\]
which is in the plane spanned by the vectors
\[
\left[\!
\begin{array}{c}
\frac{1}{\sqrt{2}} \\ 0 \\ -\frac{1}{\sqrt{2}}
\end{array}
\!\right], \quad \left[\!
\begin{array}{c}
\frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}}
\end{array}
\!\right],
\]
rotates about the line determined by the vector $\displaystyle \left[\!
\begin{array}{c}
\frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}}
\end{array}
\!\right].$ Notice that the intersection of this ellipsoid and the plane spanned by the vectors
\[
\left[\!
\begin{array}{c}
\frac{1}{\sqrt{2}} \\ 0 \\ -\frac{1}{\sqrt{2}}
\end{array}
\!\right] \quad \text{and} \quad \left[\!
\begin{array}{c}
\frac{1}{\sqrt{6}} \\ -\frac{2}{\sqrt{6}} \\ \frac{1}{\sqrt{6}}
\end{array}
\!\right]
\]
is the unit circle
\[
\left\{ (\cos\theta) \left[\!
\begin{array}{c}
\frac{1}{\sqrt{2}} \\ 0 \\ -\frac{1}{\sqrt{2}}
\end{array}
\!\right] + (\sin\theta) \left[\!
\begin{array}{c}
\frac{1}{\sqrt{6}} \\ -\frac{2}{\sqrt{6}} \\ \frac{1}{\sqrt{6}}
\end{array}
\!\right] \ : \ \theta \in [0, 2 \pi) \right\}.
\]
- T3 Since the change of coordinate matrices $\displaystyle \underset{\mathcal{E}\leftarrow\mathcal{B}}{P}$ and $\displaystyle \underset{\mathcal{B}\leftarrow\mathcal{E}}{P}$ are orthogonal matrices, we have \[ \| \mathbf{y} \|^2 = \mathbf{y}^\top \mathbf{y} = \left(\underset{\mathcal{B}\leftarrow\mathcal{E}}{P} \mathbf{x}\right)^\top \left(\underset{\mathcal{B}\leftarrow\mathcal{E}}{P} \mathbf{x}\right) = \mathbf{x}^\top \left(\underset{\mathcal{B}\leftarrow\mathcal{E}}{P}\right)^\top \underset{\mathcal{B}\leftarrow\mathcal{E}}{P} \mathbf{x} = \mathbf{x}^\top \mathbf{x} = \|\mathbf{x} \|^2. \] Therefore \[ S = \bigl\{ Q(\mathbf{x}) \, : \, \mathbf{x} \in \mathbb{R}^3, \ \| \mathbf{x} \| = 1 \bigr\} = \bigl\{y_1^2 + y_2^2 + 4 y_3^2 \, : \, \| \mathbf{y} \| = 1, \ \mathbf{y} \in\mathbb{R}^3 \bigr\}. \] Since whenever $y_1^2 + y_2^2 + y_3^2 = 1$ we have \[ 1 = y_1^2 + y_2^2 + y_3^2 \leq y_1^2 + y_2^2 + 4 y_3^2 \leq 4 y_1^2 + 4 y_2^2 + 4 y_3^2 = 4, \] we deduce that $\min S = 1$ and $\max S = 4$. The form $y_1^2 + y_2^2 + 4 y_3^2$ takes the value $1$ when $(y_1,y_2,y_3) = (\cos \theta, \sin \theta, 0)$ and the value $4$ when $(y_1,y_2,y_3) = (0,0,1)$ or $(y_1,y_2,y_3) = (0,0,-1)$. Using the change of coordinates matrix $\displaystyle \underset{\mathcal{E}\leftarrow\mathcal{B}}{P}$ we conclude that the minimum value, $1$, of $S$ is taken at the circle on the unit sphere in $\mathbb{R}^3$ given by \[ \left\{ (\cos\theta) \left[\! \begin{array}{c} \frac{1}{\sqrt{2}} \\ 0 \\ -\frac{1}{\sqrt{2}} \end{array} \!\right] + (\sin\theta) \left[\! \begin{array}{c} \frac{1}{\sqrt{6}} \\ -\frac{2}{\sqrt{6}} \\ \frac{1}{\sqrt{6}} \end{array} \!\right] \ : \ \theta \in [0, 2 \pi) \right\}. \] The situation with the maximum value $4$ of $S$ is simpler; this value is taken at two vectors: \[ \bigl\{ \mathbf{x} \in \mathbb{R}^3 \, : \, Q(\mathbf{x}) = 5 \bigr\} = \left\{\left[\! \begin{array}{c} \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \end{array} \!\right], - \left[\! \begin{array}{c} \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \end{array} \!\right] \right\}. \]
- Suggested problems for Section 7.3: 1, 3, 5, 9, 11, 12
-
When studying a specific quadratic form $Q$ we need to perform the following four Tasks:
- T1 Express the quadratic form $Q$ using a symmetric matrix $A$ as $Q(\mathbf{x}) = \mathbf{x}^\top\!A \mathbf{x}$ where $\mathbf{x}\in \mathbb{R}^n.$
- T2 Classify $Q$ based on the quadruplicity introduced in the post on Thursday, November 21, 2024: positive semidefinite, negative semidefinite, indefinite. Specify whether the form is positive definite or negative definite if applicable.
- T3 Solve the following constrained optimization problem: Consider the set of real numbers \[ S = \bigl\{ Q(\mathbf{x}) \, : \ \mathbf{x}\in \mathbb{R}^n, \ \ \| \mathbf{x} \| = 1 \bigr\}. \] Determine $\min S$ and $\max S$, and describe the sets: \[ \bigl\{ \mathbf{x} \in \mathbb{R}^n \, : \ \| \mathbf{x} \| = 1, \ \ Q(\mathbf{x}) = \min S \bigr\}, \qquad \bigl\{ \mathbf{x} \in \mathbb{R}^n \, : \ \| \mathbf{x} \| = 1, \ \ Q(\mathbf{x}) = \max S \bigr\}. \]
- T4 For \(n \in \{2,3\}\), provide a detailed description of the sets: \[ \bigl\{ \mathbf{x} \in \mathbb{R}^n \, : \, Q(\mathbf{x}) = -1 \bigr\}, \quad \bigl\{ \mathbf{x} \in \mathbb{R}^n \, : \, Q(\mathbf{x}) = 0 \bigr\}, \quad \bigl\{ \mathbf{x} \in \mathbb{R}^n \, : \, Q(\mathbf{x}) = 1 \bigr\}. \]
-
To perform Task T4 for \(n = 2\) it is essential to understand the curves: \[ x_1^2 + x_2^2 = 1, \quad x_1^2 + x_2^2 = 0, \quad x_1^2 + x_2^2 = -1, \] and \[ x_1^2 - x_2^2 = 1, \quad x_1^2 - x_2^2 = 0, \quad x_1^2 - x_2^2 = -1, \] and how these curves change when there are coefficients in front of the squares \(x_1^2\) and \(x_2^2\).
To perform Task T4 for \(n = 3\) it is essential to understand the surfaces: \[ x_1^2 + x_2^2 + x_3^2 = 1, \quad x_1^2 + x_2^2 + x_3^2 = 0, \quad x_1^2 + x_2^2 + x_3^2 = -1, \] and \[ x_1^2 + x_2^2 - x_3^2 = 1, \quad x_1^2 + x_2^2 - x_3^2 = 0, \quad x_1^2 + x_2^2 - x_3^2 = -1, \] and how these curves change when there are coefficients in front of the squares \(x_1^2\), \(x_2^2\), and \(x_3^2\); in particular when one of the coefficients is \(0\).
-
In Section 7.2 in the book the author does not discus quadratic forms with three variables. Here are some animations that might help you understand the shape of the surface which is a graph of the level surface of the quadratic form $x_1^2 + x_2^2 - x_3^2$. Here I show the surfaces in ${\mathbb R}^3$ with equations $x_1^2 + x_2^2 - x_3^2 = c$ for different values of $c$. These surfaces are called hyperboloids. You can read more at the Wikipedia
Hyperboloid page. One sheet hyperboloids are often encountered in art, see these Wikipedia pages Hyperboloid structure and list of hyperboloid structures, do not miss the Gallery at the bottom of the last Wikipedia page.
Place the cursor over the image to start the animation.
Five of the above level surfaces at different level of opacity.
- In the post on Friday, November 22, I performed Tasks T1, T2, and T3 for one of the most popular quadratic forms in \(\mathbb{R}^4\): the determinant of a \(2\times 2\) matrix \[ \det \begin{bmatrix} a & b \\ c & d \end{bmatrix} = ad - bc, \quad a,b,c,d \in \mathbb{R}. \]
- Below are two examples where I perform Tasks T1, T2, T3, and T4 for quadratic forms on \(\mathbb{R}^2\).
-
Example 1.
In this item we consider the quadratic form
\[
Q(x_1,x_2) = 6 x_1^2 - 4 x_2 x_1 + 3 x_2^2 \quad \text{where} \quad x_1, x_2 \in \mathbb{R}.
\]
- T1 We have \[ Q(x_1,x_2) = \bigl[ x_1 \ \ x_2 \bigr] \left[\! \begin{array}{cc} 6 & -2 \\ -2 & 3 \end{array} \!\right] \left[\! \begin{array}{c} x_1 \\ x_2 \end{array} \!\right] = 6 x_1^2 - 4 x_2 x_1 + 3 x_2^2 \quad \text{where} \quad x_1, x_2 \in \mathbb{R}. \]
- T2
Clearly the quadratic form $Q$ is not a zero form. To classify $Q$ as positive semidefinite, negative semidefinite, indefinite we orthogonally diagonalize the matrix of this quadratic form:
\[
\left[\!
\begin{array}{cc}
6 & -2 \\
-2 & 3
\end{array}
\!\right] = \left[\!
\begin{array}{cc}
\frac{1}{\sqrt{5}} & -\frac{2}{\sqrt{5}} \\
\frac{2}{\sqrt{5}} & \frac{1}{\sqrt{5}}
\end{array}
\!\right] \left[\!
\begin{array}{cc}
2 & 0 \\
0 & 7
\end{array}
\!\right] \left[\!
\begin{array}{cc}
\frac{1}{\sqrt{5}} & -\frac{2}{\sqrt{5}} \\
\frac{2}{\sqrt{5}} & \frac{1}{\sqrt{5}}
\end{array}
\!\right]^\top
\]
Let us introduce two bases
\[
\mathcal{B} = \left\{ \left[\!
\begin{array}{c}
\frac{1}{\sqrt{5}} \\
\frac{2}{\sqrt{5}}
\end{array}
\!\right], \left[\!
\begin{array}{c}
-\frac{2}{\sqrt{5}} \\
\frac{1}{\sqrt{5}}
\end{array}
\!\right] \right\} \qquad \text{and} \qquad
\mathcal{E} = \left\{ \left[\!
\begin{array}{c} 1 \\ 0 \end{array}
\!\right], \left[\!
\begin{array}{c}
0 \\ 1
\end{array}
\!\right] \right\}.
\]
The above orthogonal diagonalization suggests a very useful change of coordinates:
\[
\mathbf{y} = \left[\!
\begin{array}{cc}
\frac{1}{\sqrt{5}} & -\frac{2}{\sqrt{5}} \\
\frac{2}{\sqrt{5}} & \frac{1}{\sqrt{5}}
\end{array}
\!\right]^\top \mathbf{x} = \underset{\mathcal{B}\leftarrow\mathcal{E}}{P} \mathbf{x}, \qquad
\mathbf{x} = \left[\!
\begin{array}{cc}
\frac{1}{\sqrt{5}} & -\frac{2}{\sqrt{5}} \\
\frac{2}{\sqrt{5}} & \frac{1}{\sqrt{5}}
\end{array}
\!\right] \mathbf{y} = \underset{\mathcal{E}\leftarrow\mathcal{B}}{P} \mathbf{y}.
\]
The coordinates $\mathbf{y}$ are the coordinates relative to the basis $\mathcal{B}$ which consists of the blue vectors in the next image
With the change of coordinates
\[
\mathbf{y} = \underset{\mathcal{B}\leftarrow\mathcal{E}}{P} \mathbf{x}, \qquad
\mathbf{x} = \underset{\mathcal{E}\leftarrow\mathcal{B}}{P} \mathbf{y},
\]
the quadratic form $Q$ simplifies as follows
\[
6 x_1^2 - 4 x_2 x_1 + 3 x_2^2 = 2 y_1^2 + 7 y_2^2.
\]
Clearly $2 y_1^2 + 7 y_2^2 \geq 0$ for all $y_1, y_2 \in \mathbb{R}$ and $2 y_1^2 + 7 y_2^2 = 0$ if and only if $(y_1,y_2) =(0,0).$ Therefore, the given quadratic form is positive definite.
- T4
The above introduced change of coordinates yields
\[
\bigl\{ \mathbf{x} \in \mathbb{R}^2 \, : \, Q(\mathbf{x}) = -1 \bigr\} = \bigl\{ \mathbf{y} \in \mathbb{R}^2 \, : \, 2 y_1^2 + 7 y_2^2 = -1 \bigr\},
\]
(this set is clearly an empty set)
\[
\bigl\{ \mathbf{x} \in \mathbb{R}^2 \, : \, Q(\mathbf{x}) = 0 \bigr\} = \bigl\{ \mathbf{y} \in \mathbb{R}^2 \, : \, 2 y_1^2 + 7 y_2^2 = 0 \bigr\}
\]
(this set clearly consists of the zero vector only)
and
\[
\bigl\{ \mathbf{x} \in \mathbb{R}^2 \, : \, Q(\mathbf{x}) = 1 \bigr\} = \bigl\{ \mathbf{y} \in \mathbb{R}^2 \, : \, 2 y_1^2 + 7 y_2^2 = 1 \bigr\}.
\]
The set
\[
\bigl\{ (y_1,y_2) \in \mathbb{R}^2 \, : \, 2 y_1^2 + 7 y_2^2 = 1 \bigr\}
\]
is an ellipse. The vertices of this ellipse in the coordinate system relative to the basis $\mathcal{B}$ are
\[
\text{vertices}: \left(\frac{\sqrt{2}}{2}, 0 \right) , \ \left(-\frac{\sqrt{2}}{2}, 0 \right), \qquad
\text{co-vertices}: \left(0, \frac{\sqrt{7}}{7} \right) , \ \left(0, -\frac{\sqrt{7}}{7}\right).
\]
To get the coordinates of these points in the original coordinate system relative to the basis $\mathcal{E}$ we apply the change of coordinates matrix $\displaystyle \underset{\mathcal{E}\leftarrow\mathcal{B}}{P}$:
\[
\text{vertices}: \left(\frac{\sqrt{10}}{10},\frac{\sqrt{10}}{5} \right) , \ \left(-\frac{\sqrt{10}}{10}, - \frac{\sqrt{10}}{5} \right),
\]
\[
\text{co-vertices}: \left(-\frac{2 \sqrt{35}}{35}, \frac{\sqrt{35}}{35} \right) , \ \left(\frac{2 \sqrt{35}}{35}, -\frac{\sqrt{35}}{35} \right).
\]
- T3 Since the change of coordinate matrices $\displaystyle \underset{\mathcal{E}\leftarrow\mathcal{B}}{P}$ and $\displaystyle \underset{\mathcal{B}\leftarrow\mathcal{E}}{P}$ are orthogonal we have \begin{align*} \| \mathbf{y} \|^2 & = \mathbf{y}^\top \mathbf{y} \\ & = \left(\underset{\mathcal{B}\leftarrow\mathcal{E}}{P} \mathbf{x}\right)^\top \left(\underset{\mathcal{B}\leftarrow\mathcal{E}}{P} \mathbf{x}\right) \\ & = \mathbf{x}^\top \left(\underset{\mathcal{B}\leftarrow\mathcal{E}}{P}\right)^\top \underset{\mathcal{B}\leftarrow\mathcal{E}}{P} \mathbf{x} \\ & = \mathbf{x}^\top \mathbf{x} = \|\mathbf{x} \|^2. \end{align*} Therefore \begin{align*} S & = \Bigl\{ Q(\mathbf{x}) \, : \, \mathbf{x} \in \mathbb{R}^2, \ \| \mathbf{x} \| = 1 \Bigr\} \\ & = \Bigl\{ 2 y_1^2 + 7 y_2^2 \, : \, y_1^2 + y_2^2 = 1, \ y_1, y_2 \in\mathbb{R} \Bigr\}. \end{align*} Since \[ 2 = 2 y_1^2 + 2 y_2^2 \leq 2 y_1^2 + 7 y_2^2 \leq 7 y_1^2 + 7 y_2^2 = 7 \] whenever $y_1^2 + y_2^2 = 1$, we have that $\min S = 2$ and $\max S = 7$. The form $2 y_1^2 + 7 y_2^2$ takes the value $2$ when $y_1 = 1, y_2 =0$ and $y_1 = -1, y_2 =0$. Using the change of coordinates matrix $\displaystyle \underset{\mathcal{E}\leftarrow\mathcal{B}}{P}$ we conclude that \[ \bigl\{ \mathbf{x} \in \mathbb{R}^2 \, : \, Q(\mathbf{x}) = 2 \bigr\} = \left\{ \left[\! \begin{array}{c} \frac{1}{\sqrt{5}} \\ \frac{2}{\sqrt{5}} \end{array} \!\right], - \left[\! \begin{array}{c} \frac{1}{\sqrt{5}} \\ \frac{2}{\sqrt{5}} \end{array} \!\right] \right\}. \] and \[ \bigl\{ \mathbf{x} \in \mathbb{R}^2 \, : \, Q(\mathbf{x}) = 7 \bigr\} = \left\{ \left[\! \begin{array}{c} \frac{-2}{\sqrt{5}} \\ \frac{1}{\sqrt{5}} \end{array} \!\right], \left[\! \begin{array}{c} \frac{2}{\sqrt{5}} \\ \frac{-1}{\sqrt{5}} \end{array} \!\right] \right\}. \]
-
Example 2.
In this item we consider the quadratic form
\[
Q(x_1,x_2) = x_1^2 + 6 x_2 x_1 + x_2^2 \quad \text{where} \quad x_1, x_2 \in \mathbb{R}.
\]
- T1 We have \[ Q(x_1,x_2) = \bigl[ x_1 \ \ x_2 \bigr] \left[\! \begin{array}{cc} 1 & 3 \\ 3 & 1 \end{array} \!\right] \left[\! \begin{array}{c} x_1 \\ x_2 \end{array} \!\right] = x_1^2 + 6 x_2 x_1 + x_2^2 \quad \text{where} \quad x_1, x_2 \in \mathbb{R}. \]
- T2
Clearly the quadratic form $Q$ is not a zero form. To classify $Q$ as positive semidefinite, negative semidefinite, indefinite we orthogonally diagonalize the matrix of this quadratic form:
\[
\left[\!
\begin{array}{cc}
1 & 3 \\
3 & 1
\end{array}
\!\right] = \left[\!
\begin{array}{cc}
\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}
\end{array}
\!\right] \left[\!
\begin{array}{cc}
4 & 0 \\
0 & -2
\end{array}
\!\right] \left[\!
\begin{array}{cc}
\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}
\end{array}
\!\right]^\top
\]
Let us introduce two bases
\[
\mathcal{B} = \left\{ \left[\!
\begin{array}{c}
\frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}}
\end{array}
\!\right], \left[\!
\begin{array}{c}
-\frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}}
\end{array}
\!\right] \right\} \qquad \text{and} \qquad
\mathcal{E} = \left\{ \left[\!
\begin{array}{c} 1 \\ 0 \end{array}
\!\right], \left[\!
\begin{array}{c}
0 \\ 1
\end{array}
\!\right] \right\}.
\]
The above orthogonal diagonalization suggests a very useful change of coordinates:
\[
\mathbf{y} = \left[\!
\begin{array}{cc}
\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}
\end{array}
\!\right]^\top \mathbf{x} = \underset{\mathcal{B}\leftarrow\mathcal{E}}{P} \mathbf{x}, \qquad
\mathbf{x} = \left[\!
\begin{array}{cc}
\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}
\end{array}
\!\right] \mathbf{y} = \underset{\mathcal{E}\leftarrow\mathcal{B}}{P} \mathbf{y}.
\]
The coordinates $\mathbf{y}$ are the coordinates relative to the basis $\mathcal{B}$ which consists of the blue vectors in the next image
With the change of coordinates
\[
\mathbf{y} = \underset{\mathcal{B}\leftarrow\mathcal{E}}{P} \mathbf{x}, \qquad
\mathbf{x} = \underset{\mathcal{E}\leftarrow\mathcal{B}}{P} \mathbf{y},
\]
the quadratic form $Q$ simplifies as follows
\[
x_1^2 + 6 x_2 x_1 + x_2^2 = 4 y_1^2 - 2 y_2^2.
\]
Clearly $4 y_1^2 - 2 y_2^2$ is an indefinite form taking the value $4$ at $(y_1,y_2) = (1,0)$ and the value $-2$ at $(y_1,y_2) = (0,1).$
- T4
The above introduced change of coordinates yields
\[
\bigl\{ \mathbf{x} \in \mathbb{R}^2 \, : \, Q(\mathbf{x}) = -1 \bigr\} = \bigl\{ \mathbf{y} \in \mathbb{R}^2 \, : \, 4 y_1^2 - 2 y_2^2 = -1 \bigr\},
\]
\[
\bigl\{ \mathbf{x} \in \mathbb{R}^2 \, : \, Q(\mathbf{x}) = 0 \bigr\} = \bigl\{ \mathbf{y} \in \mathbb{R}^2 \, : \, 4 y_1^2 - 2 y_2^2 = 0 \bigr\}
\]
and
\[
\bigl\{ \mathbf{x} \in \mathbb{R}^2 \, : \, Q(\mathbf{x}) = 1 \bigr\} = \bigl\{ \mathbf{y} \in \mathbb{R}^2 \, : \, 4 y_1^2 - 2 y_2^2 = 1 \bigr\}.
\]
The set
\[
\bigl\{ (y_1,y_2) \in \mathbb{R}^2 \, : \, 4 y_1^2 - 2 y_2^2 = 1 \bigr\}
\]
is a hyperbola. The vertices of this hyperbola in the coordinate system relative to the basis $\mathcal{B}$ are
\[
\text{vertices}: \left(\frac{1}{2}, 0 \right) , \ \left(-\frac{1}{2}, 0 \right)
\]
and the asymptotes of this hyperbola are two lines which are determined by the vectors
\[
\left[\!
\begin{array}{c}
\frac{1}{2} \\ \frac{\sqrt{2}}{2}
\end{array}
\!\right] \qquad \text{and} \qquad \left[\!
\begin{array}{c}
\frac{1}{2} \\ -\frac{\sqrt{2}}{2}
\end{array}
\!\right].
\]
To get the coordinates of the vertices in the original coordinate system relative to the basis
$\mathcal{E}$ we apply the change of coordinates matrix $\displaystyle \underset{\mathcal{E}\leftarrow\mathcal{B}}{P}$:
\[
\text{vertices}: \left(\frac{\sqrt{2}}{4},\frac{\sqrt{2}}{4} \right) , \ \left(-\frac{\sqrt{2}}{4},-\frac{\sqrt{2}}{4} \right),
\]
and the asymptotes are determined by the vectors
\[
\left[\!
\begin{array}{c}
\frac{-2+\sqrt{2}}{4} \\ \frac{2+\sqrt{2}}{4}
\end{array}
\!\right] \qquad \text{and} \qquad \left[\!
\begin{array}{c}
\frac{2+\sqrt{2}}{4} \\ \frac{-2+\sqrt{2}}{4}
\end{array}
\!\right].
\]
The set
\[
\bigl\{ (y_1,y_2) \in \mathbb{R}^2 \, : \, 4 y_1^2 - 2 y_2^2 = 0 \bigr\}
\]
is a union of two lines that go through the origin. These lines are the asymptotes of the preceding hyperbola and are determined by the vectors
\[
\left[\!
\begin{array}{c}
\frac{1}{2} \\ \frac{\sqrt{2}}{2}
\end{array}
\!\right] \qquad \text{and} \qquad \left[\!
\begin{array}{c}
\frac{1}{2} \\ -\frac{\sqrt{2}}{2}
\end{array}
\!\right],
\]
in the coordinates relative to the basis $\mathcal{B}$. To get the coordinates in the original coordinate system relative to the basis $\mathcal{E}$ we apply the change of coordinates matrix $\displaystyle \underset{\mathcal{E}\leftarrow\mathcal{B}}{P}$:
\[
\left[\!
\begin{array}{c}
\frac{-2+\sqrt{2}}{4} \\ \frac{2+\sqrt{2}}{4}
\end{array}
\!\right] \qquad \text{and} \qquad \left[\!
\begin{array}{c}
\frac{2+\sqrt{2}}{4} \\ \frac{-2+\sqrt{2}}{4}
\end{array}
\!\right].
\]
The set
\[
\bigl\{ (y_1,y_2) \in \mathbb{R}^2 \, : \, 4 y_1^2 - 2 y_2^2 = -1 \bigr\}
\]
is a hyperbola. The vertices of this hyperbola in the coordinate system relative to the basis $\mathcal{B}$ are
\[
\text{vertices}: \left(0, \frac{\sqrt{2}}{2}\right) , \ \left(0, -\frac{\sqrt{2}}{2} \right)
\]
and the asymptotes of this hyperbola are two lines which are determined by the vectors
\[
\left[\!
\begin{array}{c}
\frac{1}{2} \\ \frac{\sqrt{2}}{2}
\end{array}
\!\right] \qquad \text{and} \qquad \left[\!
\begin{array}{c}
\frac{1}{2} \\ -\frac{\sqrt{2}}{2}
\end{array}
\!\right].
\]
To get the coordinates of the vertices in the original coordinate system relative to the basis $\mathcal{E}$ we apply the change of coordinates matrix $\displaystyle \underset{\mathcal{E}\leftarrow\mathcal{B}}{P}$:
\[
\text{vertices}: \left(-\frac{1}{2},\frac{1}{2} \right) , \ \left(\frac{1}{2}, -\frac{1}{2} \right),
\]
and the asymptotes are determined by the vectors
\[
\left[\!
\begin{array}{c}
\frac{-2+\sqrt{2}}{4} \\ \frac{2+\sqrt{2}}{4}
\end{array}
\!\right] \qquad \text{and} \qquad \left[\!
\begin{array}{c}
\frac{2+\sqrt{2}}{4} \\ \frac{-2+\sqrt{2}}{4}
\end{array}
\!\right].
\]
- T3 Since the change of coordinate matrices $\displaystyle \underset{\mathcal{E}\leftarrow\mathcal{B}}{P}$ and $\displaystyle \underset{\mathcal{B}\leftarrow\mathcal{E}}{P}$ are orthogonal we have \begin{align*} \| \mathbf{y} \|^2 &= \mathbf{y}^\top \mathbf{y} \\ & = \left(\underset{\mathcal{B}\leftarrow\mathcal{E}}{P} \mathbf{x}\right)^\top \left(\underset{\mathcal{B}\leftarrow\mathcal{E}}{P} \mathbf{x}\right) \\ & = \mathbf{x}^\top \left(\underset{\mathcal{B}\leftarrow\mathcal{E}}{P}\right)^\top \underset{\mathcal{B}\leftarrow\mathcal{E}}{P} \mathbf{x} \\ & = \mathbf{x}^\top \mathbf{x} = \|\mathbf{x} \|^2. \end{align*} Therefore \begin{align*} S &= \Bigl\{ Q(\mathbf{x}) \, : \, \mathbf{x} \in \mathbb{R}^2, \ \| \mathbf{x} \| = 1 \Bigr\} \\ & = \Bigl\{ 4 y_1^2 - 2 y_2^2 \, : \, y_1^2 + y_2^2 = 1, \ y_1, y_2 \in\mathbb{R} \Bigr\}. \end{align*} Since \[ -2 = -2 y_1^2 - 2 y_2^2 \leq 4 y_1^2 - 2 y_2^2 \leq 4 y_1^2 + 4 y_2^2 = 4 \] whenever $y_1^2 + y_2^2 = 1$, we have that $\min S = -2$ and $\max S = 4$. The form $4 y_1^2 - 2 y_2^2$ takes the value $-2$ when $y_1 = 0, y_2 = 1$ or $y_1 = 0, y_2 =-1$ and the value $4$ when $y_1 = 1, y_2 = 0$ or $y_1 = -1, y_2 =0$. Using the change of coordinates matrix $\displaystyle \underset{\mathcal{E}\leftarrow\mathcal{B}}{P}$ we conclude that \[ \bigl\{ \mathbf{x} \in \mathbb{R}^2 \, : \, Q(\mathbf{x}) = -2 \bigr\} = \left\{ \left[\! \begin{array}{c} -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{array} \!\right], \left[\! \begin{array}{c} \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} \end{array} \!\right] \right\}. \] and \[ \bigl\{ \mathbf{x} \in \mathbb{R}^2 \, : \, Q(\mathbf{x}) = 4 \bigr\} = \left\{ \left[\! \begin{array}{c} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{array} \!\right], \left[\! \begin{array}{c} -\frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} \end{array} \!\right] \right\}. \] This is illustrated in the picture below.
- By far the most popular quadratic form in any dimension \(n \in \mathbb{N}\) is the "dot square": \[ Q(\mathbf{x}) = x_1x_1 + \cdots + x_n x_n, \quad \mathbf{x} = \left[\!\begin{array}{c} x_1 \\ \vdots \\ x_n \end{array}\!\right] \in \mathbb{R}^n. \]
-
The second most popular quadratic form in dimension \(4\) is the determinant of \(2\times 2\) matrices. Here we identify a matrix \(\displaystyle \left[\!\begin{array}{cc} a & b \\ c & d \end{array}\!\right]\) with the vector \(\displaystyle \left[\!\begin{array}{c} a \\ b \\ c \\ d \end{array}\!\right]\) in \(\mathbb{R}^4\). The corresponding form is
\[
Q(\mathbf{x}) = x_1x_4 - x_2 x_3, \quad \mathbf{x} = \left[\!\begin{array}{c} x_1 \\ x_2 \\ x_3 \\ x_4 \end{array}\!\right] \in \mathbb{R}^n.
\]
- Let us analyze this form. First, identify the \(4\times 4\) matrix \(A\) such that \[ x_1x_4 - x_2 x_3 = \left[\!\begin{array}{c} x_1 \\ x_2 \\ x_3 \\ x_4 \end{array} \!\right]^\top \left[\!\begin{array}{cccc} \Box & \Box & \Box & \Box \\ \Box & \Box & \Box & \Box \\ \Box & \Box & \Box & \Box \\ \Box & \Box & \Box & \Box \end{array} \!\right] \left[\!\begin{array}{c} x_1 \\ x_2 \\ x_3 \\ x_4 \end{array} \!\right]. \] It is quite clear that \[ x_1x_4 - x_2 x_3 = \left[\!\begin{array}{c} x_1 \\ x_2 \\ x_3 \\ x_4 \end{array} \!\right]^\top \left[\!\begin{array}{cccc} 0 & 0 & 0 & \tfrac{1}{2} \\ 0 & 0 & -\tfrac{1}{2} & 0 \\ 0 & -\tfrac{1}{2} & 0 & 0 \\ \tfrac{1}{2} & 0 & 0 & 0 \end{array} \!\right] \left[\!\begin{array}{c} x_1 \\ x_2 \\ x_3 \\ x_4 \end{array} \!\right]. \]
- Thus \[ A = \left[\!\begin{array}{cccc} 0 & 0 & 0 & \tfrac{1}{2} \\ 0 & 0 & -\tfrac{1}{2} & 0 \\ 0 & -\tfrac{1}{2} & 0 & 0 \\ \tfrac{1}{2} & 0 & 0 & 0 \end{array} \!\right] = \frac{1}{2} \left[\!\begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \\ 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{array} \!\right]. \] The characteristic polynomial of \(A\) is \begin{align*} \left|\!\begin{array}{cccc} -\lambda & 0 & 0 & \tfrac{1}{2} \\ 0 & -\lambda & -\tfrac{1}{2} & 0 \\ 0 & -\tfrac{1}{2} & -\lambda & 0 \\ \tfrac{1}{2} & 0 & 0 & -\lambda \end{array} \!\right| & = \left|\!\begin{array}{cccc} 0 & 0 & 0 & \tfrac{1}{2} - 2 \lambda^2 \\ 0 & -\lambda & -\tfrac{1}{2} & 0 \\ 0 & -\tfrac{1}{2} & -\lambda & 0 \\ \tfrac{1}{2} & 0 & 0 & -\lambda \end{array} \!\right| \\ & = - \left( \frac{1}{2} - 2 \lambda^2\right) \left|\!\begin{array}{ccc} 0 & -\lambda & -\tfrac{1}{2} \\ 0 & -\tfrac{1}{2} & -\lambda \\ \tfrac{1}{2} & 0 & 0 \end{array} \!\right| \\ & = - \frac{1}{2} \left( \frac{1}{2} - 2 \lambda^2\right) \left|\!\begin{array}{cc} -\lambda & -\tfrac{1}{2} \\ -\tfrac{1}{2} & -\lambda \end{array} \!\right| \\ & = - \left( \frac{1}{4} -\lambda^2\right) \left( \lambda^2 - \frac{1}{4}\right) \\ & = \left( \lambda^2 - \frac{1}{4}\right)^2. \end{align*} Hence the eigenvalues of \(A\) are \[ \lambda_1 = -\frac{1}{2}, \quad \lambda_2 = \frac{1}{2}, \] both of multiplicity \(2\).
- Let us find the eigenvectors of \(A\). We use the method of guessing. Eigenvectors corresponding to \(\lambda_2 = \frac{1}{2}\): \[ \frac{1}{2} \left[\!\begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \\ 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{array} \!\right] \left[\!\begin{array}{c}1 \\ 0 \\ 0 \\ 1 \end{array} \!\right] = \frac{1}{2} \left[\!\begin{array}{c} 1 \\ 0 \\ 0 \\ 1 \end{array} \!\right], \quad \frac{1}{2} \left[\!\begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \\ 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{array} \!\right] \left[\!\begin{array}{c} 0 \\ -1 \\ 1 \\ 0 \end{array} \!\right] = \frac{1}{2} \left[\!\begin{array}{c} 0 \\ -1 \\ 1 \\ 0 \end{array} \!\right]. \] Eigenvectors corresponding to \(\lambda_2 = -\frac{1}{2}\): \[ \frac{1}{2} \left[\!\begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \\ 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{array} \!\right] \left[\!\begin{array}{c} 0 \\ 1 \\ 1 \\ 0 \end{array} \!\right] = -\frac{1}{2} \left[\!\begin{array}{c} 0 \\ 1 \\ 1 \\ 0 \end{array} \!\right], \quad \frac{1}{2} \left[\!\begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \\ 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{array} \!\right] \left[\!\begin{array}{c}1 \\ 0 \\ 0 \\ -1 \end{array} \!\right] = -\frac{1}{2} \left[\!\begin{array}{c} 1 \\ 0 \\ 0 \\ -1 \end{array} \!\right]. \]
- Thus, an orthonormal basis that consists of eigenvectors of \(A\) is \[ \mathcal{B} = \left\{ \left[\!\begin{array}{c} \frac{1}{\sqrt{2}} \\ 0 \\ 0 \\ \frac{1}{\sqrt{2}} \end{array} \!\right], \quad \left[\!\begin{array}{c} 0 \\ -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \\ 0 \end{array} \!\right], \quad \left[\!\begin{array}{c} 0 \\ \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \\ 0 \end{array} \!\right], \quad \left[\!\begin{array}{c} \frac{1}{\sqrt{2}} \\ 0 \\ 0 \\ -\frac{1}{\sqrt{2}} \end{array} \!\right]\right\}. \] The first two vectors correspond to the eigenvalue \(1/2\), the second two to \(-1/2\).
- An orthogonal diagonalization of the matrix \(A\) is \[\small \left[\!\begin{array}{cccc} 0 & 0 & 0 & \tfrac{1}{2} \\ 0 & 0 & -\tfrac{1}{2} & 0 \\ 0 & -\tfrac{1}{2} & 0 & 0 \\ \tfrac{1}{2} & 0 & 0 & 0 \end{array} \!\right] = \left[\!\begin{array}{cccc} \frac{1}{\sqrt{2}} & 0 & 0 & \frac{1}{\sqrt{2}} \\ 0 & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{2}} & 0 & 0 & -\frac{1}{\sqrt{2}} \end{array} \!\right] \left[\!\begin{array}{cccc} \tfrac{1}{2} & 0 & 0 & 0 \\ 0 & \tfrac{1}{2} & 0 & 0 \\ 0 & 0 & -\tfrac{1}{2} & 0 \\ 0 & 0 & 0 & -\tfrac{1}{2} \end{array} \!\right] \left[\!\begin{array}{cccc} \frac{1}{\sqrt{2}} & 0 & 0 & \frac{1}{\sqrt{2}} \\ 0 & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{2}} & 0 & 0 & -\frac{1}{\sqrt{2}} \end{array} \!\right]. \] We denote by \(U\) the orthogonal matrix in the above diagonalization. That is \[ U = \left[\!\begin{array}{cccc} \frac{1}{\sqrt{2}} & 0 & 0 & \frac{1}{\sqrt{2}} \\ 0 & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{2}} & 0 & 0 & -\frac{1}{\sqrt{2}} \end{array} \!\right]. \] Denote by \(\mathcal{E}\) the standard basis of \(\mathbb{R}^4\). Then, for the change of coordinates matrices we have: \[ \underset{\mathcal{E}\leftarrow\mathcal{B}}{P} = U, \quad \text{and} \quad \underset{\mathcal{B}\leftarrow\mathcal{E}}{P} = U^\top = U^{-1}. \]
- Let us calculate the determinant of a \(2\times 2\) matrix \(M\) given by its coordinates relative to the basis \(\mathcal{B}\): \[ \bigl[ M \bigr]_{\mathcal{B}} = \left[\!\begin{array}{c} y_1 \\ y_2 \\ y_3 \\ y_4 \end{array} \!\right]. \] That is, the vector representing the matrix \(M\) is \[ y_1 \left[\!\begin{array}{c} \frac{1}{\sqrt{2}} \\ 0 \\ 0 \\ \frac{1}{\sqrt{2}} \end{array} \!\right] + y_2 \left[\!\begin{array}{c} 0 \\ -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \\ 0 \end{array} \!\right] + y_3 \left[\!\begin{array}{c} 0 \\ \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \\ 0 \end{array} \!\right] + y_4 \left[\!\begin{array}{c} \frac{1}{\sqrt{2}} \\ 0 \\ 0 \\ -\frac{1}{\sqrt{2}} \end{array} \!\right] = \left[\!\begin{array}{c} \frac{y_1}{\sqrt{2}}+ \frac{y_4}{\sqrt{2}} \\ \frac{y_3}{\sqrt{2}} -\frac{y_2}{\sqrt{2}} \\ \frac{y_3}{\sqrt{2}} +\frac{y_2}{\sqrt{2}} \\ \frac{y_1}{\sqrt{2}} -\frac{y_4}{\sqrt{2}} \end{array} \!\right] \] The last vector represents the matrix \[ M = \left[\!\begin{array}{cc} \frac{y_1}{\sqrt{2}}+ \frac{y_4}{\sqrt{2}} & \frac{y_3}{\sqrt{2}} -\frac{y_2}{\sqrt{2}} \\[7pt] \frac{y_3}{\sqrt{2}} + \frac{y_2}{\sqrt{2}} & \frac{y_1}{\sqrt{2}} - \frac{y_4}{\sqrt{2}} \end{array}\!\right]. \] The determinant of the last matrix is \[ \frac{1}{2} y_1^2 - \frac{1}{2} y_4^2 -\left( \frac{1}{2} y_3^2 - \frac{1}{2} y_2^2\right) = \frac{1}{2} y_1^2 + \frac{1}{2} y_2^2 - \frac{1}{2} y_3^2 - \frac{1}{2} y_4^2. \]
-
The significance of the last formula for the determinant lies in its application to solving constrained optimization problems such as the following:
Problem. Find the maximum and minimum values of the determinant \(ad - bc\) of the matrix \(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\) subject to the constraint \(a^2 + b^2 + c^2 + d^2 = 1\). Additionally, identify all matrices at which these extrema occur.
-
The problem stated in the preceding item is much easier to solve in the coordinates relative to the basis \(\mathcal{B}\):
Problem. Find the maximum and minimum values of the determinant \[ \frac{1}{2} y_1^2 + \frac{1}{2} y_2^2 - \frac{1}{2} y_3^2 - \frac{1}{2} y_4^2 \] of the matrix \[ \left[\!\begin{array}{cc} \frac{y_1}{\sqrt{2}}+ \frac{y_4}{\sqrt{2}} & \frac{y_3}{\sqrt{2}} -\frac{y_2}{\sqrt{2}} \\[7pt] \frac{y_3}{\sqrt{2}} + \frac{y_2}{\sqrt{2}} & \frac{y_1}{\sqrt{2}} - \frac{y_4}{\sqrt{2}} \end{array}\!\right] \] subject to the constraint \begin{align*} 1 & = \left(\frac{y_1}{\sqrt{2}}+ \frac{y_4}{\sqrt{2}}\right)^2 + \left(\frac{y_3}{\sqrt{2}} -\frac{y_2}{\sqrt{2}}\right)^2 + \left(\frac{y_3}{\sqrt{2}} + \frac{y_2}{\sqrt{2}}\right)^2 +\left(\frac{y_1}{\sqrt{2}} - \frac{y_4}{\sqrt{2}}\right)^2 \\ & = y_1^2 + y_2^2 + y_3^2 + y_4^2 \end{align*} Additionally, identify all matrices at which these extrema occur.
- Thus, we need to find the maximum and the minimum of \[ \frac{1}{2} y_1^2 + \frac{1}{2} y_2^2 - \frac{1}{2} y_3^2 - \frac{1}{2} y_4^2 = \frac{1}{2} (y_1^2 + y_2^2) - \frac{1}{2} ( y_3^2 + y_4^2) \] subject to the constraint \[ y_1^2 + y_2^2 + y_3^2 + y_4^2 = 1. \] It follows from the constrain that \( 0 \leq y_1^2 + y_2^2 \leq 1 \) and \( 0 \leq y_3^2 + y_4^2 \leq 1\). Therefore \[ -\frac{1}{2} \leq - \frac{1}{2} ( y_3^2 + y_4^2) \leq \frac{1}{2} y_1^2 + \frac{1}{2} y_2^2 - \frac{1}{2} y_3^2 - \frac{1}{2} y_4^2 \leq \frac{1}{2} (y_1^2 + y_2^2) \leq \frac{1}{2}. \] The preceding inequalities solve the problem. The maximum is \(1/2\) and it occurs whenever \(y_1^2 + y_2^2=1\) and \(y_3=y_4=0\). The minimum is \(-1/2\) and it occurs whenever \(y_3^2 + y_4^2=1\) and \(y_1=y_2=0\).
-
The final answer:
The maximum value of the determinant subject to the given constrain is \(1/2\). Since \(y_1^2 + y_2^2=1\) if and only if there exists \(\theta \in [0,2\pi)\) such that \(y_1 = \cos \theta, y_2 = \sin\theta\), the set of all matrices satisfying the given constrain with determinant \(1/2\) is given by \[ \frac{1}{\sqrt{2}} \left[\!\begin{array}{cc} \cos \theta & -\sin \theta \\[4pt] \sin \theta & \cos \theta \end{array}\!\right]. \]
The minimum value of the determinant subject to the given constrain is \(-1/2\). Since \(y_3^2 + y_4^2=1\) if and only if there exists \(\theta \in [0,2\pi)\) such that \(y_3 = \sin \theta, y_4 = \cos\theta\), the set of all matrices satisfying the given constrain with determinant \(-1/2\) is given by \[ \frac{1}{\sqrt{2}} \left[\!\begin{array}{cc} \cos \theta & \sin \theta \\[4pt] \sin \theta & -\cos \theta \end{array}\!\right]. \]
Verify that the given matrices satisfy the given constrain and the determinants evaluate to \(1/2\) and \(-1/2\), respectively.
- Recall that the matrix \[ \left[\!\begin{array}{cc} \cos \theta & -\sin \theta \\[4pt] \sin \theta & \cos \theta \end{array}\!\right] \] is the standard matrix for the counterclockwise rotation about the origin by the angle \(\theta\) and that the matrix \[ \left[\!\begin{array}{cc} \cos \theta & \sin \theta \\[4pt] \sin \theta & -\cos \theta \end{array}\!\right] \] is the standard matrix for the reflection across the line through the origin spanned by the vector \(\left[\!\begin{array}{c} \cos(\theta/2) \\[3pt] \sin(\theta/2) \end{array}\!\right]\).
- I need to write an introduction to this topic. One way to start the topic that is discussed in Section 7.2: Quadratic Forms is to talk about congruence of graphs. It is almost a high school math topic. For example, consider two functions \(f:\mathbb{R}_+ \to \mathbb{R}\) where \(\mathbb{R}_+\) denotes the set of all positive integers and \(g:\mathbb{R} \to \mathbb{R}\) given by the formulas \[ f(x) = \dfrac{1}{2x} \qquad g(x) = \sqrt{1 + x^2}. \] I hope that some high school students might be interested in proving that the graphs of these two graphs are congruent; using the definition of congruent that is used in Euclidean geometry.
- Suggested problems for Section 7.2: Quadratic Forms: 1, 3, 5, 7, 9, 13, 17, 19, 20, 31, 32, 33, 34, 35. (5th ed: 1, 3, 5, 7, 9, 13, 17, 19, 20, 23, 24, 25, 26, 27).
- In Sections 7.2 and 7.3 we study quadratic forms.
-
A quadratic form in $n$ variables is a special kind of function $Q:\mathbb{R}^n \to \mathbb{R}.$ Below are few examples of quadratic forms
- Below are three specific quadratic forms in two variables: \[ Q(x_1,x_2) = 6 x_1^2 - 4 x_1 x_2 + 3 x_2^2, \qquad (x_1,x_2) \in \mathbb{R}^2 \] \[ Q(x_1,x_2) = x_1^2 + 6 x_1 x_2 + x_2^2, \qquad (x_1,x_2) \in \mathbb{R}^2 \] \[ Q(x_1,x_2) = 4 x_1^2 + 4 x_1 x_2 + x_2^2, \qquad (x_1,x_2) \in \mathbb{R}^2 \] In general, a quadratic form $Q$ in two variables $x_1,x_2$ is a function defined on $\mathbb{R}^2$ with the values in $\mathbb{R}$ which can be expressed as \[ Q(x_1,x_2) = a\, x_1x_1 + b\, x_1x_2 + c\, x_2x_2, \qquad (x_1,x_2) \in \mathbb{R}^2, \] where $a, b, c$ are real coefficients.
- Below are three specific quadratic forms in three variables: \[ Q(x_1,x_2,x_3) = x_1^2 -4x_1 x_2 +4 x_2 x_3 - x_3^2, \qquad (x_1,x_2,x_3) \in \mathbb{R}^3, \] \[ Q(x_1,x_2,x_3) = 4x_1 x_2 + 2 x_1 x_3 + 3 x_2^2 + 4 x_2 x_3, \qquad (x_1,x_2,x_3) \in \mathbb{R}^3, \] \[ Q(x_1,x_2,x_3) = 2 x_1^2 + 2 x_1 x_2 + 2 x_1 x_3 + 2 x_2^2 + 2 x_2 x_3 + 2 x_3^2, \qquad (x_1,x_2,x_3) \in \mathbb{R}^3, \] In general, a quadratic form $Q$ in three variables $x_1,x_2,x_3$ is a function defined on $\mathbb{R}^3$ with the values in $\mathbb{R}$ which can be expressed as \[ Q(x_1,x_2,x_3) = a\, x_1x_1 + b\, x_1x_2 + c\, x_1x_3 + d\, x_2 x_2 + e\, x_2 x_3 + f\, x_3 x_3, \quad (x_1,x_2,x_3) \in \mathbb{R}^3, \] where $a, b, c, d, e, f$ are real coefficients.
- A quadratic form $Q$ in four variables $x_1,x_2,x_3,x_4$ is a function defined on $\mathbb{R}^4$ with the values in $\mathbb{R}$ which is a linear combination of the following ten terms \[ x_1x_1, \quad x_1x_2, \quad x_1x_3, \quad x_1 x_4, \quad x_2 x_2, \quad x_2 x_3, \quad x_2x_4, \quad x_3 x_3, \quad x_3 x_4, \quad x_4 x_4. \] In other words, a quadratic form in four variables is a polynomial in four variables which contains only terms of degree $2.$
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In general, a quadratic form in $n$ variables is a polynomial in $n$ variables which contains only terms of degree $2.$ To be more specific, for $j, k \in \{1,\ldots,n\}$ with $j \leq k$ let us define the functions $q_{jk}:\mathbb{R}^n \to \mathbb{R}$ by \[ q_{jk}(\mathbf{x}) = x_j x_k, \qquad \mathbf{x} = (x_1,\ldots,x_n) \in \mathbb{R}^n. \] Notice that there are $\binom{n+1}{2} = \frac{n(n+1)}{2}$ such functions.
A linear combination of the functions $q_{jk}(\mathbf{x})$ with $j, k \in \{1,\ldots,n\}$ with $j \leq k$, is called a quadratic form in $n$ variables.
In other words, a quadratic form in $n$ variables is a function $Q:\mathbb{R}^n \to \mathbb{R}$ given by the formula \[ Q(\mathbf{x}) = \sum_{\substack{ j \leq k \\ j,k \in \{1,\ldots,n\} }} a_{jk} x_j x_k \quad \text{where} \quad \mathbf{x} = \left[\!\begin{array}{c} x_1 \\ \vdots \\ x_n \end{array}\!\right] \in \mathbb{R}^n, \] where \(a_{jk} \in \mathbb{R}\) for all \(j,k \in \{1,\ldots,n\}\) such that \(j \leq k\).
- For us, the most important fact about quadratic forms is that for each quadratic form $Q$ in $n$ variables there exists a unique symmetric $n\!\times\!n$ matrix $A$ such that \[ Q(\mathbf{x}) = \mathbf{x} \cdot (A\mathbf{x}) = \mathbf{x}^\top A\mathbf{x} \quad \text{for all} \quad \mathbf{x} \in \mathbb{R}^n. \] Such matrix $A$ is called the matrix of a quadratic form.
- In the above example, for all $(x_1,x_2) \in \mathbb{R}^2$ we have \[ Q(x_1,x_2) = a\, x_1x_1 + b\, x_1x_2 + c\, x_2x_2 = \left[\! \begin{array}{c} x_1 \\ x_2 \end{array} \!\right]^\top \left(\left[\! \begin{array}{cc} a & b/2 \\ b/2 & c \end{array} \!\right] \left[\! \begin{array}{c} x_1 \\ x_2 \end{array} \!\right] \right) \] And for all $(x_1,x_2,x_3) \in \mathbb{R}^3$ we have \begin{align*} Q(x_1,x_2,x_3) &= a\, x_1x_1 + b\, x_1x_2 + c\, x_1x_3 + d\, x_2 x_2 + e\, x_2 x_3 + f\, x_3 x_3 \\ & = \left[\! \begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array} \!\right]^\top \left(\left[\! \begin{array}{ccc} a & b/2 & c/2 \\ b/2 & d & e/2 \\ c/2 & e/2 & f \end{array} \!\right] \left[\! \begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array} \!\right] \right) \end{align*}
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In this item I will write about polychotomies in mathematics. A polychotomy is a partition of a given set of mathematical objects into disjoint classes which are all given distinct names.
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A dichotomy is a partition of a given set of mathematical objects into two disjoint classes each of which is given a name. The following are examples of dichotomies.
- The most important dichotomy for numbers is the partition of numbers into the singleton set $\{0\}$ consisting of only zero and the set of all nonzero numbers. Further, dichotomy for the nonzero real numbers is the partition of the nonzero real numbers into positive real numbers and negative real numbers.
- An important dichotomy for the set of real numbers is the partition into rational and irrational numbers.
- A useful dichotomy for complex numbers is the partition of the complex numbers into the real and nonreal numbers. A complex number $z$ is said to be nonreal if the imaginary part of $z$ is nonzero.
- Consider the set of all square matrices. A square matrix $M$ is said to be singular if $\det M = 0.$ A square matrix $M$ is said to be nonsingular if $\det M \neq 0.$ You also learned that a square matrix is invertible if and only if it is nonsingular. Thus, singular-invertible is a dichotomy for square matrices.
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A trichotomy is a partition of a given set of mathematical objects into three disjoint classes each of which is given a name. The following are examples of trichotomies.
- The most important trichotomy for the set of real numbers is the partition of numbers into singleton set $\{0\}$ consisting of only zero, the set of positive real numbers and the set of negative real numbers. As we mention before this trichotomy arrises as two dichotomies.
- In high school you learned about the trichotomy involving quadratic equations $a x^2 + b x + c = 0$ with $a\neq 0.$ Such equation can have: no solutions, exactly one solution, and exactly two solutions.
- A quadruplicity is a partition of a given set of mathematical objects into four disjoint classes each of which is given a name. I started writing about polychotomies because of the quadruplicity which arises with quadratic forms. I define that quadruplicity in the next item.
-
A dichotomy is a partition of a given set of mathematical objects into two disjoint classes each of which is given a name. The following are examples of dichotomies.
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Let $Q : \mathbb R^n \to \mathbb R$ be a quadratic form. We distinguish the following four types of quadratic forms:
- $Q$ is said to be a zero quadratic form if $Q(\mathbf x) = 0$ for all $\mathbf x \in \mathbb R^n.$
- $Q$ is said to be a positive semidefinite quadratic form if $Q(\mathbf x) \geq 0$ for all $\mathbf x \in \mathbb R^n$ and there exists $\mathbf v \in \mathbb R^n$ such that $Q(\mathbf v) \gt 0.$
- $Q$ is said to be a negative semidefinite quadratic form if $Q(\mathbf x) \leq 0$ for all $\mathbf x \in \mathbb R^n$ and there exists $\mathbf v \in \mathbb R^n$ such that $Q(\mathbf v) \lt 0.$
- $Q$ is said to be an indefinite quadratic form if there exists $\mathbf v \in \mathbb R^n$ such that $Q(\mathbf v) \gt 0$ and there exists $\mathbf u \in \mathbb R^n$ such that $Q(\mathbf u) \lt 0.$
- $Q$ is said to be a positive definite quadratic form if $Q(\mathbf x) \gt 0$ for all $\mathbf x \in \mathbb R^n\!\setminus\!\{\mathbf 0\}.$
- $Q$ is said to be a negative definite quadratic form if $Q(\mathbf x) \lt 0$ for all $\mathbf x \in \mathbb R^n\!\setminus\!\{\mathbf 0\}.$
In the image below I give a graphical representation of the above quadruplicity. The red dot represents the zero quadratic form, the green region represents the positive semidefinite quadratic forms, the blue region represents the negative semidefinite quadratic forms and the cyan region represents the indefinite quadratic forms.
In the image above, the dark green region represents the positive definite quadratic forms and the dark blue region represents the negative definite quadratic forms. These two regions are not parts of the above quadruplicity.
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We started Section 7.1: Diagonalization of Symmetric Matrices.
Suggested problems are 3, 4, 9, 10, 15, 19, 23, 24, 25, 33, 35, 36, 39, 41.
(5th edition 3, 4, 9, 10, 15, 19, 23, 24, 25(a), 27, 29, 30, 33, 35).
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I want to emphasize the importance of Theorem 2 in Section 7.1. The "only if" part of the theorem below is part (d) of Theorem 3. The "if" part of the theorem below is Exercise 25
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Theorem 3(d) says: Let $n$ be a positive integer and let $A$ be an $n\times n$ matrix. Then: If $A$ is symmetric, then $A$ is orthogonally diagonalizable.
Exercise 25 says: Let $n$ be a positive integer and let $A$ be an $n\times n$ matrix. Then: If $A$ is orthogonally diagonalizable, then $A$ is symmetric.
To prove Exercise 25, we proceed as follows. Assume that $A$ is orthogonally diagonalizable. Then there exist an orthogonal matrix $U$ and a diagonal matrix $D$ such that $A=UDU^\top$. To prove that $A$ is symmetric we calculate $A^\top$: \[ A^\top = \bigl( UDU^\top \bigr)^\top = \bigl(U^\top\bigr)^\top D^\top U^\top = UDU^\top. \] In the last calculation, we used the property of the transpose that $(XY)^\top = Y^\top X^\top$ and consequently \[ (XYZ)^\top = \bigl((XY)Z\bigr)^\top = Z^\top (XY)^\top = Z^\top Y^\top X^\top. \] We also used that every diagonal matrix is symmetric, $D^\top = D$ and repeated transpose does not change a matrix $\bigl(X^\top\bigr)^\top = X$.
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The theorem in the first item above is essential for solving Exercise 33:
Suppose \( A \) and \( B \) are both orthogonally diagonalizable and \( AB = BA \). Explain why \( AB \) is also orthogonally diagonalizable.
In fact, more mathematically, the exercise should have been stated:
If \( A \) and \( B \) are both orthogonally diagonalizable and \( AB = BA \), prove that \( AB \) is also orthogonally diagonalizable.
Proof goes like this. Assume that \( A \) and \( B \) are both orthogonally diagonalizable and \( AB = BA \). By the "if" part of the theorem stated in the first item (that is by Exercise 25) we conclude that \( A \) and \( B \) are both symmetric. Now we will use the fact that $A=A^\top$ and $B=B^\top$ and the fact that $A$ and $B$ commute to prove that $AB$ is symmetric: \[ (AB)^\top = (BA)^\top = A^\top B^\top = AB. \] Thus, $AB$ is symmetric. Now the "only if" part of the theorem stated in the first item (that is Theorem 3(d)) implies that $AB$ is orthogonally diagonalizable.
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Example. Let us find a spectral decomposition of the matrix
\[
A = \left[\!
\begin{array}{ccc}
3 & 4 & 2 \\
4 & 3 & 2 \\
2 & 2 & 0
\end{array}
\!\right].
\]
- To find the characteristic polynomial of this matrix we calculate \begin{align*} \left| \begin{array}{ccc} 3 - \lambda & 4 & 2 \\ 4 & 3 - \lambda & 2 \\ 2 & 2 & -\lambda \end{array} \right| & = \left| \begin{array}{ccc} 3 - \lambda & 4 & 2 \\ 0 & -1 - \lambda & 2 + 2 \lambda \\ 2 & 2 & -\lambda \end{array} \right| \\ & = (1+\lambda) \left| \begin{array}{ccc} 3 - \lambda & 4 & 2 \\ 0 & -1 & 2 \\ 2 & 2 & -\lambda \end{array} \right| \\ & = (1+\lambda) \left| \begin{array}{ccc} 3 - \lambda & 0 & 10 \\ 0 & -1 & 2 \\ 2 & 0 & 4-\lambda \end{array} \right| \\ & = (-1) (1+\lambda) \left| \begin{array}{cc} 3 - \lambda & 10 \\ 2 & 4-\lambda \end{array} \right| \\ & = (-1) (1+\lambda) \bigl(\lambda^2 - 7 \lambda - 8 \bigr)\\ & = (-1) (\lambda+1)^2 (\lambda - 8 \bigr)\\ & = -\lambda^3 + 6 \lambda^2 + 15 \lambda + 8. \end{align*} Thus the roots of the characteristic polynomial, that is the eigenvalues of $A$, are $-1$ and $8.$ The algebraic multiplicity of $-1$ is $2$ and the algebraic multiplicity of $8$ is $1.$ Therefore we expect that the eigenspace corresponding to $-1$ will be two-dimensional and the eigenspace corresponding to $8$ will be one-dimensional and the
- Next we find the corresponding eigenspaces: \begin{align*} \operatorname{Nul}(A-(-1)I_3) & = \operatorname{Span} \left\{ \left[ \begin{array}{r} -1 \\ 1 \\ 0 \end{array} \right], \left[ \begin{array}{r} -1 \\ 0 \\ 2 \end{array} \right] \right\}, \\ \operatorname{Nul}(A-8I_3) & = \operatorname{Span} \left\{ \left[ \begin{array}{r} 2 \\ 2 \\ 1 \end{array} \right] \right\} \end{align*}
- To find an orthogonal diagonalization of $A$ we need unit eigenvectors which are orthogonal to each other. The eigenvector $\bigl[ 2 \ 2 \ 1 \bigr]^\top$ is "nice" since its length is integer $3$ (that is, it does not involve square-root). However neither of the eigenvectors in the basis of the eigenspace corresponding to $-1$ has this property. But, if we add the vectors of the basis of the eigenspace corresponding to $-1$ we get $\bigl[ -2 \ 1 \ 2 \bigr]^\top$ and the length of this vector is also $3$. Now we need to find an eigenvector corresponding to $-1$ orthogonal to $\bigl[ -2 \ 1 \ 2 \bigr]^\top.$ To do that we use $\bigl[ -1 \ 1 \ 0 \bigr]^\top$ and apply the Gram-Schmidt orthogonalization to two vectors. \begin{align*} \mathbf{v}_1 & = \left[ \begin{array}{r} -2 \\ 1 \\ 2 \end{array} \right], \\ \mathbf{v}_2 & = \left[ \begin{array}{r} -1 \\ 1 \\ 0 \end{array} \right] - \frac{3}{9} \left[ \begin{array}{r} -2 \\ 1 \\ 2 \end{array} \right] = \left[ \begin{array}{r} -1/3 \\ 2/3 \\ -2/3 \end{array} \right] \quad \text{take the opposite vector, fewer - signs!} \end{align*} Thus, three orthogonal unit eigenvectors of $A$ are \[ \frac{1}{3}\left[ \begin{array}{r} -2 \\ 1 \\ 2 \end{array} \right], \quad \frac{1}{3}\left[ \begin{array}{r} 1 \\ -2 \\ 2 \end{array} \right], \quad \frac{1}{3}\left[ \begin{array}{r} 2 \\ 2 \\ 1 \end{array} \right]. \]
- Thus, the orthogonal diagonalization of $A$ is \[ \left[ \begin{array}{ccc} 3 & 4 & 2 \\ 4 & 3 & 2 \\ 2 & 2 & 0 \end{array} \right] = \left[ \begin{array}{ccc} -\frac{2}{3} & \frac{1}{3} & \frac{2}{3} \\ \frac{1}{3} & -\frac{2}{3} & \frac{2}{3} \\ \frac{2}{3} & \frac{2}{3} & \frac{1}{3} \end{array} \right] \left[ \begin{array}{rrr} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 8 \end{array} \right] \left[ \begin{array}{ccc} -\frac{2}{3} & \frac{1}{3} & \frac{2}{3} \\ \frac{1}{3} & -\frac{2}{3} & \frac{2}{3} \\ \frac{2}{3} & \frac{2}{3} & \frac{1}{3} \end{array} \right]. \]
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Next we will develop an alternative way of writing matrix $A$ as a linear combination of orthogonal projections onto the eigenspaces of $A$.
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The columns of \[ U = \left[ \begin{array}{ccc} -\frac{2}{3} & \frac{1}{3} & \frac{2}{3} \\ \frac{1}{3} & -\frac{2}{3} & \frac{2}{3} \\ \frac{2}{3} & \frac{2}{3} & \frac{1}{3} \end{array} \right] \] form an orthonormal basis for $\mathbb{R}^3$ which consists of unit eigenvectors of $A.$ To verify that calculate \[ U^\top U = \left[ \begin{array}{ccc} -\frac{2}{3} & \frac{1}{3} & \frac{2}{3} \\ \frac{1}{3} & -\frac{2}{3} & \frac{2}{3} \\ \frac{2}{3} & \frac{2}{3} & \frac{1}{3} \end{array} \right] \left[ \begin{array}{ccc} -\frac{2}{3} & \frac{1}{3} & \frac{2}{3} \\ \frac{1}{3} & -\frac{2}{3} & \frac{2}{3} \\ \frac{2}{3} & \frac{2}{3} & \frac{1}{3} \end{array} \right] = \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] = I_3 \] Interestingly, here \(U\) is also a symmetric matrix. Therefore \(U^\top = U\).
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The first two columns of \(U\). That is, the matrix \(Q\) (I call it \(Q\) since we studied matrices like this when we talked about the \(QR\)-factorization of a matrix with linearly independent columns.) \[ Q = \left[ \begin{array}{cc} -\frac{2}{3} & \frac{1}{3} \\ \frac{1}{3} & -\frac{2}{3} \\ \frac{2}{3} & \frac{2}{3} \end{array} \right] \] form an orthonormal basis for the eigenspace of $A$ corresponding to $-1.$ The last column \[ \left[ \begin{array}{c} \frac{2}{3} \\ \frac{2}{3} \\ \frac{1}{3} \end{array} \right] \] is an orhonormal basis for the eigenspace of $A$ corresponding to $8.$
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The orthogonal projection matrix (we call this matrix \(P_{-1}\)) onto the eigenspace of $A$ corresponding to $-1$ is \[ P_{-1} = Q Q^\top \left[ \begin{array}{cc} -\frac{2}{3} & \frac{1}{3} \\ \frac{1}{3} & -\frac{2}{3} \\ \frac{2}{3} & \frac{2}{3} \end{array} \right] \left[ \begin{array}{ccc} -\frac{2}{3} & \frac{1}{3} & \frac{2}{3} \\ \frac{1}{3} & -\frac{2}{3} & \frac{2}{3} \end{array} \right] = \frac{1}{9} \left[ \begin{array}{rrr} 5 & -4 & -2 \\ -4 & 5 & -2 \\ -2 & -2 & 8 \end{array} \right] \]
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The orthogonal projection matrix onto the eigenspace of $A$ corresponding to $8$ is \[ P_8 = \left[ \begin{array}{c} \frac{2}{3} \\ \frac{2}{3} \\ \frac{1}{3} \end{array} \right] \left[ \begin{array}{ccc} \frac{2}{3} & \frac{2}{3} & \frac{1}{3} \end{array} \right] = \frac{1}{9} \left[ \begin{array}{rrr} 4 & 4 & 2 \\ 4 & 4 & 2 \\ 2 & 2 & 1 \end{array} \right]. \]
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Since the eigenvectors that we used above form a basis for $\mathbb{R}^3$ we have \[ P_{-1} + P_8 = \frac{1}{9} \left[ \begin{array}{rrr} 5 & -4 & -2 \\ -4 & 5 & -2 \\ -2 & -2 & 8 \end{array} \right] + \frac{1}{9} \left[ \begin{array}{rrr} 4 & 4 & 2 \\ 4 & 4 & 2 \\ 2 & 2 & 1 \end{array} \right] = \left[ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] = I_3. \]
- The equality in the preceding item means that for every $\mathbf{v} \in \mathbb{R}^3$ we have \[ \mathbf{v} = P_{-1} \mathbf{v} + P_8 \mathbf{v}. \] Since $P_{-1}$ is the orthogonal projection onto the eigenspace of $A$ corresponding to $-1$ we have \[ A P_{-1} \mathbf{v} = (-1) P_{-1} \mathbf{v}. \] Similarly, since $P_{8}$ is the orthogonal projection onto the eigenspace of $A$ corresponding to $8$ we have \[ A P_{8} \mathbf{v} = 8 P_{8} \mathbf{v}. \] Therefore \[ A\mathbf{v} = A P_{-1} \mathbf{v} + A P_8 \mathbf{v} = (-1)P_{-1} \mathbf{v} + 8 P_{8} \mathbf{v} = \bigl((-1)P_{-1} + 8 P_{8}\bigr) \mathbf{v}. \] Since the last equality holds for all $\mathbf{v} \in \mathbb{R}^3$, we proved that \[ A = (-1)P_{-1} + 8 P_{8}, \] or, with matrices \[ \left[ \begin{array}{ccc} 3 & 4 & 2 \\ 4 & 3 & 2 \\ 2 & 2 & 0 \end{array} \right] = (-1) \left[ \begin{array}{ccc} \frac{5}{9} & - \frac{4}{9} & - \frac{2}{9} \\ - \frac{4}{9} & \frac{5}{9} & - \frac{2}{9} \\ - \frac{2}{9} & - \frac{2}{9} & \frac{8}{9} \end{array} \right] + 8 \left[ \begin{array}{ccc} \frac{4}{9} & \frac{4}{9} & \frac{2}{9} \\ \frac{4}{9} & \frac{4}{9} & \frac{2}{9} \\ \frac{2}{9} & \frac{2}{9} & \frac{1}{9} \end{array} \right] . \] The last two equalities are called spectral decomposition of $A.$ Although it sounds mathematically surprising, the diagonalization of $A$ is equivalent to the preceding equality.
- The orthogonal projection matrix $P_{-1}$ onto $\operatorname{Nul}(A-(-1)I_3)$ and the orthogonal projection matrix $P_8$ onto $\operatorname{Nul}(A-8I_3)$ have the following properties: \[ (P_{-1})^2 = P_{-1}, \quad P_{-1}^\top = P_{-1}, \quad I_3 - P_{-1} = P_8, \quad P_{-1} P_8 = 0, \] \[ (P_{8})^2 = P_{8}, \quad P_{8}^\top = P_{8}, \quad I_3 - P_{8} = P_{-1}, \quad P_{8} P_{-1} = 0. \]
- Please enjoy: \[ I_3 = \left[ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] = \left[ \begin{array}{ccc} \frac{5}{9} & - \frac{4}{9} & - \frac{2}{9} \\ - \frac{4}{9} & \frac{5}{9} & - \frac{2}{9} \\ - \frac{2}{9} & - \frac{2}{9} & \frac{8}{9} \end{array} \right] + \left[ \begin{array}{ccc} \frac{4}{9} & \frac{4}{9} & \frac{2}{9} \\ \frac{4}{9} & \frac{4}{9} & \frac{2}{9} \\ \frac{2}{9} & \frac{2}{9} & \frac{1}{9} \end{array} \right]. \] And then just by scaling the projections with the eigenvalues we get \[ A = \left[ \begin{array}{ccc} 3 & 4 & 2 \\ 4 & 3 & 2 \\ 2 & 2 & 0 \end{array} \right] = (-1) \left[ \begin{array}{ccc} \frac{5}{9} & - \frac{4}{9} & - \frac{2}{9} \\ - \frac{4}{9} & \frac{5}{9} & - \frac{2}{9} \\ - \frac{2}{9} & - \frac{2}{9} & \frac{8}{9} \end{array} \right] + 8 \left[ \begin{array}{ccc} \frac{4}{9} & \frac{4}{9} & \frac{2}{9} \\ \frac{4}{9} & \frac{4}{9} & \frac{2}{9} \\ \frac{2}{9} & \frac{2}{9} & \frac{1}{9} \end{array} \right] . \] We can modify the eigenvalues and define a new matrix: \[ B = \frac{1}{3} \left[ \begin{array}{ccc} 1 & 4 & 2 \\ 4 & 1 & 2 \\ 2 & 2 & -2 \end{array} \right] =(-1) \left[ \begin{array}{ccc} \frac{5}{9} & - \frac{4}{9} & - \frac{2}{9} \\ - \frac{4}{9} & \frac{5}{9} & - \frac{2}{9} \\ - \frac{2}{9} & - \frac{2}{9} & \frac{8}{9} \end{array} \right] + 2 \left[ \begin{array}{ccc} \frac{4}{9} & \frac{4}{9} & \frac{2}{9} \\ \frac{4}{9} & \frac{4}{9} & \frac{2}{9} \\ \frac{2}{9} & \frac{2}{9} & \frac{1}{9} \end{array} \right] . \] Pay attention to the changes that I made and you can tell (or guess) without any calculations what is the relationship between the matrices $B$ and $A.$
- We continued with Section 7.1: Diagonalization of Symmetric Matrices doing mostly proofs. Suggested problems are 3, 4, 9, 10, 15, 19, 23, 24, 25, 33, 35, 36, 39, 41. (5th edition 3, 4, 9, 10, 15, 19, 23, 24, 25(a), 27, 29, 30, 33, 35).
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In this item, we allow matrices to have complex entries. If you need a review, I have written a short summary on
Complex Numbers.
Theorem. All eigenvalues of a symmetric matrix are real.
Background Knowledge. Since we are working with complex numbers, in the background knowledge we will review basics operations with complex numbers and matrices with complex entries.
BK1. Let $n\in\mathbb{N}$. An $n\times n$ matrix $A$ is said to be real if all entries of $A$ are real numbers and $A^\top = A$.
BK2. In this item we state properties of the transpose operation for matrices with complex entries. We are familiar with these properties for matrices with real entries. The same properties hold for matrices with complex entries. Let $k,l,m,n \in\mathbb{N}$. Let $X$ be a $k\times l$ matrix, let $Y$ be an $l\times m$ matrix, and $Z$ be an $m\times n$ matrix.
(a) Then \[ (XY)^\top = Y^\top X^\top \quad \text{and} \quad (XYZ)^\top = Z^\top Y^\top X^\top. \] (b) If $\alpha$ is a complex number, then we have \[ (\alpha X)^\top = \alpha X^\top, \quad (\alpha X) Y = X (\alpha Y) = \alpha (XY). \]
BK3. As you can see from my summary on Complex Numbers, the conjugation is an important operation with complex numbers. Recall, first, the imaginary unit $i$ is the complex number such that $i^2 = -1$. Second, for a complex number \[ z = x + y\mkern 2mu i \quad \text{where} \quad x, y \in \mathbb{R}, \] the conjugate of $z$, denoted by $\overline{z}$, is the complex number \[ \overline{z} = x - y\mkern 2mu i. \] Notice that the complex numbers $z$ and its conjugate $\overline{z}$ are reflections of each other across the real axis, see the first picture in Complex Numbers.
Now, three important features of the conjugate.
(a) product of a complex number and its conjugate \[ z \kern 2mu \overline{z} = (x + y\mkern 2mu i) (x - y\mkern 2mu i ) = x^2 + y^2 = |z|^2 \geq 0 \] is nonnegative. In the last equality, $|z|$ is the modulus of the complex number $z$. By definition \[ |z|=\sqrt{x^2 + y^2}. \] (b) We have $z \kern 1mu \overline{z} = |z|^2 = 0$ if and only if $z = 0$.
(c) We have $z = \overline{z}$ if and only if $z \in\mathbb{R}$, that is if $z$ is a real number.
BK4. The operation of complex conjugation extends to matrices. For a matrix $X$, by $\overline{X}$ we denote the matrix in which all the entries of $X$ are replaced by their conjugates. Let $k,l,m \in\mathbb{N}$, let $X$ be a $k\times l$ matrix, let $Y$ be an $l\times m$ matrix, and let $\alpha$ be a complex number.
(a) The following algebraic rules hold for the operation of conjugation \[ \overline{\alpha \mkern 1.5mu X} = \overline{\alpha} \mkern 2mu \overline{X} \quad \text{and} \quad \overline{X Y} = \overline{X} \mkern 2mu \overline{Y}. \] (b) All the entries of a matrix $X$ are real if and only if $X = \overline{X}$.
(c) Let $n\in\mathbb{N}$ and let $\mathbf{v} \in \mathbb{C}$. Then for $\mathbf{v}$ and its conjugate $\overline{\mathbf{v}}$ we have \[ \mathbf{v} = \left[\!\begin{array}{c} v_1 \\ v_2 \\ \vdots \\ v_n \end{array}\!\right], \qquad \overline{\mathbf{v}} = \left[\!\begin{array}{c} \overline{v_1} \\ \overline{v_2} \\ \vdots \\ \overline{v_n} \end{array}\!\right], \] and for the product of the $1\times n$ matrix $\mathbf{v}^\top$ and the $n\times 1$ matrix $\overline{\mathbf{v}}$ we have \begin{align*} \mathbf{v}^\top \overline{\mathbf{v}} & = \bigl[\!\begin{array}{cccc} v_1 & v_2 & \cdots & v_n \end{array}\!\bigr] \left[\!\begin{array}{c} \overline{v_1} \\ \overline{v_2} \\ \vdots \\ \overline{v_n} \end{array}\!\right] \\ & = v_1 \overline{v_1} + v_2 \overline{v_2} + \cdots + v_n \overline{v_n} \\ & = |v_1 |^2 + |v_2 |^2 + \cdots + | v_n |^2 \geq 0. \end{align*} If $\mathbf{v}$ is a nonzero vector, then at least one of the nonnegative numbers \[ |v_1 |, \ |v_2 |, \ \ldots , | v_n |^2 \] is positive. Therefore, if $\mathbf{v}$ is a nonzero vector, then $\mathbf{v}^\top \overline{\mathbf{v}} \gt 0$.
Proof. Let $n\in\mathbb{N}$ and let $A$ be an $n\times n$ symmetric matrix. Then $A = A^\top$ and all entries of $A$ are real. Let $\lambda$ be an eigenvalue of $A$ and let $\mathbf{v}$ be a corresponding eigenvector. Then, $\mathbf{v}$ is a nonzero vector and \[ A \mathbf{v} = \lambda \mathbf{v}. \] By BK4(a) we have \[ \overline{A} \overline{\mathbf{v}} = \overline{\lambda} \overline{\mathbf{v}}. \] Since $A$ is a real matrix, by BK4(b), we have $A = \overline{A}$. Therefore, the preceding displayed equality becomes \[ A \overline{\mathbf{v}} = \overline{\lambda} \overline{\mathbf{v}}. \] Let us now calculate, using BK2(a), associativity of matrix multiplication, and the fact that $A = A^\top$, \begin{equation*} (A \mathbf{v})^\top \overline{\mathbf{v}} = (\mathbf{v}^\top A^\top) \overline{\mathbf{v}} = \mathbf{v}^\top ( A^\top \overline{\mathbf{v}}) = \mathbf{v}^\top ( A \overline{\mathbf{v}}). \end{equation*} Thus $(A \mathbf{v})^\top \overline{\mathbf{v}} = \mathbf{v}^\top ( A \overline{\mathbf{v}})$. Substituting here the equalities $A \mathbf{v} = \lambda \mathbf{v}$ and $A \overline{\mathbf{v}} = \overline{\lambda} \overline{\mathbf{v}}$, we obtain \[ (\lambda \mathbf{v})^\top \overline{\mathbf{v}} = \mathbf{v}^\top \bigl( \overline{\lambda} \overline{\mathbf{v}}\bigr). \] Applying BK2(b) to the last equality we get \[ \lambda (\mathbf{v}^\top \overline{\mathbf{v}}) = \overline{\lambda} (\mathbf{v}^\top \overline{\mathbf{v}}). \] Since $\mathbf{v}$ is a nonzero vector, by BK4(c) we have that $\mathbf{v}^\top \overline{\mathbf{v}} \gt 0$. Therefore, the last equality yields \[ \lambda = \overline{\lambda} . \] By BK3(c), we deduce that $\lambda$ is real. QED.
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The proof in the preceding item involves a lot of algebra with complex numbers. Here I present a proof that the eigenvalues of a $2\!\times\!2$ symmetric matrix are real. This proof is based on the quadratic formula.
Let $A = \begin{bmatrix} a & b \\ b & d \end{bmatrix}$ be an arbitrary $2\!\times\!2$ symmetric matrix. To calculate the eigenvalues of $A$ we solve the equation $\det(A-\lambda I) =0$ for $\lambda$. That is, we solve: \[ 0 = \left| \begin{matrix} a - \lambda & b \\ b & d -\lambda \end{matrix} \right| = (a-\lambda)(d-\lambda) - b^2 = \lambda^2 -(a+d)\lambda + ad -b^2. \] The discriminant of the preceding quadratic equation is: \begin{align*} (a+d)^2 -4(ad-b^2) &= a^2 + 2ad + d^2 - 4ad +4 b^2 \\ & = a^2 - 2ad + d^2 +4 b^2 \\ & = (a-d)^2+4b^2 \geq 0. \end{align*} Solving for $\lambda$ we get two solutions: \begin{align*} \lambda_{1} & = \frac{1}{2} \Bigl( a+d - \sqrt{(a+d)^2 - 4 b^2} \Bigr) = \frac{1}{2} \Bigl( a+d - \sqrt{(a-d)^2 + 4 b^2} \Bigr) \\ \lambda_{2} & = \frac{1}{2} \Bigl( a+d + \sqrt{(a+d)^2 - 4 b^2} \Bigr) = \frac{1}{2} \Bigl( a+d + \sqrt{(a-d)^2 + 4 b^2} \Bigr) \end{align*} Since the discriminant is nonnegative, that is $(a-d)^2 + 4 b^2 \geq 0$, both eigenvalues are real. In fact, if $(a-d)^2 + 4 b^2 = 0$, then $b = 0$ and $a=d$, so our matrix is a multiple of an identity matrix. Otherwise, that is if $(a-d)^2 + 4 b^2 \gt 0$, the symmetric matrix has two distinct eigenvalues \[ \lambda_1 = \frac{1}{2} \Bigl( a+d - \sqrt{(a-d)^2 + 4 b^2} \Bigr) \lt \lambda_2 = \frac{1}{2} \Bigl( a+d + \sqrt{(a-d)^2 + 4 b^2} \Bigr). \]
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Since we proved that eigenvalues of a symmetric matrix are real, we do not need to deal with complex numbers when studying symmetric matrices.
Theorem. Eigenspaces of a symmetric matrix which correspond to distinct eigenvalues are orthogonal.
Proof. Let $A$ be a symmetric $n\!\times\!n$ matrix. Let $\lambda$ and $\mu$ be an eigenvalues of $A$ and let $\mathbf{u}$ and $\mathbf{v}$ be a corresponding eigenvector. Then $\mathbf{u} \neq \mathbf{0},$ $\mathbf{v} \neq \mathbf{0}$ and \[ A \mathbf{u} = \lambda \mathbf{u} \quad \text{and} \quad A \mathbf{v} = \mu \mathbf{v}. \] Assume that \[ \lambda \neq \mu. \] Next we calculate the same dot product in two different ways; here we use the fact that $A^\top = A$ and algebra of the dot product. The first calculation: \[ (A \mathbf{u})\cdot \mathbf{v} = (\lambda \mathbf{u})\cdot \mathbf{v} = \lambda (\mathbf{u}\cdot\mathbf{v}) \] The second calculation: \begin{align*} (A \mathbf{u})\cdot \mathbf{v} & = (A \mathbf{u})^\top \mathbf{v} = \mathbf{u}^\top A^\top \mathbf{v} = \mathbf{u} \cdot \bigl(A^\top \mathbf{v} \bigr) = \mathbf{u} \cdot \bigl(A \mathbf{v} \bigr) \\ & = \mathbf{u} \cdot (\mu \mathbf{v} ) = \mu ( \mathbf{u} \cdot \mathbf{v}) \end{align*} Since, \[ (A \mathbf{u})\cdot \mathbf{v} = \lambda (\mathbf{u}\cdot\mathbf{v}) \quad \text{and} \quad (A \mathbf{u})\cdot \mathbf{v} = \mu (\mathbf{u}\cdot\mathbf{v}) \] we conclude that \[ \lambda (\mathbf{u}\cdot\mathbf{v}) - \mu (\mathbf{u}\cdot\mathbf{v}) = 0. \] Therefore \[ ( \lambda - \mu ) (\mathbf{u}\cdot\mathbf{v}) = 0. \] Since we assume that $ \lambda - \mu \neq 0,$ the previous displayed equality yields \[ \mathbf{u}\cdot\mathbf{v} = 0. \] This proves that any two eigenvectors corresponding to distinct eigenvalues are orthogonal. Thus, the eigenspaces corresponding to distinct eigenvalues are orthogonal. QED.
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Theorem. A symmetric $2\!\times\!2$ matrix is orthogonally diagonalizable.
Proof. Let $A = \begin{bmatrix} a & b \\ b & d \end{bmatrix}$ be an arbitrary $2\!\times\!2$ be a symmetric matrix. We need to prove that there exists an orthogonal $2\!\times\!2$ matrix $U$ and a diagonal $2\!\times\!2$ matrix $D$ such that $A = UDU^\top.$ The eigenvalues of $A$ are \[ \lambda_1 = \frac{1}{2} \Bigl( a+d - \sqrt{(a-d)^2 + 4 b^2} \Bigr), \quad \lambda_2 = \frac{1}{2} \Bigl( a+d + \sqrt{(a-d)^2 + 4 b^2} \Bigr) \] Since clearly \[ (a-d)^2 + 4 b^2 \geq 0, \] the eigenvalues $\lambda_1$ and $\lambda_2$ are real numbers.
If $\lambda_1 = \lambda_2$, then $(a-d)^2 + 4 b^2 = 0$, and consequently $b= 0$ and $a=d$; that is $A = \begin{bmatrix} a & 0 \\ 0 & a \end{bmatrix}$. Hence $A = UDU^\top$ holds with $U=I_2$ and $D = A$.
Now assume that $\lambda_1 \neq \lambda_2$. Let $\mathbf{u}_1$ be a unit eigenvector corresponding to $\lambda_1$ and let $\mathbf{u}_2$ be a unit eigenvector corresponding to $\lambda_2$. We proved that eigenvectors corresponding to distinct eigenvalues of a symmetric matrix are orthogonal. Since $A$ is symmetric, $\mathbf{u}_1$ and $\mathbf{u}_2$ are orthogonal, that is the matrix $U = \begin{bmatrix} \mathbf{u}_1 & \mathbf{u}_2 \end{bmatrix}$ is orthogonal. Since $\mathbf{u}_1$ and $\mathbf{u}_2$ are eigenvectors of $A$ we have \[ AU = U \begin{bmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{bmatrix} = UD. \] Therefore $A=UDU^\top.$ This proves that $A$ is orthogonally diagonalizable. QED.
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Theorem. For every positive integer $n$, a symmetric $n\!\times\!n$ matrix is orthogonally diagonalizable.
Proof. (You can skip this proof.) This statement can be proved by Mathematical Induction. The base case $n = 1$ is trivial. The case $n=2$ is proved above. To get a feel how mathematical induction proceeds we will prove the theorem for $n=3.$
Let $A$ be a $3\!\times\!3$ symmetric matrix. Then $A$ has an eigenvalue, which must be real. Denote this eigenvalue by $\lambda_1$ and let $\mathbf{u}_1$ be a corresponding unit eigenvector. Let $\mathbf{v}_1$ and $\mathbf{v}_2$ be unit vectors such that the vectors $\mathbf{u}_1,$ Let $\mathbf{v}_1$ and $\mathbf{v}_2$ form an orthonormal basis for $\mathbb R^3.$ Then the matrix $V_1 = \bigl[\mathbf{u}_1 \ \ \mathbf{v}_1\ \ \mathbf{v}_2\bigr]$ is an orthogonal matrix and we have \[ V_1^\top A V_1 = \begin{bmatrix} \mathbf{u}_1^\top A \mathbf{u}_1 & \mathbf{u}_1^\top A \mathbf{v}_1 & \mathbf{u}_1^\top A \mathbf{v}_2 \\[5pt] \mathbf{v}_1^\top A \mathbf{u}_1 & \mathbf{v}_1^\top A \mathbf{v}_1 & \mathbf{v}_1^\top A \mathbf{v}_2 \\[5pt] \mathbf{v}_2^\top A \mathbf{u}_1 & \mathbf{v}_2^\top A \mathbf{v}_1 & \mathbf{v}_2^\top A \mathbf{v}_2 \\\end{bmatrix}. \] Since $A = A^\top$, $A\mathbf{u}_1 = \lambda_1 \mathbf{u}_1$ and since $\mathbf{u}_1$ is orthogonal to both $\mathbf{v}_1$ and $\mathbf{v}_2$ we have \[ \mathbf{u}_1^\top A \mathbf{u}_1 = \lambda_1, \quad \mathbf{v}_j^\top A \mathbf{u}_1 = \lambda_1 \mathbf{v}_j^\top \mathbf{u}_1 = 0, \quad \mathbf{u}_1^\top A \mathbf{v}_j = \bigl(A \mathbf{u}_1\bigr)^\top \mathbf{v}_j = 0, \quad \quad j \in \{1,2\}, \] and \[ \mathbf{v}_2^\top A \mathbf{v}_1 = \bigl(\mathbf{v}_2^\top A \mathbf{v}_1\bigr)^\top = \mathbf{v}_1^\top A^\top \mathbf{v}_2 = \mathbf{v}_1^\top A \mathbf{v}_2. \] Hence, \[ \tag{**} V_1^\top A V_1 = \begin{bmatrix} \lambda_1 & 0 & 0 \\[5pt] 0 & \mathbf{v}_1^\top A \mathbf{v}_1 & \mathbf{v}_1^\top A \mathbf{v}_2 \\[5pt] 0 & \mathbf{v}_2^\top A \mathbf{v}_1 & \mathbf{v}_2^\top A \mathbf{v}_2 \\\end{bmatrix}. \] By the already proved theorem for $2\!\times\!2$ symmetric matrix there exists an orthogonal matrix $\begin{bmatrix} u_{11} & u_{12} \\[5pt] u_{21} & u_{22} \end{bmatrix}$ and a diagonal matrix $\begin{bmatrix} \lambda_2 & 0 \\[5pt] 0 & \lambda_3 \end{bmatrix}$ such that \[ \begin{bmatrix} \mathbf{v}_1^\top A \mathbf{v}_1 & \mathbf{v}_1^\top A \mathbf{v}_2 \\[5pt] \mathbf{v}_2^\top A \mathbf{v}_1 & \mathbf{v}_2^\top A \mathbf{v}_2 \end{bmatrix} = \begin{bmatrix} u_{11} & u_{12} \\[5pt] u_{21} & u_{22} \end{bmatrix} \begin{bmatrix} \lambda_2 & 0 \\[5pt] 0 & \lambda_3 \end{bmatrix} \begin{bmatrix} u_{11} & u_{12} \\[5pt] u_{21} & u_{22} \end{bmatrix}^\top. \] Substituting this equality in (**) and using some matrix algebra we get \[ V_1^\top A V_1 = \begin{bmatrix} 1 & 0 & 0 \\[5pt] 0 & u_{11} & u_{12} \\[5pt] 0 & u_{21} & u_{22} \end{bmatrix} % \begin{bmatrix} \lambda_1 & 0 & 0 \\[5pt] 0 & \lambda_2 & 0 \\[5pt] 0 & 0 & \lambda_3 \end{bmatrix} % \begin{bmatrix} 1 & 0 & 0 \\[5pt] 0 & u_{11} & u_{12} \\[5pt] 0 & u_{21} & u_{22} \end{bmatrix}^\top \] Setting \[ U = V_1 \begin{bmatrix} 1 & 0 & 0 \\[5pt] 0 & u_{11} & u_{12} \\[5pt] 0 & u_{21} & u_{22} \end{bmatrix} \quad \text{and} \quad D = \begin{bmatrix} \lambda_1 & 0 & 0 \\[5pt] 0 & \lambda_2 & 0 \\[5pt] 0 & 0 & \lambda_3 \end{bmatrix} \] we have that $U$ is an orthogonal matrix, $D$ is a diagonal matrix and $A = UDU^\top.$ This proves that $A$ is orthogonally diagonalizable. QED.
- On Friday, we talked about the second part of Section 6.8: Fourier Series. Suggested problems: 5, 6, 7, 8, 9, 11, 12. The concept of orthogonality is essential in each inner product space. Let $\mathcal{V}$ be an inner product space with the inner product $\langle\,\cdot\,,\cdot\,\rangle.$ For completeness we state the definition of an orthogonal set of vectors in an inner product space.
- For all $j,k \in \{1,\ldots,m\}$ such that $j\neq k$ we have $\langle u_j, u_k \rangle = 0$.
- For all $k \in \{1,\ldots,m\}$ we have $\langle u_k, u_k \rangle \gt 0$. (This in fact means that all the vectors in this set are nonzero vectors.)
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Here we state three important properties of an orthogonal set of vectors. Assume that $\{ u_1, \ldots, u_m \}$ is an orthogonal set of vectors in $\mathcal{V}.$
- The set $\{ {u}_1, \ldots, {u}_m \}$ is linearly independent.
- If $v \in \operatorname{Span} \{ u_1, \ldots, u_m \}$, then the solution of the vector equation \[ \require{bbox} \color{red}{\alpha_1} \color{#008000}{{u}_1} + \cdots + \color{red}{\alpha_m} \color{#008000}{{u}_m} = \color{#008000}{v} \] is given by \[ \forall \mkern+1mu k\in \{1,\ldots,m\} \quad \color{red}{\alpha_k} = \frac{\langle \color{#008000}{v}, \color{#008000}{u_k} \rangle}{\langle \color{#008000}{u_k}, \color{#008000}{u_k} \rangle}. \] Notice a beautiful separation of colors. This property I call "easy solving of a vector equation." This property is essential for answering question (b) in Problem 5 on Assignment 3.
- Let $v \in \mathcal{V}$ and let \[ \mathcal W = \operatorname{Span} \{ u_1, \ldots, u_m \}. \] Then \[ \operatorname{Proj}_{\mathcal W}({v}) = \left(\frac{\langle v, u_1 \rangle}{\langle u_1, u_1 \rangle} \right) {u}_1 + \left(\frac{\langle v, u_2 \rangle}{\langle u_2, u_2 \rangle} \right) {u}_2 + \cdots + \left( \frac{\langle v, u_m \rangle}{\langle u_m, u_m \rangle} \right) {u}_m \] This property I call "easy orthogonal projection." I derived this formula in class when we studies the dot product. The reasoning with a general inner product is essentially the same as with the dot product. In the next item below I will prove that the vector on the right-hand side of the preceding equality satisfies properties ① and ② in the definition of the orthogonal projection posted on October 13.
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In this item I will prove the third property stated in the preceding item. I will prove that the vector $\mathbf{w}$ defined as
\[
{w} =
\left(\frac{\langle v, u_1 \rangle}{\langle u_1, u_1 \rangle} \right) {u}_1
+ \left(\frac{\langle v, u_2 \rangle}{\langle u_2, u_2 \rangle} \right) {u}_2
+ \cdots + \left( \frac{\langle v, u_m \rangle}{\langle u_m, u_m \rangle} \right) {u}_m.
\]
is the projection of ${v}$ onto $\mathcal{W}.$ That is, we will prove that ${w}$ is the projection $\operatorname{Proj}_{\mathcal W}({v}).$ For the proof, we need to prove properties ① and ② in the definition of the orthogonal projection posted on October 13. That is, we need to prove:
\[
w \in \mathcal{W} \qquad \text{and} \qquad (v - w) \perp \mathcal{W}.
\]
- Since $\mathcal W$ is the span of the vectors $u_1, \ldots, u_m$ and ${w}$ is a linear combination of the vectors $u_1, \ldots, u_m$ we have that ${w} \in \mathcal{W}.$ This proves ①.
- To prove ②, that is to prove \[ ( v - w) \perp \mathcal{W} \] we recall the equivalence \[ (v - w) \perp \mathcal{W} \qquad \Leftrightarrow \qquad \forall\mkern+1mu k\in\{1,\ldots,m\} \quad \bigl\langle (v - w), u_k \bigr\rangle = 0. \]
- The proof of the right-hand side in the preceding equivalence follows. Let $k\in\{1,\ldots,m\}$ be arbitrary. We calculate \begin{align*} \bigl\langle (v - w), u_k \bigr\rangle & = \bigl\langle v , u_k \bigr\rangle - \bigl\langle w, u_k \bigr\rangle \\ &= \bigl\langle v , u_k \bigr\rangle - \left\langle \sum_{j=1}^m \frac{\langle v, u_j \rangle}{\langle u_j, u_j \rangle} \, {u}_j , {u}_k \right\rangle \\ & = \bigl\langle v , u_k \bigr\rangle - \sum_{j=1}^m \frac{\langle v, u_j \rangle}{\langle u_j, u_j \rangle} \langle u_j, u_k \rangle \\ & = \bigl\langle v , u_k \bigr\rangle - \bigl\langle v , u_k \bigr\rangle \\ & = 0. \end{align*} Since $k\in\{1,\ldots,m\}$ was arbitrary, this proves that for all $k\in\{1,\ldots,m\}$ we have $\bigl\langle (v - w), u_k \bigr\rangle = 0.$ By the equivalence in the preceding item, property ② is proved.
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Below we will demonstrate how to use the orthogonal projection formula presented above to find Fourier approximations of functions on the interval $[-\pi,\pi].$ I recall the orthogonal projection formula first.
- Let $v \in \mathcal{V},$ let $\{ u_1, \ldots, u_m \}$ be an orthogonal set of vectors, and let \[ \mathcal W = \operatorname{Span} \{ u_1, \ldots, u_m \}. \] Then \[ \operatorname{Proj}_{\mathcal W}({v}) = \left(\frac{\langle v, u_1 \rangle}{\langle u_1, u_1 \rangle} \right) {u}_1 + \left(\frac{\langle v, u_2 \rangle}{\langle u_2, u_2 \rangle} \right) {u}_2 + \cdots + \left( \frac{\langle v, u_m \rangle}{\langle u_m, u_m \rangle} \right) {u}_m \]
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To apply the above orthogonal projection formula we need a vector space with an inner product.
We consider the vector space of continuous functions on the interval is $[-\pi,\pi].$ The notation for this vector space is $C[-\pi,\pi].$ The inner product in this space is given by \[ \bigl\langle f, g \bigr\rangle = \int_{-\pi}^{\pi} f(x) g(x) dx \qquad \text{where} \quad f, g \in C[-\pi,\pi]. \] When we do not have a specific names for functions that we are considering we will write functions using the variable. For example, we write \[ \bigl\langle x^2, \cos(n x) \bigr\rangle \] for the inner product of the square function and the cosine function of frequency $n.$
- The set of functions \[ 1, \cos(t), \sin(x), \cos(2 x), \sin(2 x), \cos(3 x), \sin(3 x), \ \ldots, \ \cos(n x), \sin(n x), \ \ldots \] is an orthogonal set of functions. To prove this claim we calculate \begin{align*} \bigl\langle 1, \cos(n x) \bigr\rangle &= \int_{-\pi}^{\pi} 1 \, \cos(n x) dx = 0, \\ \bigl\langle 1, \sin(n x) \bigr\rangle &= \int_{-\pi}^{\pi} 1 \, \sin(n x) dx = 0, \\ \bigl\langle \cos(m x), \sin(n x) \bigr\rangle &= \int_{-\pi}^{\pi} \cos(m x) \, \sin(n x) dx = 0, \\ \bigl\langle \cos(m x), \cos(n x) \bigr\rangle &= \int_{-\pi}^{\pi} \cos(m x) \, \cos(n x) dx = 0 \quad \text{whenever} \quad m\neq n, \\ \bigl\langle \sin(m x), \sin(n x) \bigr\rangle &= \int_{-\pi}^{\pi} \sin(m x) \, \sin(n x) dx = 0 \quad \text{whenever} \quad m\neq n, \\ \bigl\langle 1, 1 \bigr\rangle &= \int_{-\pi}^{\pi} 1^2 dx = 2 \pi, \\ \bigl\langle \cos(n x), \cos(n x) \bigr\rangle &= \int_{-\pi}^{\pi} \bigl(\cos(n x) \bigr)^2 dx = \pi, \\ \bigl\langle \sin(n x), \sin(n x) \bigr\rangle &= \int_{-\pi}^{\pi} \bigl(\sin(n x) \bigr)^2 dx = \pi. \end{align*} These integrals are probably done in Math 125. Some of these integrals are not difficult, some require product-to-sum trigonometric identities.
- As you can see to calculate inner products in the vector space $C[-\pi,\pi]$ we have to calculate many definite integrals. Doing these calculations "by hand" is an interesting, but time consuming process. It is much more convenient to calculate these integrals using technology. There is another reason why using technology is inevitable in this setting: When a formula for a Fourier approximation is found, the only way to visually check how well it approximates a given function is to plot both the given function and the approximation that is found. My technology tool is Wolfram Mathematica, an amazing piece of software. To get you started with Mathematica I created my Getting Started with Mathematica website.
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As an alternative to integral calculations presented in Math 125, here is one example of Mathematica code that will confirm what you did in your Calculus integration class
Mathematica respondsIntegrate[Cos[m x]*Cos[n x], {x, -Pi, Pi}]
We immediately see that the above formula does not hold for $m=n.$ Next, we exercise our knowledge that for
$m, n \in \mathbb{N}$ we have $\sin(m \pi) = 0$ and $\sin(n \pi) = 0$ to verify that
\[
\int_{-\pi}^{\pi} \cos(m x) \, \cos(n x) dx = 0 \quad \text{whenever} \quad m\neq n.
\]
Warning: Mathematica has powerful commands Simplify[] and
FullSimplify[] in which we can place assumptions and ask Mathematica to algebraically
simplify mathematical expressions. For example,
Unfortunately, Mathematica response to this command is 0. This is clearly wrong when m and n are equal; as shown by evaluatingFullSimplify[ Integrate[Cos[m x]*Cos[n x], {x, -Pi, Pi}], And[n \[Element] Integers, m \[Element] Integers] ]
So, Mathematica is powerful, but one has to exercise critical thinking.FullSimplify[ Integrate[Cos[n x]*Cos[n x], {x, -Pi, Pi}], And[n \[Element] Integers] ]
- Next, I will present a calculation of the projection of the projection of the square function $x^2$ onto the span \[ \mathcal{W} = \operatorname{Span} \{ 1, \cos(x), \sin(x), \cos(2 x), \sin(2 x), \cos(3x), \sin(3x), \ \ldots, \ \cos(n x), \sin(n x) \}, \] where $n\in\mathbb{N}.$
- Recall the projection formula \[ \operatorname{Proj}_{\mathcal W}({v}) = \left(\frac{\langle v, u_1 \rangle}{\langle u_1, u_1 \rangle} \right) {u}_1 + \left(\frac{\langle v, u_2 \rangle}{\langle u_2, u_2 \rangle} \right) {u}_2 + \cdots + \left( \frac{\langle v, u_m \rangle}{\langle u_m, u_m \rangle} \right) {u}_m \] Here the role of the vector $v$ is played by the function $x^2$ defined on the interval $[-\pi,\pi]$. The role of the vectors $u_1,\ldots,u_m$ is played by the functions \[ 1, \cos(x), \sin(x), \cos(2 x), \sin(2 x), \cos(3x), \sin(3x), \ \ldots \] In this setting the coefficients $\displaystyle \frac{\langle v, u_k \rangle}{\langle u_k, u_k \rangle}$ are called the Fourier coefficients.
- To calculate the Fourier coefficients for the function $x^2$ we need to calculate the integrals \begin{align*} \bigl\langle x^2, 1 \bigr\rangle & = \int_{-\pi}^{\pi} x^2 dx = \frac{2}{3} \pi^3, \\ \bigl\langle x^2, \sin(k x) \bigr\rangle &= \int_{-\pi}^{\pi} x^2 \sin(k x) dx = 0, \\ \bigl\langle x^2, \cos(k x) \bigr\rangle &= \int_{-\pi}^{\pi} x^2 \cos(k x) dx = (-1)^k\frac{4}{k} \pi , \\ \end{align*} where $k \in \mathbb{N}.$ Hence, the Fourier coefficients for $x^2$ are \begin{align*} \frac{\bigl\langle x^2, 1 \bigr\rangle}{\bigl\langle 1, 1 \bigr\rangle} & = \frac{\frac{2}{3} \pi^3}{2\pi} = \frac{1}{3} \pi^2, \\ \frac{\bigl\langle x^2, \sin(k x) \bigr\rangle}{\bigl\langle \sin(k x), \sin(k x) \bigr\rangle} &= \frac{0}{\pi} = 0, \\ \frac{\bigl\langle x^2, \cos(k x) \bigr\rangle}{\bigl\langle \cos(k x), \cos(k x) \bigr\rangle} &= \frac{ (-1)^k\frac{4}{k} \pi}{\pi} = (-1)^k \frac{4}{k}. \\ \end{align*}
- Let $n\in \mathbb{N}$ be a fixed number. The projection (or the best approximation) of the function $x^2$ onto the span of the trigonometric functions with integer frequencies \[ 1, \cos(x), \sin(x), \cos(2 x), \sin(2 x), \ \ldots, \ \cos(nx), \sin(nx) \] is \[ \frac{1}{3} \pi^2 + \sum_{k=1}^n (-1)^k \frac{4}{k} \cos(k x). \]
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The power of the approximation found in the preceding item is illustrated with a plot of the function
$x^2$ and its approximation with a specific value of $n.$ For example, with $n=3$ the approximation of $x^2$ is
\[
\frac{1}{3}\pi ^2 - 4 \cos (x) + \cos (2 x) - \frac{4}{9} \cos (3 x).
\]
To illustrate this in Mathematica execute this command
Changing the value of nn in the above Mathematica expression one gets a better approximation.nn = 3; Plot[ {x^2, \[Pi]^2/3*1 + Sum[(4 (-1)^k)/k^2*Cos[k x], {k, 1, nn}]}, {x, -3 Pi, 3 Pi}, PlotPoints -> {100, 200}, PlotStyle -> { {RGBColor[0, 0, 0.5], Thickness[0.01]}, {RGBColor[1, 0, 0], Thickness[0.005]} }, Ticks -> {Range[-2 Pi, 2 Pi, Pi/2], Range[-14, 14, 2]}, PlotRange -> {{-Pi - 0.1, Pi + 0.1}, {-1, Pi^2 + 0.2}}, AspectRatio -> 1/GoldenRatio ]
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Since the trigonometric functions used above are periodic with period $2\pi$ it is common to consider the
periodic extension of the square function from the interval $[-\pi,\pi]$ to the entire real line.
To plot the periodic extension of functions in Mathematica I use the Mathematica function
Mod[x,2 Pi, -Pi]. To plot of Mod[x,2 Pi, -Pi]
I use Mathematica code
with the following outputPlot[ {Mod[x, 2 Pi, -Pi]}, {x, -3 Pi, 3 Pi}, PlotStyle -> {{RGBColor[0, 0, 0], Thickness[0.005]}}, Ticks -> {Range[-5 Pi, 5 Pi, Pi/2], Range[-5 Pi, 5 Pi, Pi/2]}, PlotRange -> {{-3 Pi - 0.1, 3 Pi + 0.1}, {-Pi - 1, Pi + 1}}, GridLines -> {Range[-5 Pi, 5 Pi, Pi/4], Range[-5 Pi, 5 Pi, Pi/4]}, AspectRatio -> Automatic, ImageSize -> 600 ]
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The periodic extension of the square function is just the square of the preceding Mod[x,2 Pi, -Pi].
The periodic extension and its approximation are illustrated as follows
nn = 10; Plot[ {Mod[x, 2 Pi, -Pi]^2, \[Pi]^2/3*1 + Sum[(4 (-1)^k)/k^2*Cos[k x], {k, 1, nn}]}, {x, -4 Pi, 4 Pi}, PlotPoints -> {100, 200}, PlotStyle -> {{RGBColor[0, 0, 0.5], Thickness[0.01]}, {RGBColor[1, 0, 0], Thickness[0.005]}}, Ticks -> {Range[-4 Pi, 4 Pi, Pi/2], Range[-14, 14, 2]}, PlotRange -> {{-3 Pi - 0.1, 3 Pi + 0.1}, {-1, Pi^2 + 1}}, AspectRatio -> Automatic, ImageSize -> 600 ]
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Repeating the above procedure to the linear function $x$ defined on the interval $[-\pi,\pi],$
we get the following approximation with 10 sine functions of integer frequencies:
\begin{equation*}
2 \sin(x)-\sin (2 x) + \frac{2}{3} \sin(3 x) - \frac{1}{2} \sin(4 x) + \frac{2}{5} \sin (5 x)
- \frac{1}{3} \sin(6 x)
+ \frac{2}{7} \sin(7 x) - \frac{1}{4} \sin(8 x)+\frac{2}{9} \sin(9 x) - \frac{1}{5} \sin(10 x)
\end{equation*}
nn = 10; Plot[ {Mod[x, 2 Pi, -Pi], Sum[-((2 (-1)^k)/k)*Sin[k x], {k, 1, nn}]}, {x, -4 Pi, 4 Pi}, PlotPoints -> {100, 200}, PlotStyle -> {{RGBColor[0, 0, 0.5], Thickness[0.01]}, {RGBColor[1, 0, 0], Thickness[0.005]}}, Ticks -> {Range[-4 Pi, 4 Pi, Pi/2], Range[-4 Pi, 4 Pi, Pi/2]}, GridLines -> {Range[-4 Pi, 4 Pi, Pi/4], Range[-4 Pi, 4 Pi, Pi/4]}, PlotRange -> {{-3 Pi - 0.5, 3 Pi + 0.5}, {-Pi - 1, Pi + 1}}, AspectRatio -> Automatic, ImageSize -> 600 ]
- Here is the Mathematica notebook that I used for the above calculations.
Definition. A set of vectors $\{ u_1, \ldots, u_m \}$ in the inner product space $\mathcal{V}$ is said to be an orthogonal set of vectors if it has the following two properties:
- Here I will write more about the famous sequence of orthogonal polynomials called Legendre polynomials. To be specific I will focus on the ten dimensional space of polynomials: \[ \mathbb{P}_{9} = \operatorname{Span}\bigl\{1, x, x^2, x^3, x^4, x^5, x^6, x^7, x^8, x^9\bigr\}. \] We consider $\mathbb{P}_{9}$ as a subspace of all the (piecewise) continuous functions defined on $[-1,1]$, that is \[ \mathbb{P}_{9} \subset C[-1,1]. \] And, on $C[-1,1]$ we introduce the inner product as follows: \[ \text{for all} \quad f,g \in C[-1,1] \qquad \text{we set} \qquad \langle f, g \rangle = \int_{-1}^1 f(x)g(x) dx \]
- Let us introduce the following notation for the monomials which span $\mathbb{P}_9$; \begin{alignat*}{5} \phi_0(x) &= 1, \quad & \phi_1(x) &= x, \quad & \phi_2(x) &= x^2, \quad & \phi_3(x) &= x^3, \quad & \phi_4(x) &= x^4, \\ \phi_5(x) &= x^5, \quad & \phi_6(x) &= x^6, \quad & \phi_7(x) &= x^7, \quad & \phi_8(x) &= x^8, \quad & \phi_9(x) &= x^9. \end{alignat*} The inner products of the above monomials are: for $j,k \in \{0,1,2,3,4,5,6,7,8,9\}$ \begin{alignat*}{2} \langle \phi_j , \phi_k \rangle & = \phantom{333} 0 \quad &&\text{if} \quad j + k \quad \text{is odd}, \\ \langle \phi_j , \phi_k \rangle & = \frac{2}{j+k+1} \quad &&\text{if} \quad j + k \quad \text{is even}. \end{alignat*} Below is the table of these inner products: \[ \begin{array}{cc|cccccccccc} & & \phi_0 & \phi_1 & \phi_2 & \phi_3 & \phi_4 & \phi_5 & \phi_6 & \phi_7 & \phi_8 & \phi_9 \\ & & 1 & x & x^2 & x^3 & x^4 & x^5 & x^6 & x^7 & x^8 & x^9 \\ \hline \phi_0 & 1 & 2 & 0 & \frac{2}{3} & 0 & \frac{2}{5} & 0 & \frac{2}{7} & 0 & \frac{2}{9} & 0 \\ \phi_1 & x & 0 & \frac{2}{3} & 0 & \frac{2}{5} & 0 & \frac{2}{7} & 0 & \frac{2}{9} & 0 & \frac{2}{11} \\ \phi_2 & x^2 & \frac{2}{3} & 0 & \frac{2}{5} & 0 & \frac{2}{7} & 0 & \frac{2}{9} & 0 & \frac{2}{11} & 0 \\ \phi_3 & x^3 & 0 & \frac{2}{5} & 0 & \frac{2}{7} & 0 & \frac{2}{9} & 0 & \frac{2}{11} & 0 & \frac{2}{13} \\ \phi_4 & x^4 & \frac{2}{5} & 0 & \frac{2}{7} & 0 & \frac{2}{9} & 0 & \frac{2}{11} & 0 & \frac{2}{13} & 0 \\ \phi_5 & x^5 & 0 & \frac{2}{7} & 0 & \frac{2}{9} & 0 & \frac{2}{11} & 0 & \frac{2}{13} & 0 & \frac{2}{15} \\ \phi_6 & x^6 & \frac{2}{7} & 0 & \frac{2}{9} & 0 & \frac{2}{11} & 0 & \frac{2}{13} & 0 & \frac{2}{15} & 0 \\ \phi_7 & x^7 & 0 & \frac{2}{9} & 0 & \frac{2}{11} & 0 & \frac{2}{13} & 0 & \frac{2}{15} & 0 & \frac{2}{17} \\ \phi_8 & x^8 & \frac{2}{9} & 0 & \frac{2}{11} & 0 & \frac{2}{13} & 0 & \frac{2}{15} & 0 & \frac{2}{17} & 0 \\ \phi_9 & x^9 & 0 & \frac{2}{11} & 0 & \frac{2}{13} & 0 & \frac{2}{15} & 0 & \frac{2}{17} & 0 & \frac{2}{19} \\ \end{array} \]
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The first order of business here is to find an orthogonal basis for $\mathbb{P}_{9}$. That is done by the Gram-Schmidt Orthogonalization Algorithm. I presented this at the end of the post on Friday. On Friday we found an orthogonal basis for $\mathbb{P}_{4}\subset \mathbb{P}_{9}$. Now we will find additional five orthogonal polynomials. The polynomials which are obtained by the Gram-Schmidt orthogonalization algorithm which is presented below the gray divider in the post on May 4 are denoted by $q_k(x)$:
In the calculations below we repeatedly use the fact that the monomials of odd order are orthogonal to the monomials of even powers.
\begin{align*} q_0(x) &= 1 \\ q_1(x) &= x \\ q_2(x) &= x^2 - \frac{\langle\phi_2, \phi_0 \rangle}{\langle\phi_0, \phi_0 \rangle} \phi_0(x) \\ & = x^2 -\frac{1}{3} \\ q_3(x) &= x^3 - \frac{\langle\phi_3, \phi_1 \rangle}{\langle\phi_1, \phi_1 \rangle} \phi_1(x) \\ & = x^3 - \frac{3}{5} x \\ q_4(x) &= x^4 - \frac{\langle\phi_4, \phi_0 \rangle}{\langle\phi_0, \phi_0 \rangle} \phi_0(x) - \frac{\langle\phi_4, q_2 \rangle}{\langle q_2, q_2 \rangle} q_2(x) \\ &= x^4-\frac{6}{7} x^2 +\frac{3}{35} \\ q_5(x) &= x^5 - \frac{\langle\phi_5, \phi_1 \rangle}{\langle\phi_1, \phi_1 \rangle} \phi_1(x) - \frac{\langle\phi_5, q_3 \rangle}{\langle q_3, q_3 \rangle} q_3(x) \\ &= x^5-\frac{10}{9} x^3 +\frac{5}{21} x \\ q_6(x) &= x^6 - \frac{\langle\phi_6, \phi_0 \rangle}{\langle\phi_0, \phi_0 \rangle} \phi_0(x) - \frac{\langle\phi_6, q_2 \rangle}{\langle q_2, q_2 \rangle} q_2(x) - \frac{\langle\phi_6, q_4 \rangle}{\langle q_4, q_4 \rangle} q_4(x) \\ &= x^6-\frac{15}{11} x^4 +\frac{5}{11} x^2 -\frac{5}{231} \\ q_7(x) &= x^7 - \frac{\langle\phi_7, \phi_1 \rangle}{\langle\phi_1, \phi_1 \rangle} \phi_1(x) - \frac{\langle\phi_7, q_3 \rangle}{\langle q_3, q_3 \rangle} q_3(x) - \frac{\langle\phi_7, q_5 \rangle}{\langle q_5, q_5 \rangle} q_5(x) \\ &= x^7-\frac{21}{13} x^5+\frac{105}{143} x^3-\frac{35 x}{429} \\ q_8(x) &= x^8 - \frac{\langle\phi_8, \phi_0 \rangle}{\langle\phi_0, \phi_0 \rangle} \phi_0(x) - \frac{\langle\phi_8, q_2 \rangle}{\langle q_2, q_2 \rangle} q_2(x) - \frac{\langle\phi_8, q_4 \rangle}{\langle q_4, q_4 \rangle} q_4(x) - \frac{\langle\phi_8, q_6 \rangle}{\langle q_6, q_6 \rangle} q_6(x)\\ &= x^8-\frac{28}{15} x^6+\frac{14}{13} x^4 -\frac{28}{143} x^2+\frac{7}{1287} \\ q_9(x) &= x^9 - \frac{\langle\phi_9, \phi_1 \rangle}{\langle\phi_1, \phi_1 \rangle} \phi_1(x) - \frac{\langle\phi_9, q_3 \rangle}{\langle q_3, q_3 \rangle} q_3(x) - \frac{\langle\phi_9, q_5 \rangle}{\langle q_5, q_5 \rangle} q_5(x) - \frac{\langle\phi_9, q_7 \rangle}{\langle q_7, q_7 \rangle} q_7(x)\\ &= x^9-\frac{36}{17} x^7+\frac{126}{85} x^5 -\frac{84}{221} x^3 +\frac{63}{2431} x \\ \end{align*} - It is traditional to "normalize" the orthogonal polynomials from the preceding item in such a way that they have value $1$ at $x=1$. That is, each of the polynomials in the preceding item needs to be multiplied by the value $\displaystyle \frac{1}{q_k(1)}$. Performing this multiplication we obtain the first 10 Legendre polynomials: \begin{align*} P_0(x) &= 1 \\ P_1(x) &= x \\ P_2(x) &= \frac{1}{2} \left(3 x^2-1\right) \\ P_3(x) &= \frac{1}{2} \left(5 x^3-3 x\right) \\ P_4(x) &=\frac{1}{8} \left(35 x^4-30 x^2+3\right) \\ P_5(x) &= \frac{1}{8} \left(63 x^5-70 x^3+15 x\right) \\ P_6(x) &= \frac{1}{16} \left(231 x^6-315 x^4+105 x^2-5\right)\\ P_7(x) &= \frac{1}{16} \left(429 x^7-693 x^5+315 x^3-35 x\right) \\ P_8(x) &= \frac{1}{128} \left(6435 x^8-12012 x^6+6930 x^4-1260 x^2+35\right) \\ P_9(x) &= \frac{1}{128} \left(12155 x^9-25740 x^7+18018 x^5-4620 x^3+315 x\right) \\ \end{align*}
- Orthogonal basis for $\mathbb{P}_{9}$ will be utilized to find orthogonal projections of various functions onto $\mathbb{P}_{9}$, that is to find the best approximations of various functions by polynomials of degree less than 10. Here the best approximation is in the sense of the norm induced by the above inner product: the distance between two continuous functions $f, g \in C[-1,1]$ is given by \[ \| f - g \| = \sqrt{\int_{-1}^{1} \bigl| f(x) - g(x) \bigr|^2 dx }. \]
- The calculations above involve calculating a lot of integrals. Although, integrals involving polynomials are relatively simple, the amount of arithmetic can be overwhelming. Therefore, it is convenient to use technology to help us out with calculations and with plotting of many functions that are involved. In the next item I share the Wolfram Mathematica notebook that I created for explorations of Legendre Polynomials. You can use the notebook in Wolfram Mathematica Computer Algebra System, or you can view it in free Wolfram Mathematica Reader, or you can view the pdf printout that I created.
- I created this Wolfram Mathematica notebook Gram-Schmidt Orthogonalization of Monomials in which I use Mathematica commands do demonstrate calculations related to Legendre Polynomials. I put special effort that this notebook can be used in the free Wolfram Mathematica Player which can be downloaded from Wolfram Mathematica website. You can also explore this notebook in Wolfram Mathematica Computer Algebra System which is available on the computers in the computer lab in BH 215 and the Math Center BH 209. I also created a pdf printout of the same notebook, so that you can view how Mathematica commands work without having access to Mathematica. The advantage of downloading Wolfram Mathematica Player is that the dynamic content of my notebook works in the Player and the pdf file contains just still images.
- Today we discussed Section 6.7 Inner Product Spaces. Suggested problems for Section 6.7: 1, 2, 3, 5, 7, 9, 10, 13, 16, 17, 19, 20, 21, 23, 25
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In this item, I recall the definition of an abstract inner product. In the definition below $\times$ denotes the Cartesian product between two sets.
Definition. Let $\mathcal{V}$ be a vector space over $\mathbb R.$ Below we list axioms of an inner product \(\langle\,\cdot\,,\cdot\,\rangle\) on $\mathcal{V}$:
IPE. \(\langle\,\cdot\,,\cdot\,\rangle\) is a function defined on \(\mathcal{V}\times\mathcal{V}\) with values in \(\mathbb{R}\). That is, \[ \langle\,\cdot\,,\cdot\,\rangle : \mathcal{V}\times\mathcal{V} \to \mathbb{R}. \]IPA. For all $u, v, w \in \mathcal{V}$ we have $\langle u + v, w \rangle = \langle u,w\rangle + \langle v,w\rangle$.IPS. For all $u, v \in \mathcal{V}$ and all $\alpha \in \mathbb{R}$ we have $\langle \alpha u , v \rangle = \alpha \langle u, v \rangle$.IPC. For all $u, v \in \mathcal{V}$ we have $\langle u,v\rangle = \langle v,u\rangle$.IPP. For all $v \in \mathcal{V}$ we have \[ \langle v , v \rangle \geq 0. \]IPD. For all $v \in \mathcal{V}$ we have \[ \langle v , v \rangle = 0 \quad \Rightarrow \quad v = 0_{\mathcal{V}}. \]Explanation of the abbreviations: IPE--inner product exists (it is a function), IPC--inner product is commutative, IPA--inner product respects addition of vectors, IPS--inner product respects scaling of vectors, IPP--inner product is positive, IPD--inner product is definite, in the sense that the only vector whose "inner square" is zero is the zero vector. The abbreviations are made up by me as a cute mnemonic tool.
- The preceding definition means that for every two vectors $u \in \mathcal{V}$ and $v \in \mathcal{V}$ there exists a unique real number $\langle u,v\rangle \in \mathbb{R}$ which is called the inner product of $u$ and $v$ and the inner product $\langle u,v\rangle$ has the algebraic properties IPC (inner product is commutative), IPA (inner product respects addition), IPS (inner product respects scaling), IPP (inner product is positive definite).
- The most important abstract inner products are inner products given by the Riemann integral in vector spaces of functions. I illustrate this with the inner product \[ \langle p, q \rangle = \int_{-1}^{1} p(x) q(x) dx \] in the vector space of polynomials. We can restrict ourselves to the space $\mathbb{P}_4.$ The standard basis in $\mathbb{P}_4$ is the basis which consists of monomials: \[ \phi_0(x) = 1, \quad \phi_1(x) = x, \quad \phi_2(x) = x^2, \quad \phi_3(x) = x^3, \quad \phi_4(x) = x^4. \] Set \[4 \mathcal M = \bigl\{ \phi_0, \phi_1, \phi_2, \phi_3, \phi_4 \bigr\}. \] $\mathcal M$ is not an orthogonal basis for $\mathbb{P}_5.$ In fact it is useful to calculate \begin{alignat*}{5} \langle \phi_0, \phi_0 \rangle & = 2, & \quad \langle \phi_0, \phi_1 \rangle & = 0, & \quad \langle \phi_0, \phi_2 \rangle & = \frac{2}{3}, & \quad \langle \phi_0, \phi_3 \rangle & = 0, & \quad \langle \phi_0, \phi_4 \rangle & = \frac{2}{5}, \\ \langle \phi_1, \phi_0 \rangle & = 0, & \quad \langle \phi_1, \phi_1 \rangle & = \frac{2}{3}, & \quad \langle \phi_1, \phi_2 \rangle & = 0, & \quad \langle \phi_1, \phi_3 \rangle & = \frac{2}{5}, & \quad \langle \phi_1, \phi_4 \rangle & = 0, \\ \langle \phi_2, \phi_0 \rangle & = \frac{2}{3}, & \quad \langle \phi_2, \phi_1 \rangle & = 0, & \quad \langle \phi_2, \phi_2 \rangle & = \frac{2}{5}, & \quad \langle \phi_2, \phi_3 \rangle & = 0, & \quad \langle \phi_2, \phi_4 \rangle & = \frac{2}{7}, \\ \langle \phi_3, \phi_0 \rangle & = 0, & \quad \langle \phi_3, \phi_1 \rangle & = \frac{2}{5}, & \quad \langle \phi_3, \phi_2 \rangle & = 0, & \quad \langle \phi_3, \phi_3 \rangle & = \frac{2}{7}, & \quad \langle \phi_3, \phi_4 \rangle & = 0, \\ \langle \phi_4, \phi_0 \rangle & = \frac{2}{5}, & \quad \langle \phi_4, \phi_1 \rangle & = 0, & \quad \langle \phi_4, \phi_2 \rangle & = \frac{2}{7}, & \quad \langle \phi_4, \phi_3 \rangle & = 0, & \quad \langle \phi_4, \phi_4 \rangle & = \frac{2}{9}. \\ \end{alignat*}
- One conclusion from the above table is that monomials of even degree are orthogonal to the monomials of odd degree. We will use this fact in the calculations in the next item.
- We can apply the Gram-Schmidt orthogonalization algorithm to the basis $\mathcal M$ obtain an orthogonal basis \[ \mathcal A = \bigl\{q_0, q_1, q_2, q_3, q_4 \bigr\} \] for $\mathbb{P}_4:$ \begin{align*} q_0(t) & = 1 \\ q_1(t) & = x \\ q_2(t) & = x^2 - \frac{\langle \phi_2, q_0 \rangle }{\langle q_0, q_0 \rangle} 1 = x^2 - \frac{1}{3} \\ q_3(t) & = x^3 - \frac{\langle \phi_3, q_1 \rangle }{\langle q_1, q_1 \rangle} x = x^3 - \frac{3}{5} x \\ q_4(t) & = x^4 - \frac{\langle \phi_4, q_0 \rangle }{\langle q_0, q_0 \rangle} 1 - \frac{\langle \phi_4, q_2 \rangle }{\langle q_2, q_2 \rangle} \left(x^2 - \frac{1}{3}\right) = x^4 - \frac{6}{7}x^2 + \frac{3}{35} \\ \end{align*} In the above calculation we used that \[ \langle q_0, q_0 \rangle = 2, \quad \langle q_1, q_1 \rangle =\frac{2}{3}, \quad \langle q_2, q_2 \rangle = \frac{8}{45} \] and \[ \langle \phi_2, q_0 \rangle = \frac{2}{3}, \quad \langle \phi_3, q_1 \rangle = \frac{2}{5}, \quad \langle \phi_4, q_0 \rangle = \frac{2}{5}, \quad \langle \phi_4, q_2 \rangle = \frac{16}{105}. \]
- It is common to normalize the polynomials $q_0, q_1, q_2, q_3, q_4$ so that they have values $1$ at $t=1.$ First calculate \[ q_0(1) = 1, \quad q_1(1) = 1, \quad q_2(1) = \frac{2}{3}, \quad q_3(1) = \frac{2}{5}, \quad q_4(1) = \frac{8}{35}. \] The polynomials \begin{alignat*}{2} P_0(x) & = 1 & & \\ P_1(x) & = x & & \\ P_2(x) & = \frac{1}{2} \left( 3x^2 -1 \right) & & = \frac{3}{2} q_2(x) \\ P_3(x) & = \frac{1}{2} \left( 5 x^3 -3 x \right) & & = \frac{5}{2} q_3(x) \\ P_4(x) & = \frac{1}{8} \left( 35 x^4 - 30 x^2 + 3 \right) & & = \frac{35}{8} q_4(x) \\ \end{alignat*} The polynomials $P_0, P_1, P_2, P_3, P_4$ are the first five of the sequence of orthogonal polynomials called Legendre polynomials.
- There are many examples of other sequences of orthogonal polynomials. Legendre polynomials is just one example which is presented here since the inner product in which they are orthogonal is particularly simple.
- I post several pictures related to Problem 4 on Assignment 3.
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In the pictures below, the positive direction of the vertical axes points downward for aesthetic reasons.
The intersection of the canonical rotated paraboloid $z=x^2+y^2$ and a plane $z=ax+by+c$ is an ellipse, provided that $a^2+b^2+4c>0$. The projection of this ellipse onto the $xy$-plane is always a circle. Always! I find this amazing!
Notice that the picture below is "upside-down,"
with the positive direction of the $z$-axis pointing downwards.
How would we determine the intersection of the paraboloid and the plane? Recall, the paraboloid is the set of points $(x,y, x^2 + y^2)$, while the plane is the set of points $(x,y, ax+by+c)$. For a point $(x,y,z)$ to be both, on the paraboloid, and on the plane we must have \[ x^2 + y^2 = a x + b y + c. \] Which points $(x,y)$ in the $xy$-plane satisfy the preceding equation? Rewrite the equation as \[ x^2 - a x + y^2 - b y = c, \] and completing the squares comes to the rescue, so we obtain the following equation: \[ \boxed{\left(x - \frac{a}{2} \right)^2 + \left(y-\frac{b}{2}\right)^2 = \frac{a^2}{4} + \frac{b^2}{4} + c.} \] The boxed equation is the equation of a circle in $xy$-plane centered at the point $(a/2,b/2)$ with the radius $\displaystyle\frac{1}{2}\sqrt{a^2+b^2+4c}$. Above this circle, in the plane $z=ax+by+c$ is the ellipse which is also on the paraboloid $z=x^2+y^2$. The circle whose equation is boxed, we will call the circle determined by the paraboloid $z=x^2+y^2$ and the plane $z=ax+by+c$.
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The picture below illustrates Problem 4(a). The orange points are the points
$P_1$, $P_2$, $P_3$, $P_4$. The points $Q_1$, $Q_2$, $Q_3$, $Q_4$ are the corresponding points on
the rotated paraboloid. I colored them red, but these red points don't show well on the paraboloid.
The gray line segments are $P_1Q_1$, $P_2Q_2$, $P_3Q_3$, $P_4Q_4$.
These line segments connect the points $P_1$, $P_2$, $P_3$, $P_4$ in the $xy$-plane and the corresponding
points $Q_1$, $Q_2$, $Q_3$, $Q_4$ on the rotated paraboloid.
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The picture below illustrates the solution of Problem 4(a).
The yellow plane is the least-squares fit plane that best fits the points $Q_1$, $Q_2$, $Q_3$, $Q_4$.
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The pictures below illustrate the solution of Problem 4(c).
The black circle in $xy$-plane is, in some sense, the best fit circle for the points $P_1$, $P_2$, $P_3$, $P_4$.
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Below I will describe in more details the method of finding the best fit circle to a given set of points. The method is identical to finding the least-squares fit plane to a set of given points. In this case all the given points lie on the canonical rotated paraboloid $z = x^2+y^2.$
- The standard equation for a circle in $\mathbb{R}^2$ centered at the point $(a,b)$ with the radius $r \gt 0$ is \[ (x-a)^2 + (y-b)^2 = r^2. \] Expending the squares and grouping terms, the preceding equality is equivalent to \[ x^2 + y^2 - (2a) x - (2b) y - (r^2 - a^2 - b^2) = 0. \] Substituting \[ \beta_0 = r^2 - a^2 - b^2, \quad \beta_1 = 2a, \quad \beta_2 = 2b, \] we rewrite the preceding equality as \[ \require{bbox} \bbox[5px, #FFFF00, border: 1pt solid #888800]{\beta_0 + \beta_1 x + \beta_2 y = x^2 + y^2}. \] Although the last equation does not look like an equation of a circle, it is an equation of a circle, provided that \[ \beta_0 + \bigl(\beta_1/2\bigr)^2 + \bigl(\beta_2/2\bigr)^2 \gt 0. \]
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Let $n$ be a positive integer greater than $2.$ Assume that we are given $n$ noncollinear points in $\mathbb{R}^2$: \[ (x_1, y_1), \ \ (x_2, y_2), \ \ \ldots, \ (x_n, y_n). \]
- We want to find an equation of a circle which fits these points the best. We use the yellow highlighted equation of a circle. The unknown quantities are $\color{red}{\beta_0},$ $\color{red}{\beta_1},$ $\color{red}{\beta_2}.$ The linear equations that we need to solve are \begin{align*} \color{red}{\beta_0}\cdot 1 + \color{red}{\beta_1} x_1 + \color{red}{\beta_2} y_1 &= (x_1)^2 + (y_1)^2 \\ \color{red}{\beta_0}\cdot 1 + \color{red}{\beta_1} x_2 + \color{red}{\beta_2} y_2 &= (x_2)^2 + (y_2)^2 \\ & \ \ \vdots \\ \color{red}{\beta_0}\cdot 1 + \color{red}{\beta_1} x_n + \color{red}{\beta_2} y_n &= (x_n)^2 + (y_n)^2 \end{align*}
- In matrix form the above system is: \[ \left[\begin{array}{ccc} 1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ \vdots & \vdots & \vdots \\ 1 & x_n & y_n \end{array} \right] \left[\begin{array}{c} \color{red}{\beta_0} \\ \color{red}{\beta_1} \\ \color{red}{\beta_2} \end{array} \right] = \left[\begin{array}{c} (x_1)^2 + (y_1)^2 \\ (x_2)^2 + (y_2)^2 \\ \vdots \\ (x_n)^2 + (y_n)^2 \end{array} \right]. \] This system is identical to the system for finding the least-squares plane that best fits the points \[ \bigl(x_1, y_1, (x_1)^2 + (y_1)^2 \bigr), \ \ \bigl(x_2, y_2, (x_2)^2 + (y_2)^2\bigr), \ \ \ldots, \ \bigl(x_n, y_n, (x_n)^2 + (y_n)^2\bigr). \] These points are all on the canonical rotated paraboloid $z=x^2+y^2,$ as explained at the beginning of today's post.
The normal equations for the system from the preceding item are \[ \left[\begin{array}{cccc} 1 & 1 & \cdots & 1 \\[-3pt] x_1 & x_2 & \cdots & x_n \\ y_1 & y_2 & \cdots & y_n \end{array} \right] \left[\begin{array}{ccc} 1 & x_1 & y_1 \\ 1 & x_2 & y_2 \\ \vdots & \vdots & \vdots \\ 1 & x_n & y_n \end{array} \right] \left[\begin{array}{c} \color{red}{\beta_0} \\ \color{red}{\beta_1} \\ \color{red}{\beta_2} \end{array} \right] = \left[\begin{array}{cccc} 1 & 1 & \cdots & 1 \\[-3pt] x_1 & x_2 & \cdots & x_n \\ y_1 & y_2 & \cdots & y_n \end{array} \right] \left[\begin{array}{c} (x_1)^2 + (y_1)^2 \\ (x_2)^2 + (y_2)^2 \\ \vdots \\ (x_n)^2 + (y_n)^2 \end{array} \right]. \]
- In the items below I will present the Mathematica code that I wrote to automate finding the best fit circle. The argument of this function is a given set of points, and the output of this function is the center and the radius of the best fit circle.
- You do not need Mathematica to solve Problem 4 on Assignment 3. I am posting the Mathematica code here for two reasons: first, some readers might want to engage with the code; second, I find keeping a record of what I do to be useful.
- To get you started with Mathematica I created my Mathematica website. At the end of the Mathematica website you will find a section about Linear Algebra in Mathematica.
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Now that we have a linear algebra method for finding the best fit circle to a set of points,
I wrote the Mathematica command BestCir[] to automate finding of the best fit circle:
You can copy-paste this command to a Mathematica notebook and test it on a set of points. The output of the command is a pair of the best circle's center and the best circle's radius.Clear[BestCir, gpts, mX, vY, abc]; BestCir[gpts_] := Module[ {mX, vY, abc}, mX = Transpose[Append[Transpose[gpts], Array[1 &, Length[gpts]]]]; vY = (#[[1]]^2 + #[[2]]^2) & /@ gpts; abc = Last[ Transpose[ RowReduce[ Transpose[ Append[Transpose[Transpose[mX] . mX], Transpose[mX] . vY] ] ] ] ]; {{abc[[1]]/2, abc[[2]]/2}, Sqrt[abc[[3]] + (abc[[1]]/2)^2 + (abc[[2]]/2)^2]} ]
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Below is an example for the above command. First I name a set of points mypts,
then I use the above command to find the best circles's center and the radius, which are named
cir
You can copy-paste the above code in a Mathematica cell and execute it. The result will be the following image:mypts = {{5, 2}, {-1, 5}, {3, -2}, {3, 4.5}, {-5/2, 3}, {1, 5}, {4, 3}, {-3, 1}, {-3/2, 4}, {1, -3}, {-2, -1}, {4, -1}}; cir = N[BestCir[mypts]]; Graphics[{ {PointSize[0.015], RGBColor[1, 0.5, 0], Point[#] & /@ mypts}, {RGBColor[0, 0, 0.5], PointSize[0.015], Point[cir[[1]]], Thickness[0.007], Circle[cir[[1]], cir[[2]]]} }, GridLines -> {Range[-20, 20, 1/2], Range[-20, 20, 1/2]}, GridLinesStyle -> {{GrayLevel[0.75]}, {GrayLevel[0.75]}}, Axes -> True, Ticks -> {Range[-7, 7], Range[-7, 7]}, Frame -> False, PlotRange -> {{-5.75, 5.75}, {-5.75, 5.75}}, ImageSize -> 600]
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In this example, we use our command to obtain the exact circle through three noncollinear points.
First I name a set of points mypts, then I use the above command to find the best circles's
center and the radius, which are named cir
You can copy-paste the above code in a Mathematica cell and execute it. The result will be the following image:mypts = {{3, 1}, {2, -4}, {-2, 3}}; cir = N[BestCir[mypts]]; Graphics[{ {PointSize[0.015], RGBColor[1, 0.5, 0], Point[#] & /@ mypts}, {RGBColor[0, 0, 0.5], PointSize[0.015], Point[cir[[1]]], Thickness[0.007], Circle[cir[[1]], cir[[2]]]} }, GridLines -> {Range[-20, 20, 1/2], Range[-20, 20, 1/2]}, GridLinesStyle -> {{GrayLevel[0.75]}, {GrayLevel[0.75]}}, Axes -> True, Ticks -> {Range[-7, 7], Range[-7, 7]}, Frame -> False, PlotRange -> {{-5.75, 5.75}, {-5.75, 5.75}}, ImageSize -> 600]
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In this example, we randomly generate 100 points in close to the circle centered at the origin with radius 4.
Then we use our command to obtain the best fit circle. We name a set of hundred points
mypts, then we use the above command to find the best circles's center and
the radius, which are named cir
You can copy-paste the above code in a Mathematica cell and execute it. The result will be the following image:mypts = ((4 {Cos[2 Pi #[[1]]], Sin[2 Pi #[[1]]]} + 1/70 {#[[2]], #[[3]]}) & /@ ((RandomReal[#, 3]) & /@ Range[100])); cir = N[BestCir[mypts]]; Graphics[{ {PointSize[0.015], RGBColor[1, 0.5, 0], Point[#] & /@ mypts}, {RGBColor[0, 0, 0.5], PointSize[0.015], Point[cir[[1]]], Thickness[0.007], Circle[cir[[1]], cir[[2]]]} }, GridLines -> {Range[-20, 20, 1/2], Range[-20, 20, 1/2]}, GridLinesStyle -> {{GrayLevel[0.75]}, {GrayLevel[0.75]}}, Axes -> True, Ticks -> {Range[-7, 7], Range[-7, 7]}, Frame -> False, PlotRange -> {{-5.75, 5.75}, {-5.75, 5.75}}, ImageSize -> 600]
- We discussed Section 6.6: Linear Models Suggested problems for Section 6.6: 1, 2, 3, 4, 11, 12, 13, 14, 15, 20 (5th edition: 1, 2, 3, 4, 5, 6, 7, 8, 9, 14)
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Exercise 4 in Section 6.6 is an interesting problem.
In this exercise we are given four data points
\[
( 2,3), \ \ (3,2), \ \ (5,1), \ \ (6,0),
\]
and we are asked to find the least-squares line that best fits the given data points. (We will call this line simply the
least-squares line.)
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Notice that these four points form a very narrow parallelogram. A characterizing property of a parallelogram is that its diagonals share the midpoint. For this parallelogram, the coordinates of the common midpoint of the diagonals are \[ \overline{x} = \frac{1}{4}(2+3+5+6) = 4, \quad \overline{y} = \frac{1}{4}(3+2+1+0) = 3/2. \] The long sides of this parallelogram are on the parallel lines \[ y = -\frac{2}{3}x +4 \quad \text{and} \quad y = -\frac{2}{3}x + \frac{13}{3}. \] It is natural to guess that the least square line is the line which is parallel to these two lines and half-way between them. That is the line \[ y = -\frac{2}{3}x + \frac{25}{6}. \] This line is the red line in the picture below. Clearly this line goes through the point $(4,3/2),$ the intersection of the diagonals of the parallelogram.
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The only way to verify the guess from the preceding item is to calculate the least-squares line for these four points. Let us find the least-squares solution of the equation \[ \left[\begin{array}{cc} 1 & 2 \\ 1 & 3 \\ 1 & 5 \\ 1 & 6 \end{array} \right] \left[\begin{array}{c} \beta_0 \\ \beta_1 \end{array} \right] = \left[\begin{array}{c} 3 \\ 2 \\ 1 \\ 0 \end{array} \right]. \] To get to the corresponding normal equations we multiply both sides by $X^\top$ \[ \left[\begin{array}{cccc} 1 & 1 & 1 & 1 \\ 2 & 3 & 5 & 6 \end{array} \right] \left[\begin{array}{cc} 1 & 2 \\ 1 & 3 \\ 1 & 5 \\ 1 & 6 \end{array} \right] \left[\begin{array}{c} \beta_0 \\ \beta_1 \end{array} \right] = \left[\begin{array}{cccc} 1 & 1 & 1 & 1 \\ 2 & 3 & 5 & 6 \end{array} \right] \left[\begin{array}{c} 3 \\ 2 \\ 1 \\ 0 \end{array} \right]. \] The corresponding normal equations are \[ \left[\begin{array}{cc} 4 & 16 \\ 16 & 74 \end{array} \right] \left[\begin{array}{c} \beta_0 \\ \beta_1 \end{array} \right] = \left[\begin{array}{c} 6 \\ 17 \end{array} \right]. \] Since the inverse of the above $2\!\times\!2$ matrix is \[ \left[\begin{array}{cc} 4 & 16 \\ 16 & 74 \end{array} \right]^{-1} = \frac{1}{40} \left[\begin{array}{cc} 74 & -16 \\ -16 & 4 \end{array} \right], \] and the solution of the normal equations is unique and it is given by \[ \left[\begin{array}{c} \beta_0 \\ \beta_1 \end{array} \right] = \frac{1}{40} \left[\begin{array}{cc} 74 & -16 \\ -16 & 4 \end{array} \right] \left[\begin{array}{c} 6 \\ 17 \end{array} \right] = \left[\begin{array}{c} \frac{43}{10} \\ -\frac{7}{10} \end{array} \right] \]
- Hence, the least-squares line for the given data points is \[ y = -\frac{7}{10}x + \frac{43}{10}. \] This line is the blue line in the picture below. The picture below strongly indicates that the blue line also goes through the point $(4,3/2).$ This is easily confirmed: \[ \frac{3}{2} = -\frac{7}{10}4 + \frac{43}{10}. \]
- Finally, let us calculate the least-squares error for the line which we guessed at the beginning of this problem and the least-squares line that we calculated above, that is \[ y = -\frac{2}{3}x + \frac{25}{6} \quad \text{and} \quad y = -\frac{7}{10}x + \frac{43}{10}. \] Below I calculate the residual vectors for the guessed line and for the least-squares line are as follows: \[ \left[\begin{array}{c} 3 \\ 2 \\ 1 \\ 0 \end{array} \right] - \left[\begin{array}{r} \frac{17}{6} \\ \frac{13}{6} \\ \frac{5}{6} \\ \frac{1}{6} \end{array} \right] = \left[\begin{array}{r} \frac{1}{6} \\ -\frac{1}{6} \\ \frac{1}{6} \\ -\frac{1}{6} \end{array} \right] \quad \text{and} \quad \left[\begin{array}{c} 3 \\ 2 \\ 1 \\ 0 \end{array} \right] - \left[\begin{array}{r} \frac{29}{10} \\ \frac{11}{5} \\ \frac{4}{5} \\ \frac{1}{10} \end{array} \right] = \left[\begin{array}{r} \frac{1}{10} \\ -\frac{1}{5} \\ \frac{1}{5} \\ -\frac{1}{10} \end{array} \right], \] with the corresponding norms \[ \frac{1}{3} \approx 0.333333 \qquad \text{and} \qquad \frac{1}{\sqrt{10}} \approx 0.316228 \] As it should be, the least-squares error for the least-squares line is smaller than the lease-squares error for the line guessed at the beginning of the problem.
In the image below the forest green points are the given data points.
The red line is the line which I guessed could be the least-squares line.
The blue line is the true least-squares line.
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It is amazing that what we observed in the preceding example is universal.
Proposition. If the line $y = \beta_0 + \beta_1 x$ is the least-squares line for the data points \[ (x_1,y_1), \ldots, (x_n,y_n), \] then $\overline{y} = \beta_0 + \beta_1 \overline{x}$, where \[ \overline{x} = \frac{1}{n}(x_1+\cdots+x_n), \quad \overline{y} = \frac{1}{n}(y_1+\dots+y_n). \]
The above proposition is Exercise 20 (5th edition Exercise 14) in Section 6.6.Proof. Let \[ (x_1,y_1), \ldots, (x_n,y_n), \] be given data points and set \[ \overline{x} = \frac{1}{n}(x_1+\cdots+x_n), \quad \overline{y} = \frac{1}{n}(y_1+\dots+y_n). \] Let $y = \beta_0 + \beta_1 x$ be the least-squares line for the given data points. Then the vector $\left[\begin{array}{c} \beta_0 \\ \beta_1 \end{array} \right]$ satisfies the normal equation \[ \left[\begin{array}{cccc} 1 & 1 & \cdots & 1 \\ x_1 & x_2 & \cdots & x_n \end{array} \right] \left[\begin{array}{cc} 1 & x_1 \\ 1 & x_2 \\ \vdots & \vdots \\ 1 & x_n \end{array} \right] \left[\begin{array}{c} \beta_0 \\ \beta_1 \end{array} \right] = \left[\begin{array}{cccc} 1 & 1 & \cdots & 1 \\ x_1 & x_2 & \cdots & x_n \end{array} \right] \left[\begin{array}{c} y_1 \\ y_2 \\ \vdots \\ y_n \end{array} \right]. \] Multiplying the second matrix on the left-hand side and the third vector we get \[ \left[\begin{array}{cccc} 1 & 1 & \cdots & 1 \\ x_1 & x_2 & \cdots & x_n \end{array} \right] \left[\begin{array}{c} \beta_0 + \beta_1 x_1 \\ \beta_0 + \beta_1 x_2 \\ \vdots \\ \beta_0 + \beta_1 x_n \end{array} \right] = \left[\begin{array}{cccc} 1 & 1 & \cdots & 1 \\ x_1 & x_2 & \cdots & x_n \end{array} \right] \left[\begin{array}{c} y_1 \\ y_2 \\ \vdots \\ y_n \end{array} \right]. \] The above equality is an equality of vectors with two components. The top components of these vectors are equal: \[ (\beta_0 + \beta_1 x_1) + (\beta_0 + \beta_1 x_2) + \cdots + (\beta_0 + \beta_1 x_n) = y_1 + y_2 + \cdots + y_n. \] Therefore \[ n \beta_0 + \beta_1 (x_1+x_3 + \cdots + x_n) = y_1 + y_2 + \cdots + y_n. \] Dividing by $n$ we get \[ \beta_0 + \beta_1 \frac{1}{n} (x_1+x_3 + \cdots + x_n) = \frac{1}{n}( y_1 + y_2 + \cdots + y_n). \] Hence \[ \overline{y} = \beta_0 + \beta_1 \overline{x}. \] QED.
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Do the following problem: Consider the following four data points
\[
( 0, 0, 5), \ \ (3, 0, 6), \ \ (3, 3, 14), \ \ (0, 3, 9).
\]
- Find the equation $z = \beta_0 + \beta_1 x +\beta_2 y$ of the least-squares plane that best fits the data points.
- Find the coordinates of the dark green points and the teal points in the picture below.
- Calculate the residual vector and the least-squares error.
- Find the equation of the plane through the data points \[ ( 0, 0, 5), \ \ (3, 0, 6), \ \ (0, 3, 9). \] Show that the least-squares error is larger for this plane than the error for the least-squares plane.
In the image below the navy blue points are the given data points and the light blue plane is the least-squares plane that best fits these data points. The dark green points are their projections onto the $xy$-plane. The teal points are the corresponding points in the least-square plane.
- We discussed Section 6.5 today. Suggested problems for Section 6.5: 1, 3, 6, 7, 9, 11, 12, 13, 16, 17, 18, 19, 20, 21, 27, 28, 29, 30, 31, 32.
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Exercise 27 in Section 6.5 (6th edition) is the following theorem:
Theorem. Let $m$ and $n$ be positive integers and let $A$ be an $n\!\times\!m$ matrix. Then \[ \operatorname{Nul}(A) = \operatorname{Nul}\bigl(A^\top\!A\bigr). \]
Background Knowledge established in Math 204 is as follows.
BK1. Matrix multiplication is associative. Let $j, k, l, m$ be positive integers. Let $X$ be a $j\times k$ matrix, $Y$ be a $k \times l$ matrix, and $Z$ be an $l \times m$ matrix. Then $XY$ is $j\times l$ matrix, $YZ$ is $k\times m$ and $(XY)Z = X(YZ)$.
BK2. How to calculate the transpose of a product of two matrices. Let $j, k, l$ be positive integers. Let $X$ be a $j\times k$ matrix, and let $Y$ be a $k \times l$ matrix. Then $(XY)^\top = Y^\top\mkern -2mu X^\top$. A mnemonic tool to internalize this rule, is to write the sizes of all the matrices that are involved in this equality: $XY$ is a $j\times l$ matrix, $(XY)^\top$ is an $l \times j$ matrix, $X^\top$ is a $k\times j$ matrix, $Y^\top$ is an $l\times k$ matrix. We need a formula for $l \times j$ matrix $(XY)^\top$ using a $k\times j$ matrix $X^\top$ and a $l\times k$ matrix $Y^\top$. The only way to multiply $X^\top$ and $Y^\top$ is $Y^\top\mkern -2mu X^\top$ and in this way we get an $l \times j$ matrix. This is not a proof, just a mnemonic tool. For proof, one needs to compare the entries of both matrices in the equality.
BK3. Let $n$ be a positive integer. The only vector in $\mathbb{R}^n$ whose norm is zero is the zero vector. That is for $\mathbf{v} \in \mathbb{R}^n$ we have $\|\mathbf{v}\| = 0$ if and only if $\mathbf{v} = \mathbf{0}_n$.
Proof. The set equality $\operatorname{Nul}(A^\top\!\! A ) = \operatorname{Nul}(A)$ means \[ \mathbf{x} \in \operatorname{Nul}(A^\top\!\! A ) \quad \text{if and only if} \quad \mathbf{x} \in \operatorname{Nul}(A). \] As with all equivalences, we prove this equivalence in two steps.
Step 1. Assume that $\mathbf{x} \in \operatorname{Nul}(A)$. Then $A\mathbf{x} = \mathbf{0}$. Consequently, using BK1, we have \[ (A^\top\!A)\mathbf{x} = A^\top ( \!A\mathbf{x}) = A^\top\mathbf{0} = \mathbf{0}. \] Hence, $(A^\top\!A)\mathbf{x}= \mathbf{0}$, and therefore $\mathbf{x} \in \operatorname{Nul}(A^\top\!\! A )$. Thus, we proved the implication \[ \mathbf{x} \in \operatorname{Nul}(A) \quad \Rightarrow \quad \mathbf{x} \in \operatorname{Nul}(A^\top\!\! A ). \]
Step 2. In this step, we prove the the converse: \[ \tag{*} \mathbf{x} \in \operatorname{Nul}(A^\top\!\! A ) \quad \Rightarrow \quad \mathbf{x} \in \operatorname{Nul}(A). \] Assume, $\mathbf{x} \in \operatorname{Nul}(A^\top\!\! A )$. Then, $(A^\top\!\!A) \mathbf{x} = \mathbf{0}$. Multiplying the last equality by $\mathbf{x}^\top$ we get $\mathbf{x}^\top\! (A^\top\!\! A \mathbf{x}) = 0$. Using the associativity of the matrix multiplication, that is BK1, we obtain $(\mathbf{x}^\top\!\! A^\top) (A \mathbf{x}) = 0$. Using the Linear Algebra with the transpose operation, that is BK2, we get $(A \mathbf{x})^\top\! (A \mathbf{x}) = 0$. Now recall that for every vector $\mathbf{v}$ we have $\mathbf{v}^\top \mathbf{v} = \|\mathbf{v}\|^2$. Thus, we have proved that $\|A\mathbf{x}\|^2 = 0$. Now recall BK3, that the only vector whose norm is $0$ is the zero vector, to conclude that $A\mathbf{x} = \mathbf{0}_n$. This means $\mathbf{x} \in \operatorname{Nul}(A)$. This completes the proof of implication (*). The theorem is proved. QED.
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(It is customary to end mathematical proofs with the abbreviation QED, see its Wikipedia entry.)
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In Step 2 of the preceding proof, the idea introduced in the sentence which started with the highlighted text, is a truly brilliant idea. It is a pleasure to share these brilliant mathematical vignettes with you.
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The preceding theorem has an important corollary.
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Corollary 1 is stated in
Exercises 28 and 29 in Section 6.5
Corollary 1. Let $m$ and $n$ be positive integers and let $A$ be an $n\!\times\!m$ matrix. Then the columns of $A$ are linearly independent if and only if the matrix $A^\top\!A$ is invertible.
Background Knowledge established in Math 204 is as follows.
BK1. The columns of $A$ are linearly independent if and only if $\operatorname{Nul}(A)=\{\mathbf{0}_m\}$.
BK2. Notice that the matrix $A^\top\!\! A$ is a square $m\!\times\!m$ matrix. By The Invertible Matrix Theorem The matrix $A^\top\!\! A$ is invertible if and only if $\operatorname{Nul}(A^\top\!\! A)=\{\mathbf{0}_m\}$.
Proof. By BK1 the columns of $A$ are linearly independent if and only if $\operatorname{Nul}(A)=\{\mathbf{0}_m\}$. By the Theorem we have $\operatorname{Nul}(A) = \operatorname{Nul}(A^\top\!\! A )$. Therefore $\operatorname{Nul}(A)=\{\mathbf{0}_m\}$ if and only if $\operatorname{Nul}(A^\top\!\! A)=\{\mathbf{0}_m\}$. By BK2 $\operatorname{Nul}(A^\top\!\! A)=\{\mathbf{0}_m\}$ if and only if the matrix $A^\top\!\! A$.
Based on the three equivalences stated in the preceding paragraph, we have proved the corollary. QED.
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Corollary 2 is implicitly stated in Theorem 13 in Section 6.5.
Corollary 2. Let $A$ be an $n\!\times\!m$ matrix. Then $\operatorname{Col}(A^\top\!\! A ) = \operatorname{Col}(A^\top)$.
Proof 1. The following equalities we established earlier: \begin{align*} \operatorname{Col}(A^\top\!\! A ) & = \operatorname{Row}(A^\top\!\! A ) = \bigl( \operatorname{Nul}(A^\top\!\! A ) \bigr)^\perp, \\ \operatorname{Col}(A^\top) & = \operatorname{Row}(A) = \bigl( \operatorname{Nul}(A) \bigr)^\perp \end{align*} In the above Theorem we proved that the following subspaces are equal \[ \operatorname{Nul}(A^\top\!\! A ) = \operatorname{Nul}(A). \] Equal subspaces have equal orthogonal complements. Therefore \[ \bigl(\operatorname{Nul}(A^\top\!\! A )\bigr)^\perp = \bigl( \operatorname{Nul}(A) \bigr)^\perp. \] Since earlier we proved \[ \operatorname{Col}(A^\top\!\! A ) = \bigl( \operatorname{Nul}(A^\top\!\! A ) \bigr)^\perp \quad \text{and} \quad \operatorname{Col}(A^\top) = \bigl( \operatorname{Nul}(A) \bigr)^\perp, \] the last three equalities imply \[ \operatorname{Col}(A^\top\!\! A ) = \operatorname{Col}(A^\top). \] The corollary is proved. QED.
Proof 2. (This is a direct proof. It does not use the above Theorem. It uses the existence of an orthogonal projection onto the column space of $A$.) The set equality $\operatorname{Col}(A^\top\!\! A ) = \operatorname{Col}(A^\top)$ means \[ \mathbf{x} \in \operatorname{Col}(A^\top\!\! A ) \quad \text{if and only if} \quad \mathbf{x} \in \operatorname{Col}(A^\top). \] We will prove this equivalence in two steps.
Step 1. Assume that $\mathbf{x} \in \operatorname{Col}(A^\top\!\!A).$ Then there exists $\mathbf{v} \in \mathbb{R}^m$ such that $\mathbf{x} = (A^\top\!\!A)\mathbf{v}.$ Since by the definition of matrix multiplication we have $(A^\top\!\!A)\mathbf{v} = A^\top\!(A\mathbf{v})$, we have $\mathbf{x} = A^\top\!(A\mathbf{v}).$ Consequently, $\mathbf{x} \in \operatorname{Col}(A^\top).$ Thus, we proved the implication \[ \mathbf{x} \in \operatorname{Col}(A^\top\!\!A) \quad \Rightarrow \quad \mathbf{x} \in \operatorname{Col}(A^\top). \]
Step 2. Now we prove the converse: \[ \mathbf{x} \in \operatorname{Col}(A^\top) \quad \Rightarrow \quad \mathbf{x} \in \operatorname{Col}(A^\top\!\!A). \] Assume, $\mathbf{x} \in \operatorname{Col}(A^\top).$ By the definition of the column space of $A^\top$, there exists $\mathbf{y} \in \mathbb{R}^n$ such that $\mathbf{x} = A^\top\!\mathbf{y}.$ Let $\widehat{\mathbf{y}}$ be the orthogonal projection of $\mathbf{y}$ onto $\operatorname{Col}(A).$ That is $\widehat{\mathbf{y}} \in \operatorname{Col}(A)$ and $\mathbf{y} - \widehat{\mathbf{y}} \in \bigl(\operatorname{Col}(A)\bigr)^{\perp}.$ Since $\widehat{\mathbf{y}} \in \operatorname{Col}(A),$ there exists $\mathbf{v} \in \mathbb{R}^m$ such that $\widehat{\mathbf{y}} = A\mathbf{v}.$ Since $\bigl(\operatorname{Col}(A)\bigr)^{\perp} = \operatorname{Nul}(A^\top),$ the relationship $\mathbf{y} - \widehat{\mathbf{y}} \in \bigl(\operatorname{Col}(A)\bigr)^{\perp}$ yields $A^\top\bigl(\mathbf{y} - \widehat{\mathbf{y}}\bigr) = \mathbf{0}.$ Consequently, since $\widehat{\mathbf{y}} = A\mathbf{v},$ we deduce $A^\top\bigl(\mathbf{y} - A\mathbf{v}\bigr) = \mathbf{0}.$ Hence \[ \mathbf{x} = A^\top\mathbf{y} = \bigl(A^\top\!\!A\bigr) \mathbf{v} \quad \text{with} \quad \mathbf{v} \in \mathbb{R}^m. \] This proves that $\mathbf{x} \in \operatorname{Col}(A^\top\!\!A).$ Thus, the implication \[ \mathbf{x} \in \operatorname{Col}(A^\top) \quad \Rightarrow \quad \mathbf{x} \in \operatorname{Col}(A^\top\!\!A) \] is proved. The corollary is proved. QED.
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Corollary 3 item is stated in Exercise 30 in Section 6.5. Corollary 3 is a consequence of Corollary 2. The hint given in Exercise 30 will result in a different proof of Corollary 3.
Corollary 3. Let $A$ be an $n\!\times\!m$ matrix. The matrices $A$, $A^\top$, $A^\top\!\! A$ and $A A^\top$ have the same rank.
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A few years ago, a student's insightful question prompted me to explore the connection between the $QR$-factorization of an $n\times m$ matrix $A$ and the $m$-dimensional volume of the parallelepiped spanned by the columns of $A$ in the Euclidean space $\mathbb{R}^n$. While I don't remember the question's exact wording, I provide a paraphrased version below.
Question. Given two vectors in $\mathbb{R}^3$ we learned how to calculate the area of the parallelogram spanned by these two vectors. In a multivariable calculus class we used the cross product for this calculation. Given three vectors in $\mathbb{R}^3$ we learned how to calculate the volume of the parallelepiped spanned by these three vectors. We used the determinant of the $3\times 3$ matrix whose columns are the given three vectors. If we are given three vectors in $\mathbb{R}^4$, is there a way to calculate the volume of the three-dimensional parallelepiped spanned by these three vectors?
- My answer to the above question is in the items below. It took me a while to come up with this answer. Later, I will add a picture that I used in class to motivate the reasoning below.
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Let us find the $QR$-factorization of the following matrix:
\[
A = \left[
\begin{array}{rrr}
4 & -1 & -7 \\
2 & 8 & 7 \\
2 & 4 & -8 \\
1 & 5 & 5 \\
\end{array}
\right] = \bigl[ \mathbf{a}_1 \ \ \mathbf{a}_2 \ \ \mathbf{a}_3 \bigr].
\]
The goal here is to use the $QR$-factorization to calculate the volume of the $3$-dimensional parallelepiped spanned in $\mathbb{R}^4$ by the vectors $\mathbf{a}_1$, $\mathbf{a}_2$, and $\mathbf{a}_3$. Notice that
\[
\| \mathbf{a}_1 \| = 5, \quad \|\mathbf{a}_2\| = \sqrt{106}, \quad \|\mathbf{a}_3\| = \sqrt{187}.
\]
Thus, the volume we seek is certainly smaller or equal to the product of these norms:
\[
5 \sqrt{19822} \approx 703.953.
\]
- We first perform the Gram-Schmidt orthogonalization algorithm on the vectors $\mathbf{a}_1$, $\mathbf{a}_2$, and $\mathbf{a}_3$. \begin{align*} \mathbf{v}_1 & = \mathbf{a}_1 = \left[\!\begin{array}{r} 4 \\ 2 \\ 2 \\ 1 \\ \end{array}\! \right] \\ \mathbf{v}_2 & = \mathbf{a}_2 - \left(\frac{\mathbf{a}_2\cdot \mathbf{v}_1}{\mathbf{v}_1\cdot \mathbf{v}_1}\right) \mathbf{v}_1 = \left[\!\begin{array}{r} -1 \\ 8 \\ 4 \\ 5 \\ \end{array}\! \right] - \frac{25}{25} \left[\!\begin{array}{r} 4 \\ 2 \\ 2 \\ 1 \\ \end{array}\! \right] = \left[\!\begin{array}{r} -5 \\ 6 \\ 2 \\ 4 \\ \end{array}\! \right] \\ \mathbf{v}_3 & = \mathbf{a}_3 - \left(\frac{\mathbf{a}_3\cdot \mathbf{v}_1}{\mathbf{v}_1\cdot \mathbf{v}_1}\right) \mathbf{v}_1 - \left(\frac{\mathbf{a}_3\cdot \mathbf{v}_2}{\mathbf{v}_2\cdot \mathbf{v}_2}\right) \mathbf{v}_2 \\ & = \left[\!\begin{array}{r} -7 \\ 7 \\ -8 \\ 5 \\ \end{array}\! \right] - \frac{-25}{25} \left[\!\begin{array}{r} 4 \\ 2 \\ 2 \\ 1 \\ \end{array}\! \right] - \frac{81}{81} \left[\!\begin{array}{r} -5 \\ 6 \\ 2 \\ 4 \\ \end{array}\! \right] = \left[\!\begin{array}{r} 2 \\ 3 \\ -8 \\ 2 \\ \end{array}\! \right] \end{align*}
- Since the vectors $\mathbf{v}_1$, $\mathbf{v}_2$, and $\mathbf{v}_3$ form an orthogonal set of vectors, the volume of the cuboid spanned by the vectors $\mathbf{v}_1$, $\mathbf{v}_2$, and $\mathbf{v}_3$ is \[ \| \mathbf{v}_1 \| \times \| \mathbf{v}_2 \| \times \| \mathbf{v}_3 \| = 5\times 9\times 9 = 405. \] Since this cuboid is a shear transformation of the parallelepiped spanned by the vectors $\mathbf{a}_1$, $\mathbf{a}_2$, and $\mathbf{a}_3$, the volume of the parallelepiped is also $405$. So, the Gram-Schmidt orthogonalization gives the volume that we are asked to calculate. The volume is the product of the lengths of the vectors $\mathbf{v}_1$, $\mathbf{v}_2$, and $\mathbf{v}_3$.
- To get the $QR$-factorization of $A$, we normalize the vectors $\mathbf{v}_1$, $\mathbf{v}_2$, and $\mathbf{v}_3$: \[ \mathbf{u}_1 = \frac{1}{5}\left[\!\begin{array}{r} 4 \\ 2 \\ 2 \\ 1 \\ \end{array}\! \right], \quad \mathbf{u}_2 = \frac{1}{9}\left[\!\begin{array}{r} -5 \\ 6 \\ 2 \\ 4 \\ \end{array}\! \right], \quad \mathbf{u}_3 = \frac{1}{9} \left[\!\begin{array}{r} 2 \\ 3 \\ -8 \\ 2 \\ \end{array}\! \right]. \]
- Next, we can write the $QR$-factorization of $A$: \begin{align*} Q & = \left[ \begin{array}{rrr} \frac{4}{5} & -\frac{5}{9} & \frac{2}{9} \\ \frac{2}{5} & \frac{2}{3} & \frac{1}{3} \\ \frac{2}{5} & \frac{2}{9} & -\frac{8}{9} \\ \frac{1}{5} & \frac{4}{9} & \frac{2}{9} \\ \end{array} \right], \\ R & = Q^\top A = \left[ \begin{array}{cccc} \frac{4}{5} & \frac{2}{5} & \frac{2}{5} & \frac{1}{5} \\ -\frac{5}{9} & \frac{2}{3} & \frac{2}{9} & \frac{4}{9} \\ \frac{2}{9} & \frac{1}{3} & -\frac{8}{9} & \frac{2}{9} \\ \end{array} \right]\left[ \begin{array}{rrr} 4 & -1 & -7 \\ 2 & 8 & 7 \\ 2 & 4 & -8 \\ 1 & 5 & 5 \\ \end{array} \right] = \left[ \begin{array}{ccc} 5 & 5 & -5 \\ 0 & 9 & 9 \\ 0 & 0 & 9 \\ \end{array} \right]. \end{align*} It is important to observe that $R$ can also be read from the Gram-Schmidt orthogonalization. The columns of $R$ are the coefficients of the columns of $A$ as written as linear combinations of the columns of $\mathbf{u}_1$, $\mathbf{u}_2$, and $\mathbf{u}_3$. Importantly, the entries on the diagonal of $R$ are the lengths of the vectors $\mathbf{v}_1$, $\mathbf{v}_2$, and $\mathbf{v}_3$.
- Thus, the $QR$-factorization of $A$ is \[ \left[\begin{array}{rrr} 4 & -1 & -7 \\ 2 & 8 & 7 \\ 2 & 4 & -8 \\ 1 & 5 & 5 \\ \end{array} \right] = \left[ \begin{array}{rrr} \frac{4}{5} & -\frac{5}{9} & \frac{2}{9} \\ \frac{2}{5} & \frac{2}{3} & \frac{1}{3} \\ \frac{2}{5} & \frac{2}{9} & -\frac{8}{9} \\ \frac{1}{5} & \frac{4}{9} & \frac{2}{9} \\ \end{array} \right] \left[ \begin{array}{ccc} 5 & 5 & -5 \\ 0 & 9 & 9 \\ 0 & 0 & 9 \\ \end{array} \right]. \]
- The volume of the $3$-dimensional parallelepiped spanned in $\mathbb{R}^4$ by the vectors $\mathbf{a}_1$, $\mathbf{a}_2$, and $\mathbf{a}_3$ is \[ \det(R) = \left| \begin{array}{ccc} 5 & 5 & -5 \\ 0 & 9 & 9 \\ 0 & 0 & 9 \\ \end{array} \right| = 405. \]
- A natural question occurs: Is it possible to calculate this volume directly from the matrix $A$ (for a general $n\times m$ matrix $A$ with linearly independent columns) without doing the Gram-Schmidt orthogonalization? The answer is yes, and it follows from the calculations below: \begin{align*} A & = QR \\ A^\top A &= R^\top Q^\top \\ A^\top A &= R^\top Q^\top QR = R^\top I_m R = R^\top R \\ \det(R^\top) & = \det(R) \\ \det(A^\top A) & = \det(R^\top R) = \det(R^\top) \det(R) =\bigl(\det R\bigr)^2. \end{align*} Consequently \[ \det(R) = \sqrt{\det(A^\top A)}. \] For our specific matrix $A$: \[ A^\top A = \left[ \begin{array}{rrrr} 4 & 2 & 2 & 1 \\ -1 & 8 & 4 & 5 \\ -7 & 7 & -8 & 5 \\ \end{array} \right] \left[ \begin{array}{rrr} 4 & -1 & -7 \\ 2 & 8 & 7 \\ 2 & 4 & -8 \\ 1 & 5 & 5 \\ \end{array}\right] = \left[ \begin{array}{rrr} 25 & 25 & -25 \\ 25 & 106 & 56 \\ -25 & 56 & 187 \\ \end{array} \right] \] and \begin{align*} \left| \begin{array}{rrr} 25 & 25 & -25 \\ 25 & 106 & 56 \\ -25 & 56 & 187 \\ \end{array} \right| &=\left| \begin{array}{rrr} 25 & 25 & -25 \\ 0 & 81 & 81 \\ 0 & 81 & 162 \\ \end{array} \right| \\ & = \left| \begin{array}{rrr} 25 & 25 & -25 \\ 0 & 81 & 81 \\ 0 & 0 & 81 \\ \end{array} \right| \\ & = 25\times 81 \times 81 \\ & = (5\times 9 \times 9)^2 \\ &= 405^2. \end{align*}
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We are studying three vectors in \(\mathbb{R}^4\), but we have captured these three vectors in special three orthonormal vectors which in the picture below are colored purple and plotted thick. These vectors play role of the vectors \(\left[\begin{array}{c} 1\\ 0\\ 0\end{array}\right]\), \(\left[\begin{array}{c} 0\\ 1\\ 0\end{array}\right]\), \(\left[\begin{array}{c} 0\\ 0\\ 1\end{array}\right]\). So, although what is given is in \(\mathbb{R}^4\), we can pretend that we are in \(\mathbb{R}^3\) and draw the picture below to visualize the given vectors with their coordinates relative to the basis
\[
\color{purple}{\mathcal{U}}
= \bigl\{\color{purple}{\mathbf{u}_1}, \color{purple}{\mathbf{u}_2}, \color{purple}{\mathbf{u}_3}\bigr\}.
\]
From the matrix \(R\) calculated above we have
\[
\bigl[ \color{green}{\mathbf{a}_1} \bigr]_{\color{purple}{\mathcal{U}}} =
\left[ \begin{array}{c} 5 \\ 0 \\ 0 \end{array} \right], \qquad
\bigl[ \color{green}{\mathbf{a}_2} \bigr]_{\color{purple}{\mathcal{U}}} =
\left[ \begin{array}{c} 5 \\ 9 \\ 0 \end{array} \right], \qquad
\bigl[ \color{green}{\mathbf{a}_3} \bigr]_{\color{purple}{\mathcal{U}}} =
\left[ \begin{array}{c} -5 \\ 9 \\ 9 \end{array} \right].
\]
To make the pictures nicer, I had to shorten the vectors \(\color{green}{\mathbf{a}_1}\), \(\color{green}{\mathbf{a}_2}\), and \(\color{green}{\mathbf{a}_3}\). In the pictures below I present
\[
\left[ \tfrac{3}{8} \color{green}{\mathbf{a}_1} \right]_{\color{purple}{\mathcal{U}}} =
\frac{3}{8} \left[ \begin{array}{c} 5 \\ 0 \\ 0 \end{array} \right], \qquad
\left[ \tfrac{3}{8} \color{green}{\mathbf{a}_2} \right]_{\color{purple}{\mathcal{U}}} =
\frac{3}{8}\left[ \begin{array}{c} 5 \\ 9 \\ 0 \end{array} \right], \qquad
\left[ \tfrac{3}{8} \color{green}{\mathbf{a}_3} \right]_{\color{purple}{\mathcal{U}}} =
\frac{3}{8} \left[ \begin{array}{c} -5 \\ 9 \\ 9 \end{array} \right].
\]
The point of the post today was to calculate the volume of the parallelepiped spanned by the vectors \(\color{green}{\mathbf{a}_1}\), \(\color{green}{\mathbf{a}_2}\), and \(\color{green}{\mathbf{a}_3}\). In the picture below that parallelepiped is colored green.
The base of the green parallelepiped is a parallelogram. The volume of the parallelepiped will not change if I transform the base into the parallelogram with the same area. In the picture below I show in purple the parallelepiped with a rectangular base spanned by two orthogonal vectors \(\color{green}{\mathbf{a}_1}\) and the projection of \(\color{green}{\mathbf{a}_2}\) onto \(\color{purple}{\mathbf{u}_2}\). This rectangle has the same area as the parallelogram which is the base of the green parallelepiped. Therefore the purple parallelepiped with the rectangular base has the same volume as the green parallelepiped.
Finally, in the picture below, I show the rectangular prism (I like to call such a shape, cuboid) which has the same base as the parallelepiped in the picture immediately above, and the same height as the green parallelepiped. Therefore, the cuboid pictured below has the sam volume as the green parallelepiped that we started with. When we are multiplying the diagonal entries of the matrix \(R\), we are calculating the volume of the cuboid below.
Multiplying the diagonal entries of the matrix \(R\) seems like a simple rule how to calculate the volume. But, I wanted to show the nice geometry behind that multiplication. That is why I made these picutes. I will try to make even better ones in time.
- The topic today was Section 6.4: Gram-Schmidt Orthogonalization Algorithm. I made up this name. It seems that it is an appropriate modern name. But, I think that mathematically the most fitting name would be: Gram-Schmidt Orthogonalization Recursion. The Gram-Schmidt Orthogonalization is a recursive formula. The formula for the next vector is given in terms of the previously calculated vectors. Suggested problems for Section 6.4: 2, 3, 5, 7, 9, 13, 15, 17, 19, 20
- The presentation of the $QR$ factorization in the textbook somewhat obscures the direct connection between the Gram-Schmidt orthogonalization algorithm and the $QR$ factorization. Below I will demonstrate the connection.
-
Let $m, n \in \mathbb{N}$. Let $\mathbf{a}_1, \mathbf{a}_2, \ldots, \mathbf{a}_m$ be linearly independent vectors in $\mathbb{R}^n$.
- The Gram-Schmidt orthogonalization algorithm produces the mutually orthogonal vectors $\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_m$ defined as follows: \begin{align*} \mathbf{v}_1 & = \mathbf{a}_1 \\ \mathbf{v}_2 & = \mathbf{a}_2 - \left(\frac{\mathbf{a}_2\cdot \mathbf{v}_1}{\mathbf{v}_1 \cdot \mathbf{v}_1}\right) \mathbf{v}_1 \\ \mathbf{v}_3 & = \mathbf{a}_3 - \left(\frac{\mathbf{a}_3\cdot \mathbf{v}_1}{\mathbf{v}_1 \cdot \mathbf{v}_1}\right) \mathbf{v}_1 - \left(\frac{\mathbf{a}_3\cdot \mathbf{v}_2}{\mathbf{v}_2 \cdot \mathbf{v}_2}\right) \mathbf{v}_2 \\ & \ \ \vdots \\ \mathbf{v}_m & = \mathbf{a}_m - \left(\frac{\mathbf{a}_m\cdot \mathbf{v}_1}{\mathbf{v}_1 \cdot \mathbf{v}_1} \right) \mathbf{v}_1 - \cdots - \left(\frac{\mathbf{a}_m\cdot \mathbf{v}_{m-1}}{\mathbf{v}_{m-1} \cdot \mathbf{v}_{m-1}} \right) \mathbf{v}_{m-1} \\ \end{align*}
- We can rewrite the above vector equations as \begin{align*} \mathbf{a}_1 & = \mathbf{v}_1 \\ \mathbf{a}_2 & = \left(\frac{\mathbf{a}_2\cdot \mathbf{v}_{1}}{\mathbf{v}_{1} \cdot \mathbf{v}_{1}} \right) \mathbf{v}_1 + \mathbf{v}_2 \\ \mathbf{a}_3 & = \left( \frac{\mathbf{a}_3\cdot \mathbf{v}_{1}}{\mathbf{v}_{1} \cdot \mathbf{v}_{1}}\right) \mathbf{v}_1 + \left( \frac{\mathbf{a}_3\cdot \mathbf{v}_{2}}{\mathbf{v}_{2} \cdot \mathbf{v}_{2}} \right) \mathbf{v}_2 + \mathbf{v}_3 \\ & \ \ \vdots \\ \mathbf{a}_m & = \left( \frac{\mathbf{a}_m\cdot \mathbf{v}_{1}}{\mathbf{v}_{1} \cdot \mathbf{v}_{1}} \right) \mathbf{v}_1 + \cdots + \left( \frac{\mathbf{a}_m\cdot \mathbf{v}_{m-1}}{\mathbf{v}_{m-1} \cdot \mathbf{v}_{m-1}} \right) \mathbf{v}_{m-1} + \mathbf{v}_m \\ \end{align*}
- Now we normalize the vectors $\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_m$. That is, we introduce the unit vectors $\mathbf{u}_1, \mathbf{u}_2, \ldots, \mathbf{u}_m$ defined as follows: \[ \mathbf{u}_k = \frac{1}{\|\mathbf{v}_k\|} \mathbf{v}_k \quad \text{for all} \quad k \in \{1,\ldots,m\}. \] We use the fact that $\mathbf{v}_k \cdot \mathbf{v}_k = \|\mathbf{v}_k\|^2$ to rewrite the formulas for the vectors $\mathbf{a}_1,\dots, \mathbf{a}_m$ in terms of the orthonormal vectors $\mathbf{u}_1,\ldots,\mathbf{u}_m$ as follows: \begin{align*} \mathbf{a}_1 & = \|\mathbf{v}_1\| \, \mathbf{u}_1 \\ \mathbf{a}_2 & = \left( \mathbf{a}_2\cdot \mathbf{u}_{1} \right) \mathbf{u}_1 + \|\mathbf{v}_2\| \, \mathbf{u}_2 \\ \mathbf{a}_3 & = \left( \mathbf{a}_3\cdot \mathbf{u}_{1} \right) \mathbf{u}_1 + \left( \mathbf{a}_3\cdot \mathbf{u}_{2} \right) \mathbf{u}_2 + \|\mathbf{v}_3\| \, \mathbf{u}_3 \\ & \ \ \vdots \\ \mathbf{a}_m & = \left( \mathbf{a}_m\cdot \mathbf{u}_{1} \right) \mathbf{u}_1 + \cdots + \left( \mathbf{a}_m\cdot \mathbf{u}_{m-1} \right) \mathbf{u}_{m-1} + \|\mathbf{v}_m\| \, \mathbf{u}_m \end{align*}
- Now set \[ \alpha_{j,k} = \mathbf{a}_k\cdot \mathbf{u}_{j} \quad \text{for} \quad j \in \{1,\ldots,k-1\}, \ \ k \in \{2,\ldots,m\} \] and the above equations can be rewritten as \begin{align*} \mathbf{a}_1 & = \|\mathbf{v}_1\| \, \mathbf{u}_1 \\ \mathbf{a}_2 & = \alpha_{1,2} \, \mathbf{u}_1 + \|\mathbf{v}_2\| \, \mathbf{u}_2 \\ \mathbf{a}_3 & = \alpha_{1,3} \, \mathbf{u}_1 + \alpha_{2,3} \, \mathbf{u}_2 + \|\mathbf{v}_3\| \, \mathbf{u}_3 \\ & \ \ \vdots \\ \mathbf{a}_m & = \alpha_{1,m} \, \mathbf{u}_1 + \cdots + \alpha_{m-1,m} \, \mathbf{u}_{m-1} + \|\mathbf{v}_m\| \, \mathbf{u}_m \\ \end{align*}
- The preceding vector equations can be written in matrix form as follows \[ \Bigl[\begin{array}{ccccc} \mathbf{a}_1 & \mathbf{a}_2 & \mathbf{a}_3 & \cdots & \mathbf{a}_m \end{array} \Bigr] = \Bigl[\begin{array}{ccccc} \mathbf{u}_1 & \mathbf{u}_2 & \mathbf{u}_3 & \cdots & \mathbf{u}_m \end{array} \Bigr] \left[\begin{array}{ccccc} \|\mathbf{v}_1\| & \alpha_{1,2} & \alpha_{1,3} & \cdots & \alpha_{1,m} \\ 0 & \|\mathbf{v}_2\| & \alpha_{2,3} & \cdots & \alpha_{2,m} \\ 0 & 0 & \|\mathbf{v}_3\| & \cdots & \alpha_{3,m} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & \|\mathbf{v}_m\| \\ \end{array} \right] \]
- The preceding matrix equation is the $QR$ factorization of $A$: \[ A = QR \] with \begin{equation*} A = \left[\begin{array}{ccccc} \mathbf{a}_1 & \mathbf{a}_2 & \mathbf{a}_3 & \cdots & \mathbf{a}_m \end{array} \right] \quad \text{and} \quad Q = \left[\begin{array}{ccccc} \mathbf{u}_1 & \mathbf{u}_2 & \mathbf{u}_3 & \cdots & \mathbf{u}_m \end{array} \right] \end{equation*} and the matrix $R$ is an upper triangular matrix with positive terms on the diagonal. Since the vectors $\mathbf{u}_1, \mathbf{u}_2,\ldots, \mathbf{u}_m$ are orthonormal, we have $Q^{\top} Q = I_m$. Therefore the upper triangular $m\!\times\!m$ matrix $R$ can be calculated as \[ R = Q^{\top} A. \]
-
Next I will state the $QR$ factorization of a matrix with linearly independent columns as a theorem.
Theorem. Every $n\times m$ matrix $A$ with linearly independent columns can be written as a product $A = QR$ where $Q$ is an $n\times m$ matrix whose columns form an orthonormal basis for the column space of $A$ and $R$ is an $m\times m$ upper triangular invertible matrix with positive entries on its diagonal.
-
The $QR$ factorization of a matrix is just the Gram-Schmidt orthogonalization algorithm for the columns of $A$ written in matrix form. The only difference is that the Gram-Schmidt orthogonalization algorithm produces orthogonal vectors which we have to normalize to obtain the matrix $ Q$ with orthonormal columns.
-
Example. A nice example is given by calculating $QR$ factorization of the $3\!\times\!2$ matrix \[ A = \left[ \begin{array}{rr} 1 & 1 \\[2pt] 2 & 4 \\[2pt] 2 & 3 \end{array}\right]. \]
- Denote the columns of $A$ by $\mathbf{a}_1$ and $\mathbf{a}_2$.
- The Gram-Schmidt orthogonalization of the vectors $\mathbf{a}_1$ and $\mathbf{a}_2$ leads to vectors \[ \mathbf{v}_1 = \left[ \begin{array}{r} 1 \\[2pt] 2 \\[2pt] 2 \end{array}\right], \quad \mathbf{v}_2 = \left[ \begin{array}{r} -2/3 \\[2pt] 2/3 \\[2pt] 1/3 \end{array}\right]. \] These vectors are calculated as \[ \tag{*} \mathbf{v}_1 = \mathbf{a}_1, \quad \mathbf{v}_2 = \mathbf{a}_2 - \left(\frac{\mathbf{a}_2 \cdot \mathbf{v}_1}{\mathbf{v}_1 \cdot \mathbf{v}_1} \right) \mathbf{v}_1 = \mathbf{a}_2 - \frac{5}{3} \mathbf{v}_1. \]
- Next we normalize the vectors $\mathbf{v}_1$ and $\mathbf{v}_2$. The norm of vector $\mathbf{v}_1$ is $3$ and the norm of $\mathbf{v}_2$ is 1. Hence the following vectors are orthonormal: \[ \mathbf{u}_1 = \frac{1}{3} \left[ \begin{array}{r} 1 \\[2pt] 2 \\[2pt] 2 \end{array}\right], \quad \mathbf{u}_2 = \frac{1}{3} \left[ \begin{array}{r} -2 \\[2pt] 2 \\[2pt] 1 \end{array}\right]. \]
- We can rewrite equalities (*) using the vectors $\mathbf{u}_1$ and $\mathbf{u}_2$ as follows \[ \tag{**} \mathbf{a}_1 = 3\, \mathbf{u}_1 = 3\, \mathbf{u}_1 + 0\, \mathbf{u}_2, \quad \mathbf{a}_2 = \frac{5}{3} \, 3\, \mathbf{u}_1 + \mathbf{u}_2 = 5\,\mathbf{u}_1 + \mathbf{u}_2. \] In the matrix form the equalities (**) can be written as \[ \left[ \begin{array}{rr} 1 & 1 \\[2pt] 2 & 4 \\[2pt] 2 & 3 \end{array}\right] = \frac{1}{3} \left[ \begin{array}{rr} 1 & -2 \\[2pt] 2 & 2 \\[2pt] 2 & 1 \end{array}\right] \left[ \begin{array}{rr} 3 & 5 \\[2pt] 0 & 1 \end{array}\right]. \] where \[ Q = \frac{1}{3} \left[ \begin{array}{rr} 1 & -2 \\[2pt] 2 & 2 \\[2pt] 2 & 1 \end{array}\right] \] is a matrix with orthonormal columns and its column space is identical to the columns space of the matrix $A$. Here \[ R = \left[ \begin{array}{rr} 3 & 5 \\[2pt] 0 & 1 \end{array}\right]. \]
- Notice that on the diagonal of the matrix $R$ are the norms of the vectors $\mathbf{v}_1$ and $\mathbf{v}_2$ which we obtained by the Gram-Schmidt orthogonalization algorithm. Since the matrix $Q$ has orthonormal columns we have $Q^\top Q = I_2$. Therefore the matrix $R$ can be calculated as \[ R = Q^\top A. \] That is \[ \left[ \begin{array}{rr} 3 & 5 \\[2pt] 0 & 1 \end{array}\right] = \frac{1}{3} \left[ \begin{array}{rrr} 1 & 2 & 2 \\[2pt] -2 & 2 & 1 \end{array}\right] \left[ \begin{array}{rr} 1 & 1 \\[2pt] 2 & 4 \\[2pt] 2 & 3 \end{array}\right]. \] The last calculation of $R$ might be simpler than making adjustments to the coefficients of the Gram-Schmidt orthogonalization algorithm as we did in this above example. However, it is good to know that $R$ is closely related to the coefficients that appear in the Gram-Schmidt orthogonalization algorithm.
-
In the next example, I will demonstrate a useful simplification strategy when calculating the vectors $\mathbf{v}_1$, $\mathbf{v}_2$, $\mathbf{v}_3$, and so on. Following the given formulas, calculation the vectors $\mathbf{v}$s will frequently involve fractions and make the arithmetic of the subsequent calculations more difficult. Recall, the objective here is to produce orthogonal set of vectors keeping the running spans equal.
To simplify the arithmetic, at each step of the Gram-Schmidt algorithm, we can replace a vector $\mathbf{v}_k$ by its scaled version $\alpha \mathbf{v}_k$ with a conveniently chosen $\alpha \gt 0$.
In this way we can avoid fractions in vectors $\mathbf{v}_1$, $\mathbf{v}_2$, $\mathbf{v}_3.$ In the next item I present an example.
-
Example.
Calculate the $QR$-factorization of the matrix:
\[
\left[\!
\begin{array}{ccc}
6 & 6 & 1 \\
3 & 6 & 1 \\
2 & 1 & 1
\end{array}\!
\right].
\]
- We first apply the Gram-Schmidt algorithm to the vectors \[ \left[\! \begin{array}{c} 6 \\ 3 \\ 2 \end{array}\! \right], \quad \left[\! \begin{array}{c} 6 \\ 6 \\ 1 \end{array}\! \right], \quad \left[\! \begin{array}{c} 1 \\ 1 \\ 1 \end{array}\! \right]. \]
- We first calculate \begin{align*} \mathbf{v}_1 & = \left[\! \begin{array}{c} 6 \\ 3 \\ 2 \end{array}\!\right], \\ \mathbf{v}_2 & = \left[\! \begin{array}{c} 6 \\ 6 \\ 1 \end{array}\!\right] - \frac{8}{7} \left[\!\begin{array}{c} 6 \\ 3 \\ 2 \end{array}\!\right] = \left[\!\begin{array}{c} -6/7 \\ 18/7 \\ -9/7 \end{array}\!\right] = \frac{3}{7} \left[\!\begin{array}{r} -2 \\ 6 \\ -3 \end{array}\!\right], \quad \\ & \phantom{\hspace{1cm}} \text{continue with the scaled version} \ \mathbf{v}_2 = \left[\!\begin{array}{r} -2 \\ 6 \\ -3 \end{array}\!\right] \\ \mathbf{v}_3 & = \left[\!\begin{array}{c} 1\\ 1\\ 1 \end{array}\!\right] - \frac{11}{49}\left[\! \begin{array}{c} 6 \\ 3 \\ 2 \end{array}\!\right] - \frac{1}{49} \left[\!\begin{array}{r} -2 \\ 6 \\ -3 \end{array}\!\right] = \frac{1}{49} \left[\!\begin{array}{r} 49 - 66 + 2 \\ 49 - 33 - 6 \\ 49 - 22 + 3 \end{array}\!\right] = \frac{5}{49} \left[\!\begin{array}{r} -3 \\ 2 \\ 6 \end{array}\!\right], \\ & \phantom{\hspace{4cm}} \text{continue with the scaled version} \ \mathbf{v}_3 = \left[\!\begin{array}{r} -3 \\ 2 \\ 6 \end{array}\!\right]. \end{align*}
- Next we normalize the vectors $\mathbf{v}_1$, $\mathbf{v}_2$, $\mathbf{v}_3$: \[ \frac{1}{7} \left[\! \begin{array}{r} 6 \\ 3 \\ 2 \end{array}\! \right], \quad \frac{1}{7} \left[\! \begin{array}{r} -2 \\ 6 \\ -3 \end{array}\! \right], \quad \frac{1}{7} \left[\! \begin{array}{r} -3 \\ 2 \\ 6 \end{array}\! \right]. \] The preceding unit vectors are the columns of $Q$.
- Next we calculate $R = Q^\top A$: \[ \frac{1}{7} \left[\! \begin{array}{rrr} 6 & 3 & 2 \\ -2 & 6 & -3 \\ -3 & 2 & 6 \end{array}\! \right]\left[\! \begin{array}{ccc} 6 & 6 & 1 \\ 3 & 6 & 1 \\ 2 & 1 & 1 \end{array}\! \right] = \left[\! \begin{array}{rrr} 7 & 8 & 11/7 \\ 0 & 3 & 1/7 \\ 0 & 0 & 5/7 \end{array}\! \right] \]
- Thus \[ \left[\! \begin{array}{ccc} 6 & 6 & 1 \\ 3 & 6 & 1 \\ 2 & 1 & 1 \end{array}\! \right] = \frac{1}{7} \left[\! \begin{array}{rrr} 6 & -2 & -3 \\ 3 & 6 & 2 \\ 2 & -3 & 6 \end{array}\! \right] \left[\! \begin{array}{rrr} 7 & 8 & 11/7 \\ 0 & 3 & 1/7 \\ 0 & 0 & 5/7 \end{array}\! \right] \]
- The matrix \[ \frac{1}{7} \left[\! \begin{array}{rrr} 6 & -2 & -3 \\ 3 & 6 & 2 \\ 2 & -3 & 6 \end{array}\! \right] = \left[\! \begin{array}{rrr} 6/7 & -2/7 & -3/7 \\ 3/7 & 6/7 & 2/7 \\ 2/7 & -3/7 & 6/7 \end{array}\! \right] \] is a $3\!\times\!3$ matrix with orthonormal columns. A square matrix with orthonormal columns is called an orthogonal matrix. An orthogonal matrix is very special since its inverse equals its transpose. Verify that for the above matrix $Q$: \[ \left(\frac{1}{7} \left[\! \begin{array}{rrr} 6 & 3 & 2 \\ -2 & 6 & -3 \\ -3 & 2 & 6 \end{array}\! \right]\right) \left(\frac{1}{7} \left[\! \begin{array}{rrr} 6 & -2 & -3 \\ 3 & 6 & 2 \\ 2 & -3 & 6 \end{array}\! \right]\right) = \frac{1}{49} \left[\! \begin{array}{rrr} 49 & 0 & 0 \\ 0 & 49 & 0 \\ 0 & 0 & 49 \end{array}\! \right] \]
- For practice, find $QR$ factorizations of the following matrices \[ \left[ \begin{array}{ccc} -1 & -1 & 3 \\ 1 & 5 & -1 \\ 1 & 1 & 3 \\ -1 & -5 & 7 \end{array} \right] \quad \left[ \begin{array}{ccc} 6 & 8 & 7 \\ 3 & 6 & 0 \\ 2 & 2 & 0 \end{array} \right] \quad \left[ \begin{array}{ccc} 2 & 2 & 1 \\ 1 & 2 & 8 \\ 2 & 3 & 1 \end{array} \right] \quad \left[ \begin{array}{ccc} 4 & -1 & -7 \\ 2 & 8 & 7 \\ 2 & 4 & -8 \\ 1 & 5 & 5 \end{array} \right] \] \[ \left[ \begin{array}{ccc} 2 & -6 & 4 \\ -5 & 9 & 1 \\ 4 & 4 & 9 \\ 2 & -4 & 5 \end{array} \right] \]
- Today I reviewed the content that I posted yesterday. The concepts and theorems introduced in Section 6.3 are important, and I think accessible, so I want to help you to understand them. Similar, or better, almost identical concepts and theorems appear in different branches of Mathematics. Pay attention to The Orthogonal Decomposition Theorem (Theorem 8) and The Best Approximation Theorem (Theorem 9). and Theorem 10 which I will call The Standard Matrix of an Orthogonal Projection.
- We discussed Section 6.3 today. Suggested problems for Section 6.3 are: 1, 2, 4, 5, 7, 10, 11, 13, 15, 16 17, 19, 20, 21, 23.
-
The main concept in this section is the following definition:
Definition. Let $\mathcal{V}$ be a subspace of $\mathbb{R}^n$ and let $\mathbf{y} \in \mathbb{R}^n.$ The vector $\widehat{\mathbf{y}} \in \mathbb{R}^n$ is called the orthogonal projection of $\mathbf{y}$ onto $\mathcal{V}$ if the vector $\widehat{\mathbf{y}}$ has the following two properties:
- ① $\widehat{\mathbf{y}} \in \mathcal{V},$
- ② $\mathbf{y} - \widehat{\mathbf{y}} \in \mathcal{V}^{\perp}.$
The notation for the orthogonal projection $\widehat{\mathbf{y}}$ of $\mathbf{y} \in \mathbb{R}^n$ onto $\mathcal{V}$ is: \[ \widehat{\mathbf{y}} = \operatorname{Proj}_{\mathcal{V}}(\mathbf{y}). \] The transformation $\operatorname{Proj}_{\mathcal{V}}: \mathbb{R}^n \to \mathcal{V}$ is called the orthogonal projection of $\mathbb{R}^n$ onto $\mathcal{V}$.
- There are two important theorems in Section 6.3 are The Orthogonal Decomposition Theorem (Theorem 8) and The Best Approximation Theorem (Theorem 9).
-
Another important theorem in this section is Theorem 10 which I will call The Standard Matrix of an Orthogonal Projection.
My presentation here is slightly different from the book. The difference is that we do not assume that we are given unit vectors. Before we proceed with the theorem let us review an important property of a matrix with orthogonal nonzero columns.
- Let \(\bigl\{ \mathbf{v}_1 \, \cdots \, \mathbf{v}_m \bigr\}\) be an orthogonal set of nonzero vectors in \(\mathbb{R}^n\). This means that \[ \forall \mkern 1mu j,k \in \{1,\ldots,m\} \quad \text{we have:} \quad \left\{ \begin{array}{ll} \text{if} \ \ j\neq k, \ \ \mathbf{v}_j \cdot \mathbf{v}_k = 0, \\ \text{if} \ \ j = k, \ \ \mathbf{v}_k \cdot \mathbf{v}_k \gt 0, \\ \end{array} \right. \] Set \[ V = \bigl[ \mathbf{v}_1 \, \cdots \, \mathbf{v}_m \bigr]. \]
- Then \(V\) is \(n\times m\) matrix. Its transpose \(V^\top\) is \(m\times n\) matrix given by \[ V^\top = \left[ \begin{array}{c} \mathbf{v}_1^\top \\ \vdots \\ \mathbf{v}_m^\top \end{array} \right]. \]
- Matrix multiplication \(V^\top V\) results in the \(m\times m\) diagonal matrix with positive entries on the diagonal: \begin{align*} V^\top V & = \left[ \begin{array}{c} \mathbf{v}_1^\top \\ \vdots \\ \mathbf{v}_m^\top \end{array} \right] \bigl[ \mathbf{v}_1 \, \cdots \, \mathbf{v}_m \bigr] \\[5pt] & = \left[ \begin{array}{cccc} \mathbf{v}_1^\top \mathbf{v}_1 & \mathbf{v}_1^\top \mathbf{v}_2 & \cdots & \mathbf{v}_1^\top \mathbf{v}_m \\[5pt] \mathbf{v}_2^\top \mathbf{v}_1 & \mathbf{v}_2^\top \mathbf{v}_2 & \cdots & \mathbf{v}_2^\top \mathbf{v}_m \\[3pt] \vdots & \vdots & \ddots & \vdots \\[7pt] \mathbf{v}_m^\top \mathbf{v}_1 & \mathbf{v}_m^\top \mathbf{v}_2 & \cdots & \mathbf{v}_m^\top \mathbf{v}_m \end{array} \right] \\[5pt] & = \left[ \begin{array}{cccc} \mathbf{v}_1 \cdot \mathbf{v}_1 & \mathbf{v}_1\cdot \mathbf{v}_2 & \cdots & \mathbf{v}_1\cdot \mathbf{v}_m \\[5pt] \mathbf{v}_2\cdot \mathbf{v}_1 & \mathbf{v}_2\cdot \mathbf{v}_2 & \cdots & \mathbf{v}_2\cdot \mathbf{v}_m \\[3pt] \vdots & \vdots & \ddots & \vdots \\[7pt] \mathbf{v}_m\cdot \mathbf{v}_1 & \mathbf{v}_m\cdot \mathbf{v}_2 & \cdots & \mathbf{v}_m\cdot \mathbf{v}_m \end{array} \right] \\[5pt] & = \left[ \begin{array}{cccc} \mathbf{v}_1 \cdot \mathbf{v}_1 & 0 & \cdots & 0 \\[5pt] 0 & \mathbf{v}_2\cdot \mathbf{v}_2 & \cdots & 0 \\[3pt] \vdots & \vdots & \ddots & \vdots \\[7pt] 0 & 0 & \cdots & \mathbf{v}_m\cdot \mathbf{v}_m \end{array} \right] \end{align*}
- Since all diagonal entries of \(V^\top V\) are positive, the above diagonal matrix is invertible: \[ (V^\top V)^{-1} = \left[ \begin{array}{cccc} \frac{1}{\mathbf{v}_1 \cdot \mathbf{v}_1} & 0 & \cdots & 0 \\[5pt] 0 & \frac{1}{\mathbf{v}_2\cdot \mathbf{v}_2} & \cdots & 0 \\[3pt] \vdots & \vdots & \ddots & \vdots \\[7pt] 0 & 0 & \cdots & \frac{1}{\mathbf{v}_m\cdot \mathbf{v}_m} \end{array} \right] \]
- Assumption 1. $\mathcal{V}$ is a subspace of $\mathbb{R}^n,$ $\mathbf{v}_1, \ldots, \mathbf{v}_m \in \mathcal{V},$ and \[ \mathcal{V} = \operatorname{span} \{ \mathbf{v}_1, \ldots, \mathbf{v}_m \} \]
- Assumption 2. The set $\{ \mathbf{v}_1, \ldots, \mathbf{v}_m \}$ is an orthogonal set.
- Claim. For every $\mathbf y \in \mathbb R^n$ we have \[ \operatorname{Proj}_{\mathcal{V}}(\mathbf{y}) = V(V^\top V)^{-1}V^\top\!\mathbf{y}, \] where \[ V = \bigl[ \mathbf{v}_1 \, \cdots \, \mathbf{v}_m \bigr] \]
- BK 1. Since by Assumption 2. the set $\{ \mathbf{v}_1, \ldots, \mathbf{v}_m \}$ is an orthogonal set, the matrix $V$ is an $n\!\times\!m$ matrix with orthogonal columns. Therefore, as explained before the theorem, we have that $V^\top V$ is a diagonal matrix with positive entries on the diagonal; therefore invertible.
- BK 2. Since \[ V = \bigl[ \mathbf{v}_1 \, \cdots \, \mathbf{v}_m \bigr], \] by the definition of $\operatorname{Col}(V)$ we have \[ \operatorname{Col}(V) = \operatorname{span} \{ \mathbf{v}_1, \ldots, \mathbf{v}_m \} = \mathcal{V}. \]
- BK 3. By Theorem 3 in Section 6.1 we have that the orthogonal complement of the column space of \(V\) is the null space of \(V^\top\). That is \[ \bigl(\operatorname{Col}(V)\bigr)^\perp = \operatorname{Nul}(V^\top). \] Since by BK 2 we have \[ \mathcal{V} = \operatorname{Col}(V). \] we conclude that \[ \mathcal{V}^\perp = \bigl(\operatorname{Col}(V)\bigr)^\perp = \operatorname{Nul}(V^\top). \] Consequently, we have $\mathbf{x} \in \mathcal{V}^\perp$ if and only if $V^\top\!\mathbf{x} = \mathbf{0}$.
The proof of Theorem 10 given in the book is deceptively simple. Please do understand the proof in the book. Below I will give another proof of this theorem.
Theorem
Assumptions
In the proof of the theorem we use the following Background Knowledge, abbreviated as BK. Background Knowledge consists of the facts that we use in proof and that have been proved elsewhere.
The proof starts here.
Let $\mathbf{y} \in \mathbb R^n$ be arbitrary. By the definition of orthogonal projection, to prove that \[ V(V^\top V)^{-1}V^\top\! \mathbf y = \operatorname{Proj}_{\mathcal{V}}(\mathbf{y}) \] we have to prove two claims: The first \[ V(V^\top V)^{-1}V^\top\!\mathbf{y} \in \mathcal{V} \] and, the second, \[ \mathbf{y} - V(V^\top V)^{-1}V^\top\!\mathbf{y} \in \mathcal{V}^\perp. \] Since every vector of the form $V\mathbf{v}$ belongs to $\operatorname{Col}(V)$, we have that \[ V(V^\top V)^{-1}V^\top\! \mathbf{y} = V\bigl((V^\top V)^{-1}V^\top\! \mathbf{y}\bigr) \in \operatorname{Col}(V). \] By BK 2 we have $\operatorname{Col}(V) = \mathcal{V}.$ Therefore $V(V^\top V)^{-1}V^\top \mathbf{y} \in \mathcal{V}$ is proved. This proves the first claim.
To prove the second claim, we use BK 3. By BK 3: $\mathbf{x} \in \mathcal{V}^\perp$ if and only if $V^\top\!\mathbf{x} = \mathbf{0}$. Therefore, to prove \[ \mathbf{y} - V(V^\top V)^{-1}V^\top\!\mathbf{y} \in \mathcal{V}^\perp, \] we need to prove \[ V^\top \! \bigl( \mathbf{y} - V(V^\top V)^{-1}V^\top\!\mathbf{y} \bigr) = \mathbf{0}. \] Let us calculate $V^\top \! \bigl( \mathbf{y} - V(V^\top V)^{-1}V^\top\!\mathbf{y} \bigr)$ below \begin{align*} V^\top \! \bigl( \mathbf{y} - V(V^\top V)^{-1}V^\top\!\mathbf{y} \bigr) & = V^\top \! \mathbf{y} - V^\top \! \bigl(V(V^\top V)^{-1}V^\top\!\mathbf{y} \bigr) \\ &= V^\top \! \mathbf{y} - \bigl( V^\top \! V\bigr) (V^\top V)^{-1} \bigl(V^\top\!\mathbf{y}\bigr) \\ & = V^\top \! \mathbf{y} - I_m \bigl(V^\top\!\mathbf{y}\bigr) \\ & = V^\top \! \mathbf{y} - V^\top\!\mathbf{y} \\ & = \mathbf{0} \end{align*} In the preceding sequence of equalities, the first equality holds since the matrix multiplication is distributive, the second equality holds since the matrix multiplication is associative, the third equality follows from BK 1, the fourth equality follows from the definition of the matrix $I_m$, and the fifth equality holds by the definition of the opposite vector.
In conclusion, for every $\mathbf{y} \in \mathbb R^n$ we proved that the vector $V(V^\top V)^{-1}V^\top\!\mathbf{y}$ has the properties ① and ② in the definition of orthogonal projection onto $\mathcal{V}$. This proves \[ \operatorname{Proj}_{\mathcal{V}}(\mathbf{y}) = V(V^\top V)^{-1}V^\top\!\mathbf{y}. \]
The proof ends here. - In the preceding item we proved that the orthogonal projection of \(\mathbf{y} \in \mathbb{R}^n\) onto the column space of \(V\) is given by the formula \[ \operatorname{Proj}_{\mathcal{V}}(\mathbf{y}) = V(V^\top V)^{-1}V^\top\!\mathbf{y}. \] Next we will rewrite the preceding matrix formula as a linear combination of the columns of \(V\). Recall that \[ V = \bigl[ \mathbf{v}_1 \, \cdots \, \mathbf{v}_m \bigr], \qquad V^\top = \left[ \begin{array}{c} \mathbf{v}_1^\top \\ \vdots \\ \mathbf{v}_m^\top \end{array} \right] . \] Therefore \[ V^\top\!\mathbf{y} = \left[ \begin{array}{c} \mathbf{v}_1^\top \\ \mathbf{v}_2^\top \\ \vdots \\ \mathbf{v}_m^\top \end{array} \right] \mathbf{y} = \left[ \begin{array}{c} \mathbf{v}_1^\top \mathbf{y} \\ \vdots \\ \mathbf{v}_m^\top \mathbf{y} \end{array} \right] = \left[ \begin{array}{c} \mathbf{y} \cdot \mathbf{v}_1 \\ \mathbf{y} \cdot \mathbf{v}_2 \\ \vdots \\ \mathbf{y} \cdot \mathbf{v}_m \end{array} \right]. \] Further, \[ (V^\top V)^{-1}V^\top\!\mathbf{y} = \left[ \begin{array}{cccc} \frac{1}{\mathbf{v}_1 \cdot \mathbf{v}_1} & 0 & \cdots & 0 \\[5pt] 0 & \frac{1}{\mathbf{v}_2\cdot \mathbf{v}_2} & \cdots & 0 \\[3pt] \vdots & \vdots & \ddots & \vdots \\[7pt] 0 & 0 & \cdots & \frac{1}{\mathbf{v}_m\cdot \mathbf{v}_m} \end{array} \right] \left[ \begin{array}{c} \mathbf{y} \cdot \mathbf{v}_1 \\ \mathbf{y} \cdot \mathbf{v}_2 \\ \vdots \\ \mathbf{y} \cdot \mathbf{v}_m \end{array} \right] = \left[ \begin{array}{c} \frac{\mathbf{y} \cdot \mathbf{v}_1}{\mathbf{v}_1 \cdot \mathbf{v}_1} \\ \frac{\mathbf{y} \cdot \mathbf{v}_2}{\mathbf{v}_2 \cdot \mathbf{v}_2} \\ \vdots \\ \frac{\mathbf{y} \cdot \mathbf{v}_m}{\mathbf{v}_m \cdot \mathbf{v}_m} \end{array} \right]. \] Finally, \begin{align*} \operatorname{Proj}_{\mathcal{V}}(\mathbf{y}) & = V(V^\top V)^{-1}V^\top\!\mathbf{y} \\[5pt] & = V \left[ \begin{array}{c} \frac{\mathbf{y} \cdot \mathbf{v}_1}{\mathbf{v}_1 \cdot \mathbf{v}_1} \\ \frac{\mathbf{y} \cdot \mathbf{v}_2}{\mathbf{v}_2 \cdot \mathbf{v}_2} \\ \vdots \\ \frac{\mathbf{y} \cdot \mathbf{v}_m}{\mathbf{v}_m \cdot \mathbf{v}_m} \end{array} \right] \\[5pt] & = \left(\frac{\mathbf{y} \cdot \mathbf{v}_1}{\mathbf{v}_1 \cdot \mathbf{v}_1}\right) \mathbf{v}_1 + \left(\frac{\mathbf{y} \cdot \mathbf{v}_2}{\mathbf{v}_2 \cdot \mathbf{v}_2} \right) \mathbf{v}_2 + \cdots + \left(\frac{\mathbf{y} \cdot \mathbf{v}_m}{\mathbf{v}_m \cdot \mathbf{v}_m} \right) \mathbf{v}_m . \end{align*}
-
I should have taken a picture of the blackboard I created on Tuesday, where I proved the orthogonal decomposition theorem in two dimensions and derived the Cauchy-Bunyakovsky-Schwarz inequality from it. This pair of results is truly a mathematical gem! To preserve it for posterity on my website, I recreated Tuesday’s blackboard in the Goodnotes app on my iPad. Here it is for you to enjoy.
- We started Section 6.2: Orthogonal sets of vectors today. Suggested problems are: 2, 3, 5, 8, 9, 11, 13, 15, 17, 19, 21, 23, 25, 26, 27, 29. Below I present the main points and their relation to important theorem in Section 6.2: Orthogonal Projections
-
Let $m, n \in \mathbb{N}$. Let $\mathbf{v}_1, \ldots, \mathbf{v}_m \in \mathbb{R}^n$ be an orthogonal set of nonzero vectors. What the last phrase means is the following
- For all $j,k \in \{1,\ldots,m\}$ we have $\mathbf{v}_j \cdot \mathbf{v}_k = 0$ whenever $j \neq k$.
- For all $k \in \{1,\ldots,m\}$ we have $\mathbf{v}_k \cdot \mathbf{v}_k \gt 0$.
-
It is truly important that we always connect vectors with matrices formed by those vectors. Let $\mathbf{v}_1, \ldots, \mathbf{v}_m \in \mathbb{R}^n$ be an orthogonal set of nonzero vectors.
Let us form the $n\times m$ matrix $M$ whose columns are the vectors $\mathbf{v}_1, \ldots, \mathbf{v}_m$: \[ M = \left[\!\begin{array}{cccc} \mathbf{v}_1 & \mathbf{v}_2 & \cdots & \mathbf{v}_m \end{array}\!\!\right]. \] Recall that $M^\top$ is an $m\times n$ matrix. Now calculate the product of $M^\top$ and $M$ as follows: \begin{align*} M^\top M & = \left[\!\begin{array}{c} \mathbf{v}_1^\top \\ \mathbf{v}_2^\top \\ \vdots \\ \mathbf{v}_m^\top \end{array}\!\!\right] \left[\!\begin{array}{cccc} \mathbf{v}_1 & \mathbf{v}_2 & \cdots & \mathbf{v}_m \end{array}\!\!\right] \\[6pt] & = \left[\!\begin{array}{cccc} \mathbf{v}_1\!\!\cdot\!\mathbf{v}_1 & \mathbf{v}_1\!\!\cdot\!\mathbf{v}_2 & \cdots & \mathbf{v}_1\!\!\cdot\!\mathbf{v}_m \\ \mathbf{v}_2\!\!\cdot\!\mathbf{v}_1 & \mathbf{v}_2\!\!\cdot\!\mathbf{v}_2 & \cdots & \mathbf{v}_2\!\!\cdot\!\mathbf{v}_m \\ \vdots & \vdots & \ddots & \vdots \\ \mathbf{v}_m\!\!\cdot\!\mathbf{v}_1 & \mathbf{v}_m\!\!\cdot\!\mathbf{v}_2 & \cdots & \mathbf{v}_m\!\!\cdot\!\mathbf{v}_m \end{array}\!\!\right] \\[6pt] & = \left[\!\begin{array}{cccc} \mathbf{v}_1\!\!\cdot\!\mathbf{v}_1 & 0 & \cdots & 0 \\ 0 & \mathbf{v}_2\!\!\cdot\!\mathbf{v}_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \mathbf{v}_m\!\!\cdot\!\mathbf{v}_m \end{array}\!\!\right] \end{align*} Thus $M^\top\! M$ is a diagonal $m\times m$ matrix.
The above matrix calculation shows that the vectors $\mathbf{v}_1, \ldots, \mathbf{v}_m \in \mathbb{R}^n$ form an orthogonal set of nonzero vectors if and only if the matrix $M^\top\! M$ is a diagonal $m\times m$ with nonzero entries on the diagonal.
-
Why are orthogonal sets of nonzero vectors in $\mathbb{R}^n$ important? My answer to that question are the following three properties, for which I use the adjective straightforward. Look at the proofs in the textbook, the proofs of these properties are straightforward using some standard algebraic manipulations with linear equations.
-
Straightforward linear independence. The vectors in an orthogonal set of nonzero vectors $\mathbf{v}_1, \ldots, \mathbf{v}_m$ are linearly independent. This is Theorem 4 in Section 6.2.
-
Straightforward linear combinations. If \[ \mathbf{y} = \alpha_1 \mathbf{v}_1 + \alpha_2 \mathbf{v}_1 + \cdots + \alpha_m \mathbf{v}_m, \] then \[ \alpha_1 = \frac{\mathbf{y}\cdot \mathbf{v}_1}{\mathbf{v}_1 \cdot \mathbf{v}_1}, \quad \alpha_2 = \frac{\mathbf{y}\cdot \mathbf{v}_2}{\mathbf{v}_2 \cdot \mathbf{v}_2}, \quad \ldots, \quad \alpha_m = \frac{\mathbf{y}\cdot \mathbf{v}_m}{\mathbf{v}_m \cdot \mathbf{v}_m}. \]
In other words:
If $\mathbf{y} \in \operatorname{Span}\bigl\{\mathbf{v}_1, \ldots, \mathbf{v}_m\bigr\}$, then \[ \require{bbox} \bbox[#FFFF00, 8px, border: 2px solid #808000]{ \mathbf{y} = \left(\frac{\mathbf{y}\cdot \mathbf{v}_1}{\mathbf{v}_1 \cdot \mathbf{v}_1} \right) \mathbf{v}_1 + \left(\frac{\mathbf{y}\cdot \mathbf{v}_2}{\mathbf{v}_2 \cdot \mathbf{v}_2}\right) \mathbf{v}_2+ \cdots + \left(\frac{\mathbf{y}\cdot \mathbf{v}_m}{\mathbf{v}_m \cdot \mathbf{v}_m}\right) \mathbf{v}_m.} \]
This is Theorem 5 in Section 6.2.
-
Straightforward projections onto the span. Let $\mathcal{W} = \operatorname{Span}\bigl\{\mathbf{v}_1, \ldots, \mathbf{v}_m\bigr\}$ and let $\mathbf{y} \in \mathbb{R}^n$. Then the orthogonal projection of $\mathbf{y}$ onto $\mathcal{W}$ is given by \[ \require{bbox} \bbox[#FFFF00, 8px, border: 2px solid #808000]{ \widehat{\mathbf{y}} = \operatorname{Proj}_{\mathcal{W}}(\mathbf{y}) = \left(\frac{\mathbf{y}\cdot \mathbf{v}_1}{\mathbf{v}_1 \cdot \mathbf{v}_1}\right) \mathbf{v}_1 + \left(\frac{\mathbf{y}\cdot \mathbf{v}_2}{\mathbf{v}_2 \cdot \mathbf{v}_2}\right) \mathbf{v}_2 + \cdots + \left(\frac{\mathbf{y}\cdot \mathbf{v}_m}{\mathbf{v}_m \cdot \mathbf{v}_m}\right) \mathbf{v}_m.} \]
This is Theorem 8 in Section 6.3.
-
-
Examples.
-
Consider the vectors \[ \left[\! \begin{array}{r} -8 \\ 1 \\ 4 \end{array} \!\right], \quad \left[\! \begin{array}{r} 1 \\ -8 \\ 4 \end{array} \!\right], \quad \left[\! \begin{array}{r} 4 \\ 4 \\ 7 \end{array} \!\right]. \]
These vectors form an orthogonal sets of nonzero vectors. To verify this claim, it is sufficient to calculate \[ \left[\! \begin{array}{r} -8 \\ 1 \\ 4 \end{array} \!\right] \cdot \left[\! \begin{array}{r} 1 \\ -8 \\ 4 \end{array} \!\right] = 0, \quad \left[\! \begin{array}{r} -8 \\ 1 \\ 4 \end{array} \!\right] \cdot \left[\! \begin{array}{r} 4 \\ 4 \\ 7 \end{array} \!\right] = 0, \quad \left[\! \begin{array}{r} 1 \\ -8 \\ 4 \end{array} \!\right] \cdot \left[\! \begin{array}{r} 4 \\ 4 \\ 7 \end{array} \!\right] = 0. \]
By Theorem 4 in Section 6.2 these vectors are linearly independent. This is an example of straightforward linear independence.
-
As shown in the preceding item the vectors \[ \left[\! \begin{array}{r} -8 \\ 1 \\ 4 \end{array} \!\right], \quad \left[\! \begin{array}{r} 1 \\ -8 \\ 4 \end{array} \!\right], \quad \left[\! \begin{array}{r} 4 \\ 4 \\ 7 \end{array} \!\right], \] form an orthogonal set of nonzero vectors. By Theorem 4 in Section 6.2 they are linearly independent. Consequently they form a basis for $\mathbb{R}^3$.
To express the vector $\left[\! \begin{array}{r} 1 \\ 2 \\ 3 \end{array} \!\right]$ as a linear combination of the vectors \[ \left[\! \begin{array}{r} -8 \\ 1 \\ 4 \end{array} \!\right], \quad \left[\! \begin{array}{r} 1 \\ -8 \\ 4 \end{array} \!\right], \quad \left[\! \begin{array}{r} 4 \\ 4 \\ 7 \end{array} \!\right], \] we use the straightforward linear combinations, that is Theorem 5 in Section 6.2.
We calculate \[ \left[\! \begin{array}{r} 1 \\ 2 \\ 3 \end{array} \!\right] \cdot \left[\! \begin{array}{r} -8 \\ 1 \\ 4 \end{array} \!\right] = 6, \quad \left[\! \begin{array}{r} 1 \\ 2 \\ 3 \end{array} \!\right] \cdot \left[\! \begin{array}{r} 1 \\ -8 \\ 4 \end{array} \!\right] = -3, \quad \left[\! \begin{array}{r} 1 \\ 2 \\ 3 \end{array} \!\right] \cdot \left[\! \begin{array}{r} 4 \\ 4 \\ 7 \end{array} \!\right] = 33, \] and \[ \left[\! \begin{array}{r} -8 \\ 1 \\ 4 \end{array} \!\right] \cdot \left[\! \begin{array}{r} -8 \\ 1 \\ 4 \end{array} \!\right] = 81, \quad \left[\! \begin{array}{r} 1 \\ -8 \\ 4 \end{array} \!\right] \cdot \left[\! \begin{array}{r} 1 \\ -8 \\ 4 \end{array} \!\right] = 81, \quad \left[\! \begin{array}{r} 4 \\ 4 \\ 7 \end{array} \!\right] \cdot \left[\! \begin{array}{r} 4 \\ 4 \\ 7 \end{array} \!\right] = 81. \] Then \begin{align*} \left[\! \begin{array}{r} 1 \\ 2 \\ 3 \end{array} \!\right] & = \frac{6}{81} \left[\! \begin{array}{r} -8 \\ 1 \\ 4 \end{array} \!\right] + \frac{-3}{81} \left[\! \begin{array}{r} 1 \\ -8 \\ 4 \end{array} \!\right] + \frac{33}{81} \left[\! \begin{array}{r} 4 \\ 4 \\ 7 \end{array} \!\right] \\ & = \left[\! \begin{array}{r} -\frac{16}{27} \\ \frac{2}{27} \\ \frac{8}{27} \end{array} \!\right] + \left[\! \begin{array}{r} -\frac{1}{27} \\ \frac{8}{27} \\ -\frac{4}{27} \end{array} \!\right] + \left[\! \begin{array}{r} \frac{44}{27} \\ \frac{44}{27} \\ \frac{77}{27} \end{array} \!\right] \end{align*}
-
Consider the vectors \[ \left[\! \begin{array}{r} 1 \\ 1 \\ 5 \\ 3 \end{array} \!\right], \quad \left[\! \begin{array}{r} 1 \\ -1 \\ -3 \\ 5 \end{array} \!\right]. \] These two vectors form an orthogonal set of nonzero vectors in $\mathbb{R}^4$.
To calculate the orthogonal projection of the vector $\mathbf{y} = \left[\! \begin{array}{r} 1 \\ 2 \\ 3 \\ 4 \end{array} \!\right]$ onto the linear span of the given two vectors, that is \[ \mathcal{W} = \operatorname{Span}\left\{ \left[\! \begin{array}{r} 1 \\ 1 \\ 5 \\ 3 \end{array} \!\right], \left[\! \begin{array}{r} 1 \\ -1 \\ -3 \\ 5 \end{array} \!\right] \right\} \] we use straightforward projections onto the span, that is Theorem 8 in Section 6.3.
To calculate the projection, we need to calculate \[ \left[\! \begin{array}{r} 1 \\ 2 \\ 3 \\ 4 \end{array} \!\right] \cdot \left[\! \begin{array}{r} 1 \\ 1 \\ 5 \\ 3 \end{array} \!\right] = 30, \quad \left[\! \begin{array}{r} 1 \\ 2 \\ 3 \\ 4 \end{array} \!\right] \cdot \left[\! \begin{array}{r} 1 \\ -1 \\ -3 \\ 5 \end{array} \!\right] = 10 \] and \[ \left[\! \begin{array}{r} 1 \\ 1 \\ 5 \\ 3 \end{array} \!\right] \cdot \left[\! \begin{array}{r} 1 \\ 1 \\ 5 \\ 3 \end{array} \!\right] = 36, \quad \left[\! \begin{array}{r} 1 \\ -1 \\ -3 \\ 5 \end{array} \!\right] \cdot \left[\! \begin{array}{r} 1 \\ -1 \\ -3 \\ 5 \end{array} \!\right] = 36. \] Then the projection is: \begin{align*} \widehat{\mathbf{y}} = \operatorname{Proj}_{\mathcal{W}} \left[\! \begin{array}{r} 1 \\ 2 \\ 3 \\ 4 \end{array} \!\right] & = \frac{30}{36} \left[\! \begin{array}{r} 1 \\ 1 \\ 5 \\ 3 \end{array} \!\right] + \frac{10}{36} \left[\! \begin{array}{r} 1 \\ -1 \\ -3 \\ 5 \end{array} \!\right] \\ & = \left[\! \begin{array}{c} \frac{5}{6} \\ \frac{5}{6} \\ \frac{25}{6} \\ \frac{15}{6} \end{array} \!\right] + \left[\! \begin{array}{r} \frac{5}{18} \\ -\frac{5}{18} \\ -\frac{15}{18} \\ \frac{25}{18} \end{array} \!\right] \\ & = \left[\! \begin{array}{c} \frac{10}{9} \\ \frac{5}{9} \\ \frac{30}{9} \\ \frac{35}{9} \end{array} \!\right]. \end{align*} To verify that this calculation is correct we calculate \[ \mathbf{y} - \widehat{\mathbf{y}} = \left[\! \begin{array}{r} 1 \\ 2 \\ 3 \\ 4 \end{array} \!\right] - \left[\! \begin{array}{c} \frac{10}{9} \\ \frac{5}{9} \\ \frac{30}{9} \\ \frac{35}{9} \end{array} \!\right] = \left[\! \begin{array}{r} -\frac{1}{9} \\ \frac{13}{9} \\ -\frac{3}{9} \\ \frac{1}{9} \end{array} \!\right]. \] This vector should be orthogonal to the both given vectors: \[ \frac{1}{9} \left[\! \begin{array}{r} -1 \\ 13 \\ -3 \\ 1 \end{array} \!\right] \cdot \left[\! \begin{array}{r} 1 \\ 1 \\ 5 \\ 3 \end{array} \!\right] = 0, \quad \frac{1}{9} \left[\! \begin{array}{r} -1 \\ 13 \\ -3 \\ 1 \end{array} \!\right] \cdot \left[\! \begin{array}{r} 1 \\ -1 \\ -3 \\ 5 \end{array} \!\right] = 0 \] Finally, to celebrate our calculation we write \[ \left[\! \begin{array}{r} 1 \\ 2 \\ 3 \\ 4 \end{array} \!\right] = \left[\! \begin{array}{c} \frac{10}{9} \\ \frac{5}{9} \\ \frac{30}{9} \\ \frac{35}{9} \end{array} \!\right] + \left[\! \begin{array}{r} -\frac{1}{9} \\ \frac{13}{9} \\ -\frac{3}{9} \\ \frac{1}{9} \end{array} \!\right], \] and we point out that two vectors in the sum are orthogonal to each other. Moreover, the first vector in the sum is in the subspace $\mathcal{W}$ and the second vector in the sum is in the subspace $\mathcal{W}^\perp.$
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- Here is a summary of some important points from Section 6.1 that I did not emphasize enough on Friday. Please pay particular attention to the calculations with matrices.
-
The following concepts emerge from dot product in $\mathbb{R}^n$:
- the length of a vector,
- the distance between two vectors,
- the concept of orthogonality among vectors,
- the concept of orthogonal complement,
- the Pythagorean theorem for orthogonal vectors,
- the abstract concept of the angle between vectors.
- Here is a proof of the Law of Cosines and its connection to dot product.
- Here is a proof of the classical Pythagorean Theorem.
- One of the most important theorems in Section 6.1 is Theorem 3. I did not talk about it in class (I will briefly mention it on Thursday, please study it from the textbook and here. Theorem 3 states: Let $m$ and $n$ be positive integers. Let $A$ be an $m\!\times\!n$ matrix. Then \[ \require{bbox} \bbox[#FFFF00, 8px, border: 3px solid #808000]{ \bigl(\operatorname{Row}(A)\bigr)^\perp = \operatorname{Nul}(A) } \quad \text{and} \quad \bbox[#FFFF00, 8px, border: 3px solid #808000]{ \bigl(\operatorname{Col}(A)\bigr)^\perp = \operatorname{Nul}(A^\top)}. \] And also \[ \bbox[#FFFF00, 8px, border: 3px solid #808000]{ \bigl(\operatorname{Nul}(A)\bigr)^\perp = \operatorname{Row}(A)} \quad \text{and} \quad \bbox[#FFFF00, 8px, border: 3px solid #808000]{\bigl(\operatorname{Nul}(A^\top)\bigr)^\perp = \operatorname{Col}(A)}. \]
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The theorem from the previous item is useful for a problem like this: Find the orthogonal complement of the following span:
\[
\operatorname{Span}\left\{ \left[\! \begin{array}{c}
1 \\ 1 \\ 1 \\ 1
\end{array} \!\right],
\left[\! \begin{array}{c}
-1 \\ 1 \\ 1 \\ -1
\end{array} \!\right]\right\}.
\]
- We first observe that \[ \operatorname{Span}\left\{ \left[\! \begin{array}{c} 1 \\ 1 \\ 1 \\ 1 \end{array} \!\right], \left[\! \begin{array}{c} -1 \\ 1 \\ 1 \\ -1 \end{array} \!\right]\right\} = \operatorname{Row}\left( \left[\! \begin{array}{cccc} 1 & 1 & 1 & 1 \\ -1 & 1 & 1 & -1 \end{array} \!\right] \right). \] Therefore, by Theorem 3 in Section 6.1 we deduce \[ \left( \operatorname{Span}\left\{ \left[\! \begin{array}{c} 1 \\ 1 \\ 1 \\ 1 \end{array} \!\right], \left[\! \begin{array}{c} -1 \\ 1 \\ 1 \\ -1 \end{array} \!\right]\right\} \right)^\perp = \operatorname{Nul}\left( \left[\! \begin{array}{cccc} 1 & 1 & 1 & 1 \\ -1 & 1 & 1 & -1 \end{array} \!\right] \right). \]
- To find the preceding null space Row Reduce to RREF: \[ \left[\! \begin{array}{cccc} 1 & 1 & 1 & 1 \\ -1 & 1 & 1 & -1 \end{array} \!\right] \ \ \sim \quad \cdots \quad \sim \ \ \left[\! \begin{array}{cccc} 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \end{array} \!\right]. \] From the last RREF matrix we deduce that \[ \operatorname{Nul}\left( \left[\! \begin{array}{cccc} 1 & 1 & 1 & 1 \\ -1 & 1 & 1 & -1 \end{array} \!\right] \right) = \operatorname{Span}\left\{ \left[\! \begin{array}{c} -1 \\ 0 \\ 0 \\ 1 \end{array} \!\right], \left[\! \begin{array}{c} 0 \\ -1 \\ 1 \\ 0 \end{array} \!\right]\right\} \]
- Therefore \[ \left( \operatorname{Span}\left\{ \left[\! \begin{array}{c} 1 \\ 1 \\ 1 \\ 1 \end{array} \!\right], \left[\! \begin{array}{c} -1 \\ 1 \\ 1 \\ -1 \end{array} \!\right]\right\} \right)^\perp = \operatorname{Span}\left\{ \left[\! \begin{array}{c} -1 \\ 0 \\ 0 \\ 1 \end{array} \!\right], \left[\! \begin{array}{c} 0 \\ -1 \\ 1 \\ 0 \end{array} \!\right]\right\} \]
- It is a curious fact that \[ \operatorname{Span}\left\{ \left[\! \begin{array}{c} -1 \\ 0 \\ 0 \\ 1 \end{array} \!\right], \left[\! \begin{array}{c} 0 \\ -1 \\ 1 \\ 0 \end{array} \!\right]\right\} = \operatorname{Span}\left\{ \left[\! \begin{array}{c} -1 \\ -1 \\ 1 \\ 1 \end{array} \!\right], \left[\! \begin{array}{c} -1 \\ 1 \\ - 1 \\ 1 \end{array} \!\right]\right\}. \] The preceding claim is a consequence of the fact that \[ \operatorname{Span} \bigl\{ \mathbf{u}, \mathbf{v} \bigr\} = \operatorname{Span} \bigl\{ \mathbf{u} + \mathbf{v}, \mathbf{u} - \mathbf{v} \bigr\}, \] and the fact that $\mathbf{u}$ and $\mathbf{v}$ are linearly independent if and only if $\mathbf{u}+\mathbf{v}$ and $\mathbf{u}-\mathbf{v}$ are linearly independent. (Prove these facts for exercise.)
- As a consequence of the preceding two items we have \[ \left( \operatorname{Span}\left\{ \left[\! \begin{array}{c} 1 \\ 1 \\ 1 \\ 1 \end{array} \!\right], \left[\! \begin{array}{c} -1 \\ 1 \\ 1 \\ -1 \end{array} \!\right]\right\} \right)^\perp = \operatorname{Span}\left\{ \left[\! \begin{array}{c} -1 \\ -1 \\ 1 \\ 1 \end{array} \!\right], \left[\! \begin{array}{c} -1 \\ 1 \\ -1 \\ 1 \end{array} \!\right]\right\}. \]
- The four vectors that we used in the preceding item are remarkable four vectors in $\mathbb{R}^4.$ One can see how remarkable they are if we put these four vectors as the columns of a $4\!\times\!4$ matrix \[ U = \left[\! \begin{array}{cccc} 1 & -1 & -1 & -1 \\ 1 & 1 & -1 & 1 \\ 1 & 1 & 1 & -1 \\ 1 & -1 & 1 & 1 \end{array} \!\right] \] and calculate \[ U^\top U = \left[\! \begin{array}{cccc} 1 & 1 & 1 & 1 \\ -1 & 1 & 1 & -1 \\ -1 & -1 & 1 & 1 \\ -1 & 1 & -1 & 1 \end{array} \!\right] \left[\! \begin{array}{cccc} 1 & -1 & -1 & -1 \\ 1 & 1 & -1 & 1 \\ 1 & 1 & 1 & -1 \\ 1 & -1 & 1 & 1 \end{array} \!\right] = \left[\! \begin{array}{cccc} 4 & 0 & 0 & 0 \\ 0 & 4 & 0 & 0 \\ 0 & 0 & 4 & 0 \\ 0 & 0 & 0 & 4 \end{array} \!\right] \] The preceding matrix equality tells us in a matrix form that the columns of the matrix $U$ are orthogonal to each other and each column as a vector in $\mathbb{R}^4$ has norm (length) equal to $2.$
- We started Section 6.1 today. Suggested problems: 1, 5, 7, 8, 9-12, 13, 15-18, 20, 22, 24, 25, 26, 27, 28, 29, 30, 31, 32 (do this problem by hand), 33 (do this problem by hand).
-
The following concepts emerge from dot product in $\mathbb{R}^n$:
- the length of a vector,
- the distance between two vectors,
- the concept of orthogonality among vectors,
- the concept of orthogonal complement,
- the Pythagorean theorem for orthogonal vectors,
- the abstract concept of the angle between vectors.
-
In the animation below I illustrate the dot product in $\mathbb{R}^2$ with approximately $100$ pairs of vectors.
- It is important for you to internalize that we have been working with the dot product all along when multiplying matrices. Let $k,m$ and $n$ be positive integers and let $A$ be a $k\!\times\!m$ matrix and $B$ be a $m\!\times\!n$. Then $A$ has $k$ rows and each row of $A$ is a transpose of a vector in $\mathbb{R}^m$. Similarly, $B$ has $n$ columns and each column of $B$ is a vector in $\mathbb{R}^m$. Now introduce the notation: \[ \mathbf{r}_1, \mathbf{r}_2, \ldots, \mathbf{r}_k \in \mathbb{R}^m \quad \text{are the transposes of the rows of} \quad A \] \[ \mathbf{c}_1, \mathbf{c}_2, \ldots, \mathbf{c}_n \in \mathbb{R}^m \quad \text{are the columns of} \quad B \] So, I can write the matrices $A$ and $B$ as \[ A = \left[\!\begin{array}{c} \mathbf{r}_1^\top \\ \mathbf{r}_2^\top \\ \vdots \\ \mathbf{r}_k^\top \end{array}\!\!\right], \qquad B = \left[\!\begin{array}{cccc} \mathbf{c}_1 & \mathbf{c}_2 & \cdots & \mathbf{c}_n \end{array}\!\!\right]. \] Now we calculate the product of $A$ and $B$ as follows: \[ A B = \left[\!\begin{array}{c} \mathbf{r}_1^\top \\ \mathbf{r}_2^\top \\ \vdots \\ \mathbf{r}_k^\top \end{array}\!\!\right] \left[\!\begin{array}{cccc} \mathbf{c}_1 & \mathbf{c}_2 & \cdots & \mathbf{c}_n \end{array}\!\!\right] = \left[\!\begin{array}{cccc} \mathbf{r}_1\!\!\cdot\!\mathbf{c}_1 & \mathbf{r}_1\!\!\cdot\!\mathbf{c}_2 & \cdots & \mathbf{r}_1\!\!\cdot\!\mathbf{c}_n \\ \mathbf{r}_2\!\!\cdot\!\mathbf{c}_1 & \mathbf{r}_2\!\!\cdot\!\mathbf{c}_2 & \cdots & \mathbf{r}_2\!\!\cdot\!\mathbf{c}_n \\ \vdots & \vdots & \ddots & \vdots \\ \mathbf{r}_k\!\!\cdot\!\mathbf{c}_1 & \mathbf{r}_k\!\!\cdot\!\mathbf{c}_2 & \cdots & \mathbf{r}_k\!\!\cdot\!\mathbf{c}_n \end{array}\!\!\right] \]
-
As an example, I want to recall the matrix multiplication related to the Reduced Row Echelon Form of a matrix. Below is a $4\times 5$ matrix and its reduced row echelon form:
\[
A =
\left[\! \begin{array}{rrrrr}
1 & 3 & 2 & 2 & 2 \\
2 & 0 & -2 & 1 & 1 \\
2 & 1 & -1 & 1 & 2 \\
1 & 4 & 3 & 2 & 3
\end{array} \!\right] \ \sim \quad \cdots \quad \sim \
\left[\!
\begin{array}{rrrrr}
1 & 0 & -1 & 0 & 1 \\
0 & 1 & 1 & 0 & 1 \\
0 & 0 & 0 & 1 & -1 \\
0 & 0 & 0 & 0 & 0 \\
\end{array}
\!\right]
\]
- Whenever the reduced row echelon form (RREF) is found we should observe the following remarkable identity: The matrix product of the $4\times 3$ matrix consisting of the pivot columns of $A$ and the $3\times 5$ matrix consisting of the nonzero rows of the RREF of $A$ results in the matrix $A$: \[ \require{bbox} \bbox[#7FBFBF, 8px, border: 3px solid teal]{ \left[\! \begin{array}{rrr} 1 & 3 & 2 \\ 2 & 0 & 1 \\ 2 & 1 & 1 \\ 1 & 4 & 2 \end{array} \!\right] \left[\! \begin{array}{rrrrr} 1 & 0 & -1 & 0 & 1 \\ 0 & 1 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 & -1 \\ \end{array} \!\right] = \left[\! \begin{array}{rrrrr} 1 & 3 & 2 & 2 & 2 \\ 2 & 0 & -2 & 1 & 1 \\ 2 & 1 & -1 & 1 & 2 \\ 1 & 4 & 3 & 2 & 3 \end{array} \!\right] = A }. \] I decided to color this identity in teal since it is beautealful.
- Notice that the columns in the $4\!\times\!3$ matrix in the teal identity are the basis vectors for $\operatorname{Col}(A).$ Similarly, the rows in the $3\!\times\!5$ matrix in the teal identity are the basis vectors for $\operatorname{Row}(A).$
- By the rules of matrix multiplication the teal identity tells us that the columns of the matrix $A$ are linear combinations of the columns of the $4\!\times\!3$ matrix in the teal identity. The coefficients in these linear combinations are the columns of $3\!\times\!5$ matrix in the teal identity.
- By the rules of matrix multiplication the teal identity tells us that the rows of the matrix $A$ are linear combinations of the rows of the $3\!\times\!5$ matrix in the teal identity. The coefficients in these linear combinations are the rows of $4\!\times\!3$ matrix in the teal identity.
- In this item I will illustrate the multiplication of two matrices below by emphasising the dot products between $\mathbb{R}^3$ vectors: \[ \require{bbox} \bbox[#7FBFBF, 8px, border: 3px solid teal]{ \left[\! \begin{array}{rrr} 1 & 3 & 2 \\ 2 & 0 & 1 \\ 2 & 1 & 1 \\ 1 & 4 & 2 \end{array} \!\right] \left[\! \begin{array}{rrrrr} 1 & 0 & -1 & 0 & 1 \\ 0 & 1 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 & -1 \\ \end{array} \!\right] = \left[\! \begin{array}{rrrrr} 1 & 3 & 2 & 2 & 2 \\ 2 & 0 & -2 & 1 & 1 \\ 2 & 1 & -1 & 1 & 2 \\ 1 & 4 & 3 & 2 & 3 \end{array} \!\right] }. \] Here the left matrix in the product is a $4\times\!3$ matrix with the rows: \[ \left[\! \begin{array}{r} 1 \\ 3 \\ 2 \end{array} \!\right], \qquad \left[\! \begin{array}{r} 2 \\ 0 \\ 1 \end{array} \!\right], \qquad \left[\! \begin{array}{r} 2 \\ 1 \\ 1 \end{array} \!\right], \qquad \left[\! \begin{array}{r} 1 \\ 4 \\ 2 \end{array} \!\right], \] and right matrix in the product is a $3\times\!5$ matrix with the columns: \[ \left[\! \begin{array}{r} 1 \\ 0 \\ 0 \end{array} \!\right], \qquad \left[\! \begin{array}{r} 0 \\ 1 \\ 0 \end{array} \!\right], \qquad \left[\! \begin{array}{r} -1 \\ 1 \\ 0 \end{array} \!\right], \qquad \left[\! \begin{array}{r} 0 \\ 0 \\ 1 \end{array} \!\right], \qquad \left[\! \begin{array}{r} 1 \\ 1 \\ -1 \end{array} \!\right]. \] It is useful to think of the matrix multiplication as follows (as you read through the matrix-vector arithmetic below always look for dot products in different forms, always): \begin{align*} & \left[\!\! \begin{array}{r} \phantom{\biggl|\biggr.} \bigl[\!\begin{array}{ccc} 1 & 3 & 2 \end{array} \!\bigr] \\ \bigl[\!\begin{array}{ccc} 2 & 0 & 1 \end{array} \!\bigr] \\ \bigl[\!\begin{array}{ccc} 2 & 1 & 1 \end{array} \!\bigr] \\ \bigl[\!\begin{array}{ccc} 1 & 4 & 2 \end{array} \!\bigr] \end{array} \!\right] \left[\! \begin{array}{rrrrr} \left[\! \begin{array}{r} 1 \\ 0 \\ 0 \end{array} \!\right] & \left[\! \begin{array}{r} 0 \\ 1 \\ 0 \end{array} \!\right] & \left[\! \begin{array}{r} -1 \\ 1 \\ 0 \end{array} \!\right] & \left[\! \begin{array}{r} 0 \\ 0 \\ 1 \end{array} \!\right] & \left[\! \begin{array}{r} 1 \\ 1 \\ -1 \end{array} \!\right] \end{array} \!\right] \\ & = \left[\! \begin{array}{rrrrr} \bigl[\!\begin{array}{ccc} 1 & 3 & 2 \end{array} \!\bigr]\!\left[\! \begin{array}{r} 1 \\ 0 \\ 0 \end{array} \!\right] & \bigl[\!\begin{array}{ccc} 1 & 3 & 2 \end{array} \!\bigr]\!\left[\! \begin{array}{r} 0 \\ 1 \\ 0 \end{array} \!\right] & \bigl[\!\begin{array}{ccc} 1 & 3 & 2 \end{array} \!\bigr]\!\left[\! \begin{array}{r} -1 \\ 1 \\ 0 \end{array} \!\right] & \bigl[\!\begin{array}{ccc} 1 & 3 & 2 \end{array} \!\bigr]\!\left[\! \begin{array}{r} 0 \\ 0 \\ 1 \end{array} \!\right] & \bigl[\!\begin{array}{ccc} 1 & 3 & 2 \end{array} \!\bigr]\!\left[\! \begin{array}{r} 1 \\ 1 \\ -1 \end{array} \!\right] \\ \bigl[\!\begin{array}{ccc} 2 & 0 & 1 \end{array} \!\bigr]\!\left[\! \begin{array}{r} 1 \\ 0 \\ 0 \end{array} \!\right] & \bigl[\!\begin{array}{ccc} 2 & 0 & 1 \end{array} \!\bigr]\!\left[\! \begin{array}{r} 0 \\ 1 \\ 0 \end{array} \!\right] & \bigl[\!\begin{array}{ccc} 2 & 0 & 1 \end{array} \!\bigr]\!\left[\! \begin{array}{r} -1 \\ 1 \\ 0 \end{array} \!\right] & \bigl[\!\begin{array}{ccc} 2 & 0 & 1 \end{array} \!\bigr]\!\left[\! \begin{array}{r} 0 \\ 0 \\ 1 \end{array} \!\right] & \bigl[\!\begin{array}{ccc} 2 & 0 & 1 \end{array} \!\bigr]\!\left[\! \begin{array}{r} 1 \\ 1 \\ -1 \end{array} \!\right] \\ \bigl[\!\begin{array}{ccc} 2 & 1 & 1 \end{array} \!\bigr]\!\left[\! \begin{array}{r} 1 \\ 0 \\ 0 \end{array} \!\right] & \bigl[\!\begin{array}{ccc} 2 & 1 & 1 \end{array} \!\bigr]\!\left[\! \begin{array}{r} 0 \\ 1 \\ 0 \end{array} \!\right] & \bigl[\!\begin{array}{ccc} 2 & 1 & 1 \end{array} \!\bigr]\!\left[\! \begin{array}{r} -1 \\ 1 \\ 0 \end{array} \!\right] & \bigl[\!\begin{array}{ccc} 2 & 1 & 1 \end{array} \!\bigr]\!\left[\! \begin{array}{r} 0 \\ 0 \\ 1 \end{array} \!\right] & \bigl[\!\begin{array}{ccc} 2 & 1 & 1 \end{array} \!\bigr]\!\left[\! \begin{array}{r} 1 \\ 1 \\ -1 \end{array} \!\right] \\ \bigl[\!\begin{array}{ccc} 1 & 4 & 2 \end{array} \!\bigr]\!\left[\! \begin{array}{r} 1 \\ 0 \\ 0 \end{array} \!\right] & \bigl[\!\begin{array}{ccc} 1 & 4 & 2 \end{array} \!\bigr]\!\left[\! \begin{array}{r} 0 \\ 1 \\ 0 \end{array} \!\right] & \bigl[\!\begin{array}{ccc} 1 & 4 & 2 \end{array} \!\bigr]\!\left[\! \begin{array}{r} -1 \\ 1 \\ 0 \end{array} \!\right] & \bigl[\!\begin{array}{ccc} 1 & 4 & 2 \end{array} \!\bigr]\!\left[\! \begin{array}{r} 0 \\ 0 \\ 1 \end{array} \!\right] & \bigl[\!\begin{array}{ccc} 1 & 4 & 2 \end{array} \!\bigr]\!\left[\! \begin{array}{r} 1 \\ 1 \\ -1 \end{array} \!\right] \\ \end{array} \!\right] \\ & = \left[\! \begin{array}{ccccc} 1\!\cdot\!1 + 3\!\cdot\!0 + 2\!\cdot\!0 & 1\!\cdot\!0 + 3\!\cdot\!1 + 2\!\cdot\!0 & 1\!\cdot\!(-1) + 3\!\cdot\!1 + 2\!\cdot\!0 & 1\!\cdot\!0 + 3\!\cdot\!0 + 2\!\cdot\!1 & 1\!\cdot\!1 + 3\!\cdot\!1 + 2\!\cdot\!(-1) \\ 2\!\cdot\!1 + 0\!\cdot\!0 + 1\!\cdot\!0 & 2\!\cdot\!0 + 0\!\cdot\!1 + 1\!\cdot\!0 & 2\!\cdot\!(-1) + 0\!\cdot\!1 + 1\!\cdot\!0 & 2\!\cdot\!0 + 0\!\cdot\!0 + 1\!\cdot\!1 & 2\!\cdot\!1 + 0\!\cdot\!1 + 1\!\cdot\!(-1) \\ 2\!\cdot\!1 + 1\!\cdot\!0 + 1\!\cdot\!0 & 2\!\cdot\!0 + 1\!\cdot\!1 + 1\!\cdot\!0 & 2\!\cdot\!(-1) + 1\!\cdot\!1 + 1\!\cdot\!0 & 2\!\cdot\!0 + 1\!\cdot\!0 + 1\!\cdot\!1 & 2\!\cdot\!1 + 1\!\cdot\!1 + 1\!\cdot\!(-1) \\ 1\!\cdot\!1 + 4\!\cdot\!0 + 2\!\cdot\!0 & 1\!\cdot\!0 + 4\!\cdot\!1 + 2\!\cdot\!0 & 1\!\cdot\!(-1) + 4\!\cdot\!1 + 2\!\cdot\!0 & 1\!\cdot\!0 + 4\!\cdot\!0 + 2\!\cdot\!1 & 1\!\cdot\!1 + 4\!\cdot\!1 + 2\!\cdot\!(-1) \end{array} \!\right] \\ &= \left[\! \begin{array}{rrrrr} 1 & 3 & 2 & 2 & 2 \\ 2 & 0 & -2 & 1 & 1 \\ 2 & 1 & -1 & 1 & 2 \\ 1 & 4 & 3 & 2 & 3 \end{array} \!\right]. \end{align*}
- Today in class I illustrated action of 2x2 matrix with complex eigenvalues geometrically. I created small Wolfram Mathematica notebook 20241024.nb. You can download this notebook and start using Mathematica.
- To get you started with Mathematica I created my Mathematica website. At the end of the Mathematica website you will find a section about Linear Algebra in Mathematica.
-
In this item I will work out the details how the Hidden Rotation-Dilation Theorem works for a real $3\times 3$ matrix with complex eigenvalues.
Consider the $3\times 3$ matrix
\[
A = \left[\! \begin{array}{ccc}
0 & 0 & 1\\ 1 & 0 & 0 \\ 0 & 1 & 0
\end{array} \!\right].
\]
You might be able to figure out the geometric action of this matrix by looking at what this matrix does to the coordinate vectors.
However, I want to demonstrate how to use the eigenvalues and eigenvectors to figure this action out.
- First calculate the eigenvalues by finding the characteristic polynomial of $A$ \begin{align*} \det\left( A - \lambda I_3 \right) & = \left| \begin{array}{ccc} -\lambda & 0 & 1\\ 1 & -\lambda & 0 \\ 0 & 1 & -\lambda \end{array} \right| \\ & = (-\lambda) \left| \begin{array}{cc} -\lambda & 0 \\ 1 & -\lambda \end{array} \right| + \left| \begin{array}{cc} 1 & -\lambda \\ 0 & 1 \end{array} \right| \\ & = -\lambda^3 + 1. \end{align*} To find all the roots of the cubic equation $\lambda^3 - 1 = 0$, we factor it \[ \lambda^3 - 1 = (\lambda -1) (\lambda^2 + \lambda + 1), \] and find out that \[ \lambda_1 = 1, \quad \lambda_2 = -\frac{1}{2} - i \frac{\sqrt{3}}{2}, \quad \lambda_3 = -\frac{1}{2} + i \frac{\sqrt{3}}{2} \] are the eigenvalues of $A$.
- To calculate an eigenvector corresponding to the eigenvalue $\displaystyle \lambda_2 = -\frac{1}{2} - i \frac{\sqrt{3}}{2}$ we need to find the nullspace of the matrix \[ A - \lambda_2 I_3 = \left[\!\begin{array}{ccc} \frac{1}{2} + i \frac{\sqrt{3}}{2} & 0 & 1 \\ 1 & \frac{1}{2} + i \frac{\sqrt{3}}{2} & 0 \\ 0 & 1 & \frac{1}{2} + i \frac{\sqrt{3}}{2} \end{array}\right] \] It is convenient to introduce the notation \[ z = \frac{1}{2} + i \frac{\sqrt{3}}{2} \] and calculate \begin{align*} \overline{z} & = \frac{1}{2} - i \frac{\sqrt{3}}{2} \\ z^2 & = \frac{1}{4} + i \frac{\sqrt{3}}{2} - \frac{3}{4} = -\frac{1}{4} + i \frac{\sqrt{3}}{2} = - \overline{z} \\ -z^2 & = \overline{z} \\ z \mkern 1.5mu \overline{z} & = \left(\frac{1}{2} + i \frac{\sqrt{3}}{2}\right) \left(\frac{1}{2} - i \frac{\sqrt{3}}{2}\right) = \frac{1}{4} - (-1) \frac{3}{4} = 1 \\ -z \mkern 1.5mu \overline{z} & =-1 \end{align*} Now we calculate the Reduced Row Echelon Form of $A - \lambda_2 I_3$ as follows (at the 1st step we swap the first two rows; at the 2nd step we replace the second row with $(-z)\text{Row1} + \text{Row2}$; at the 3rd step we swap the last two rows; at the 4th step we replace the third row with $(-\overline{z})\text{Row2} + \text{Row3}$) \begin{alignat*}{2} \text{1st} \ & \text{step} \quad \left[\!\begin{array}{ccc} z & 0 & 1 \\ 1 & z & 0 \\ 0 & 1 & z \end{array}\right] && \sim \left[\!\begin{array}{ccc} 1 & z & 0 \\ z & 0 & 1 \\ 0 & 1 & z \end{array}\right] \\ \text{2nd} \ & \text{step} && \sim \left[\!\begin{array}{ccc} 1 & z & 0 \\ 0 & \overline{z} & 1 \\ 0 & 1 & z \end{array}\right] \\ \text{3rd} \ & \text{step} && \sim \left[\!\begin{array}{ccc} 1 & 0 & \overline{z} \\ 0 & 1 & z \\ 0 & \overline{z} & 1 \end{array}\right] \\ \text{4th} \ & \text{step} && \sim \left[\!\begin{array}{ccc} 1 & 0 & \overline{z} \\ 0 & 1 & z \\ 0 & 0 & 0 \end{array}\right] \\ \end{alignat*} To find the eigenspace corresponding to $\displaystyle \lambda_2 = -\frac{1}{2} - i \frac{\sqrt{3}}{2}$ we solve the system $x_1 + \overline{z} x_3 = 0$ and $x_2 + z x_3 = 0$. Here $x_3$ is a free variable, call it $s$, and for $x_1$ and $x_2$ we have \begin{align*} x_1 & = - s \overline{z} = - s \left(\frac{1}{2} - i \frac{\sqrt{3}}{2}\right) \\ x_2 &= - s z = - s \left(\frac{1}{2} + i \frac{\sqrt{3}}{2}\right) \\ x_3 & = s \end{align*} It is convenient to take $s= 2$. Then we get that a corresponding eigenvector is \[ \mathbf{v}_2 = \left[ \begin{array}{c} -1 \\ -1 \\ 2 \end{array} \right] + i \left[ \begin{array}{c} \sqrt{3} \\ -\sqrt{3} \\ 0 \end{array} \right]. \]
- Corresponding eigenvectors : \begin{alignat*}{3} \require{bbox} \lambda_1 & = \bbox[#FF8000]{1} \quad && \text{a corresponding eigenvector is} \quad \mathbf{v}_1 &&= \left[ \bbox[#FF8000]{\begin{array}{c} 1 \\ 1 \\ 1\end{array}} \right] \\ \lambda_2 & = \bbox[#00FF00]{-\frac{1}{2}} - i \bbox[#8080FF]{\frac{\sqrt{3}}{2}} \quad && \text{a corresponding eigenvector is} \quad \mathbf{v}_2 && = \left[ \bbox[#00FF00]{\begin{array}{c} -1 \\ -1 \\ 2 \end{array}} \right] + i \left[ \bbox[#8080FF]{\begin{array}{c} \sqrt{3} \\ -\sqrt{3} \\ 0 \end{array}} \right] \\ \lambda_3 & = \bbox[#00FF00]{-\frac{1}{2}} + i \bbox[#8080FF]{\frac{\sqrt{3}}{2}} \quad && \text{a corresponding eigenvector is} \quad \mathbf{v}_3 && = \left[\bbox[#00FF00]{\begin{array}{c} -1 \\ -1 \\ 2 \end{array}} \right] - i \left[ \bbox[#8080FF]{\begin{array}{c} \sqrt{3} \\ -\sqrt{3} \\ 0\end{array}} \right] \end{alignat*}
- Homework: Verify that the above calculated eigenvalues and eigenvectors are correct.
- The significance of the eigenvector $\left[\! \begin{array}{c} 1 \\ 1 \\ 1\end{array} \!\right]$ is that it remains unchanged under $A$. Not only that, but any scalar multiple of this eigenvector remains unchanged under $A$. For all $x\in \mathbb{R}$ we have: \[ \left[\! \begin{array}{ccc} 0 & 0 & 1\\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{array} \!\right] \left[\! \begin{array}{c} x \\ x \\ x\end{array} \!\right] = \left[\! \begin{array}{c} x \\ x \\ x \end{array} \!\right] \]
- Below we will discover the significance of the real and the imaginary part of the eigenvector $\mathbb{v}_2$: \[ \left[ \bbox[#00FF00]{\begin{array}{c} -1 \\ -1 \\ 2 \end{array}} \right], \quad \left[ \bbox[#8080FF]{\begin{array}{c} \sqrt{3} \\ -\sqrt{3} \\ 0 \end{array}} \right]. \]
- Recall the following equivalences that we established yesterday: \[ A (\bbox[#00FF00]{\mathbf{x}} + i \mkern 1mu \bbox[#8080FF]{\mathbf{y}}) = (\bbox[#00FF00]{a}-i \mkern 1mu \bbox[#8080FF]{b}) (\bbox[#00FF00]{\mathbf{x}} + i \mkern 1mu \bbox[#8080FF]{\mathbf{y}}) \quad \Leftrightarrow \quad \begin{array}{l} A\bbox[#00FF00]{\mathbf{x}} = \phantom{-} \bbox[#00FF00]{a} \mkern 1mu \bbox[#00FF00]{\mathbf{x}} +\bbox[#8080FF]{b} \mkern 1mu \bbox[#8080FF]{\mathbf{y}} \\ A \bbox[#00FF00]{\mathbf{x}} = - \bbox[#8080FF]{b} \bbox[#00FF00]{\mathbf{x}} + \bbox[#00FF00]{a} \bbox[#8080FF]{\mathbf{y}} \end{array} \quad \Leftrightarrow \quad A \left[\! \begin{array}{cc} \bbox[#00FF00]{\mathbf{x}} & \bbox[#8080FF]{\mathbf{y}} \end{array} \!\right] = \left[\! \begin{array}{cc} \bbox[#00FF00]{\mathbf{x}} & \bbox[#8080FF]{\mathbf{y}} \end{array} \!\right] \left[\! \begin{array}{cc} \bbox[#00FF00]{a} & -\bbox[#8080FF]{b} \\ \bbox[#8080FF]{b} & \bbox[#00FF00]{a} \end{array} \!\right]. \]
- Applying the preceding equivalence to the complex eigenvalue $\lambda_2$ and its eigenvector we found above, we get \[ \left[\! \begin{array}{ccc} 0 & 0 & 1\\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{array} \!\right] \left[\! \begin{array}{cc} \bbox[#00FF00]{\begin{array}{c} -1 \\ -1 \\ 2 \end{array}} & \bbox[#8080FF]{\begin{array}{c} \sqrt{3} \\ -\sqrt{3} \\ 0 \end{array}} \end{array}\!\right] = \left[\! \begin{array}{cc} \bbox[#00FF00]{\begin{array}{c} -1 \\ -1 \\ 2 \end{array}} & \bbox[#8080FF]{\begin{array}{c} \sqrt{3} \\ -\sqrt{3} \\ 0 \end{array}} \end{array}\!\right] \left[\! \begin{array}{cc} \bbox[#00FF00]{-\frac{1}{2}} & -\bbox[#8080FF]{\frac{\sqrt{3}}{2}} \\ \bbox[#8080FF]{\frac{\sqrt{3}}{2}} & \bbox[#00FF00]{-\frac{1}{2}} \end{array} \!\right]. \] Since \[ \bbox[#00FF00]{\cos\bigl(2\pi/3\bigr)} = \bbox[#00FF00]{-\frac{1}{2}} \qquad \text{and} \qquad \bbox[#8080FF]{\sin\bigl(2\pi/3\bigr)}= \bbox[#8080FF]{\frac{\sqrt{3}}{2}}, \] the matrix \[ \left[\! \begin{array}{cc} \bbox[#00FF00]{-\frac{1}{2}} & -\bbox[#8080FF]{\frac{\sqrt{3}}{2}} \\ \bbox[#8080FF]{\frac{\sqrt{3}}{2}} & \bbox[#00FF00]{-\frac{1}{2}} \end{array} \!\right] \] represents the counterclockwise rotation by $\theta = 2 \pi/3.$
- The identity that we discovered in the preceding item tells us that the matrix $A$ acts as the rotation by $\theta = 2 \pi/3$ in the plane spanned by the vectors \[ \left[ \bbox[#00FF00]{\begin{array}{c} -1 \\ -1 \\ 2 \end{array}} \right], \quad \left[\bbox[#8080FF]{\begin{array}{c} \sqrt{3} \\ -\sqrt{3} \\ 0 \end{array}} \right] \]
- Now the question arises, how to implement the eigenvalue $\lambda_1 = \bbox[#FF8000]{1}$ and its eigenvector $\left[\bbox[#FF8000]{\begin{array}{c} 1 \\ 1 \\ 1\end{array}} \right]$ to discover an identity corresponding to hidden dilation-rotation for a $3\times 3$ matrix. The following identity holds: \[ \left[\! \begin{array}{ccc} 0 & 0 & 1\\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{array} \!\right] \left[\! \begin{array}{ccc} \bbox[#FF8000]{\begin{array}{c} 1 \\ 1 \\ 1\end{array}} & \bbox[#00FF00]{\begin{array}{c} -1 \\ -1 \\ 2 \end{array}} & \bbox[#8080FF]{\begin{array}{c} \sqrt{3} \\ -\sqrt{3} \\ 0 \end{array}} \end{array} \!\right] = \left[\! \begin{array}{ccc} \bbox[#FF8000]{\begin{array}{c} 1 \\ 1 \\ 1\end{array}} & \bbox[#00FF00]{\begin{array}{c} -1 \\ -1 \\ 2 \end{array}} & \bbox[#8080FF]{\begin{array}{c} \sqrt{3} \\ -\sqrt{3} \\ 0 \end{array}} \end{array} \!\right] \left[\! \begin{array}{ccc} \bbox[#FF8000]{1} & 0 & 0 \\ 0 & \bbox[#00FF00]{-\frac{1}{2}} & -\bbox[#8080FF]{\frac{\sqrt{3}}{2}} \\ 0 & \bbox[#8080FF]{\frac{\sqrt{3}}{2}} & \bbox[#00FF00]{-\frac{1}{2}} \end{array} \!\right]. \] Now calculate \[ \left[\! \begin{array}{ccc} \bbox[#FF8000]{\begin{array}{c} 1 \\ 1 \\ 1\end{array}} & \bbox[#00FF00]{\begin{array}{c} -1 \\ -1 \\ 2 \end{array}} & \bbox[#8080FF]{\begin{array}{c} \sqrt{3} \\ -\sqrt{3} \\ 0 \end{array}} \end{array} \!\right]^{-1} = \frac{1}{6} \left[\! \begin{array}{ccc} 2 & 2 & 2 \\ -1 & -1 & 2 \\ \sqrt{3} & -\sqrt{3} & 0 \end{array} \!\right] \] Thus, the hidden dilation-rotation formula for the $3\times 3$ matrix $A$ is \[ \left[\! \begin{array}{ccc} 0 & 0 & 1\\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{array} \!\right] = \left[\! \begin{array}{ccc} \bbox[#FF8000]{\begin{array}{c} 1 \\ 1 \\ 1\end{array}} & \bbox[#00FF00]{\begin{array}{c} -1 \\ -1 \\ 2 \end{array}} & \bbox[#8080FF]{\begin{array}{c} \sqrt{3} \\ -\sqrt{3} \\ 0 \end{array}} \end{array} \!\right] \left[\! \begin{array}{ccc} \bbox[#FF8000]{1} & 0 & 0 \\ 0 & \bbox[#00FF00]{-\frac{1}{2}} & -\bbox[#8080FF]{\frac{\sqrt{3}}{2}} \\ 0 & \bbox[#8080FF]{\frac{\sqrt{3}}{2}} & \bbox[#00FF00]{-\frac{1}{2}} \end{array} \!\right] \left[\! \begin{array}{ccc} \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\ -\frac{1}{6} & -\frac{1}{6} & \frac{1}{3} \\ \frac{\sqrt{3}}{6} & -\frac{\sqrt{3}}{6} & 0 \end{array} \!\right] . \]
- Homework: Explain the meaning of the formula in the preceding item using the concept of change of coordinates.
- Today we started Section 5.5. Suggested problems for Section 5.5: 1-6, 7-12, 13, 16, 17, 18, 21, 25, 26.
- I rewrote my introduction to Complex Numbers. I added a colored picture of the Complex Plane in which I introduce the conjugation of complex numbers as the reflection across the real axis.
-
We work with complex numbers in this section. Recall the properties of complex conjugation. If $z = a + i b,$ with $a, b \in \mathbb{R},$ then the complex conjugate $\overline{z}$ of $z$ is the complex number $\overline{z} = a - i b.$ The following algebra holds for complex conjugation. For complex numbers $z \in \mathbb{C}$ and $w \in \mathbb{C}$ we have
\[
\overline{z + w} = \overline{z} + \overline{w} \qquad \text{and} \qquad \overline{z \, w} = \overline{z}\mkern 4mu \overline{w}.
\]
To verify these algebraic properties of conjugation we set $z = a + i b$ and $w = c + i d$ with $a, b, c, d \in \mathbb{R}.$
- We calculate \begin{align*} z + w & = (a+c) + i (b + d), \\ \overline{z + w} & = (a+c) - i (b + d), \\ \overline{z} & = a - i b, \\ \overline{w} & = c - i d, \\ \overline{z} + \overline{w} & = (a+c) + i (-b - d) = (a+c) - i (b + d). \end{align*} Comparing the second equality and the last equality above, we deduce $\overline{z + w} = \overline{z} + \overline{w}.$
- We calculate \begin{align*} z \, w & = (a+ib) (c +i d) = (ac - bd) + i (ad + bc) \\ \overline{z \, w} & = (ac - bd) - i (ad + bc) \\ \overline{z} & = a - i b, \\ \overline{w} & = c - i d, \\ \overline{z} \mkern 4mu \overline{w} & = (a - i b) (c - i d) = (ac - bd) + i (-ad-bc) = (ac - bd) - i (ad+bc). \end{align*} Comparing the second equality and the last equality above, we deduce $\overline{z \, w} = \overline{z} \mkern 3mu \overline{w}.$
- Let $z = a + i b,$ with $a, b \in \mathbb{R}.$ The $\overline{z} = z$ if and only if $a - i b = a + i b.$ The last equality is equivalent to $2 i b = 0$ and this equality is equivalent to $b=0.$ And $b=0$ means $z = a \in \mathbb{R}.$ Therefore, for $z \in \mathbb{C}$ we have $\overline{z} = z$ if and only if $z \in \mathbb{R}.$
- A summary of Section 5.5 is as follows. Let $A$ be a real $n\!\times\!n$ matrix. Assume that $A$ has a complex eigenvalue $\lambda = a - i b$, where $a,b \in \mathbb{R}$ with $b\neq0$, and that a corresponding eigenvector is \[ \mathbf{v} = \mathbf{x} + i \mathbf{y} \in\mathbb{C}^n \quad \text{where} \quad \mathbf{x}, \mathbf{y} \in \mathbb{R}^n. \] That is we assume \[ A \mathbf{v} = \lambda \mathbf{v}, \quad \mathbf{v} \neq \mathbf{0}, \quad \operatorname{Im}(\lambda) = -b \neq 0. \] Next we write the preceding equality with the real and imaginary components: \[ A (\mathbf{x} + i \mathbf{y}) = (a-i b) (\mathbf{x} + i \mathbf{y}). \] Now taking the conjugate of both sides of this equation we get \[ \overline{A (\mathbf{x} + i \mathbf{y})} = \overline{(a-i b) (\mathbf{x} + i \mathbf{y})}. \] Next, using the rules for the conjugation we get \[ \overline{A} \, \overline{(\mathbf{x} + i \mathbf{y})} = \overline{(a-i b)} \, \overline{(\mathbf{x} + i \mathbf{y})}. \] Since $A$ consists of real numbers we have $\overline{A} = A.$ Consequently, the preceding equality reads \[ A (\mathbf{x} - i \mathbf{y}) = (a + i b) (\mathbf{x} - i \mathbf{y}). \] The last equality means that $\overline{\lambda} = a + i b$ is an eigenvalue of $A$ and its corresponding eigenvector is $\overline{\mathbf{v}} = \mathbf{x} - i \mathbf{y}.$
- The conclusion from the previous item is: If $\lambda = a - i b$, where $a,b \in \mathbb{R}$ with $b\neq0$, is a complex eigenvalue of a real matrix $A$ and \[ \mathbf{v} = \mathbf{x} + i \mathbf{y} \quad \text{where} \quad \mathbf{x}, \mathbf{y} \in \mathbb{R}^n, \] is a corresponding eigenvector, then the conjugate $\overline{\lambda} = a + i b$ is also an eigenvalue of $A$ and its corresponding eigenvector is the conjugate vector of $\mathbf{v},$ that is $\overline{\mathbf{v}} = \mathbf{x} - i \mathbf{y}.$
- Assume that a real matrix $A$ has a complex eigenvalue $\lambda = a - i b$, where $a,b \in \mathbb{R}$ with $b\neq0$, and that a corresponding eigenvector is \[ \mathbf{v} = \mathbf{x} + i \mathbf{y} \quad \text{where} \quad \mathbf{x}, \mathbf{y} \in \mathbb{R}^n. \] This means that \[ A \bigl(\mathbf{x} + i \mathbf{y}\bigr) = ( a - i b) \bigl(\mathbf{x} + i \mathbf{y}\bigr). \] Using the linearity of the matrix-vector multiplication and algebra with vectors we can rewrite the preceding equality as \[ A \mathbf{x} + i A \mathbf{y} = (a\mkern 1mu \mathbf{x} + b\mkern 1mu \mathbf{y}) + i (- b\mkern 1mu \mathbf{x} + a\mkern 1mu \mathbf{y}). \] Since the vectors $A \mathbf{x}$, $A \mathbf{y}$ and $a\mkern 1mu \mathbf{x} + b\mkern 1mu \mathbf{y}$, $- b\mkern 1mu \mathbf{x} + a\mkern 1mu \mathbf{y}$ are vectors with real entries the preceding equality implies that \begin{align*} A\mkern 1mu \mathbf{x} & = \ \ a\mkern 1mu \mathbf{x} + b\mkern 1mu \mathbf{y} \\ A\mkern 1mu \mathbf{y} & = - b\mkern 1mu \mathbf{x} + a\mkern 1mu \mathbf{y} . \end{align*} The last two vector equalities can be rewritten as one matrix equality \[ A \bigl[ \mathbf{x} \ \ \mathbf{y} \bigr] = \bigl[ A\mkern 1mu\mathbf{x} \ \ A\mkern 1mu\mathbf{y} \bigr] = \bigl[ a\mkern 1mu \mathbf{x} + b\mkern 1mu \mathbf{y} \ \ \ - b\mkern 1mu \mathbf{x} + a\mkern 1mu \mathbf{y} \bigr]. \] The last matrix can be factored as \[ \bigl[ a\mkern 1mu \mathbf{x} + b\mkern 1mu \mathbf{y} \ \ \ - b\mkern 1mu \mathbf{x} + a\mkern 1mu \mathbf{y} \bigr] = \bigl[ \mathbf{x} \ \ \mathbf{y} \bigr] \left[\! \begin{array}{rr} a & -b \\ b & a \end{array} \!\right]. \] Finally, the last two equalities yield \[ A \bigl[ \mathbf{x} \ \ \mathbf{y} \bigr] = \bigl[ \mathbf{x} \ \ \mathbf{y} \bigr] \left[\! \begin{array}{rr} a & -b \\ b & a \end{array} \!\right]. \] Notice that the matrices in the preceding equality are $n\!\times\!2$ matrices.
-
Let $A$ be a real $n\!\times\!n$ matrix. Assume that $A$ has a complex eigenvalue $\lambda = a - i b$, where $a,b \in \mathbb{R}$ with $b\neq0$, and that a corresponding eigenvector is
\[
\mathbf{v} = \mathbf{x} + i \mathbf{y} \in \mathbb{C}^n \quad \text{where} \quad \mathbf{x}, \mathbf{y} \in \mathbb{R}^n.
\]
In class we will prove the following implication: If
\[
A (\mathbf{x} + i \mathbf{y}) = (a - i b) (\mathbf{x} + i \mathbf{y}), \quad \mathbf{x} + i \mathbf{y} \neq \mathbf{0}, \quad b \neq 0,
\]
than
\[
\mathbf{x} \quad \text{and} \quad \mathbf{y} \qquad \text{are linearly independent.}
\]
We will prove this statement in two steps.
Theorem. Let $A$ be a real $n\!\times\!n$ matrix. Let $\lambda = a + i b$ with \(b \neq 0\) be a complex eigenvalue of $A$ with a corresponding eigenvector \(\mathbf{v} \neq \mathbf{0}\), that is \(A \mathbf{v} = \lambda \mathbf{v}\). Then the vectors \(\mathbf{v}\) and its conjugate \(\overline{\mathbf{v}}\) are linearly independent.
Proof. We proved earlier that, since \(A\) is a real matrix, we have \(A \overline{\mathbf{v}} = \overline{\lambda} \overline{\mathbf{v}}\). To prove that \(\mathbf{v}\) and \(\overline{\mathbf{v}}\) are linearly independent we need to prove the implication: For \(\alpha, \beta \in \mathbb{C}\) we have \[ \alpha \mathbf{v} + \beta\mkern 1mu \overline{\mathbf{v}} = \mathbf{0} \quad \Rightarrow \quad \alpha = 0 \ \ \text{and} \ \ \beta = 0. \] Assume \(\alpha \mathbf{v} + \beta\mkern 1mu \overline{\mathbf{v}} = \mathbf{0}\). Apply \(A\) to both sides of the assumed equality and get \begin{equation*} \mathbf{0} = A \mathbf{0} = A \bigl(\alpha \mathbf{v} + \beta\mkern 1mu \overline{\mathbf{v}} \bigr) = \alpha A \mathbf{v} + \beta A \overline{\mathbf{v}} = \alpha \lambda \mathbf{v} + \beta\mkern 1mu \overline{\lambda} \overline{\mathbf{v}}. \end{equation*} Thus, \[ \alpha \lambda \mathbf{v} + \beta\mkern 1mu \overline{\lambda} \overline{\mathbf{v}} = \mathbf{0}. \] Multiplying both sides of the equality \(\alpha \mathbf{v} + \beta\mkern 1mu \overline{\mathbf{v}} = \mathbf{0}\) by \( \overline{\lambda}\) we get \[ \alpha \overline{\lambda} \mathbf{v} + \beta\mkern 1mu \overline{\lambda} \overline{\mathbf{v}} = \mathbf{0}. \] Subtracting the last two displayed equalities we get \[ \mathbf{0} = \alpha \lambda \mathbf{v} - \alpha \overline{\lambda} \mathbf{v} = \alpha \bigl(\lambda - \overline{\lambda} \bigr) \mathbf{v}. \] Since \(\lambda - \overline{\lambda} = 2 i b \neq 0\) and \(\mathbf{v} \neq \mathbf{0}\), the last displayed equality implies \(\alpha = 0\). Substituting \(\alpha = 0\) in \(\alpha \mathbf{v} + \beta\mkern 1mu \overline{\mathbf{v}} = \mathbf{0}\), we get \(\beta\mkern 1mu \overline{\mathbf{v}} = \mathbf{0}\). Since \(\overline{\mathbf{v}} \neq \mathbf{0}\), we deduce that \(\beta = 0\). QED.Theorem. Let $A$ be a real $n\!\times\!n$ matrix. Let $\lambda = a + i b$ with \(b \neq 0\) be a complex eigenvalue of $A$ with a corresponding eigenvector \(\mathbf{v} \neq \mathbf{0}\) and let \(\mathbf{v} = \mathbf{x} + i \mathbf{y}\), with \(\mathbf{x}, \mathbf{y} \in \mathbb{R}^n\). Then \(\mathbf{x}\) and \(\mathbf{y}\) are linearly independent.
Proof. To prove that \(\mathbf{x}\) and \(\mathbf{y}\) are linearly independent we need to prove the implication: For \(\alpha, \beta \in \mathbb{R}\) we have \[ \alpha \mathbf{x} + \beta\mkern 1mu \mathbf{y} = \mathbf{0} \quad \Rightarrow \quad \alpha = 0 \ \ \text{and} \ \ \beta = 0. \] Assume \(\alpha\mkern 1mu \mathbf{x} + \beta\mkern 1mu \mathbf{y} = \mathbf{0}\), recall that \(\overline{\mathbf{v}} = \mathbf{x} - i \mathbf{y}\), and calculate \begin{align*} (\alpha - i \beta) \mathbf{v} + (\alpha + i \beta) \overline{\mathbf{v}} & = (\alpha - i \beta) (\mathbf{x} + i \mathbf{y}) + (\alpha + i \beta) ((\mathbf{x} - i \mathbf{y})) \\ & = \alpha \mathbf{x} + \beta \mathbf{y} + i \alpha \mathbf{y} - i \beta \mathbf{x} + \alpha \mathbf{x} + \beta \mathbf{y} + i \beta \mathbf{x} - i \alpha \mathbf{y} \\ & = 2 (\alpha \mathbf{x} + \beta \mathbf{y}) \\ & = \mathbf{0}. \end{align*} Thus, the assumption \(\alpha\mkern 1mu \mathbf{x} + \beta\mkern 1mu \mathbf{y} = \mathbf{0}\) implies \[ (\alpha - i \beta) \mathbf{v} + (\alpha + i \beta) \overline{\mathbf{v}} = \mathbf{0}. \] Since by the preceding theorem the vectors \( \mathbf{v}\) and \( \overline{\mathbf{v}}\) are linearly independent, we deduce that \[ \alpha - i \beta = 0 \quad \text{and} \quad \alpha + i \beta = 0. \] Therefore \(\alpha = 0\) and \(\beta = 0\). QED. - Now assume that $A$ is a real $2\!\times\!2$ matrix. In this case, since the real vectors $\mathbf{x}$ and $\mathbf{y}$ are linearly independent, the $2\!\times\!2$ matrix $\bigl[ \mathbf{x} \ \ \mathbf{y} \bigr]$ is invertible. Therefore, the equality \[ A \bigl[ \mathbf{x} \ \ \mathbf{y} \bigr] = \bigl[ \mathbf{x} \ \ \mathbf{y} \bigr] \left[\! \begin{array}{rr} a & -b \\ b & a \end{array} \!\right] \] can be rewritten as \[ A = \bigl[ \mathbf{x} \ \ \mathbf{y} \bigr] \left[\! \begin{array}{rr} a & -b \\ b & a \end{array} \!\right] \bigl[ \mathbf{x} \ \ \mathbf{y} \bigr]^{-1}. \] The matrix \[ \left[\! \begin{array}{rr} a & -b \\ b & a \end{array} \!\right] \] is a composition of a scaling and a rotation. To see that factor out $\sqrt{a^2+b^2}$ from the preceding matrix: \[ \sqrt{a^2+b^2} \left[\! \begin{array}{rr} \frac{a}{\sqrt{a^2+b^2}} & -\frac{b}{\sqrt{a^2+b^2}} \\ \frac{b}{\sqrt{a^2+b^2}} & \frac{a}{\sqrt{a^2+b^2}} \end{array} \!\right]. \] Since \[ \left( \frac{a}{\sqrt{a^2+b^2}} \right)^2 + \left( \frac{b}{\sqrt{a^2+b^2}} \right)^2 = 1, \] there exists an angle $\theta \in [0,2\pi)$ such that \[ \cos \theta = \frac{a}{\sqrt{a^2+b^2}} \quad \text{and} \quad \sin \theta = \frac{b}{\sqrt{a^2+b^2}} . \] Thus, with $\alpha = \sqrt{a^2+b^2}$ we have \[ \left[\! \begin{array}{rr} a & -b \\ b & a \end{array} \!\right] = \alpha \left[\! \begin{array}{rr} \cos\theta & -\sin \theta \\ \sin\theta & \cos\theta \end{array} \!\right]. \]
-
The above considerations are summarized in the following theorem which is called the Hidden Rotation-Dilation Theorem:
In the above theorem the matrix \[ \alpha \left[\! \begin{array}{rr} \cos\theta & -\sin \theta \\ \sin\theta & \cos\theta \end{array} \!\right] = \left[\! \begin{array}{rr} a & -b \\ b & a \end{array} \!\right] \] is the Hidden Rotation-Dilation, $\alpha$ dilates and $\theta$ rotates.Theorem. Let $A$ be a real $2\!\times\!2$ matrix with a nonreal eigenvalue $a-i b$ and a corresponding eigenvector $\mathbf{x} + i \mathbf{y}$. Here $a, b \in \mathbb{R},$ $b\neq 0$ and $\mathbf{x}, \mathbf{y}\in \mathbb{R}^2.$ Then the $2\!\times\!2$ matrix \[ P = \bigl[ \mathbf{x} \ \ \mathbf{y} \bigr] \] is invertible and \[ A = \alpha P \left[\! \begin{array}{rr} \cos\theta & -\sin \theta \\ \sin\theta & \cos\theta \end{array} \!\right] P^{-1}, \] where $\alpha = \sqrt{a^2 + b^2}$ and $\theta \in [0, 2\pi)$ is such that \[ \cos \theta = \frac{a}{\sqrt{a^2 + b^2}}, \quad \sin \theta = \frac{b}{\sqrt{a^2 + b^2}}. \]
- Here is an example of the above procedure. Consider the matrix \[ \left[\! \begin{array}{rr} 1 & -3 \\ 6 & 7 \end{array} \!\right]. \] The eigenvalues of this matrix are \[ 4 - 3i \qquad \text{and} \qquad 4+3i. \] Corresponding eigenvectors are \[ \left[\! \begin{array}{r} 1 \\ -1 \end{array} \!\right] + i \left[\! \begin{array}{r} 0 \\ 1 \end{array} \!\right] \qquad \text{and} \qquad \left[\! \begin{array}{r} 1 \\ -1 \end{array} \!\right] - i \left[\! \begin{array}{r} 0 \\ 1 \end{array} \!\right] \] One of the identities for the matrix $\left[\! \begin{array}{rr} 1 & -3 \\ 6 & 7 \end{array} \!\right]$ that we established in the previous item is \[ \left[\! \begin{array}{rr} 1 & -3 \\ 6 & 7 \end{array} \!\right] \left[\! \begin{array}{rr} 1 & 0 \\ -1 & 1 \end{array} \!\right] = \left[\! \begin{array}{rr} 1 & 0 \\ -1 & 1 \end{array} \!\right] \left[\! \begin{array}{rr} 4 & -3 \\ 3 & 4 \end{array} \!\right]. \] Since $\sqrt{4^2+3^2} = 5$ we have \[ \left[\! \begin{array}{rr} 4 & -3 \\ 3 & 4 \end{array} \!\right] = 5 \left[\! \begin{array}{rr} \frac{4}{5} & -\frac{3}{5} \\ \frac{3}{5} & \frac{4}{5} \end{array} \!\right] = 5 \left[\! \begin{array}{rr} \cos\theta & -\sin \theta \\ \sin\theta & \cos\theta \end{array} \!\right], \quad \text{where} \quad \theta = \arccos \frac{4}{5} \approx 0.643501. \] Thus \[ \left[\! \begin{array}{rr} 1 & -3 \\ 6 & 7 \end{array} \!\right] = 5 \left[\! \begin{array}{rr} 1 & 0 \\ -1 & 1 \end{array} \!\right] \left[\! \begin{array}{rr} \frac{4}{5} & -\frac{3}{5} \\ \frac{3}{5} & \frac{4}{5} \end{array} \!\right] \left[\! \begin{array}{rr} 1 & 0 \\ 1 & 1 \end{array} \!\right] \]
- Here is another example of the above procedure. Consider the matrix \[ \left[\! \begin{array}{rr} -1 & 2 \\ -1 & 1 \end{array} \!\right]. \] For this matrix it is interesting to calculate its square \[ \left[\! \begin{array}{rr} -1 & 2 \\ -1 & 1 \end{array} \!\right] \left[\! \begin{array}{rr} -1 & 2 \\ -1 & 1 \end{array} \!\right] = \left[\! \begin{array}{rr} -1 & 0 \\ 0 & -1 \end{array} \!\right] \] and then \[ \left[\! \begin{array}{rr} -1 & 2 \\ -1 & 1 \end{array} \!\right]^4 = \left[\! \begin{array}{rr} -1 & 0 \\ 0 & -1 \end{array} \!\right] \left[\! \begin{array}{rr} -1 & 0 \\ 0 & -1 \end{array} \!\right] = \left[\! \begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array} \!\right]. \] Explain why the fourth power of the given matrix is the identity matrix by using the method presented in this post.
- Today talked about complex numbers, see Appendix B Complex Numbers in the textbook. I wrote my own introduction to complex numbers. There are 11 exercises at the end of my notes.
- Today we went through the details of the example (starting with "Considered the linear mapping" ending with "gave us the inverse of \(M\)") posted yesterday.
- We also reviewed the example (starting with "In Problem 2 on Assignment 1" ending with the end of the post) posted yesterday.
- Today we continued with Section 5.4. Suggested problems are: 1, 3, 4, 5, 6, 7, 9, 15, 17, 27, 28.
-
The content of Section 5.4 can be used to obtain the
standard matrix of a reflection.
- In the picture below we show the reflection across the green line. The unit vector along the green line and a vector orthogonal to it are colored teal: \[ \color{teal}{\mathbf{u}_1} = \color{teal}{\left[\! \begin{array}{c} \cos \theta \\ \sin \theta \end{array} \!\right]}, \ \color{teal}{\mathbf{u}_2} = \color{teal}{\left[\! \begin{array}{r} - \sin \theta \\ \cos \theta \end{array} \!\right]}. \]
-
In the picture below we denote the reflection of a vector $\color{purple}{\mathbf{v}}$ by $T\color{purple}{\mathbf{v}}$.
An illustration of a Reflection across the green line
- Let \[ \color{teal}{\mathcal B} = \bigl\{ \color{teal}{\mathbf{u}_1}, \color{teal}{\mathbf{u}_2} \bigr\} \quad \text{and} \quad \mathcal E = \bigl\{ \mathbf{e}_1 , \mathbf{e}_2 \bigr\}. \] As we learned in Chapter 4 the change of coordinate matrices are \[ \underset{\mathcal{E}\leftarrow\mathcal{B}}{P} = \left[\! \begin{array}{rr} \cos \theta & - \sin \theta \\ \sin \theta & \cos \theta \end{array} \!\right] \quad \text{and} \quad \underset{\mathcal{B}\leftarrow\mathcal{E}}{P} = \left[\! \begin{array}{rr} \cos \theta & \sin \theta \\ - \sin \theta & \cos \theta \end{array} \!\right] \]
- It is clear that the matrix of the reflection $T$ relative to the teal basis $\color{teal}{\mathcal B}$ is \[ \bigl[ T \bigr]_{\color{teal}{\mathcal B}} = \left[\! \begin{array}{rr} 1 & 0 \\ 0 & -1 \end{array} \!\right]. \] This means that if we have the coordinates of a vector $\color{purple}{\mathbf{v}}$ relative to the teal basis $\color{teal}{\mathcal B}$ then we can easily calculate the coordinates of its reflection $T\color{purple}{\mathbf{v}}$ relative to the teal basis $\color{teal}{\mathcal B}:$ \[ \bigl[ T \color{purple}{\mathbf{v}} \bigr]_{\color{teal}{\mathcal B}} = \left[\! \begin{array}{rr} 1 & 0 \\ 0 & -1 \end{array} \!\right] \bigl[ \color{purple}{\mathbf{v}} \bigr]_{\color{teal}{\mathcal B}}. \]
- Now recall the power of the change of coordinates matrix: \[ \bigl[ \color{purple}{\mathbf{v}} \bigr]_{\color{teal}{\mathcal B}} = \underset{\mathcal{B}\leftarrow\mathcal{E}}{P} \bigl[ \color{purple}{\mathbf{v}} \bigr]_{\mathcal E} = \underset{\mathcal{B}\leftarrow\mathcal{E}}{P} \color{purple}{\mathbf{v}} \] and that \[ T \color{purple}{\mathbf{v}} = \bigl[ T \color{purple}{\mathbf{v}} \bigr]_{\mathcal E} = \underset{\mathcal{E}\leftarrow\mathcal{B}}{P} \bigl[ T \color{purple}{\mathbf{v}} \bigr]_{\color{teal}{\mathcal B}} \]
- Now put together the preceding three displayed relations: \[ T \color{purple}{\mathbf{v}} = \underset{\mathcal{E}\leftarrow\mathcal{B}}{P} \bigl[ T \color{purple}{\mathbf{v}} \bigr]_{\color{teal}{\mathcal B}} = \underset{\mathcal{E}\leftarrow\mathcal{B}}{P}\left[\! \begin{array}{rr} 1 & 0 \\ 0 & -1 \end{array} \!\right] \bigl[ \color{purple}{\mathbf{v}} \bigr]_{\color{teal}{\mathcal B}} = \underset{\mathcal{E}\leftarrow\mathcal{B}}{P}\left[\! \begin{array}{rr} 1 & 0 \\ 0 & -1 \end{array} \!\right] \underset{\mathcal{B}\leftarrow\mathcal{E}}{P} \color{purple}{\mathbf{v}}. \] Thus, the matrix of the reflection is \begin{align*} \underset{\mathcal{E}\leftarrow\mathcal{B}}{P}\left[\! \begin{array}{rr} 1 & 0 \\ 0 & -1 \end{array} \!\right] \underset{\mathcal{B}\leftarrow\mathcal{E}}{P} &= \left[\! \begin{array}{rr} \cos \theta & - \sin \theta \\ \sin \theta & \cos \theta \end{array} \!\right] \left[\! \begin{array}{rr} 1 & 0 \\ 0 & -1 \end{array} \!\right] \left[\! \begin{array}{rr} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array} \!\right] \\ & = \left[\! \begin{array}{rr} \cos \theta & - \sin \theta \\ \sin \theta & \cos \theta \end{array} \!\right] \left[\! \begin{array}{rr} \cos \theta & \sin \theta \\ \sin \theta & -\cos \theta \end{array} \!\right] \\ & = \left[\! \begin{array}{cc} (\cos \theta)^2 - (\sin\theta)^2 & 2 (\sin \theta)(\cos\theta) \\ 2 (\sin \theta)(\cos\theta) & (\sin\theta)^2 - (\cos \theta)^2 \end{array} \!\right] \\ & = \color{blue}{\left[\! \begin{array}{rr} \cos(2\theta) & \sin(2\theta) \\ \sin(2\theta) & -\cos(2\theta) \end{array} \!\right]} \end{align*}
- Thus, the standard matrix of the reflection across the green line which makes the angle \(\theta\) with the positive \(x\)-axis is \[ \color{blue}{\left[\! \begin{array}{rr} \cos(2\theta) & \sin(2\theta) \\ \sin(2\theta) & -\cos(2\theta) \end{array} \!\right]}. \]
-
As an exercise consider a similar setting in $\mathbb{R}^3.$ Consider the plane in $\mathbb{R}^3$ given as
\[
\operatorname{Span} \left\{
\left[\!
\begin{array}{r}
1 \\
- 1 \\
0
\end{array}
\!\right],
\left[\!
\begin{array}{r}
0 \\
1 \\
-1
\end{array}
\!\right]
\right\}.
\]
Your task is to find the matrix of the reflection in $\mathbb R^3$ across this plane.
- To help you out I will point out that the vector \[ \left[\! \begin{array}{c} 1 \\ 1 \\ 1 \end{array} \!\right] \] is orthogonal to the given plane. Thus, the reflection across the given plane, call it \(R,\) leaves the vectors \(\left[\! \begin{array}{r} 1 \\ - 1 \\ 0 \end{array} \!\right]\) and \( \left[\! \begin{array}{r} 0 \\ 1 \\ -1 \end{array} \!\right]\) unchanged and reflects the vector \(\left[\! \begin{array}{c} 1 \\ 1 \\ 1 \end{array} \!\right]\) to its opposite vector \(-\left[\! \begin{array}{c} 1 \\ 1 \\ 1 \end{array} \!\right].\) That is \[ R \left[\! \begin{array}{r} 1 \\ - 1 \\ 0 \end{array} \!\right] = \left[\! \begin{array}{r} 1 \\ - 1 \\ 0 \end{array} \!\right], \quad R\left[\! \begin{array}{r} 0 \\ 1 \\ -1 \end{array} \!\right] = \left[\! \begin{array}{r} 0 \\ 1 \\ -1 \end{array} \!\right], \quad R\left[\! \begin{array}{c} 1 \\ 1 \\ 1 \end{array} \!\right] = -\left[\! \begin{array}{c} 1 \\ 1 \\ 1 \end{array} \!\right]. \]
- The problem presented below involves an evaluation mapping for polynomials. This problem is similar to Problems 9 and 10 in Section 5.4.
-
Considered the linear mapping \(T: \mathbb{P}_3 \to \mathbb{R}^4\) defined by
\[
T\bigl(p\bigr) = \left[\! \begin{array}{c} p(-1) \\ p(0) \\ p(1) \\ p(2) \end{array} \!\right] \quad \text{for all} \quad p(x) \in \mathbb{P}_3.
\]
A mapping like this is sometimes called an evaluation mapping. The goal in this item is to find a matrix representation for $T$ relative to the standard bases for \(\mathbb{P}_3\) and \(\mathbb{R}^4.\)
- Recall that the standard basis in \(\mathbb{P}_3\) is the set of all monomials \(\mathcal{M} = \bigl\{1,x,x^2,x^3\bigr\}\) and the standard basis for \(\mathbb{R}^4\) is the set of the columns of the identity matrix \(I_4.\) We denote this basis by \(\mathcal{E}.\)
- The matrix representation for \(T\) relative to the basis \(\mathcal{M}\) of \(\mathbb{P}_3\) and the basis \(\mathcal{E}\) of \(\mathbb{R}^4\) is the matrix \(M\) with the following property \[ M \bigl[p\bigr]_{\mathcal{M}} = \bigl[Tp]_{\mathcal{E}} \quad \text{for all} \quad p(x) \in \mathbb{P}_3. \] In this case we can figure out the matrix \(M\) directly, based on the definition.
- Let \(p(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3.\) Then \[ \bigl[Tp]_{\mathcal{E}} = Tp = \left[\! \begin{array}{c} p(-1) \\ p(0) \\ p(1) \\ p(2) \end{array} \!\right] = \left[\!\begin{array}{l} a_0 - a_1 + a_2 - a_3 \\ a_0 \\ a_0 + a_1 + a_2 + a_3 \\ a_0 + 2 a_1 + 4 a_2 + 8 a_3 \end{array} \!\right]. \] Thus we need a \(4\times 4\) matrix \(M\) such that \[ \left[\!\begin{array}{cccc} \Box & \Box & \Box & \Box \\ \Box & \Box & \Box & \Box \\ \Box & \Box & \Box & \Box \\ \Box & \Box & \Box & \Box \end{array} \!\right] \left[\!\begin{array}{c} a_0 \\ a_1 \\ a_2 \\ a_3 \end{array} \!\right] = \left[\! \begin{array}{l} a_0 - a_1 + a_2 - a_3 \\ a_0 \\ a_0 + a_1 + a_2 + a_3 \\ a_0 + 2 a_1 + 4 a_2 + 8 a_3 \end{array} \!\right]. \] By the definition of the action of a matrix on a vector we can reconstruct the matrix \(M\) \[ \left[\!\begin{array}{cccc} 1 & -1 & 1 & -1 \\ 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 \\ 1 & 2 & 4 & 8 \end{array} \!\right]\left[\!\begin{array}{c} a_0 \\ a_1 \\ a_2 \\ a_3 \end{array} \!\right] = \left[\! \begin{array}{l} a_0 - a_1 + a_2 - a_3 \\ a_0 \\ a_0 + a_1 + a_2 + a_3 \\ a_0 + 2 a_1 + 4 a_2 + 8 a_3 \end{array} \!\right]. \]
- Let us introduce a notation for the monomials in \(\mathcal{M}\): \[ q_0(x) = 1, \quad q_1(x) = x, \quad q_2(x) = x^2, \quad q_3(x) = x^3, \qquad x\in \mathbb{R}. \] By formula (4) in Section 5.4 we have \[ M = \Bigl[ \bigl[ Tq_0\bigr]_{\mathcal{E}} \ \, \bigl[Tq_1\bigr]_{\mathcal{E}} \ \, \bigl[Tq_2\bigr]_{\mathcal{E}} \ \, \bigl[ Tq_3 \bigr]_{\mathcal{E}} \ \, \Bigr] = \left[\!\begin{array}{cccc} 1 & -1 & 1 & -1 \\ 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 \\ 1 & 2 & 4 & 8 \end{array} \!\right] \] which, of course, coincides with what we got above.
- The determinant of \(M\) is equal to \(12.\) Therefore \(M,\) and so \(T,\) is invertible. Let us calculate the matrix representation for \(T^{-1}.\) We will do that not by inverting \(M,\) but by considering polynomials.
- We will use formula (4) in Section 5.4 to determine \(M^{-1}.\) For convenience we use the standard notation for the columns of the identity matrix \(I_4\) \[ \mathbf{e}_1 = \left[\!\begin{array}{c} 1 \\ 0 \\ 0 \\ 0 \end{array} \!\right], \quad \mathbf{e}_2 = \left[\!\begin{array}{c} 0 \\ 1 \\ 0 \\ 0 \end{array} \!\right], \quad\mathbf{e}_3 = \left[\!\begin{array}{c} 0 \\ 0 \\ 1 \\ 0 \end{array} \!\right], \quad\mathbf{e}_4 = \left[\!\begin{array}{c} 0 \\ 0 \\ 0 \\ 1 \end{array} \!\right]. \] Then \[ M^{-1} = \Biggl[ \Bigl[ T^{-1} \mathbf{e}_1 \Bigr]_{\mathcal{M}} \quad \Bigl[ T^{-1} \mathbf{e}_2 \Bigr]_{\mathcal{M}} \quad \Bigl[ T^{-1} \mathbf{e}_3 \Bigr]_{\mathcal{M}} \quad \Bigl[ T^{-1} \mathbf{e}_4 \Bigr]_{\mathcal{M}} \Biggr] \] So, to find \(M^{-1}\) we need to calculate the polynomials \[ T^{-1} \left[\!\begin{array}{c} 1 \\ 0 \\ 0 \\ 0 \end{array} \!\right], \quad T^{-1} \left[\!\begin{array}{c} 0 \\ 1 \\ 0 \\ 0 \end{array} \!\right], \quad T^{-1} \left[\!\begin{array}{c} 0 \\ 0 \\ 1 \\ 0 \end{array} \!\right], \quad T^{-1} \left[\!\begin{array}{c} 0 \\ 0 \\ 0 \\ 1 \end{array} \!\right]. \]
- What does \[ p(x) = T^{-1} \left[\!\begin{array}{c} 1 \\ 0 \\ 0 \\ 0 \end{array} \!\right] \] mean? This means that \[ \left[\! \begin{array}{c} p(-1) \\ p(0) \\ p(1) \\ p(2) \end{array} \!\right] = \left[\!\begin{array}{c} 1 \\ 0 \\ 0 \\ 0 \end{array} \!\right]. \] What we learned about polynomials makes it "easy" to find \(p(x)\) such that \[ p(0) = 0, \quad p(1) = 0, \quad p(2) = 0. \] A possible \(p(x)\) is \[ p(x) = x(x-1)(x-2). \] However, for this \(p(x)\) we have \(p(-1) = -6.\) Since we need \(p(-1) = 1,\) the needed \(p(x)\) 1s \[ p(x) = - \frac{1}{6} x(x-1)(x-2) = 0\cdot 1 - \frac{1}{3} x + \frac{1}{2} x^2 - \frac{1}{6} x^3. \]
- What does \[ p(x) = T^{-1} \left[\!\begin{array}{c} 0 \\ 1 \\ 0 \\ 0 \end{array} \!\right] \] mean? This means that \[ \left[\! \begin{array}{c} p(-1) \\ p(0) \\ p(1) \\ p(2) \end{array} \!\right] = \left[\!\begin{array}{c} 0 \\ 1 \\ 0 \\ 0 \end{array} \!\right]. \] What we learned about polynomials makes it "easy" to find \(p(x)\) such that \[ p(-1) = 0, \quad p(1) = 0, \quad p(2) = 0. \] A possible \(p(x)\) is \[ p(x) = (x+1)(x-1)(x-2). \] However, for this \(p(x)\) we have \(p(0) = 2.\) Since we need \(p(0) = 1,\) the needed \(p(x)\) is \[ p(x) = \frac{1}{2} (x+1)(x-1)(x-2) = 1 - \frac{1}{2} x - x^2 + \frac{1}{2} x^3. \]
- What does \[ p(x) = T^{-1} \left[\!\begin{array}{c} 0 \\ 0 \\ 1 \\ 0 \end{array} \!\right] \] mean? This means that \[ \left[\! \begin{array}{c} p(-1) \\ p(0) \\ p(1) \\ p(2) \end{array} \!\right] = \left[\!\begin{array}{c} 0 \\ 0 \\ 1 \\ 0 \end{array} \!\right]. \] What we learned about polynomials makes it "easy" to find \(p(x)\) such that \[ p(-1) = 0, \quad p(0) = 0, \quad p(2) = 0. \] A possible \(p(x)\) is \[ p(x) = (x+1)x(x-2). \] However, for this \(p(x)\) we have \(p(1) = -2.\) Since we need \(p(1) = 1,\) the needed \(p(x)\) is \[ p(x) = - \frac{1}{2} (x+1)x(x-2) = 0\cdot 1 + x + \frac{1}{2} x^2 - \frac{1}{2} x^3. \]
- What does \[ p(x) = T^{-1} \left[\!\begin{array}{c} 0 \\ 0 \\ 0 \\ 1 \end{array} \!\right] \] mean? This means that \[ \left[\! \begin{array}{c} p(-1) \\ p(0) \\ p(1) \\ p(2) \end{array} \!\right] = \left[\!\begin{array}{c} 0 \\ 0 \\ 0 \\ 1 \end{array} \!\right]. \] What we learned about polynomials makes it "easy" to find \(p(x)\) such that \[ p(-1) = 0, \quad p(0) = 0, \quad p(1) = 0. \] A possible \(p(x)\) is \[ p(x) = (x+1)x(x-1). \] However, for this \(p(x)\) we have \(p(2) = 6.\) Since we need \(p(2) = 1\), the needed \(p(x)\) is \[ p(x) = \frac{1}{6} (x+1)x(x-1) = 0\cdot 1 - \frac{1}{6} x + 0 \cdot x^2 + \frac{1}{6} x^3. \]
- The last four items give us \(M^{-1}\) as follows \[ M^{-1} = \left[\!\begin{array}{cccc} 0 & 1 & 0 & 0 \\ -\frac{1}{3} & -\frac{1}{2} & 1 & -\frac{1}{6} \\ \frac{1}{2} & -1 & \frac{1}{2} & 0 \\ -\frac{1}{6} & \frac{1}{2} & -\frac{1}{2} & \frac{1}{6} \end{array}\!\right] \]
- It remains to verify: \[ M \, M^{-1} = \left[\!\begin{array}{cccc} 1 & -1 & 1 & -1 \\ 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 \\ 1 & 2 & 4 & 8 \end{array} \!\right] \left[\!\begin{array}{cccc} 0 & 1 & 0 & 0 \\ -\frac{1}{3} & -\frac{1}{2} & 1 & -\frac{1}{6} \\ \frac{1}{2} & -1 & \frac{1}{2} & 0 \\ -\frac{1}{6} & \frac{1}{2} & -\frac{1}{2} & \frac{1}{6} \end{array}\!\right] = \left[\!\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \!\right] \] It is really amazing that a calculation with polynomials gave us the inverse of \(M\).
-
Here I will present some interesting linear transformations of the vector space $\mathbb{P}_4$ of polynomials of degree $\leq 4$. In these examples I always calculate the matrix of a linear transformation with respect to the basis of this space which consists of monomials:
\[
\mathcal{M} =\bigl\{1, t, t^2, t^3,t^4 \bigr\}.
\]
Notice that this basis has 5 elements, so the space $\mathbb{P}_4$ is a five-dimensional vector space. Sometimes it is convenient to introduce notation for monomials: We set
\[
\phi_0(t) = 1, \quad \phi_1(t) = t, \quad \phi_2(t) = t^2, \quad \phi_3(t) = t^3, \quad \phi_4(t) = t^4.
\]
- Let $D: \mathbb P_4 \to \mathbb P_4$ be the linear transformation of taking the derivative with respect to $t$. That is, for every $p(t) \in \mathbb P_n$ we set $(Dp)(t) = p'(t).$ Then the matrix representation of $D$ relative to $\mathcal M$ is the following $5\!\times\!5$ matrix \[ \left[\! \begin{array}{ccccc} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 0 & 4 \\ 0 & 0 & 0 & 0 & 0 \\ \end{array}\!\right]. \]
- Let $R: \mathbb P_4 \to \mathbb P_4$ be the linear transformation defined for every $p(t) \in \mathbb P_4$ by \[ (Rp)(t) = t^4 p(1/t). \] Then the matrix representation of $R$ relative to $\mathcal M$ is the following $5\!\times\!5$ matrix \[ Z_{5} = \left[\! \begin{array}{ccccc} 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ \end{array}\!\right]. \]
-
Let $T: \mathbb P_4 \to \mathbb P_4$ be the linear transformation defined for every $p(t) \in \mathbb P_4$ by
\[
(Tp)(t) = (1-t^2)p''(t) - tp'(t).
\]
Then the matrix representation of $T$ relative to $\mathcal M$ is the following $5\!\times\!5$ matrix
\[
\left[
\begin{array}{rrrrr}
0 & 0 & 2 & 0 & 0 \\
0 & -1 & 0 & 6 & 0 \\
0 & 0 & -4 & 0 & 12 \\
0 & 0 & 0 & -9 & 0 \\
0 & 0 & 0 & 0 & -16
\end{array}
\right].
\]
The linear transformation introduced here is related to the Chebyshev differential equation and Chebyshev polynomials of the first kind, see my web-page about this topic using linear algebra.
- In Problem 2 on Assignment 1 we studied a vector space of trigonometric functions. Here is a subspace of that vector space: \[ \mathcal{H} = \operatorname{Span}\bigl\{ 1, (\cos t)^2, (\cos t)^4, (\cos t)^6 \bigr\}. \] We proved that the set \[ \mathcal{B} = \bigl\{ 1, (\cos t)^2, (\cos t)^4, (\cos t)^6 \bigr\}. \] is a basis for the vector space $\mathcal{H}.$
- In Section 5.4 we study the concept of the matrix of a linear transformation. The remarkable linear transformation in the space \[ \mathcal{H} = \operatorname{Span}\bigl\{ 1, (\cos t)^2, (\cos t)^4, (\cos t)^6 \bigr\}. \] is the second derivative: Denote by $T: \mathcal{H} \to \mathcal{H}$ the second derivative \[ \forall\mkern 0mu f \in \mathcal{H} \qquad (T f)(t) = \frac{d^2}{dt^2} f(t). \] It is not immediately clear that the second derivative of a function in $\mathcal{H}$ is again in $\mathcal{H}$. Next we will prove that by calculating the second derivatives of each function in the basis \[ \mathcal{B} = \bigl\{ 1, (\cos t)^2, (\cos t)^4, (\cos t)^6 \bigr\}. \]
- Clearly the second derivative of $1$ is $0$. Therefore, $T(1) = 0 \in \mathcal{H}$ and \[ \require{bbox} \bigl[T (1)\bigr]_{\mathcal{B}} = \left[\bbox[#20FFFF]{\begin{array}{r} 0 \\ 0 \\ 0 \\ 0 \end{array}}\right] \] Let us calculate the second derivative of $(\cos t)^2$: \begin{align*} \frac{d^2}{dt^2} \bigl( (\cos t)^2 \bigr) & = \frac{d}{dt} \bigl(2 (\cos t)(-\sin t) \bigr) \\ & = 2 (-\sin t)(-\sin t) + 2 (\cos t)(-\cos t) \\ & = 2 (\sin t)^2 - 2 (\cos t)^2 \\ & = 2 \bigl(1 - (\cos t)^2 \bigr) - 2 (\cos t)^2 \\ & = 2 - 4 (\cos t)^2. \end{align*} Thus \[ T \bigl((\cos t)^2\bigr) = 2 - 4 (\cos t)^2. \] Therefore $T \bigl((\cos t)^2\bigr) \in \mathcal{H}$ and \[ \require{bbox} \bigl[T \bigl((\cos t)^2\bigr)\bigr]_{\mathcal{B}} = \left[\bbox[#FF8000]{\begin{array}{r} 2 \\ -4 \\ 0 \\ 0 \end{array}}\right] \] Let us calculate the second derivative of $(\cos t)^4$: \begin{align*} \frac{d^2}{dt^2} \bigl( (\cos t)^4 \bigr) & = \frac{d}{dt} \bigl(4 (\cos t)^3(-\sin t) \bigr) \\ & = 4\cdot 3 (\cos t)^2 (-\sin t)^2 + 4 (\cos t)^3(-\cos t) \\ & = 12 (\cos t)^2 \bigl(1 - (\cos t)^2 \bigr) - 4 (\cos t)^4 \\ & = 12 (\cos t)^2 - 16 (\cos t)^4. \end{align*} Thus \[ T \bigl((\cos t)^4\bigr) = 12 (\cos t)^2 - 16 (\cos t)^4. \] Therefore $T \bigl((\cos t)^4\bigr) \in \mathcal{H}$ and \[ \require{bbox} \bigl[T \bigl((\cos t)^4\bigr)\bigr]_{\mathcal{B}} = \left[\bbox[#808020]{\begin{array}{r} 0 \\ 12 \\ -16 \\ 0 \end{array}}\right] \] Let us calculate the second derivative of $(\cos t)^6$: \begin{align*} \frac{d^2}{dt^2} \bigl( (\cos t)^6 \bigr) & = \frac{d}{dt} \bigl(6 (\cos t)^5(-\sin t) \bigr) \\ & = 6\cdot 5 (\cos t)^4 (-\sin t)^2 + 6 (\cos t)^5(-\cos t) \\ & = 30 (\cos t)^4 \bigl(1 - (\cos t)^2 \bigr) - 6 (\cos t)^6 \\ & = 30 (\cos t)^4 - 36 (\cos t)^6. \end{align*} Thus \[ T \bigl((\cos t)^6\bigr) = 30 (\cos t)^4 - 36 (\cos t)^6. \] Therefore $T \bigl((\cos t)^6\bigr) \in \mathcal{H}$ and \[ \require{bbox} \bigl[T \bigl((\cos t)^6\bigr)\bigr]_{\mathcal{B}} = \left[\bbox[#802080]{\begin{array}{r} 0 \\ 0 \\ 30 \\ -36 \end{array}}\right]. \]
- Based on what we calculated in the preceding item, the matrix of the second derivative relative to the basis $\mathcal{B}$ is \[ \bigl[ T \bigr]_{\mathcal{B}} = \left[\! \begin{array}{rrrr} \bbox[#20FFFF]{\begin{array}{r} 0 \\ 0 \\ 0 \\ 0 \end{array}} & \bbox[#FF8000]{\begin{array}{r} 2 \\ -4 \\ 0 \\ 0 \end{array}} & \bbox[#808020]{\begin{array}{r} 0 \\ 12 \\ -16 \\ 0 \end{array}} & \bbox[#802080]{\begin{array}{r} 0 \\ 0 \\ 30 \\ -36 \end{array}} \end{array} \!\right] = \left[\! \begin{array}{rrrr} 0 & 2 & 0 & 0 \\ 0 & -4 & 12 & 0 \\ 0 & 0 & -16 & 30 \\ 0 & 0 & 0 & -36 \end{array} \!\right] . \]
- Let us calculate the eigenvalues and the eigenvectors of the matrix \[ \left[\! \begin{array}{rrrr} 0 & 2 & 0 & 0 \\ 0 & -4 & 12 & 0 \\ 0 & 0 & -16 & 30 \\ 0 & 0 & 0 & -36 \end{array} \!\right]. \] Since this is an upper triangular matrix, its eigenvalues are $0$, $-4$, $-16$, $-36$. The eigenvectors are \begin{align*} \left[\! \begin{array}{rrrr} 0 & 2 & 0 & 0 \\ 0 & -4 & 12 & 0 \\ 0 & 0 & -16 & 30 \\ 0 & 0 & 0 & -36 \end{array} \!\right] \left[\! \begin{array}{r} 1 \\ 0 \\ 0 \\ 0 \end{array} \!\right] &= \phantom{-1}0 \left[\! \begin{array}{r} 1 \\ 0 \\ 0 \\ 0 \end{array} \!\right], \\ \left[\! \begin{array}{rrrr} 0 & 2 & 0 & 0 \\ 0 & -4 & 12 & 0 \\ 0 & 0 & -16 & 30 \\ 0 & 0 & 0 & -36 \end{array} \!\right] \left[\! \begin{array}{r} -1 \\ 2 \\ 0 \\ 0 \end{array} \!\right] & = (-4) \left[\! \begin{array}{r} -1 \\ 2 \\ 0 \\ 0 \end{array} \!\right], \\ \left[\! \begin{array}{rrrr} 0 & 2 & 0 & 0 \\ 0 & -4 & 12 & 0 \\ 0 & 0 & -16 & 30 \\ 0 & 0 & 0 & -36 \end{array} \!\right] \left[\! \begin{array}{r} 1 \\ -8 \\ 8 \\ 0 \end{array} \!\right] &= (-16) \left[\! \begin{array}{r} 1 \\ -8 \\ 8 \\ 0 \end{array} \!\right], \\ \left[\! \begin{array}{rrrr} 0 & 2 & 0 & 0 \\ 0 & -4 & 12 & 0 \\ 0 & 0 & -16 & 30 \\ 0 & 0 & 0 & -36 \end{array} \!\right] \left[\! \begin{array}{r} -1 \\ 18 \\ -48 \\ 32 \end{array} \!\right] &= (-36) \left[\! \begin{array}{r} -1 \\ 18 \\ -48 \\ 32 \end{array} \!\right]. \end{align*}
- The vectors \[ \left[\! \begin{array}{r} 1 \\ 0 \\ 0 \\ 0 \end{array} \!\right], \qquad \left[\! \begin{array}{r} -1 \\ 2 \\ 0 \\ 0 \end{array} \!\right], \qquad \left[\! \begin{array}{r} 1 \\ -8 \\ 8 \\ 0 \end{array} \!\right], \qquad \left[\! \begin{array}{r} -1 \\ 18 \\ -48 \\ 32 \end{array} \!\right], \] are the coordinate vectors relative to the basis \[ \mathcal{B} = \bigl\{ 1, (\cos t)^2, (\cos t)^4, (\cos t)^6 \bigr\} \] of the following functions: \begin{align*} & \phantom{=-}1 \\ & \phantom{=}- 1 + 2 (\cos t)^2\\ & \phantom{=-} 1 - Â 8 (\cos t)^2 + 8 (\cos t)^4 \\ & \phantom{=}- 1 + 18 (\cos t)^2 - 48 (\cos t)^4 +Â 32 (\cos t)^6 \end{align*} and since their coordinate vectors are eigenvectors of the matrix of the second derivative, the above functions are the eigenfunctions of the second derivative. But, we have encountered these functions before: \begin{align*} 1 & = \phantom{-}1 \\ \cos(2t) & = - 1 + 2 (\cos t)^2\\ \cos(4t) & = \phantom{-} 1 - Â 8 (\cos t)^2 + 8 (\cos t)^4 \\ \cos(6t) & = - 1 + 18 (\cos t)^2 - 48 (\cos t)^4 +Â 32 (\cos t)^6 \end{align*}
- Conclusion. In the preceding items, we used linear algebra to discover that the eigenfunctions of the second derivative in the vector space $\mathcal{H}$ are the function \[ \mathcal{C} = \bigl\{ 1, \cos(2 t), \cos(4 t), \cos(6 t) \bigr\}. \] In the post on April 10, we proved that these functions form a basis of $\mathcal{H}$. In this post we have discovered that the matrix of the second derivative relative to the basis $\mathcal{C}$ is a diagonal matrix: \[ \left[\begin{array}{rrrr} 0 & 0 & 0 & 0 \\ 0 & -4 & 0 & 0 \\ 0 & 0 & -16 & 0 \\ 0 & 0 & 0 & -36 \end{array}\right] \]
- Now we found two matrices (relative to two different basis) of the second derivative. What is the relationship between these two matrices? To answer this question, recall that on April 10 we calculated \[ \underset{\mathcal{B}\leftarrow\mathcal{C}}{P} = \left[\begin{array}{rrrr} 1 & -1 & 1 & -1 \\ 0 & 2 & -8 & 18 \\ 0 & 0 & 8 & -48 \\ 0 & 0 & 0 & 32 \end{array}\right] \qquad \text{and} \qquad \underset{\mathcal{C}\leftarrow\mathcal{B}}{P} = \left[\begin{array}{rrrr} 1 & -1 & 1 & -1 \\ 0 & 2 & -8 & 18 \\ 0 & 0 & 8 & -48 \\ 0 & 0 & 0 & 32 \end{array}\right]^{-1} = \left[\!\begin{array}{cccc} 1 & \frac{1}{2} & \frac{3}{8} & \frac{5}{16} \\ 0 & \frac{1}{2} & \frac{1}{2} & \frac{15}{32} \\ 0 & 0 & \frac{1}{8} & \frac{3}{16} \\ 0 & 0 & 0 & \frac{1}{32} \\ \end{array}\!\right] \] The relationship between two matrix representations of the second derivative is as follows: \[ \left[\! \begin{array}{rrrr} 0 & 2 & 0 & 0 \\ 0 & -4 & 12 & 0 \\ 0 & 0 & -16 & 30 \\ 0 & 0 & 0 & -36 \end{array} \!\right] = \underset{\mathcal{B}\leftarrow\mathcal{C}}{P} \left[\begin{array}{rrrr} 0 & 0 & 0 & 0 \\ 0 & -4 & 0 & 0 \\ 0 & 0 & -16 & 0 \\ 0 & 0 & 0 & -36 \end{array}\right] \underset{\mathcal{C}\leftarrow\mathcal{B}}{P} \]
-
Let $A$ be a $3\times 3$ matrix. Last few days we studied the following matrix initial value problem:
Find a $3\times 3$ matrix-valued function $Y(t)$ which satisfies: \[ Y'(t) = A\mkern 1mu Y(t) \quad \text{and} \quad Y(0) = I_3. \]It is a fact from differential equation that the solution of this initial value problem is unique. The unique solution of the above boxed initial value problem is called the matrix-exponential function: \[ Y(t) = {\Large e}^{A\mkern 0.75mu t}. \]
It is important to notice that from the boxed equation evaluated at $t=0$ we obtain \[ Y'(0) = A\mkern 1mu Y(0) = A\mkern 1mu I_3 = A . \] -
The exponential notation introduced in the previous item is analogous to what we learned in a calculus class. Let $a$ be a real number. The unique solution $y(t)$ of the initial \[ y'(t) = a\mkern 0.75mu y(t) \quad \text{and} \quad y(0) = 1 \] is the scalar exponential function \[ y(t) = e^{a\mkern 0.75mu t}. \]
- In the preceding item, instead of a $3\times 3$ matrix $A$ we can consider any square matrix. In the example in the next item we will work with $2\times 2$ matrix functions.
-
Recall the rotation matrix in $\mathbb{R}^2$. The $2\times 2$ rotation matrix
\[
\left[\!\begin{array}{rr}
\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta
\end{array}\!\right]
\]
is the matrix representation of the linear transformation which rotates counterclockwise points of the plane $\mathbb{R}^2$ through an angle $\theta$ about the origin.
- The rotation matrix can be viewed as a matrix-valued function of $\theta$ \[ R(\theta) = \left[\!\begin{array}{rr} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{array}\!\right]. \] The values of this function are $2\times 2$ matrices.
- Let us calculate the derivative of this matrix: \[ \frac{d}{d\theta} \left[\!\begin{array}{rr} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{array}\!\right] = \left[\!\begin{array}{cc} -\sin\theta & -\cos\theta \\ \cos\theta & -\sin\theta \end{array}\!\right]. \]
- We have \[ R(0) = \left[\!\begin{array}{rr} \cos 0 & -\sin 0 \\ \sin 0 & \cos 0 \end{array}\!\right] = \left[\!\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\!\right] \] and \[ R'(0) = \left[\!\begin{array}{rr} -\sin 0 & -\cos 0 \\ \cos 0 & -\sin 0 \end{array}\!\right] = \left[\!\begin{array}{r} 0 \\ 1\end{array} \mkern 4mu \begin{array}{r} -1 \\ 0 \end{array} \!\right] = A. \]
- Let us now verify whether $R(\theta)$ satisfies $R'(\theta) = A\mkern 1mu R(\theta)$: \[ \left[\!\begin{array}{rr} -\sin\theta & -\cos\theta \\ \cos\theta & -\sin\theta \end{array}\!\right] = \left[\!\begin{array}{r} 0 \\ 1\end{array} \mkern 4mu \begin{array}{r} -1 \\ 0 \end{array} \!\right] \left[\!\begin{array}{rr} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{array}\!\right]. \] Since the last identity holds true, we have in fact proved that \[ {\huge e}^{^{\left[\!\begin{array}{r} 0 \\ 1\end{array} \mkern 0.5mu \begin{array}{r} -1 \\ 0 \end{array} \!\right]{\Large \theta}}} = \left[\!\begin{array}{rr} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{array}\!\right]. \]
- The last identity can be rewritten as \[ {\huge e}^{^{\left[\!\begin{array}{r} 0 \\ 1\end{array} \mkern 0.5mu \begin{array}{r} -1 \\ 0 \end{array} \!\right]{\Large \theta}}} = \left[\!\begin{array}{cc} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{array}\!\right] = \left[\!\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\!\right] (\cos\theta) + \left[\!\begin{array}{r} 0 \\ 1\end{array} \mkern 4mu \begin{array}{r} -1 \\ 0 \end{array} \!\right] (\sin \theta). \]
- We will talk more about this when we discuss the complex numbers. The complex numbers are introduced by first introducing the imaginary unit $i$ such that $i^2 = -1$. Since the matrix $\displaystyle \left[\!\begin{array}{r} 0 \\ 1\end{array} \mkern 4mu \begin{array}{r} -1 \\ 0 \end{array} \!\right]$ has the property \[ \left[\!\begin{array}{r} 0 \\ 1\end{array} \mkern 4mu \begin{array}{r} -1 \\ 0 \end{array} \!\right]^2 = \left[\!\begin{array}{r} 0 \\ 1\end{array} \mkern 4mu \begin{array}{r} -1 \\ 0 \end{array} \!\right]\left[\!\begin{array}{r} 0 \\ 1\end{array} \mkern 4mu \begin{array}{r} -1 \\ 0 \end{array} \!\right] = \left[\!\begin{array}{r} -1 \\ 0\end{array} \mkern 4mu \begin{array}{r} 0 \\ -1 \end{array} \!\right] = - \left[\!\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\!\right], \] the matrix $\displaystyle \left[\!\begin{array}{r} 0 \\ 1\end{array} \mkern 4mu \begin{array}{r} -1 \\ 0 \end{array} \!\right]$ can be identified by the imaginary unit $i.$ Similarly the identity matrix $\displaystyle \left[\!\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\!\right]$ can be identified with the real number $1$.
- With this identification, the identity \[ {\huge e}^{^{\left[\!\begin{array}{r} 0 \\ 1\end{array} \mkern 4mu \begin{array}{r} -1 \\ 0 \end{array} \!\right] \Large \theta}} = \left[\!\begin{array}{rr} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{array}\!\right] = \left[\!\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\!\right] (\cos\theta) + \left[\!\begin{array}{r} 0 \\ 1\end{array} \mkern 4mu \begin{array}{r} -1 \\ 0 \end{array} \!\right] (\sin \theta). \] can be rewritten with complex numbers as follows: For all real numbers $\theta$ we have \[ {\Large e}^{i \theta} = (\cos\theta) + i (\sin \theta). \] This is Euler's identity, one of the most famous formulas in Mathematics.
-
In the preceding item we deduced Euler's identity. I want to celebrate it by carving it in a teal plate:
Even better celebration is brought by including $0$ in the formula
-
But, in today's post we established a tight connection between Euler's formula and the matrix exponential:
-
Yesterday I posted about the matrix-valued exponential function $t\mapsto e^{At}$. Here $A$ is a $3\times 3$ matrix. We defined the matrix-valued exponential function $t\mapsto e^{At}$ as to be the unique solution $Y(t)$ of the following problem:
Find a $3\times 3$ matrix-valued function $Y(t)$ which satisfies: \[ Y'(t) = A\mkern 1mu Y(t) \quad \text{and} \quad Y(0) = I_3. \] -
Let us show how to solve the above differential equation when $A$ is a diagonal matrix, call it $D$.
- With a diagonal matrix it is reasonable to expect that the solution will be a diagonal matrix. \begin{multline*} \left[\! \begin{array}{ccc} y_1'(t) & 0 & 0 \\ 0 & y_2'(t) & 0 \\ 0 & 0 & y_3'(t) \end{array} \!\right] = \left[\! \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array} \!\right] \left[\! \begin{array}{ccc} y_1(t) & 0 & 0 \\ 0 & y_2(t) & 0 \\ 0 & 0 & y_3(t) \end{array} \!\right] \\ \text{and} \qquad \left[\! \begin{array}{ccc} y_1(0) & 0 & 0 \\ 0 & y_2(0) & 0 \\ 0 & 0 & y_3(0) \end{array} \!\right] = \left[\! \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \!\right]. \end{multline*}
- But the matrix equation in the preceding item is just a very complicated way of writing three scalar equations \begin{align*} y_1'(t) & = 1\mkern 1mu y_1(t), \quad y_1(0) = 1, \\ y_2'(t) & = 2\mkern 1mu y_2(t), \quad y_2(0) = 1, \\ y_3'(t) & = 2\mkern 1mu y_3(t), \quad y_3(0) = 1. \end{align*} For each of these equations we know the solution \[ y_1(t) = e^t, \qquad y_2(t) = e^{2t}, \qquad y_3(t) = e^{2t}. \]
-
Thus, the matrix-valued function
\[
Y(t) = \left[\!
\begin{array}{ccc}
e^{t} & 0 & 0 \\
0 & e^{2t} & 0 \\
0 & 0 & e^{2t}
\end{array} \!\right]
\]
solves the problem
-
It is quite amazing that, now that we solved the problem $Y'(t) = D Y(t)$ and $Y(0) = I_3$ for a diagonal matrix, we can solve the problem $X'(t) = A X(t)$ and $X(0) = I_3$ for a diagonalizable matrix $A$. That is for a matrix $A$ for which
\[
A = PDP^{-1},
\]
where $P$ is an invertible matrix whose columns are eigenvectors of $A$. The reasoning goes as follows:
- We know the function $Y(t)$ which solves \[ Y'(t) = D Y(t) \quad \text{and} \quad Y(0) = I_3. \]
- The last two equalities we can multiply by matrices $P$ and $P^{-1}$ as follows \[ P\mkern 2mu Y'(t) P^{-1} = P DP^{-1} P\mkern 2mu Y(t) P^{-1} \quad \text{and} \quad P\mkern 2mu Y(0) P^{-1} = P\mkern 2mu I_3 P^{-1} = I_3. \] Since we know that $A = PDP^{-1}$, the last differential equation can be rewritten as \[ P\mkern 2mu Y'(t) P^{-1} = A\mkern 2mu P\mkern 2mu Y(t) P^{-1} \quad \text{and} \quad P\mkern 2mu Y(0) P^{-1} = I_3 \]
- Now set \[ X(t) = P\mkern 2mu Y(t) P^{-1}. \] Then \[ X(0) = P\mkern 2mu Y(0) P^{-1} = I_3 \] and, since $P$ and $P^{-1}$ are constant matrices \[ X'(t) = \frac{d}{dt} \bigl( P\mkern 2mu Y(t) P^{-1} \bigr) = P\mkern 2mu Y'(t) P^{-1}. \] Substituting the last derivative in \[ P\mkern 2mu Y'(t) P^{-1} = A\mkern 2mu P\mkern 2mu Y(t) P^{-1}, \] we see that \[ X'(t) = A \mkern 2mu X(t). \]
- Thus, \[ X(t) = P\mkern 2mu Y(t) P^{-1} \quad \text{solves} \quad X'(t) = A Y(t) \quad \text{and} \quad X(0) = I_3. \]
-
Now recall the diagonalization
\[
\left[\!
\begin{array}{rrr}
3 & 1 & -1 \\
1 & 3 & -1 \\
3 & 3 & -1
\end{array}
\!\right] = \left[\begin{array}{rrr}
\bbox [#FF8000]{\begin{array}{c} 1 \\ 1 \\ 3 \end{array}} &
\bbox[#8080FF]{\begin{array}{r} -1 \\ 1 \\ 0 \end{array}} &
\bbox[#80FF80]{\begin{array}{c} 1 \\ 0 \\ 1 \end{array}}
\end{array} \right]
\left[\!
\begin{array}{rrr}
\bbox [#FF8000]{1} & 0 & 0 \\
0 & \bbox[#8080FF]{2} & 0 \\
0 & 0 & \bbox[#80FF80]{2}
\end{array} \!\right]
\left[\!
\begin{array}{rrr}
-1 & -1 & 1 \\
1 & 2 & -1 \\
3 & 3 & -2
\end{array}
\!\right].
\]
-
We calculated that the matrix
\[
Y(t) = \left[\!
\begin{array}{ccc}
e^{t} & 0 & 0 \\
0 & e^{2t} & 0 \\
0 & 0 & e^{2t}
\end{array} \!\right]
\]
solves the problem
- The reasoning in the previous square (■) item shows that the matrix function \[ X(t) = \left[\begin{array}{rrr} \bbox [#FF8000]{\begin{array}{c} 1 \\ 1 \\ 3 \end{array}} & \bbox[#8080FF]{\begin{array}{r} -1 \\ 1 \\ 0 \end{array}} & \bbox[#80FF80]{\begin{array}{c} 1 \\ 0 \\ 1 \end{array}} \end{array} \right] \left[\! \begin{array}{rrr} \bbox [#FF8000]{e^{t}} & 0 & 0 \\ 0 & \bbox[#8080FF]{e^{2t}} & 0 \\ 0 & 0 & \bbox[#80FF80]{e^{2t}} \end{array} \!\right] \left[\! \begin{array}{rrr} -1 & -1 & 1 \\ 1 & 2 & -1 \\ 3 & 3 & -2 \end{array} \!\right] = \left[\! \begin{array}{rrr} 2 e^{2 t}-e^t & e^{2 t}-e^t & e^t-e^{2 t} \\ e^{2 t}-e^t & 2 e^{2 t}-e^t & e^t-e^{2 t} \\ 3 e^{2 t}-3 e^t & 3 e^{2 t}-3 e^t & 3 e^t-2 e^{2 t} \end{array} \!\right], \] or simplified \[ X(t) = e^t \mkern 1mu \left[\! \begin{array}{rrr} 2 e^{t}-1 & e^{t}-1 & 1-e^{t} \\ e^{t}-1 & 2 e^{t}-1 & 1-e^{t} \\ 3 e^{t}-3 & 3 e^{t}-3 & 3 -2 e^{t} \end{array} \!\right] \] solves the problem \[ X'(t) = \left[\! \begin{array}{rrr} 3 & 1 & -1 \\ 1 & 3 & -1 \\ 3 & 3 & -1 \end{array} \!\right] X(t) \quad \text{and} \quad X(0) = I_3. \]
- Let us verify \[ X(0) = e^0 \mkern 1mu \left[\! \begin{array}{rrr} 2 e^{0}-1 & e^{0}-1 & 1-e^{0} \\ e^{0}-1 & 2 e^{0}-1 & 1-e^{0} \\ 3 e^{0}-3 & 3 e^{0}-3 & 3 -2 e^{0} \end{array} \!\right] = \left[\! \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \!\right] \]
- Let us calculate \[ X'(t) = \frac{d}{dt} \left[\! \begin{array}{rrr} 2 e^{2 t}-e^t & e^{2 t}-e^t & e^t-e^{2 t} \\ e^{2 t}-e^t & 2 e^{2 t}-e^t & e^t-e^{2 t} \\ 3 e^{2 t}-3 e^t & 3 e^{2 t}-3 e^t & 3 e^t-2 e^{2 t} \end{array} \!\right] = \left[\! \begin{array}{rrr} 4 e^{2 t}-e^t & 2 e^{2 t}-e^t & e^t-2 e^{2 t} \\ 2 e^{2 t}-e^t & 4 e^{2 t}-e^t & e^t-2 e^{2 t} \\ 6 e^{2 t}-3 e^t & 6 e^{2 t}-3 e^t & 3 e^t-4 e^{2 t} \end{array} \!\right] \] Or, slightly simplified \[ X'(t) = e^t \mkern 1mu \left[\! \begin{array}{rrr} 4 e^{t}-1 & 2 e^{t}-1 & 1-2 e^{t} \\ 2 e^{t}-1 & 4 e^{t}-1 & 1-2 e^{t} \\ 6 e^{t}-3 & 6 e^{t}-3 & 3-4 e^{t} \end{array} \!\right] \]
- Now we need to verify the equality $X'(t) = AX(t)$, that is \[ e^t \mkern 1mu \left[\! \begin{array}{rrr} 4 e^{t}-1 & 2 e^{t}-1 & 1-2 e^{t} \\ 2 e^{t}-1 & 4 e^{t}-1 & 1-2 e^{t} \\ 6 e^{t}-3 & 6 e^{t}-3 & 3-4 e^{t} \end{array} \!\right] \stackrel{?}{=} e^{t} \left[\! \begin{array}{rrr} 3 & 1 & -1 \\ 1 & 3 & -1 \\ 3 & 3 & -1 \end{array} \!\right] \left[\! \begin{array}{rrr} 2 e^{t}-1 & e^{t}-1 & 1-e^{t} \\ e^{t}-1 & 2 e^{t}-1 & 1-e^{t} \\ 3 e^{t}-3 & 3 e^{t}-3 & 3 -2 e^{t} \end{array} \!\right]. \] Let us verify the entry in the first row and the first column: \begin{align*} 4 e^{t}-1 & \stackrel{?}{=} 3 (2 e^{t}-1) + 1 (e^{t}-1) + (-1) (3 e^{t}-3) \\ & = 6 e^t - 3 + e^t -1 - 3 e^t + 3 \\ & = 4 e^t - 1. \end{align*} Let us verify the entry in the first row and the second column: \begin{align*} 2 e^{t}-1 & \stackrel{?}{=} 3 (e^{t}-1) + 1 (2e^{t}-1) + (-1) (3 e^{t}-3) \\ & = 3 e^t - 3 + 2 e^t -1 - 3 e^t + 3 \\ & = 2 e^t - 1. \end{align*} Let us verify the entry in the first row and the third column: \begin{align*} 1 - 2 e^{t} & \stackrel{?}{=} 3 (1 - e^{t}) + 1 (1 -e^{t}) + (-1) (3 - 2 e^{t}) \\ & = 3 - 3 e^t + 1 - e^t -3 + 2 e^t \\ & = 1 - 2 e^t. \end{align*} Let us verify the entry in the second row and the first column: \begin{align*} 2 e^{t} -1 & \stackrel{?}{=} 1 (2 e^{t}-1) + 3 (e^{t}-1) + (-1) (3 e^{t}-3) \\ & = 2 e^t - 1 + 3 e^t -3 - 3 e^t + 3 \\ & = 2 e^t - 1. \end{align*} Let us verify the entry in the second row and the second column: \begin{align*} 4 e^{t}-1 & \stackrel{?}{=} 1 (e^{t}-1) + 3 (2e^{t}-1) + (-1) (3 e^{t}-3) \\ & = e^t - 1 + 6 e^t -3 - 3 e^t + 3 \\ & = 4 e^t - 1. \end{align*} Let us verify the entry in the second row and the third column: \begin{align*} 1 - 2 e^{t} & \stackrel{?}{=} 1 (1 - e^{t}) + 3 (1 -e^{t}) + (-1) (3 - 2 e^{t}) \\ & = 1 - e^t + 3 - 3 e^t -3 + 2 e^t \\ & = 1 - 2 e^t. \end{align*} Let us verify the entry in the third row and the first column: \begin{align*} 6 e^{t} -3 & \stackrel{?}{=} 3 (2 e^{t}-1) + 3 (e^{t}-1) + (-1) (3 e^{t}-3) \\ & = 6 e^t - 3 + 3 e^t -3 - 3 e^t + 3 \\ & = 6 e^t - 3. \end{align*} Let us verify the entry in the second row and the second column: \begin{align*} 6 e^{t}-3 & \stackrel{?}{=} 3 (e^{t}-1) + 3 (2e^{t}-1) + (-1) (3 e^{t}-3) \\ & = 3 e^t - 3 + 6 e^t -3 - 3 e^t + 3 \\ & = 6 e^t - 3. \end{align*} Let us verify the entry in the second row and the third column: \begin{align*} 3 - 4 e^{t} & \stackrel{?}{=} 3 (1 - e^{t}) + 3 (1 -e^{t}) + (-1) (3 - 2 e^{t}) \\ & = 3 - 3 e^t + 3 - 3 e^t -3 + 2 e^t \\ & = 3 - 4 e^t. \end{align*} So we verified all nine entries of the product $AX(t)$ and confirmed the equality $X'(t) = AX(t)$.
-
We calculated that the matrix
\[
Y(t) = \left[\!
\begin{array}{ccc}
e^{t} & 0 & 0 \\
0 & e^{2t} & 0 \\
0 & 0 & e^{2t}
\end{array} \!\right]
\]
solves the problem
- On Monday and Tuesday we studied the matrix \[ A = \left[\! \begin{array}{rrr} 3 & 1 & -1 \\ 1 & 3 & -1 \\ 3 & 3 & -1 \end{array} \!\right]. \] The most important result (posted on Thursday) was: Conclusion. Since the given matrix $A$ has three linearly independent eigenvectors, it is diagonalizable. The following equality is the diagonalization of $A.$ \[ \left[\! \begin{array}{rrr} 3 & 1 & -1 \\ 1 & 3 & -1 \\ 3 & 3 & -1 \end{array} \!\right] = \left[\begin{array}{rrr} \bbox [#FF8000]{\begin{array}{c} 1 \\ 1 \\ 3 \end{array}} & \bbox[#8080FF]{\begin{array}{r} -1 \\ 1 \\ 0 \end{array}} & \bbox[#80FF80]{\begin{array}{c} 1 \\ 0 \\ 1 \end{array}} \end{array} \right] \left[\! \begin{array}{rrr} \bbox [#FF8000]{1} & 0 & 0 \\ 0 & \bbox[#8080FF]{2} & 0 \\ 0 & 0 & \bbox[#80FF80]{2} \end{array} \!\right] \left[\! \begin{array}{rrr} -1 & -1 & 1 \\ 1 & 2 & -1 \\ 3 & 3 & -2 \end{array} \!\right]. \] Please verify the preceding equality.
- Knowing a diagonalization opens doors for so many calculations with a matrix $A$. First, we can positive integer powers of the matrix $A$. For example \begin{align*} A^{8}&= \bigl(P\mkern 2mu D P^{-1}\bigr)^{8} \\ \text{(the meaning of the 8th power) } & = \bigl(P\mkern 2mu D P^{-1}\bigr)\bigl(P\mkern 2mu D P^{-1}\bigr)\bigl(P\mkern 2mu D P^{-1}\bigr)\bigl(P\mkern 2mu D P^{-1}\bigr)\bigl(P\mkern 2mu D P^{-1}\bigr)\bigl(P\mkern 2mu D P^{-1}\bigr)\bigl(P\mkern 2mu D P^{-1}\bigr)\bigl(P\mkern 2mu D P^{-1}\bigr) \\ \text{(matrix multiplication is associative) } & = P\mkern 2mu D P^{-1} P\mkern 2mu D P^{-1} P\mkern 2mu D P^{-1} P\mkern 2mu D P^{-1} P\mkern 2mu D P^{-1} P\mkern 2mu D P^{-1} P\mkern 2mu D P^{-1} P\mkern 2mu D P^{-1} \\ P^{-1} P = I_3 \ \ & = P D I_3 D I_3 D I_3 D I_3 D I_3 D I_3 D I_3 D P^{-1} \\ D I_3 = D \ \ & = P DDDDDDDD P^{-1} \\ \text{(the meaning of the 8th power) } & = P D^8 P^{-1} \end{align*}
- Any power of a diagonal matrix is easy to calculate. For example, \[ \left[\! \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array} \!\right]^8 = \left[\! \begin{array}{rrr} 1^8 & 0 & 0 \\ 0 & 2^8 & 0 \\ 0 & 0 & 2^8 \end{array} \!\right] = \left[\! \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 256 & 0 \\ 0 & 0 & 256 \end{array} \!\right]. \] Therefore, \begin{align*} \left[\! \begin{array}{rrr} 3 & 1 & -1 \\ 1 & 3 & -1 \\ 3 & 3 & -1 \end{array} \!\right]^8 & = \left[\begin{array}{rrr} \bbox [#FF8000]{\begin{array}{c} 1 \\ 1 \\ 3 \end{array}} & \bbox[#8080FF]{\begin{array}{r} -1 \\ 1 \\ 0 \end{array}} & \bbox[#80FF80]{\begin{array}{c} 1 \\ 0 \\ 1 \end{array}} \end{array} \right] \left[\! \begin{array}{ccc} \bbox [#FF8000]{1} & 0 & 0 \\ 0 & \bbox[#8080FF]{2^8} & 0 \\ 0 & 0 & \bbox[#80FF80]{2^8} \end{array} \!\right] \left[\! \begin{array}{rrr} -1 & -1 & 1 \\ 1 & 2 & -1 \\ 3 & 3 & -2 \end{array} \!\right] \\[5pt] & = \left[\begin{array}{rrr} \bbox [#FF8000]{\begin{array}{c} 1 \\ 1 \\ 3 \end{array}} & \bbox[#8080FF]{\begin{array}{r} -1 \\ 1 \\ 0 \end{array}} & \bbox[#80FF80]{\begin{array}{c} 1 \\ 0 \\ 1 \end{array}} \end{array} \right] \left[\! \begin{array}{ccc} \bbox [#FF8000]{1} & 0 & 0 \\ 0 & \bbox[#8080FF]{256} & 0 \\ 0 & 0 & \bbox[#80FF80]{256} \end{array} \!\right] \left[\! \begin{array}{rrr} -1 & -1 & 1 \\ 1 & 2 & -1 \\ 3 & 3 & -2 \end{array} \!\right] \\[5pt] &= \left[\! \begin{array}{ccc} 2^9-1 & 2^8-1 & 1-2^8 \\[5pt] 2^8-1 & 2^9 - 1 & 1-2^8 \\[5pt] 3 (2^8-1) & 3 (2^8-1) & 3 - 2^9 \end{array} \!\right] \\[5pt] &= \left[\! \begin{array}{rrr} 511 & 255 & -255 \\[5pt] 255 & 511 & -255 \\[5pt] 765 & 765 & -509 \end{array} \!\right] \end{align*}
- This logic extends to other functions. For example, the square root. What is $\sqrt{A}$ for the given $3\times 3$ matrix $A$? We will give a precise answer to this question later. For now let us find a $3\times 3$ matrix $X$ such that \[ X^2 = A. \] There are several such matrices. You can try to calculate exactly how many. One of such matrices is \begin{align*} X & = \left[\begin{array}{rrr} \bbox [#FF8000]{\begin{array}{c} 1 \\ 1 \\ 3 \end{array}} & \bbox[#8080FF]{\begin{array}{r} -1 \\ 1 \\ 0 \end{array}} & \bbox[#80FF80]{\begin{array}{c} 1 \\ 0 \\ 1 \end{array}} \end{array} \right] \left[\! \begin{array}{rrr} \bbox [#FF8000]{1} & 0 & 0 \\ 0 & \bbox[#8080FF]{\sqrt{2}} & 0 \\ 0 & 0 & \bbox[#80FF80]{\sqrt{2}} \end{array} \!\right] \left[\! \begin{array}{rrr} -1 & -1 & 1 \\ 1 & 2 & -1 \\ 3 & 3 & -2 \end{array} \!\right] \\[5pt] &= \left[\! \begin{array}{ccc} 2 \sqrt{2}-1 & \sqrt{2}-1 & 1-\sqrt{2} \\[5pt] \sqrt{2}-1 & 2 \sqrt{2}-1 & 1-\sqrt{2} \\[5pt] 3 \bigl(\sqrt{2}-1\bigr) & 3 \bigl(\sqrt{2}-1\bigr) & 3-2 \sqrt{2} \end{array} \!\right] \end{align*}
- Most importantly, a diagonalization makes it possible do define the matrix-valued exponential function. First recall the exponential function that you studied in calculus.
-
Let $a \in \mathbb{R}.$ The importance of the function $t\mapsto e^{at}$ is that it is the unique solution $y(t)$ of the problem:
Find a function $y(t)$ which satisfies: \[ y'(t) = a y(t) \quad \text{and} \quad y(0) = 1. \]
The goal of Problem 5 on Assignment 1 is to explore analogous problem in which real number $a$ is substituted with a diagonalizable $3\times 3$ matrix $A$.
In analogy with the scalar case, we define the matrix-valued exponential function $t\mapsto e^{At}$ to be the unique solution $Y(t)$ of the following problem:
- Problem 5 on Assignment 1 deals with $3\times 3$ matrices whose entries are functions of a real variable $t$. For example, \[ \left[\!\begin{array}{ccc} 1 & t & \frac{t^2}{2} \\ 0 & 1 & t \\ 0 & 0 & 1 \end{array}\!\right], \quad \left[\!\begin{array}{ccc} \cos (t) & -\sin (t) & 0 \\[5pt] \sin (t) & \cos (t) & 0 \\[5pt] 0 & 0 & e^t \end{array}\!\right], \quad \frac{1}{6} \left[\!\begin{array}{ccc} 5 e^{-t}+1 & 1-e^{-t} & 2 e^{-t}-2 \\[5pt] 1-e^{-t} & 5 e^{-t}+1 & 2 e^{-t}-2 \\[5pt] 2 e^{-t}-2 & 2 e^{-t}-2 & 2 e^{-t}+4 \end{array} \!\right]. \] We will consider only functions which are defined for all real numbers $t$. That is we consider functions $F$ such that \[ F:\mathbb{R} \to \mathbb{R}^{3\times 3} \] ($F$ is defined for all real numbers and the values of $F$ are $3\times 3$ matrices).
- Let \[ F(t) = \left[\!\begin{array}{ccc} f_{11}(t) & f_{12}(t) & f_{13}(t) \\[5pt] f_{21}(t) & f_{22}(t) & f_{23}(t) \\[5pt] f_{31}(t) & f_{32}(t) & f_{33}(t) \end{array} \!\right], \] where all the functions $f_{jk}(t)$ with $j,k \in \{1,2,3\}$ are differentiable functions. Then the derivative $F'(t)$ is the matrix-valued function whose entries are the derivatives of the entries of $F(t)$. That is \[ F'(t) = \left[\!\begin{array}{ccc} f_{11}'(t) & f_{12}'(t) & f_{13}'(t) \\[5pt] f_{21}'(t) & f_{22}'(t) & f_{23}'(t) \\[5pt] f_{31}'(t) & f_{32}'(t) & f_{33}'(t) \end{array} \!\right]. \]
- For example, the derivatives of the three matrix-valued functions that we listed as examples are: \[ \left[\!\begin{array}{ccc} 0 & 1 & t \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array}\!\right], \quad \left[\!\begin{array}{ccc} -\sin (t) & -\cos (t) & 0 \\[5pt] \cos (t) & -\sin (t) & 0 \\[5pt] 0 & 0 & e^t \end{array}\!\right], \quad \frac{1}{6} \left[\!\begin{array}{ccc} -5 e^{-t} & e^{-t} & -2 e^{-t}\\[5pt] e^{-t} & -5 e^{-t} & -2 e^{-t} \\[5pt] -2 e^{-t} & -2 e^{-t} & - 2 e^{-t} \end{array} \!\right]. \]
- You learned many formulas for derivatives in calculus. Here we will bring up only one such formula for matrix valued functions. Let $F(t)$ be a differentiable $3\times 3$ matrix-valued function and let $A$ and $B$ be constant $3\times 3$ matrices. Then \[ \frac{d}{dt} \bigl( A \mkern 1mu F(t) \mkern 1mu B \bigr) = A \mkern 1mu F'(t) \mkern 1mu B. \]
-
Let us now consider each of the three examples of matrix-valued function separately.
- For the first example we have \[ Y(t) = \left[\!\begin{array}{ccc} 1 & t & \frac{t^2}{2} \\ 0 & 1 & t \\ 0 & 0 & 1 \\ \end{array}\!\right], \quad Y'(t) = \left[\!\begin{array}{ccc} 0 & 1 & t \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ \end{array}\!\right] \] and \[ Y(0) = \left[\!\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0\\ 0 & 0 & 1 \\ \end{array}\!\right], \quad Y'(0) = \left[\!\begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ \end{array}\!\right] = A. \] Let us now verify whether $Y'(t) = A Y(t)$ \[ \left[\!\begin{array}{ccc} 0 & 1 & t \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ \end{array}\!\right] = \left[\!\begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ \end{array}\!\right] \left[\!\begin{array}{ccc} 1 & t & \frac{t^2}{2} \\ 0 & 1 & t \\ 0 & 0 & 1 \\ \end{array}\!\right]. \] Since the above identity holds true, we have in fact proved that \[ {\huge e}^{^{\left[\!\begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ \end{array}\!\right] {\huge t}}} = \left[\!\begin{array}{ccc} 1 & t & \frac{t^2}{2} \\ 0 & 1 & t \\ 0 & 0 & 1 \\ \end{array}\!\right]. \]
- For the second example we have \[ Y(t) = \left[\!\begin{array}{ccc} \cos (t) & -\sin (t) & 0 \\ \sin (t) & \cos (t) & 0 \\ 0 & 0 & e^t \\ \end{array}\!\right], \quad Y'(t) = \left[\!\begin{array}{ccc} -\sin (t) & -\cos (t) & 0 \\[5pt] \cos (t) & -\sin (t) & 0 \\[5pt] 0 & 0 & e^t \end{array}\!\right] \] and \[ Y(0) = \left[\!\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0\\ 0 & 0 & 1 \\ \end{array}\!\right], \quad Y'(0) = \left[\!\begin{array}{ccc} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{array}\!\right] = A. \] Let us now verify whether $Y'(t) = A Y(t)$ \[ \left[\!\begin{array}{ccc} -\sin (t) & -\cos (t) & 0 \\[5pt] \cos (t) & -\sin (t) & 0 \\[5pt] 0 & 0 & e^t \end{array}\!\right] = \left[\!\begin{array}{ccc} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array}\!\right] \left[\!\begin{array}{ccc} \cos (t) & -\sin (t) & 0 \\[5pt] \sin (t) & \cos (t) & 0 \\[5pt] 0 & 0 & e^t \end{array}\!\right]. \] Since the above identity holds true, we have in fact proved that \[ {\huge e}^{^{\left[\!\begin{array}{ccc} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{array}\!\right] {\huge t}}} = \left[\!\begin{array}{ccc} \cos (t) & -\sin (t) & 0 \\[5pt] \sin (t) & \cos (t) & 0 \\[5pt] 0 & 0 & e^t \end{array}\!\right]. \]
- For the third example we have \[ Y(t) = \frac{1}{6} \left[\!\begin{array}{ccc} 5 e^{-t}+1 & 1-e^{-t} & 2 e^{-t}-2 \\[5pt] 1-e^{-t} & 5 e^{-t}+1 & 2 e^{-t}-2 \\[5pt] 2 e^{-t}-2 & 2 e^{-t}-2 & 2 e^{-t}+4 \end{array} \!\right], \quad Y'(t) = \frac{1}{6} \left[\!\begin{array}{ccc} -5 e^{-t} & e^{-t} & -2 e^{-t}\\[5pt] e^{-t} & -5 e^{-t} & -2 e^{-t} \\[5pt] -2 e^{-t} & -2 e^{-t} & - 2 e^{-t} \end{array} \!\right] \] and \[ Y(0) = \left[\!\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array}\!\right], \quad Y'(0) = \frac{1}{6} \left[\!\begin{array}{ccc} -5 & 1 & -2 \\ 1 & -5 & 2 \\ -2 & -2 & -2 \end{array}\!\right] = A. \] Let us now verify whether $Y'(t) = A Y(t)$ \[ \frac{1}{6}\left[\!\begin{array}{ccc} -5 e^{-t} & e^{-t} & -2 e^{-t}\\[5pt] e^{-t} & -5 e^{-t} & -2 e^{-t} \\[5pt] -2 e^{-t} & -2 e^{-t} & - 2 e^{-t} \end{array}\!\right] = \frac{1}{6} \left[\!\begin{array}{ccc} -5 & 1 & -2 \\ 1 & -5 & 2 \\ -2 & -2 & -2 \end{array}\!\right] \frac{1}{6}\left[\!\begin{array}{ccc} 5 e^{-t}+1 & 1-e^{-t} & 2 e^{-t}-2 \\[5pt] 1-e^{-t} & 5 e^{-t}+1 & 2 e^{-t}-2 \\[5pt] 2 e^{-t}-2 & 2 e^{-t}-2 & 2 e^{-t}+4 \end{array} \!\right]. \] Since the above identity holds true, we have in fact proved that \[ {\huge e}^{^{\frac{1}{6} \left[\!\begin{array}{ccc} -5 & 1 & -2 \\ 1 & -5 & 2 \\ -2 & -2 & -2 \\ \end{array}\!\right] {\huge t}}} = \frac{1}{6} \left[\!\begin{array}{ccc} 5 e^{-t}+1 & 1-e^{-t} & 2 e^{-t}-2 \\[5pt] 1-e^{-t} & 5 e^{-t}+1 & 2 e^{-t}-2 \\[5pt] 2 e^{-t}-2 & 2 e^{-t}-2 & 2 e^{-t}+4 \end{array} \!\right]. \]
- Read Section 5.1. Suggested problems for Section 5.1: 1, 3, 4, 5, 6, 8, 11, 15, 16, 17, 19, 20, 24-27, 29, 30, 31.
- A related Wikipedia link: Eigenvalue, eigenvector and eigenspace.
- Below are animations of different matrices in action. In each scene the navy blue vector is the image of the sea green vector under the multiplication by a matrix $A$. For easier visualization of the action the heads of vectors leave traces.
- Just looking at the movies you can guess what the eigenvalues and eigenvectors of the featured matrix are. In particular it is easy to see whether an eigenvalue is positive, negative, zero, or complex, ...
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Moreover, looking at the movies, you can also SEE what the matrix used in each movie is. This is done by using, what I called "matrix surgery":
(n-by-n matrix M) (k-th column of the n-by-n identity matrix) = (k-th column of the n-by-n matrix M).
Place the cursor over the image to start the animation.
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- For Section 5.2 do 1-8, 11, 12, 14, 15, (in all these problems you can find eigenvectors as well) 9, 13, 18, 19, 20, 21, 24, 25, 27.
- Two examples follow.
-
Example 1. In this item I will illustrate how to calculate eigenvalues and the corresponding eigenspaces of a specific $3\!\times\!3$ matrix. Consider the matrix
\[
A = \left[\!
\begin{array}{rrr}
3 & 1 & -1 \\
1 & 3 & -1 \\
3 & 3 & -1
\end{array}
\!\right] .
\]
- First we find the characteristic polynomial of this matrix. The characteristic polynomial is the determinant of the following matrix: \[ A - \lambda I_3 = \left[\! \begin{array}{rrr} 3 & 1 & -1 \\ 1 & 3 & -1 \\ 3 & 3 & -1 \end{array} \!\right] - \left[\! \begin{array}{rrr} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{array} \!\right] = \left[\! \begin{array}{ccc} 3-\lambda & 1 & -1 \\ 1 & 3-\lambda & -1 \\ 3 & 3 & -1-\lambda \end{array} \!\right] \] Next we calculate this determinant: \begin{align*} \left|\! \begin{array}{ccc} 3-\lambda & 1 & -1 \\ 1 & 3-\lambda & -1 \\ 3 & 3 & -1-\lambda \end{array} \!\right| &= \left|\! \begin{array}{ccc} 2-\lambda & -2+\lambda & 0 \\ 1 & 3-\lambda & -1 \\ 3 & 3 & -1-\lambda \end{array} \!\right| \\ &= \left|\! \begin{array}{ccc} 2-\lambda & 0 & 0 \\ 1 & 4-\lambda & -1 \\ 3 & 6 & -1-\lambda \end{array} \!\right| \\ &= (2-\lambda) \bigl( (4-\lambda)(-1-\lambda) + 6 \bigr) \\ & = (2-\lambda)\bigl(\lambda^2 - 3 \lambda + 2\bigr) \\ & = -(\lambda - 2)^2 ( \lambda - 1) \end{align*} (At the first equality sign, we subtracted the second row from the first. At the second equality sign, we added the first column to the second. These operations do not change the value of a determinant.)
- Thus the eigenvalues of the matrix $A$ are $1$ and $2.$
- Next we find the eigenspace corresponding to the eigenvalue $1.$ For that we need to find the nullspace of the matrix \[ A - 1 I_3 = \left[\! \begin{array}{rrr} 3 & 1 & -1 \\ 1 & 3 & -1 \\ 3 & 3 & -1 \end{array} \!\right] - \left[\! \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \!\right] = \left[\! \begin{array}{ccc} 2 & 1 & -1 \\ 1 & 2 & -1 \\ 3 & 3 & -2 \end{array} \!\right]. \] So, we row reduce the preceding matrix and find its nullspace: \[ \left[\! \begin{array}{ccc} 2 & 1 & -1 \\ 1 & 2 & -1 \\ 3 & 3 & -2 \end{array} \!\right] \sim \left[\! \begin{array}{ccc} 1 & 2 & -1 \\ 0 & 3 & -1 \\ 0 & 3 & -1 \end{array} \!\right] \sim \left[\! \begin{array}{ccc} 1 & 2 & -1 \\ 0 & 1 & -1/3 \\ 0 & 0 & 0 \end{array} \!\right] \sim \left[\! \begin{array}{ccc} 1 & 0 & -1/3 \\ 0 & 1 & -1/3 \\ 0 & 0 & 0 \end{array} \!\right]. \] Thus, the eigenspace corresponding to the eigenvalue $1$ is the subspace \[ \left\{ \left[\! \begin{array}{c} s/3 \\ s/3 \\ s \end{array} \!\right] \ : \ s \in \mathbb{R} \right\} = \operatorname{Span} \left\{ \left[\! \begin{array}{c} 1 \\ 1 \\ 3 \end{array} \!\right] \right\}. \] Hence one eigenvector is $\left[\! \begin{array}{c} 1 \\ 1 \\ 3 \end{array} \!\right].$
- Next we find the eigenspace corresponding to the eigenvalue $2.$ For that we need to find the nullspace of the matrix \[ A - 2 I_3 = \left[\! \begin{array}{rrr} 3 & 1 & -1 \\ 1 & 3 & -1 \\ 3 & 3 & -1 \end{array} \!\right] - \left[\! \begin{array}{rrr} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array} \!\right] = \left[\! \begin{array}{rrr} 1 & 1 & -1 \\ 1 & 1 & -1 \\ 3 & 3 & -3 \end{array} \!\right]. \] So, we row reduce the preceding matrix and find its nullspace: \[ \left[\! \begin{array}{rrr} 1 & 1 & -1 \\ 1 & 1 & -1 \\ 3 & 3 & -3 \end{array} \!\right] \sim \left[\! \begin{array}{rrr} 1 & 1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \!\right]. \] Thus, the eigenspace is the subspace \[ \left\{ \left[\! \begin{array}{c} -s + t \\ s \\ t \end{array} \!\right] \ : \ s, t \in \mathbb{R} \right\} = \operatorname{Span} \left\{ \left[\! \begin{array}{c} -1 \\ 1 \\ 0 \end{array} \!\right], \left[\! \begin{array}{c} 1 \\ 0 \\ 1 \end{array} \!\right] \right\}. \] Hence two linearly independent eigenvectors corresponding to the eigenvalue $2$ are $\left[\! \begin{array}{c} -1 \\ 1 \\ 0 \end{array} \!\right]$ and $\left[\! \begin{array}{c} 1 \\ 0 \\ 1 \end{array} \!\right].$
- The magic of what we found by now is that we found a basis of $\mathbb{R}^3$ which consists of eigenvectors of $A:$ \[ \left[\bbox[#FF8000]{\begin{array}{c} 1 \\ 1 \\ 3 \end{array}}\right], \quad \left[\bbox[#8080FF]{\begin{array}{r} -1 \\ 1 \\ 0 \end{array}}\right], \quad \left[\bbox[#80FF80]{\begin{array}{c} 1 \\ 0 \\ 1 \end{array}} \right]. \]
- It is easy to verify whether these are really eigenvectors: \begin{align*} \left[\! \begin{array}{rrr} 3 & 1 & -1 \\ 1 & 3 & -1 \\ 3 & 3 & -1 \end{array} \!\right] \left[\bbox [#FF8000]{\begin{array}{c} 1 \\ 1 \\ 3 \end{array}}\right] & = \bbox[#FF8000]{1} \left[\bbox[#FF8000]{\begin{array}{c} 1 \\ 1 \\ 3 \end{array}}\right], \\ \left[\! \begin{array}{rrr} 3 & 1 & -1 \\ 1 & 3 & -1 \\ 3 & 3 & -1 \end{array}\!\right] \left[\bbox[#8080FF]{\begin{array}{r} -1 \\ 1 \\ 0 \end{array}}\right] & = \bbox[#8080FF]{2} \left[\bbox[#8080FF]{\begin{array}{r} -1 \\ 1 \\ 0 \end{array}} \right], \\ \left[\!\begin{array}{rrr} 3 & 1 & -1 \\ 1 & 3 & -1 \\ 3 & 3 & -1 \end{array}\!\right] \left[\bbox[#80FF80]{\begin{array}{c} 1 \\ 0 \\ 1 \end{array}}\right] & = \bbox[#80FF80]{2} \left[\bbox[#80FF80]{ \begin{array}{c} 1 \\ 0 \\ 1 \end{array}}\right]. \end{align*} Yes, they are.
- Three matrix-vector equalities from the preceding item can be expressed as one matrix equality as follows: \[ \left[\! \begin{array}{rrr} 3 & 1 & -1 \\ 1 & 3 & -1 \\ 3 & 3 & -1 \end{array}\!\right] \left[\begin{array}{rrr} \bbox [#FF8000]{\begin{array}{c} 1 \\ 1 \\ 3 \end{array}} & \bbox[#8080FF]{\begin{array}{r} -1 \\ 1 \\ 0 \end{array}} & \bbox[#80FF80]{\begin{array}{c} 1 \\ 0 \\ 1 \end{array}} \end{array} \right] = \left[\begin{array}{rrr} \bbox [#FF8000]{\begin{array}{c} 1 \\ 1 \\ 3 \end{array}} & \bbox[#8080FF]{\begin{array}{r} -1 \\ 1 \\ 0 \end{array}} & \bbox[#80FF80]{\begin{array}{c} 1 \\ 0 \\ 1 \end{array}} \end{array} \right] \left[\! \begin{array}{rrr} \bbox [#FF8000]{1} & 0 & 0 \\ 0 & \bbox[#8080FF]{2} & 0 \\ 0 & 0 & \bbox[#80FF80]{2} \end{array} \!\right] . \] The preceding matrix equality is the first step towards a diagonalization of the given matrix.
- In the next item we will prove that the matrix whose columns are the eigenvectors is invertible. Therefore we can write the matrix equality from the preceding item as follows: \[ \left[\! \begin{array}{rrr} 3 & 1 & -1 \\ 1 & 3 & -1 \\ 3 & 3 & -1 \end{array}\!\right] = \left[\begin{array}{rrr} \bbox [#FF8000]{\begin{array}{c} 1 \\ 1 \\ 3 \end{array}} & \bbox[#8080FF]{\begin{array}{r} -1 \\ 1 \\ 0 \end{array}} & \bbox[#80FF80]{\begin{array}{c} 1 \\ 0 \\ 1 \end{array}} \end{array} \right] \left[\! \begin{array}{rrr} \bbox [#FF8000]{1} & 0 & 0 \\ 0 & \bbox[#8080FF]{2} & 0 \\ 0 & 0 & \bbox[#80FF80]{2} \end{array} \!\right] \left[\begin{array}{rrr} \bbox [#FF8000]{\begin{array}{c} 1 \\ 1 \\ 3 \end{array}} & \bbox[#8080FF]{\begin{array}{r} -1 \\ 1 \\ 0 \end{array}} & \bbox[#80FF80]{\begin{array}{c} 1 \\ 0 \\ 1 \end{array}} \end{array} \right]^{-1} \] The preceding equality is called the diagonalization of $A.$ Although, one more step is needed; we need to find the inverse which appears in the preceding equality. We calculate the inverse in the next item.
- I also made the claim that three eigenvectors are linearly independent. Let us verify that as well. \[ \left[\!\begin{array}{rrr|ccc} \bbox [#FF8000]{\begin{array}{c} 1 \\ 1 \\ 3 \end{array}} & \bbox[#8080FF]{\begin{array}{r} -1 \\ 1 \\ 0 \end{array}} & \bbox[#80FF80]{\begin{array}{c} 1 \\ 0 \\ 1 \end{array}} & \begin{array}{c} 1 \\ 0 \\ 0 \end{array} & \begin{array}{c} 0 \\ 1 \\ 0 \end{array} & \begin{array}{c} 0 \\ 0 \\ 1 \end{array} \end{array} \!\right] \sim \cdots \sim \left[\! \begin{array}{ccc|rrr} 1 & 0 & 0 & -1 & -1 & 1\\ 0 & 1 & 0 & 1 & 2 & -1 \\ 0 & 0 & 1 & 3 & 3 & -2 \end{array} \!\right]. \] We know that the right-hand side matrix in the Reduced Row Echelon Form is the inverse of the matrix whose columns are the eigenvectors. To verify the row reduction above, we calculate: \[ \left[\begin{array}{rrr} \bbox [#FF8000]{\begin{array}{c} 1 \\ 1 \\ 3 \end{array}} & \bbox[#8080FF]{\begin{array}{r} -1 \\ 1 \\ 0 \end{array}} & \bbox[#80FF80]{\begin{array}{c} 1 \\ 0 \\ 1 \end{array}} \end{array} \right] \left[\! \begin{array}{rrr} -1 & -1 & 1 \\ 1 & 2 & -1 \\ 3 & 3 & -2 \end{array} \!\right] = \left[\! \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \!\right]. \]
- Conclusion. Since the given matrix $A$ has three linearly independent eigenvectors, it is diagonalizable. The following equality is the diagonalization of $A.$ \[ \left[\! \begin{array}{rrr} 3 & 1 & -1 \\ 1 & 3 & -1 \\ 3 & 3 & -1 \end{array} \!\right] = \left[\begin{array}{rrr} \bbox [#FF8000]{\begin{array}{c} 1 \\ 1 \\ 3 \end{array}} & \bbox[#8080FF]{\begin{array}{r} -1 \\ 1 \\ 0 \end{array}} & \bbox[#80FF80]{\begin{array}{c} 1 \\ 0 \\ 1 \end{array}} \end{array} \right] \left[\! \begin{array}{rrr} \bbox [#FF8000]{1} & 0 & 0 \\ 0 & \bbox[#8080FF]{2} & 0 \\ 0 & 0 & \bbox[#80FF80]{2} \end{array} \!\right] \left[\! \begin{array}{rrr} -1 & -1 & 1 \\ 1 & 2 & -1 \\ 3 & 3 & -2 \end{array} \!\right] \] Please verify the preceding equality.
-
Example 2. In this item I will illustrate how to calculate eigenvalues and the corresponding eigenspaces of a specific $4\!\times\!4$ matrix. The purpose is to demonstrate a matrix that is not diagonalizable. Consider the matrix
\[
A = \left[\!
\begin{array}{rrrr}
0 & 0 & -1 & -1 \\
-1 & 0 & 0 & 0 \\
2 & 1 & 2 & 1 \\
-2 & -1 & -1 & 0
\end{array}
\!\right] .
\]
- First we find the characteristic polynomial of this matrix. The characteristic polynomial is the determinant of the following matrix: \[ A - \lambda I_4 = \left[\! \begin{array}{rrrr} 0 & 0 & -1 & -1 \\ -1 & 0 & 0 & 0 \\ 2 & 1 & 2 & 1 \\ -2 & -1 & -1 & 0 \end{array} \!\right] - \left[\! \begin{array}{rrrr} \lambda & 0 &0 & 0 \\ 0 & \lambda & 0 & 0 \\ 0 & 0 & \lambda & 0 \\ 0 & 0 & 0 & \lambda \end{array} \!\right] = \left[\! \begin{array}{cccc} -\lambda & 0 & -1 & -1 \\ -1 & -\lambda & 0 & 0 \\ 2 & 1 & 2-\lambda & 1 \\ -2 & -1 & -1 & -\lambda \end{array} \!\right] \] Next we calculate the determinant of the preceding matrix: \begin{align*} \left|\! \begin{array}{cccc} -\lambda & 0 & -1 & -1 \\ -1 & -\lambda & 0 & 0 \\ 2 & 1 & 2-\lambda & 1 \\ -2 & -1 & -1 & -\lambda \end{array} \!\right| & = \left|\!\begin{array}{cccc} -\lambda & \lambda^2 & -1 & -1 \\ -1 & 0 & 0 & 0 \\ 2 & 1 - 2 \lambda & 2-\lambda & 1 \\ -2 & -1 + 2\lambda & -1 & -\lambda \end{array}\!\right| \\[6pt] & = \left|\!\begin{array}{ccc} \lambda^2 & -1 & -1 \\ 1 - 2 \lambda & 2-\lambda & 1 \\ -1 + 2\lambda & -1 & -\lambda \end{array}\!\right| \\[6pt] & = \left|\!\begin{array}{ccc} \lambda^2 & -1 & -1 \\ 1 -2 \lambda & 2-\lambda & 1 \\ 0 & 1-\lambda & 1 -\lambda \end{array}\!\right| \\[6pt] & = (1-\lambda) \left|\!\begin{array}{ccc} \lambda^2 & -1 & -1 \\ 1 -2 \lambda & 2-\lambda & 1 \\ 0 & 1 & 1 \end{array}\!\right| \\[6pt] & = (1-\lambda) \left|\!\begin{array}{ccc} \lambda^2 & 0 & 0 \\ 1 -2 \lambda & 2-\lambda & 1 \\ 0 & 1 & 1 \end{array}\!\right| \\[6pt] & = \lambda^2 (1-\lambda) \left|\!\begin{array}{cc} 2-\lambda & 1 \\ 1 & 1 \end{array}\!\right| \\[6pt] & = \lambda^2 (1-\lambda) (1-\lambda) \end{align*} (At the first equality sign, we subtracted the first column multiplied by $-\lambda$ from the second column. At the second equality sign, we perform the cofactor expansion along the second row. At the third equality sign, we add the second row to the third. At the fourth equality sign, we factor out the common factor $(1-\lambda)$ from the third row. At the fifth equality sign, we add the third row to the first. At the sixth equality sign, we perform the cofactor expansion along the first row. At the last equality sign, we calculate the $2\!\times\!2$ determinant.)
- Thus the eigenvalues of the matrix $A$ are $0$ and $1.$ The algebraic multiplicities of both eigenvalues is $2.$ Next we calculate the geometric multiplicities of these eigenvalues.
- Next we find the eigenspace corresponding to the eigenvalue $0.$ For that we need to find the nullspace of the matrix \[ A - 0 I_4 = \left[\! \begin{array}{rrrr} 0 & 0 & -1 & -1 \\ -1 & 0 & 0 & 0 \\ 2 & 1 & 2 & 1 \\ -2 & -1 & -1 & 0 \end{array} \!\right] \]
- So, we row reduce the matrix $A:$ \[ \left[\! \begin{array}{rrrr} 0 & 0 & -1 & -1 \\ -1 & 0 & 0 & 0 \\ 2 & 1 & 2 & 1 \\ -2 & -1 & -1 & 0 \end{array} \!\right] \sim \cdots \sim \left[\! \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{array} \!\right] \] The nullspace of the preceding matrix is the eigenspace corresponding to the eigenvalue $0$; which we calculate to be \[ \operatorname{Nul}(A-0 I_4) = \left\{ \left[\! \begin{array}{c} 0 \\ s \\ -s \\ s \end{array} \!\right] \ : \ s \in \mathbb{R} \right\} = \operatorname{Span} \left\{ \left[\! \begin{array}{r} 0 \\ 1 \\ -1 \\ 1 \end{array} \!\right] \right\}. \]
- Thus, the eigenspace corresponding to the eigenvalue $0$ is one-dimensional.
- Next we find the eigenspace corresponding to the eigenvalue $1.$ For that we need to find the nullspace of the matrix \[ A - 1 I_4 = \left[\! \begin{array}{rrrr} 0 & 0 & -1 & -1 \\ -1 & 0 & 0 & 0 \\ 2 & 1 & 2 & 1 \\ -2 & -1 & -1 & 0 \end{array} \!\right] - \left[\! \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \!\right] = \left[\! \begin{array}{rrrr} -1 & 0 & -1 & -1 \\ -1 & -1 & 0 & 0 \\ 2 & 1 & 1 & 1 \\ -2 & -1 & -1 & -1 \end{array} \!\right] \]
- So, we row reduce the preceding matrix: \[ \left[\! \begin{array}{rrrr} -1 & 0 & -1 & -1 \\ -1 & -1 & 0 & 0 \\ 2 & 1 & 1 & 1 \\ -2 & -1 & -1 & -1 \end{array} \!\right] \sim \cdots \sim \left[\! \begin{array}{rrrr} 1 & 0 & 1 & 1 \\ 0 & 1 & -1 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \!\right] \] Thus, the eigenspace corresponding to the eigenvalue $1$ is the subspace \[ \operatorname{Nul}(A-1 I_4) = \left\{ \left[\! \begin{array}{c} -s-t \\ s+t \\ s \\ t \end{array} \!\right] \ : \ s, t \in \mathbb{R} \right\} = \operatorname{Span} \left\{ \left[\! \begin{array}{r} -1 \\ 1 \\ 1 \\ 0 \end{array} \!\right], \left[\! \begin{array}{r} -1 \\ 1 \\ 0 \\ 1 \end{array} \!\right] \right\}. \]
- Thus, the eigenspace corresponding to the eigenvalue $1$ is two-dimensional.
- Conclusion. Since we found all eigenspaces of the $4\!\times\!4$ matrix $A$ and these eigenspaces have dimensions $1$ and $2$, we conclude that we can have at most three linearly independent eigenvectors. Consequently, we can not have a basis for $\mathbb R^4$ which consists of eigenvectors of $A.$ This shows that the matrix $A$ is not diagonalizable.
- On Friday we discussed some aspects of the Four Problems from the textbook that relate to vector spaces of trigonometric functions. These problems are in this linked PDF file.
- There is much more about trigonometric functions then what is presented in these four problems. I wrote about it on this webpage. The linked webpage explores the relationship between the powers of the cosine function and the multiple angle cosine functions.
- In the items below I will present some methods do deal with the Four Problems related to trigonometric functions linked above.
- Today we proved that the functions in the following set are linearly independent: \[ \bigl\{ 1, \cos t, (\cos t)^2, (\cos t)^3, (\cos t)^4, (\cos t)^5, (\cos t)^6 \bigr\}. \] What does this mean? This means that the only linear combination of the functions in \(\mathcal{B}\) which gives the zero function is the linear combination with zero coefficients. That is, we have to prove \begin{eqnarray*} \forall t \in\mathbb{R} \quad \alpha_0 + \alpha_1 (\cos t) + \alpha_2 (\cos t)^2 + \alpha_3 (\cos t)^3 + \alpha_4 (\cos t)^4 + \alpha_5 (\cos t)^5 + \alpha_6 (\cos t)^6 = 0 \quad \\ \Longrightarrow \quad \alpha_0 = 0, \ \alpha_1 = 0, \ \alpha_2 = 0, \ \alpha_3 = 0 \ \alpha_4 = 0, \ \alpha_5 = 0, \ \alpha_6 = 0. \end{eqnarray*} To prove this implication we assume \[ \forall t \in\mathbb{R} \quad \alpha_0 + \alpha_1 (\cos t) + \alpha_2 (\cos t)^2 + \alpha_3 (\cos t)^3 + \alpha_4 (\cos t)^4 + \alpha_5 (\cos t)^5 + \alpha_6 (\cos t)^6 = 0. \] Following our experience with other proofs of this kind, we could choose \(7\) values of \(t\) and obtain a homogeneous system of seven linear equations with seven unknowns \(\alpha_j, j \in\{0,1,\ldots,6\}\). However, to prove that such a system has only the trivial solution is not likely to be done by hand. Therefore in the textbook there is [M] next to this problem. [M] meaning that a usage of technology is recommended. I will use Wolfram Mathematica.
- Computer algebra system Wolfram Mathematica will be very useful in this class. To get started with Mathematica see my Mathematica page. Please watch the videos that are on my Mathematica page. Watching the movies is very helpful to get started with Mathematica efficiently! Mathematica is available in the computer labs in BH 215, HH 233 and in the Math Center BH 209/211A.
- In class I created a small Mathematica notebook 20241004.nb in which I illustrated the first proof below. I also created 20231031.pdf, for you to enjoy this file without access to Mathematica.
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We will present two proofs. The first proof is using Wolfram Mathematica. The second proof uses differentiation which you learned in Calculus.
- Proof 1. Since we assume \[ \forall t \in\mathbb{R} \quad \alpha_0 + \alpha_1 (\cos t) + \alpha_2 (\cos t)^2 + \alpha_3 (\cos t)^3 + \alpha_4 (\cos t)^4 + \alpha_5 (\cos t)^5 + \alpha_6 (\cos t)^6 = 0, \] the preceding equality holds for the following seven values of \(t\): \[ 0,\quad \frac{\pi}{6},\quad \frac{\pi}{3},\quad \frac{\pi}{2}, \quad \frac{2\pi }{3}, \quad \frac{5 \pi}{6},\quad \pi. \] Substituting these seven values of \(t\) in the equation \[ \alpha_0 + \alpha_1 (\cos t) + \alpha_2 (\cos t)^2 + \alpha_3 (\cos t)^3 + \alpha_4 (\cos t)^4 + \alpha_5 (\cos t)^5 + \alpha_6 (\cos t)^6 = 0, \] we get a homogeneous system of seven linear equations with seven unknowns \[ \alpha_0, \quad \alpha_1, \quad \alpha_2, \quad \alpha_3, \quad \alpha_4, \quad \alpha_5, \quad \alpha_6. \] Mathematica gives us the matrix of this system: \[ \left[ \begin{array}{ccccccc} 1 & 1 & 1 & 1 & 1 & 1 & 1 \\[3pt] 1 & \frac{\sqrt{3}}{2} & \frac{3}{4} & \frac{3 \sqrt{3}}{8} & \frac{9}{16} & \frac{9 \sqrt{3}}{32} & \frac{27}{64} \\[3pt] 1 & \frac{1}{2} & \frac{1}{4} & \frac{1}{8} & \frac{1}{16} & \frac{1}{32} & \frac{1}{64} \\[3pt] 1 & 0 & 0 & 0 & 0 & 0 & 0 \\[3pt] 1 & -\frac{1}{2} & \frac{1}{4} & -\frac{1}{8} & \frac{1}{16} & -\frac{1}{32} & \frac{1}{64} \\[3pt] 1 & -\frac{\sqrt{3}}{2} & \frac{3}{4} & -\frac{3 \sqrt{3}}{8} & \frac{9}{16} & -\frac{9 \sqrt{3}}{32} & \frac{27}{64} \\[3pt] 1 & -1 & 1 & -1 & 1 & -1 & 1 \end{array} \right] \] and its determinant is \[ -\frac{27 \sqrt{3}}{8192}. \] Thus, the homogeneous system of seven linear equations with seven unknowns that we obtained for \[ \alpha_0, \quad \alpha_1, \quad \alpha_2, \quad \alpha_3, \quad \alpha_4, \quad \alpha_5, \quad \alpha_6. \] has only the trivial solution. This completes the proof of linear independence.
- Proof 2. We assume \[ \forall t \in\mathbb{R} \quad \alpha_0 + \alpha_1 (\cos t) + \alpha_2 (\cos t)^2 + \alpha_3 (\cos t)^3 + \alpha_4 (\cos t)^4 + \alpha_5 (\cos t)^5 + \alpha_6 (\cos t)^6 = 0. \] Since we assume that the last displayed identity holds for all $t\in\mathbb{R},$ substituting $t=\pi/2$ yields \[ \alpha_0 = 0. \] Therefore, the assumption becomes \[ \forall t \in\mathbb{R} \quad \alpha_1 (\cos t) + \alpha_2 (\cos t)^2 + \alpha_3 (\cos t)^3 + \alpha_4 (\cos t)^4 + \alpha_5 (\cos t)^5 + \alpha_6 (\cos t)^6 = 0. \] Substituting \[ s = \cos t, \] we obtain \[ \forall s \in [-1,1] \quad \alpha_1 s + \alpha_2 s^2 + \alpha_3 s^3 + \alpha_4 s^4 + \alpha_5 s^5 + \alpha_6 s^6= 0. \] Differentiating the preceding identity with respect to $s \in (-1,1)$ we obtain \[ \forall s \in (-1,1) \quad \alpha_1 + 2\alpha_2 s + 3\alpha_3 s^2 + 4\alpha_4 s^3 + 5\alpha_5 s^4 + 6\alpha_6 s^5= 0. \] Substituting \(s=0\) in the preceding identity we obtain \(\alpha_1=0\). Differentiating the preceding identity again with respect to $s \in (-1,1)$ we obtain \[ \forall s \in (-1,1) \quad 2 \alpha_2 + 6 \alpha_3 s + 12\alpha_4 s^2 + 20 \alpha_5 s^3 + 30\alpha_6 s^4 = 0. \] Substituting \(s=0\) in the preceding identity we obtain \(\alpha_2=0\). Differentiating the preceding identity with respect to $s \in (-1,1)$ we obtain \[ \forall s \in (-1,1)\quad 6 \alpha_3 + 24 \alpha_4 s + 60 \alpha_5 s^2 + 120 \alpha_6 s^3= 0. \] Substituting \(s=0\) in the preceding identity we obtain \(\alpha_3=0\). Differentiating the preceding identity with respect to $s \in (-1,1)$ we obtain \[ \forall s \in (-1,1)\quad 24 \alpha_4 + 120 \alpha_5 s + 360 \alpha_6 s^2= 0. \] Substituting \(s=0\) in the preceding identity we obtain \(\alpha_4=0\). Differentiating the preceding identity with respect to $s \in (-1,1)$ we obtain \[ \forall s \in (-1,1)\quad 120 \alpha_5 + 720 \alpha_6 s = 0. \] Substituting \(s=0\) in the preceding identity we obtain \(\alpha_5=0\). Differentiating the preceding identity with respect to $s \in (-1,1)$ we obtain \[ \forall s \in (-1,1)\quad 720 \alpha_6 = 0. \] Hence, \(\alpha_6 =0\). Thus, without use of technology we proved that the given seven functions are linearly independent. This completes the proof.
- Using Mathematical Induction, which in this case is just an example of Recursive Reasoning, the second proof can be adopted for any number of powers of the cosine function.
- Set \[ \mathcal{H} = \operatorname{Span}\bigl\{ 1, (\cos t)^2, (\cos t)^4, (\cos t)^6 \bigr\} \] and \[ \mathcal{B} = \bigl\{ 1, (\cos t)^2, (\cos t)^4, (\cos t)^6 \bigr\}. \] It can be deduced from the preceding item that $\mathcal{B}$ is a basis for $\mathcal{H}.$
- The first of the following three trigonometric identities is called the double angle identity for cosine. The second can be deduced from the first, and the third one follows from the first two and the addition identity for cosine. \begin{align*} \cos(2t) & = - 1 + 2 (\cos t)^2\\ \cos(4t) & = \phantom{-} 1 - Â 8 (\cos t)^2 + 8 (\cos t)^4 \\ \cos(6t) & = - 1 + 18 (\cos t)^2 - 48 (\cos t)^4 +Â 32 (\cos t)^6 \end{align*} Now, knowing that \(\mathcal{B}\) is a basis for \(\mathcal{H}\), in the language of linear algebra, these identities can be expressed as follows \[ \cos(2t) \in \mathcal{H}, \quad \cos(4t) \in \mathcal{H}, \quad\cos(6t) \in \mathcal{H} \] and \[ \bigl[ 1 \bigr]_{\mathcal{B}}=\left[\begin{array}{r} 1 \\ 0 \\ 0 \\ 0 \end{array}\right], \quad \bigl[\cos(2t)\bigr]_{\mathcal{B}}=\left[\begin{array}{r} -1 \\ 2 \\ 0 \\ 0 \end{array}\right], \quad \bigl[\cos(4t)\bigr]_{\mathcal{B}}=\left[\begin{array}{r} 1 \\ -8 \\ 8 \\ 0 \end{array}\right], \quad \bigl[\cos(6t)\bigr]_{\mathcal{B}}=\left[\begin{array}{r} -1 \\ 18 \\ -48 \\ 32 \end{array}\right]. \] It follows from Theorem 8 in Section 4.4 in the textbook that the functions $1,$ $\cos(2t),$ $\cos(4t),$ $\cos(6t)$ are linearly independent if and only if their coordinates are linearly independent. That is the functions $1,$ $\cos(2t),$ $\cos(4t),$ $\cos(6t)$ are linearly independent if and only if the vectors \[ \left[\begin{array}{r} 1 \\ 0 \\ 0 \\ 0 \end{array}\right], \quad \left[\begin{array}{r} -1 \\ 2 \\ 0 \\ 0 \end{array}\right], \quad \left[\begin{array}{r} 1 \\ -8 \\ 8 \\ 0 \end{array}\right], \quad \left[\begin{array}{r} -1 \\ 18 \\ -48 \\ 32 \end{array}\right] \] are linearly independent. Since \[ \det \left[\begin{array}{rrrr} 1 & -1 & 1 & -1 \\ 0 & 2 & -8 & 18 \\ 0 & 0 & 8 & -48 \\ 0 & 0 & 0 & 32 \end{array}\right] = 1\cdot 2 \cdot 8 \cdot 32 = 512 \neq 0 \] the coordinate vectors are linearly independent. Therefore, the functions $1,$ $\cos(2t),$ $\cos(4t),$ $\cos(6t)$ are linearly independent. Consequently, the functions $1,$ $\cos(2t),$ $\cos(4t),$ $\cos(6t)$ form a basis for the vector space $\mathcal{H}.$
- Set \[ \mathcal{C} = \bigl\{ 1, \cos(2 t), \cos(4 t), \cos(6 t) \bigr\}. \] Then $\mathcal{C}$ is a basis for the vector space $\mathcal{H}.$ Read the post of Tuesday, October 1, 2024. Based on that post we deduce that \[ \underset{\mathcal{B}\leftarrow\mathcal{C}}{P} = \left[\begin{array}{rrrr} 1 & -1 & 1 & -1 \\ 0 & 2 & -8 & 18 \\ 0 & 0 & 8 & -48 \\ 0 & 0 & 0 & 32 \end{array}\right]. \] Therefore \[ \underset{\mathcal{C}\leftarrow\mathcal{B}}{P} = \left[\begin{array}{rrrr} 1 & -1 & 1 & -1 \\ 0 & 2 & -8 & 18 \\ 0 & 0 & 8 & -48 \\ 0 & 0 & 0 & 32 \end{array}\right]^{-1} = \left[\!\begin{array}{cccc} 1 & \frac{1}{2} & \frac{3}{8} & \frac{5}{16} \\ 0 & \frac{1}{2} & \frac{1}{2} & \frac{15}{32} \\ 0 & 0 & \frac{1}{8} & \frac{3}{16} \\ 0 & 0 & 0 & \frac{1}{32} \\ \end{array}\!\right] = \left[ \bigl[ 1 \bigr]_{\mathcal{C}} \ \ \bigl[ (\cos t)^2 \bigr]_{\mathcal{C}} \ \ \bigl[ (\cos t)^4 \bigr]_{\mathcal{C}} \ \ \bigl[ (\cos t)^6 \bigr]_{\mathcal{C}} \right]. \] Therefore, for example, \[ (\cos t)^6 = \frac{5}{16} + \frac{15}{32} \cos (2 t)+\frac{3}{16} \cos (4 t)+\frac{1}{32} \cos (6 t). \] Thus, we have discovered a trigonometric identity using linear algebra only.
- Today we discussed the concept of a basis in the context of polynomials. In the textbook that is Section 4.4, specifically Example 5, the theory before it and Problems 13 and 14. These two problems deals with the space $\mathbb{P}_2$ of all polynomials of the degree less or equal than $2$. Below I use the space $\mathbb{P}_3$ of all polynomials of the degree less or equal than $3$ to illustrate the basic concepts.
- Here we use the definition of a linearly independent set on page 210 in the textbook, which I reviewed in yesterday's post. We also reviewed the concept of a basis from Section 4.3.
- Important examples of finite dimensional vector spaces are spaces of polynomials. For $n\in\mathbb{N}$ by $\mathbb{P}_n$ we denote the vector space of all polynomials of degree less or equal to $n.$ The most important step in understanding the vector space $\mathbb{P}_n$ is establishing that the monomials form a basis of this space. I wrote this webpage with a proof which uses only linear algebra. The proof which I give below for $\mathbb{P}_3$ uses calculus. The proof which is given in our textbook uses the Fundamental Theorem of Algebra (which is more difficult to prove).
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Below we study the polynomial space $\mathbb{P}_3$, that is the vector space of all polynomials of degree less or equal to $3.$
- $\mathbb{P}_3$ denotes the vector space of all polynomials of degree less or equal $3.$ That is, in set-builder notation, \[ \mathbb{P}_3 = \bigl\{a_0 + a_1 x + a_2 x^2 + a_3 x^3 \, : \, a_0, a_1, a_2, a_3 \in \mathbb{R} \bigr\}. \] Recall that the constant polynomial $1$ is a polynomial in $\mathbb{P}_3$. To get this polynomial in the above set-builder notation we take $a_0 = 1,$ $a_1 = 0,$ $a_2 =0,$ and $a_3 = 0.$ To get the polynomial $x$ in the above set-builder notation we take $a_0 = 0,$ $a_1 = 1,$ $a_2 = 0,$ and $a_3 = 0.$ Similarly, to get the square polynomial $x^2$ in the above set-builder notation we take $a_0 = 0,$ $a_1 = 0,$ $a_2 = 1,$ and and $a_3 = 0.$ To get the cube polynomial $x^3$ in the above set-builder notation we take $a_0 = 0,$ $a_1 = 0,$ $a_2 = 0,$ and and $a_3 = 1.$ Using the concept of the span the above expression for $\mathbb{P}_3$ in set-builder notation can be written using the concept of the span as \[ \mathbb{P}_3 = \operatorname{Span}\bigl\{ 1, x, x^2, x^3 \bigr\}. \]
- As you probably learned in Math 204 the polynomials $1, x, x^2, x^3$ are linearly independent. These polynomials are called monomials.
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Below is a proof that the monomials $1, x, x^2, x^3$ are linearly independent in the vector space ${\mathbb P}_3$. First we need to be specific what we need to prove.
Let $\alpha_1,$ $\alpha_2,$ $\alpha_3,$ and $\alpha_4$ be scalars in $\mathbb{R}.$ We need to prove the following implication: If \[ \require{bbox} \bbox[5px, #88FF88, border: 1pt solid green]{\alpha_1\cdot 1 + \alpha_2 x + \alpha_3 x^2 + \alpha_4 x^3 =0 \quad \text{for all} \quad x \in \mathbb{R}}, \] then \[ \bbox[5px, #FF4444, border: 1pt solid red]{\alpha_1 = 0, \quad \alpha_2 =0, \quad \alpha_3 = 0, \quad \alpha_4 = 0}. \] Proof.
- Assume that \[ \require{bbox} \bbox[5px, #88FF88, border: 1pt solid green]{\alpha_1\cdot 1 + \alpha_2 x + \alpha_3 x^2 + \alpha_4 x^3 =0 \quad \text{for all} \quad x \in \mathbb{R}}, \]
- Consider the left-hand side of the preceding green identity as a function of $x$ and take the derivative with respect to $x$. We obtain \[ \bbox[5px, #88FF88, border: 1pt solid green]{\alpha_2 + 2 \alpha_3 x + 3 \alpha_4 x^2 =0 \quad \text{for all} \quad x \in \mathbb{R}}. \]
- Again, consider the left-hand side of the preceding green identity as a function of $x$ and take the derivative with respect to $x$. We obtain \[ \bbox[5px, #88FF88, border: 1pt solid green]{2 \alpha_3 + 6 \alpha_4 x =0 \quad \text{for all} \quad x \in \mathbb{R}}. \]
- Again, consider the left-hand side of the preceding green identity as a function of $x$ and take the derivative with respect to $x$. We obtain \[ \bbox[5px, #88FF88, border: 1pt solid green]{6 \alpha_4 = 0 \quad \text{for all} \quad x \in \mathbb{R}}. \]
- Substituting $x=0$ in the preceding four green identities we obtain \[ \bbox[5px, #88FF88, border: 1pt solid green]{\alpha_1 = 0, \quad \alpha_2 =0, \quad 2 \alpha_3 = 0, \quad 6 \alpha_4 = 0}. \] Dividing the third equation by $2$ and the fourth equation by $6$ we obtain \[ \bbox[5px, #88FF88, border: 1pt solid green]{\alpha_1 = 0, \quad \alpha_2 =0, \quad \alpha_3 = 0, \quad \alpha_4 = 0}. \] In this way we have greenifyed the red statement. That is, we proved it.
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Here is an alternative proof that the monomials $1, x, x^2, x^3$ are linearly independent in the vector space ${\mathbb P}_3$.
- Assume that $\alpha_1,$ $\alpha_2,$ $\alpha_3,$ and $\alpha_4$ are scalars in $\mathbb{R}$ such that \[ \require{bbox} \bbox[5px, #88FF88, border: 1pt solid green]{\alpha_1\cdot 1 + \alpha_2 x + \alpha_3 x^2 + \alpha_4 x^3 =0 \quad \text{for all} \quad x \in \mathbb{R}}. \] The objective here is to prove \[ \bbox[5px, #FF4444, border: 1pt solid red]{\alpha_1 = 0, \quad \alpha_2 =0, \quad \alpha_3 = 0, \quad \alpha_4 = 0}. \]
- The above green identity holds for all $x\in\mathbb{R}.$ In particular, it holds for specific $x=-1,$ $x=0,$ $x=1,$ and $x=2.$ That is, we have \[ \bbox[5px, #88FF88, border: 1pt solid green]{ \begin{array}{lr} \alpha_1 - \phantom{2} \alpha_2 +\phantom{4}\alpha_3 -\phantom{8} \alpha_4 &=0 \\ \alpha_1 &=0 \\ \alpha_1 + \phantom{2} \alpha_2 + \phantom{4}\alpha_3 +\phantom{8} \alpha_4 &=0 \\ \alpha_1 + 2 \alpha_2 + 4 \alpha_3 + 8 \alpha_4 &=0 \\ \end{array} } \]
- The last green box contains a homogeneous system of linear equations which can be written in a matrix form as \[ \bbox[5px, #88FF88, border: 1pt solid green]{ \left[\!\begin{array}{rrrr} 1 & -1 & 1 & -1\\ 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 \\ 1 & 2 & 4 & 8 \end{array}\!\right] \left[\!\begin{array}{c} \alpha_1 \\ \alpha_2 \\ \alpha_3 \\ \alpha_4 \end{array}\!\right] = \left[\!\begin{array}{c} 0 \\ 0 \\ 0 \\ 0 \end{array}\!\right] } \]
- Let us calculate the determinant of the preceding $4\!\times\!4$ matrix: \[ \left|\!\begin{array}{rrrr} 1 & -1 & 1 & -1\\ 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 \\ 1 & 2 & 4 & 8 \end{array}\!\right| = -1 \left|\!\begin{array}{rrr} -1 & 1 & -1\\ 1 & 1 & 1 \\ 2 & 4 & 8 \end{array}\!\right| = -1 \left|\!\begin{array}{rrr} 0 & 2 & 0\\ 1 & 1 & 1 \\ 2 & 4 & 8 \end{array}\!\right| = (-1) (-2) \left|\!\begin{array}{rrr} 1 & 1 \\ 2 & 8 \end{array}\!\right| = (1)(-2) 6 = 12. \] Since the determinant of the above $4\!\times\!4$ matrix is $12$, the above homogeneous system of linear equations has only the trivial solution. That is, \[ \bbox[5px, #88FF88, border: 1pt solid green]{\alpha_1 = 0, \quad \alpha_2 =0, \quad \alpha_3 = 0, \quad \alpha_4 = 0}. \] In this way we have greenifyed the red statement. That is, we proved it.
- Hence, \[ \mathbb{P}_3 = \operatorname{Span}\bigl\{ 1, x, x^2, x^3 \bigr\} \] and $\mathcal{M} = \bigl\{ 1, x, x^2, x^3 \bigr\}$ is a basis for $\mathbb{P}_3.$ I denoted this basis by $\mathcal{M}$ since the polynomials $1,$ $x,$ $x^2$, $x^3$ are called monomials. The basis $\mathcal{M} = \bigl\{ 1, x, x^2, x^3 \bigr\}$ is called the standard basis for $\mathbb{P}_3$.
- The concept of a basis of a finite-dimensional vector space is essential for everything that we discus in this class. The definition of a basis is at the beginning of Section 4.3 in the textbook.
- In yesterday's blog I reviewed the Unique Representation Theorem (on page 218 in the textbook) and the concept of a coordinate mapping relative to a basis.
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In Problem 13 in Section 4.4, we use Corollary 1 and Corollary 2 (reviewed yesterday) to prove that the set
\[
\mathcal B = \bigl\{1+x^2, x+ x^2, 1+2 x+ x^2\bigr\}
\]
is a basis for $\mathbb{P}_2$.
- We first express coordinate vectors relative to the basis $\mathcal{M}_2=\{1,x,x^2\}$ for each of the polynomials in $\mathcal{B}$: \[ \bigl[1+x^2\bigr]_{\mathcal{M}_2} = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}, \quad \bigl[x+x^2\bigr]_{\mathcal{M}_2} = \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}, \quad \bigl[1+2 x+x^2\bigr]_{\mathcal{M}_2} = \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}. \]
- Since \[ \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 1 & 1 & 1 \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 1 & 0 \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}, \] the vectors \[ \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}, \quad \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}, \quad \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}, \] are linearly independent and span $\mathbb{R}^3$. By Corollary 1 and Corollary 2, the polynomials \[ 1+x^2, \quad x+x^2, \quad 1+2 x+x^2, \] are linearly independent and the span $\mathbb{P}_2$.
- We have the change of coordinate basis \[ \underset{\mathcal{M}_2\leftarrow\mathcal{B}}{P} = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 1 & 1 & 1 \end{bmatrix}. \] To calculate the change of coordinate basis $\displaystyle \underset{\mathcal{B}\leftarrow\mathcal{M}_2}{P}$ we calculate the inverse of $\displaystyle \underset{\mathcal{M}_2\leftarrow\mathcal{B}}{P}$: \begin{align*} \left[\!\begin{array}{ccc|ccc} 1 & 0 & 1 & 1 & 0 & 0 \\ 0 & 1 & 2 & 0 & 1 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1 \end{array}\!\right] & \sim \left[\!\begin{array}{ccc|ccc} 1 & 0 & 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & -1 & 0 & 1 \\ 0 & 1 & 2 & 0 & 1 & 0 \end{array}\!\right] \\ & \sim \left[\!\begin{array}{ccc|ccc} 1 & 0 & 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & -1 & 0 & 1 \\ 0 & 0 & 2 & 1 & 1 & -1 \end{array}\!\right] \\ & \sim \left[\!\begin{array}{ccc|ccc} 1 & 0 & 1 & \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ 0 & 1 & 0 & -1 & 0 & 1 \\ 0 & 0 & 1 & \frac{1}{2} & \frac{1}{2} & -\frac{1}{2} \end{array}\!\right] \\ \end{align*}
- Thus \[ \underset{\mathcal{B}\leftarrow\mathcal{M}_2}{P} = \begin{bmatrix} \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -1 & 0 & 1 \\ \frac{1}{2} & \frac{1}{2} & -\frac{1}{2} \end{bmatrix}. \] The practical meaning of the above matrix is as follows: \begin{align*} 1 & = \tfrac{1}{2} (1+x^2) + (-1) (x+x^2) + \tfrac{1}{2} (1 + 2x + x^2) \\ x & =\bigl(-\tfrac{1}{2} \bigr) (1+x^2) \qquad \qquad + \tfrac{1}{2} (1 + 2x + x^2) \\ x^2 & = \tfrac{1}{2} (1+x^2) + (1) (x+x^2) + \bigl(-\tfrac{1}{2} \bigr) (1 + 2x + x^2) \end{align*}
- One of the most important concepts in linear algebra is the concept of a basis of a vector space. In this post, I will review that concept. The definition of a basis is in Section 4.3.
- The definition of a linearly independent set on page 210. Next, I will restate this definition as an implication: An indexed set of vectors $\{\mathbf{v}_1,\ldots,\mathbf{v}_m\}$ in a vector space $\mathcal{V}$ is said to be linearly independent if the following implication holds \[ \alpha_1 \mathbf{v}_1 + \cdots + \alpha_m \mathbf{v}_m = \mathbf{0} \quad \text{implies} \quad \alpha_k = 0 \quad \text{for all} \quad k \in \{1,\ldots,m\}. \] There are many other equivalent ways of stating this definition. However, the above statement is the only formal definition which is easiest to use when we need to prove that certain vectors are linearly independent.
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Let $\mathcal{H}$ be a subspace of a vector space $\mathcal{V}$. An indexed set of vectors $\mathcal{B} = \{\mathbf{v}_1,\ldots,\mathbf{v}_m\}$ in a vector space $\mathcal{H}$ is said to be a basis for $\mathcal{H}$ if the following to conditions are satisfied
- $\mathcal{B} = \{\mathbf{v}_1,\ldots,\mathbf{v}_m\}$ is a linearly independent set.
- $\mathcal{H} = \operatorname{Span}\{\mathbf{v}_1,\ldots,\mathbf{v}_m\}$.
- A closely related to the concept of a basis are the Unique Representation Theorem on page 218 and a coordinate mapping relative to a basis.
- Let $\mathcal{B} = \{\mathbf{b}_1,\ldots,\mathbf{b}_n\}$ be a basis of a vector space $\mathcal{V}.$ Please understand the concept of the coordinates of a vector in $\mathcal{V}$ relative to the basis $\mathcal{B}.$ This definition is on page 218. Also understand the concept of the coordinate mapping determined by the basis $\mathcal{B}.$ Briefly, for a vector $\mathbf{v}$ in $\mathcal{V}$ we have \[ [\mathbf{v}]_{\mathcal{B}} = \left[\!\!\begin{array}{c} \alpha_1 \\ \alpha_2 \\ \vdots \\ \alpha_n \end{array}\!\!\right] \quad \text{if and only if} \quad \mathbf{v} = \alpha_1 \mathbf{b}_1 + \alpha_2 \mathbf{b}_2 + \cdots + \alpha_n \mathbf{b}_n \] The coordinate mapping relative to a basis $\mathcal{B}$ is a linear mapping with domain $\mathcal{V}$ and with the range $\mathbb{R}^n$ which is defined by \[ \mathcal{V} \ni \mathbf{v} \longmapsto [\mathbf{v}]_{\mathcal{B}} = \left[\!\!\begin{array}{c} \alpha_1 \\ \alpha_2 \\ \vdots \\ \alpha_n \end{array}\!\!\right] \in \mathbb{R}^n. \]
- The fundamental theorem about the coordinate mapping is Theorem 8 on page 221. In Theorem 8, the book uses the adjectives 'one-to-one' and 'onto' for functions. I encourage you to use different terminology for these concepts. My preferred synonym for 'one-to-one function' is injection, and for 'onto function', is surjection.
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Since the concepts of injection and surjection are basic concepts related to functions, I state their definitions here.
Definition. A function $f$ from $A$ to $B$, $f:A\to B$, is called surjection if it satisfies condition the following condition:
- For every $y \in B$ there exists $x \in A$ such that $f(x) = y$.
Definition. A function $f$ from $A$ to $B$, $f:A\to B$, is called injection if it satisfies the following condition
- For every $x_1, x_2 \in A$ we have that $f(x_1) = f(x_2)$ implies $x_1 = x_2$.
An equivalent formulation of the preceding condition is:
- For every $x_1, x_2 \in A$ we have that $x_1 \neq x_2$ implies $f(x_1) \neq f(x_2)$.
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Together with the terminology injection and surjection goes the following terminology
Definition. A function $f:A\to B$ is called bijection if it satisfies the following two conditions:
- For every $y \in B$ there exists $x \in A$ such that $f(x) = y$.
- For every $x_1, x_2 \in A$ we have that $f(x_1) = f(x_2)$ implies $x_1 = x_2$.
In other words, a function $f:A\to B$ is called bijection if it is both an injection and a surjection.
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In the context of vector spaces the following is an important definition:
Definition. Let $\mathcal V$ and $\mathcal W$ be vector spaces. A linear bijection $T: \mathcal V \to \mathcal W$ is said to be an isomorphism.
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Since I use different terminology, I will restate Theorem 8 from page 221 here.
Theorem 8. Let $n \in \mathbb{N}$. Let $\mathcal{B} = \{\mathbf{b}_1, \ldots, \mathbf{b}_n\}$ be a basis of a vector space $\mathcal V$. The coordinate mapping \[ \mathbf{v} \mapsto [\mathbf{v}]_\mathcal{B}, \qquad \mathbf{v} \in \mathcal V, \] is a linear bijection between the vector space $\mathcal V$ and the vector space $\mathbb{R}^n.$
Theorem 8. Let $n \in \mathbb{N}$. Let $\mathcal{B} = \{\mathbf{b}_1, \ldots, \mathbf{b}_n\}$ be a basis of a vector space $\mathcal V$. The coordinate mapping \[ \mathbf{v} \mapsto [\mathbf{v}]_\mathcal{B}, \qquad \mathbf{v} \in \mathcal{V}, \] is an isomorphism between the vector space $\mathcal V$ and the vector space $\mathbb{R}^n.$
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The next two corollaries of Theorem 8 are useful tools in dealing with abstract vector spaces.
Corollary 1. Let $m, n \in \mathbb{N}$. Let $\mathcal{B} = \{\mathbf{b}_1, \ldots, \mathbf{b}_n\}$ be a basis of a vector space $\mathcal V$. Then the following statements are equivalent:
- Vectors $\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_m$ are linearly independent in $\mathcal V$.
- The columns of the $n\times m$ matrix $\Bigl[ [\mathbf{v}_1]_\mathcal{B} \ [\mathbf{v}_2]_\mathcal{B} \ \cdots \ [\mathbf{v}_m]_\mathcal{B} \Bigr]$ are linearly independent.
Corollary 2. Let $m, n \in \mathbb{N}$. Let $\mathcal{B} = \{\mathbf{b}_1, \ldots, \mathbf{b}_n\}$ be a basis of a vector space $\mathcal V$. Then the following statements are equivalent:
- Vectors $\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_m$ span the space $\mathcal V$.
- The columns of the $n\times m$ matrix $\Bigl[ [\mathbf{v}_1]_\mathcal{B} \ [\mathbf{v}_2]_\mathcal{B} \ \cdots \ [\mathbf{v}_m]_\mathcal{B} \Bigr]$ span the space $\mathbb{R}^n$.
-
Consider the picture below. No numbers are given 4, just the picture. In the picture below we are given two bases of $\mathbb{R}^2$, one blue and one purple:
\[
\color{blue}{\mathcal B} = \bigl\{ \color{blue}{\mathbf{b}_1}, \color{blue}{\mathbf{b}_2} \bigr\}, \quad \color{purple}{\mathcal C} = \bigl\{ \color{purple}{\mathbf{c}_1}, \color{purple}{\mathbf{c}_2} \bigr\}
\]
- Suggested exercises for Section 4.7: Change of Basis are 2, 3, 4, 6, 8, 9, 11, 12, 19, 20.
-
A brief review of the Change of Coordinates Matrix follows. Let $m, n \in \mathbb{N}$ and $m\leq n$. Let $\mathcal{H}$ be a subspace of $\mathbb{R}^n$ and let
\[
\mathcal{A} = \bigl\{\mathbf{a}_1,\ldots,\mathbf{a}_m\bigr\}
\]
and
\[
\mathcal{B} = \bigl\{\mathbf{b}_1,\ldots,\mathbf{b}_m\bigr\}
\]
be two bases of $\mathcal{H}.$ By definition of a basis this implies that
\[
\mathcal{H} = \operatorname{Span}\bigl\{\mathbf{a}_1,\ldots,\mathbf{a}_m\bigr\} = \operatorname{Span}\bigl\{\mathbf{b}_1,\ldots,\mathbf{b}_m\bigr\}
\]
and both
\[
\mathcal{A} = \bigl\{\mathbf{a}_1,\ldots,\mathbf{a}_m\bigr\} \quad \text{and} \quad \mathcal{B} = \bigl\{\mathbf{b}_1,\ldots,\mathbf{b}_m\bigr\}
\]
are linearly independent. We proved in class that the change of coordinates matrix $\displaystyle\underset{\mathcal{B}\leftarrow\mathcal{A}}{P}$ is given by
\[
\underset{\mathcal{B}\leftarrow\mathcal{A}}{P} = \Bigl[ \bigl[\mathbf{a}_1\bigr]_{\mathcal{B}} \ \cdots \ \bigl[ \mathbf{a}_m\bigr]_{\mathcal{B}} \Bigr]
\]
and analogously
\[
\underset{\mathcal{A}\leftarrow\mathcal{B}}{P} = \Bigl[ \bigl[\mathbf{b}_1\bigr]_{\mathcal{A}} \ \cdots \ \bigl[ \mathbf{b}_m\bigr]_{\mathcal{A}} \Bigr].
\]
But, how to calculate
\[
\bigl[\mathbf{a}_1\bigr]_{\mathcal{B}}, \ldots,\bigl[ \mathbf{a}_m\bigr]_{\mathcal{B}}?
\]
Let us look at
\[
\bigl[\mathbf{a}_1\bigr]_{\mathcal{B}} = \left[\!\begin{array}{c}
x_1 \\ \vdots \\ x_m
\end{array}\!\right].
\]
To find the real numbers $x_1, \ldots, x_m$ we have to solve the nonhomogeneous vector equation
\[
x_1 \mathbf{b}_1 + x_2 \mathbf{b}_2 + \cdots + x_m \mathbf{b}_m = \mathbf{a}_1.
\]
To solve the preceding equation we row reduce
\[
\Bigl[\!\begin{array}{cccc|c}
\mathbf{b}_1 & \mathbf{b}_2 & \cdots & \mathbf{b}_m & \mathbf{a}_1\end{array}\!\Bigr].
\]
Since the vectors $\mathbf{b}_1, \mathbf{b}_2, \ldots, \mathbf{b}_m$ are linearly independent, the Reduced Row Echelon Form of the preceding augmented matrix has the following form
\[
\left[\!\begin{array}{cccc|c}
1 & 0 & \cdots & 0 & \text{the solution for} \ x_1 \\
0 & 1 & \cdots & 0 & \text{the solution for} \ x_2 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \cdots & 1 & \text{the solution for} \ x_m \\
0 & 0 & \cdots & 0 & 0 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \cdots & 0 & 0
\end{array}\!\right].
\]
Notice that in the preceding matrix the bottom zero rows are present only in the case when $n \gt m.$ If $n \gt m$, then there are exactly $n-m$ rows of zeros. Also notice that the above system must be consistent since the vector $\mathbf{a}_1$ is in the span of the vectors $\mathbf{b}_1, \mathbf{b}_2, \ldots, \mathbf{b}_m.$
To solve the nonhomogeneous vector equations \[ x_1 \mathbf{b}_1 + x_2 \mathbf{b}_2 + \cdots + x_m \mathbf{b}_m = \mathbf{a}_2, \quad \ldots, \quad x_1 \mathbf{b}_1 + x_2 \mathbf{b}_2 + \cdots + x_m \mathbf{b}_m = \mathbf{a}_m, \] we just build the bigger augmented matrix: \[ \Bigl[\!\begin{array}{cccc|cccc} \mathbf{b}_1 & \mathbf{b}_2 & \cdots & \mathbf{b}_m & \mathbf{a}_1 & \mathbf{a}_2 & \cdots & \mathbf{a}_m \end{array}\!\Bigr]. \] Since the vectors $\mathbf{b}_1, \ldots, \mathbf{b}_m$ are linearly independent, the RREF off the matrix whose columns are the vectors of $\mathcal{B}$ consists of the identity matrix $I_m$ and $n-m$ zero rows at the bottom if $n\gt m.$ Therefore \[ \Bigl[\!\begin{array}{ccc|ccc} \mathbf{b}_1 & \cdots & \mathbf{b}_m & \mathbf{a}_1 & \cdots & \mathbf{a}_m \end{array}\!\Bigr] \sim \cdots \sim \left[\! \begin{array}{c|c} I_m & \underset{\mathcal{B}\leftarrow\mathcal{A}}{P} \\ 0 & 0 \end{array} \!\right]. \] In the preceding RREF, the zero matrices at the bottom are present only if $n-m \gt 0.$ Then, if $n-m \gt 0,$ these matrices are of the size $(n-m)\!\times\!m;$ they both have $m$ columns and $n-m$ rows consisting of zeros. -
In the next example we are given two bases of a two-dimensional subspace of $\mathbb{R}^4$ and we are asked to find a change of coordinate matrices between these two bases:
\[
\mathcal{H} = \operatorname{Span}\left\{\left[\!\begin{array}{c}
1 \\ 2 \\ 1 \\ 3
\end{array}\!\right], \left[\!\begin{array}{c}
2 \\ 3 \\ 1 \\ 5
\end{array}\!\right]\right\} = \operatorname{Span}\left\{\left[\!\begin{array}{c}
2 \\ 5 \\ 3 \\ 7
\end{array}\!\right], \left[\!\begin{array}{r}
1 \\ 0 \\ -1 \\ 1
\end{array}\!\right]\right\}.
\]
Set
\[
\mathcal{A} = \left\{\left[\!\begin{array}{c}
1 \\ 2 \\ 1 \\ 3
\end{array}\!\right], \left[\!\begin{array}{c}
2 \\ 3 \\ 1 \\ 5
\end{array}\!\right]\right\}, \qquad \mathcal{B} = \left\{\left[\!\begin{array}{c}
2 \\ 5 \\ 3 \\ 7
\end{array}\!\right], \left[\!\begin{array}{r}
1 \\ 0 \\ -1 \\ 1
\end{array}\!\right]\right\}.
\]
To calculate $\displaystyle\underset{\mathcal{B}\leftarrow\mathcal{A}}{P}$ we need to row reduce the matrix
\[
\left[\!\begin{array}{cr|cc}
2 & 1 & 1 & 2 \\
5 & 0 & 2 & 3 \\
3 & -1 & 1 & 1 \\
7 & 1 & 3 & 5
\end{array}\!\right]
\]
The RREF of the preceding matrix will certainly include fractions. Therefore we rather find $\displaystyle\underset{\mathcal{A}\leftarrow\mathcal{B}}{P}$ for which we need to row reduce (without fractions)
\[
\left[\!\begin{array}{cc|cr}
1 & 2 & 2 & 1 \\
2 & 3 & 5 & 0 \\
1 & 1 & 3 & -1 \\
3 & 5 & 7 & 1
\end{array}\!\right] \sim \cdots \sim
\left[\!\begin{array}{cc|rr}
1 & 0 & 4 & -3 \\
0 & 1 & -1 & 2 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{array}\!\right].
\]
Hence
\[
\underset{\mathcal{A}\leftarrow\mathcal{B}}{P} = \left[\!\begin{array}{rr}
4 & -3 \\
-1 & 2
\end{array}\!\right].
\]
Let us verify this calculation. Is it true that:
\[
\left[\!\begin{array}{c}
2 \\ 5 \\ 3 \\ 7
\end{array}\!\right] = (4) \left[\!\begin{array}{c}
1 \\ 2 \\ 1 \\ 3
\end{array}\!\right] + (-1)\left[\!\begin{array}{c}
2 \\ 3 \\ 1 \\ 5
\end{array}\!\right], \qquad \left[\!\begin{array}{r}
1 \\ 0 \\ -1 \\ 1
\end{array}\!\right] = (-3) \left[\!\begin{array}{c}
1 \\ 2 \\ 1 \\ 3
\end{array}\!\right] + (2) \left[\!\begin{array}{c}
2 \\ 3 \\ 1 \\ 5
\end{array}\!\right]?
\]
Yes. Therefore the following equalities are correct:
\[
\bigl[ \mathbf{b}_1\bigr]_{\mathcal{A}} = \left[\!\begin{array}{r}
4 \\ -1
\end{array}\!\right], \qquad \bigl[ \mathbf{b}_2\bigr]_{\mathcal{A}} = \left[\!\begin{array}{r}
-3 \\ 2
\end{array}\!\right].
\]
Hence, $\displaystyle\underset{\mathcal{A}\leftarrow\mathcal{B}}{P}$ is correct.
We have \[ \underset{\mathcal{B}\leftarrow\mathcal{A}}{P} = \left(\underset{\mathcal{A}\leftarrow\mathcal{B}}{P}\right)^{-1} = \frac{1}{5} \left[\!\begin{array}{rr} 2 & 3 \\ 1 & 4 \end{array}\!\right]. \] Verify this: \[ \left[\!\begin{array}{c} 1 \\ 2 \\ 1 \\ 3 \end{array}\!\right] = \frac{2}{5}\left[\!\begin{array}{c} 2 \\ 5 \\ 3 \\ 7 \end{array}\!\right] + \frac{1}{5} \left[\!\begin{array}{r} 1 \\ 0 \\ -1 \\ 1 \end{array}\!\right], \qquad \left[\!\begin{array}{c} 2 \\ 3 \\ 1 \\ 5 \end{array}\!\right] = \frac{3}{5} \left[\!\begin{array}{c} 2 \\ 5 \\ 3 \\ 7 \end{array}\!\right] + \frac{4}{5} \left[\!\begin{array}{r} 1 \\ 0 \\ -1 \\ 1 \end{array}\!\right]. \] True. Therefore the following equalites are correct: \[ \bigl[ \mathbf{a}_1\bigr]_{\mathcal{B}} = \frac{1}{5}\left[\!\begin{array}{r} 2 \\ 1 \end{array}\!\right], \qquad \bigl[ \mathbf{a}_2\bigr]_{\mathcal{B}} = \frac{1}{5} \left[\!\begin{array}{r} 3 \\ 4 \end{array}\!\right]. \] Hence, $\displaystyle\underset{\mathcal{B}\leftarrow\mathcal{A}}{P}$ is correct.
-
To celebrate the concept of reduced row echelon form of a matrix and the full power of its utility I wrote the webpage
-
Consider the following $3\times 5$ matrix
\[
A = \left[\!\begin{array}{rrrrr}
1 & 2 & 0 & 1 & 2 \\
2 & 4 & 1 & 0 & 5 \\
3 & 6 & 2 & -1 & 8
\end{array}\right].
\]
- Today I announced that for a matrix named \(A\) by \(a_{ij}\) we will denote the entry of \(A\) in the \(i\)-th row and the \(j\)-th column of \(A\). That is, for the above matrix \(A\) we have \begin{alignat*}{4} a_{11} &= 1, \quad a_{12} &&= 2, \quad a_{13} &&= 0, \quad a_{14} &&=\phantom{-} 1, \quad a_{15} &&= 2, \\ a_{21} &= 2, \quad a_{22} &&= 4, \quad a_{23} &&= 1, \quad a_{24} &&=\phantom{-} 0, \quad a_{25} &&= 5, \\ a_{31} &= 3, \quad a_{32} &&= 6, \quad a_{33} &&= 2, \quad a_{34} &&= -1, \quad a_{35} &&= 8. \end{alignat*}
- Similarly, for the columns of \(A\) we will use the notation \(A_{\mkern 2mu\cdot,1}\), the first column, \(A_{\mkern 2mu\cdot,2}\), the second column, \(A_{\mkern 2mu\cdot,3}\), the third column, \(A_{\mkern 2mu\cdot,4}\), the fourth column, and \(A_{\mkern 2mu\cdot,5}\), the fifth column. That is, \[ A_{\mkern 2mu\cdot,1} = \left[\!\begin{array}{c} 1 \\ 2 \\ 3 \end{array}\right], \quad A_{\mkern 2mu\cdot,2} = \left[\!\begin{array}{c} 2 \\ 4 \\ 6 \end{array}\right], \quad A_{\mkern 2mu\cdot,3} = \left[\!\begin{array}{c} 0 \\ 1 \\ 2 \end{array}\right], \quad A_{\mkern 2mu\cdot,4} = \left[\!\begin{array}{r} 1 \\ 0 \\ -1 \end{array}\right], \quad A_{\mkern 2mu\cdot,5} = \left[\!\begin{array}{c} 1 \\ 5 \\ 8 \end{array}\right]. \]
- Similarly, for the rows of \(A\) we will use the notation \(A_{1,\mkern0.5mu\cdot\mkern 1mu}\), the first row, \(A_{2,\mkern0.5mu\cdot\mkern 1mu}\), the second row, and \(A_{3,\mkern0.5mu\cdot\mkern 1mu}\), the third row. That is, \[ A_{1,\mkern0.5mu\cdot\mkern 1mu} = \bigl[\!\begin{array}{rrrrr} 1 & 2 & 0 & 1 & 2 \end{array}\bigr], \quad A_{2,\mkern0.5mu\cdot\mkern 1mu} = \bigl[\!\begin{array}{rrrrr} 2 & 4 & 1 & 0 & 5 \end{array}\bigr], \quad A_{3,\mkern0.5mu\cdot\mkern 1mu} = \bigl[\!\begin{array}{rrrrr} 3 & 6 & 2 & -1 & 8 \end{array}\bigr] \]
-
The role of \(\mkern1mu \cdot\mkern1mu\) in this notation is that is stands for any index, in the first column, the second index is \(1\), while the second one can be any of allowed positive integers \(\{1,2,3\}\), indicated by \(\mkern1mu \cdot\mkern1mu\) as a wildcard index.
The reason I use capital Roman letters for rows and columns, and lowercase letters for entries, is that rows and columns are also matrices—special matrices, but matrices nonetheless—while the entries are scalars, specifically real numbers. It is important to make a clear distinction between these objects: matrices and scalars.
- We calculated the Reduced Echelon Form of $A$ \begin{alignat*}{2} \text{1st} \ & \text{step} \quad \left[\!\begin{array}{rrrrr} 1 & 2 & 0 & 1 & 2 \\ 2 & 4 & 1 & 0 & 5 \\ 3 & 6 & 2 & -1 & 8 \end{array}\right] && \sim \left[\!\begin{array}{rrrrr} 1 & 2 & 0 & 1 & 2 \\ 0 & 0 & 1 & -2 & 1 \\ 0 & 0 & 2 & -4 & 2 \end{array}\right] \\ \text{2nd} \ & \text{step} && \sim \left[\!\begin{array}{rrrrr} 1 & 2 & 0 & 1 & 2 \\ 0 & 0 & 1 & -2 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \\ \end{alignat*}
- Let us write \(A\) and it Reduced Row Echelon Form \(B\) nest to each other \[ A = \left[\!\begin{array}{rrrrr} 1 & 2 & 0 & 1 & 2 \\ 2 & 4 & 1 & 0 & 5 \\ 3 & 6 & 2 & -1 & 8 \end{array}\right] \, \sim \cdots \sim \, \left[\!\begin{array}{rrrrr} 1 & 2 & 0 & 1 & 2 \\ 0 & 0 & 1 & -2 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] = B. \]
-
Each step in a row reduction can be achieved by multiplication by a matrix.
Step the row operations the matrix used the matrix inverse 1st $\mkern 5mu \begin{array}{l} \sideset{_n}{_1}R \to \sideset{_o}{_1}R, \\ \sideset{_n}{_2}R \to (-2)\sideset{_o}{_1}R + \sideset{_o}{_2}R,\\ \sideset{_n}{_3}R \to (-3)\sideset{_o}{_1}R + \sideset{_o}{_3}R \end{array} $ $E_1 = \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ -2 & 1 & 0 \\ -3 & 0 & 1 \\ \end{array}\right]$ $(E_1)^{-1} = \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 0 & 1 \\ \end{array}\right]$ 2nd $\mkern 5mu \begin{array}{l} \sideset{_n}{_1}R \to \sideset{_o}{_1}R, \\ \sideset{_n}{_2}R \to \sideset{_o}{_2}R, \\ \sideset{_n}{_3}R \to (-2)\sideset{_o}{_2}R + \sideset{_o}{_3}R \end{array}$ $E_2 = \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -2 & 1 \end{array}\right]$ $(E_2)^{-1} = \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 2 & 1 \end{array}\right]$ -
The importance of the careful keeping track of each matrix is that we can calculate which single matrix performs the above Row Reduction:
\[ M = E_2 E_1 = \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -2 & 1 \end{array}\right] \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ -2 & 1 & 0 \\ -3 & 0 & 1 \\ \end{array}\right] = \left[ \begin{array}{rrr} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 1 & -2 & 1 \\ \end{array} \right] \] You can verify that \[ MA = \left[ \begin{array}{rrr} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 1 & -2 & 1 \\ \end{array} \right] \left[\!\begin{array}{rrrrr} 1 & 2 & 0 & 1 & 2 \\ 2 & 4 & 1 & 0 & 5 \\ 3 & 6 & 2 & -1 & 8 \end{array}\right] = \left[\!\begin{array}{rrrrr} 1 & 2 & 0 & 1 & 2 \\ 0 & 0 & 1 & -2 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] = B. \] -
The last matrix multiplication in the preceding item is important since it reveals the following important fact
-
But, also,
- The last six linear combinations prove that the row spaces of the of the matrix $A$ and the row space of the RREF of $A$ are identical.
-
Since the nonzero rows of the RREF of $A$ are linearly independent, we deduce that
The nonzero rows of the Reduced Row Echelon Form of $A$ form a basis for the row space of $A$.
-
For what Reduced Row Echelon Form of a matrix tells us about the Column Space of a matrix read
-
Let us work through another example of a $3\times 5$ matrix
\[
A = \left[\!\begin{array}{rrrrr}
3 & 1 & 5 & 1 & 2 \\
2 & 2 & 2 & 1 & 4 \\
1 & 2 & 0 & 1 & 5 \end{array}\right]
\].
- We calculated the Reduced Echelon Form of $A$ \begin{alignat*}{2} \text{1st} \ & \text{step} \quad \left[\!\begin{array}{rrrrr} 3 & 1 & 5 & 1 & 2 \\ 2 & 2 & 2 & 1 & 4 \\ 1 & 2 & 0 & 1 & 5 \end{array}\right] && \sim \left[\!\begin{array}{rrrrr} 1 & 2 & 0 & 1 & 5 \\ 1 & 1 & 1 & \frac{1}{2} & 2 \\ 1 & \frac{1}{3} & \frac{5}{3} & \frac{1}{3} & \frac{2}{3} \end{array}\right] \\ \text{2nd} \ & \text{step} && \sim \left[\!\begin{array}{rrrrr} 1 & 2 & 0 & 1 & 5 \\ 0 & -1 & 1 & -\frac{1}{2} & -3 \\ 0 & -\frac{5}{3} & \frac{5}{3} & -\frac{2}{3} & -\frac{13}{3} \end{array}\right] \\ \text{3rd} \ & \text{step} && \sim \left[\!\begin{array}{rrrrr} 1 & 0 & 2 & 0 & -1 \\ 0 & 1 & -1 & \frac{1}{2} & 3 \\ 0 & 0 & 0 & \frac{1}{6} & \frac{2}{3} \end{array}\right] \\ \text{4th} \ & \text{step} && \sim \left[\!\begin{array}{rrrrr} 1 & 0 & 2 & 0 & -1 \\ 0 & 1 & -1 & 0 & 1 \\ 0 & 0 & 0 & 1 & 4 \end{array}\right] \\ \end{alignat*}
-
Each step in a row reduction can be achieved by multiplication by a matrix.
Step the row operations the matrix used the matrix inverse 1st $ \mkern 5mu \begin{array}{l} \sideset{_n}{_1}R \to \sideset{_o}{_3}R, \\ \sideset{_n}{_2}R\to\frac{1}{2}\sideset{_o}{_2}R,\\ \sideset{_n}{_3}R \to \frac{1}{3} \sideset{_o}{_1}R \end{array}$ $E_1 = \left[\!\begin{array}{rrr} 0 & 0 & 1 \\ 0 & \frac{1}{2} & 0 \\ \frac{1}{3} & 0 & 0 \\ \end{array}\right]$ $(E_1)^{-1} = \left[\!\begin{array}{rrr} 0 & 0 & 3 \\ 0 & 2 & 0 \\ 1 & 0 & 0 \\ \end{array}\right]$ 2nd $\mkern 5mu \begin{array}{l} \sideset{_n}{_1}R \to \sideset{_o}{_1}R, \\ \sideset{_n}{_2}R \to (-1)\sideset{_o}{_1}R+\sideset{_o}{_2}R, \\ \sideset{_n}{_3}R \to (-1)\sideset{_o}{_1}R + \sideset{_o}{_3}R \end{array}$ $E_2 = \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right]$ $(E_2)^{-1} = \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{array}\right]$ 3rd $\mkern 5mu \begin{array}{l} \sideset{_n}{_1}R \to \sideset{_o}{_1}R + 2 \sideset{_o}{_2}R, \\ \sideset{_n}{_2}R \to (-1)\sideset{_o}{_2}R, \\ \sideset{_n}{_3}R \to -\frac{5}{3} \sideset{_o}{_2}R + \sideset{_o}{_3}R \end{array} $ $E_3 = \left[\!\begin{array}{rrr} 1 & 2 & 0 \\ 0 & -1 & 0 \\ 0 & -\frac{5}{3} & 1 \end{array}\right]$ $(E_3)^{-1} = \left[\!\begin{array}{rrr} 1 & 2 & 0 \\ 0 & -1 & 0 \\ 0 & -\frac{5}{3} & 1 \end{array}\right]$ 4th $\mkern 5mu \begin{array}{l} \sideset{_n}{_1}R \to \sideset{_o}{_1}R,\\ \sideset{_n}{_2}R \to \sideset{_o}{_2}R + (-3)\sideset{_o}{_3}R, \\ \sideset{_n}{_3}R \to 6 \sideset{_o}{_3}R \end{array}$ $E_4 = \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & -3 \\ 0 & 0 & 6 \end{array}\right]$ $(E_4)^{-1} = \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & \frac{1}{2} \\ 0 & 0 & \frac{1}{6} \end{array}\right]$ -
The importance of the careful keeping track of each matrix is that we can calculate which single matrix performs the above Row Reduction:
\[ M = E_4 E_3 E_2 E_1 = \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & -3 \\ 0 & 0 & 6 \end{array}\right]\left[\!\begin{array}{rrr} 1 & 2 & 0 \\ 0 & -1 & 0 \\ 0 & -\frac{5}{3} & 1 \end{array}\right] \left[\!\begin{array}{rrr} 1 & 0 & 0 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right] \left[\!\begin{array}{rrr} 0 & 0 & 1 \\ 0 & \frac{1}{2} & 0 \\ \frac{1}{3} & 0 & 0 \\ \end{array}\right] = \left[ \begin{array}{rrr} 0 & 1 & -1 \\ -1 & 2 & -1 \\ 2 & -5 & 4 \\ \end{array} \right] \] You can verify that \[ MA = \left[ \begin{array}{rrr} 0 & 1 & -1 \\ -1 & 2 & -1 \\ 2 & -5 & 4 \\ \end{array} \right] \left[\!\begin{array}{rrrrr} 3 & 1 & 5 & 1 & 2 \\ 2 & 2 & 2 & 1 & 4 \\ 1 & 2 & 0 & 1 & 5 \end{array}\right] = \left[\!\begin{array}{rrrrr} 1 & 0 & 2 & 0 & -1 \\ 0 & 1 & -1 & 0 & 1 \\ 0 & 0 & 0 & 1 & 4 \end{array}\right] = B. \] -
The last matrix multiplication in the preceding item is important since it reveals the following important fact
-
But, also,
It is important to observe that forming the above three linear combinations is straightforward because of the positioning of the leading $1$s in the rows of the RREF. I tried to emphasise this by
- The last six linear combinations prove that the row spaces of the of the matrix $A$ and the row space of the RREF of $A$ are identical.
-
Since the nonzero rows of the RREF of $A$ are linearly independent, we deduce that
The nonzero rows of the Reduced Row Echelon Form of $A$ form a basis for the row space of $A$.
-
For what Reduced Row Echelon Form of a matrix tells us about the Column Space of a matrix read
-
I wanted to provide a colorful beginning of the New Academic Year and the Fall Quarter. So I started the class by talking about the colors in relation to linear algebra. I love the application of vectors to COLORS so much that I wrote a webpage to celebrate it: Color Cube.
-
It is important to point out that in the red-green-blue coloring scheme, the following eighteen colors stand out. I present them in six steps with three colors in each step.
-
Step One:
- Black (vector $(0,0,0)$),
- White (vector $(1,1,1)$) and the color between them,
- Gray (vector $(1/2,1/2,1/2)$) which can be considered as dark White;
-
Step Two:
the coordinate colors,
- Red (vector $(1,0,0)$),
- Green (vector $(0,1,0)$),
- Blue (vector $(0,0,1)$);
-
Step Three:
the complementary colors to RGB which are CMY,
- Cyan (vector $(0,1,1)$) absence of Red,
- Magenta (vector $(1,0,1)$) absence of Green,
- Yellow (vector $(1,1,0)$) absence of Blue;
-
Step Four:
the dark RGB colors,
- Maroon (vector $(1/2,0,0)$) is the dark Red,
- Forest (vector $(0,1/2,0)$) is the dark Green,
- Navy (vector $(0,0,1/2)$) is the dark Blue;
-
Step Five:
the dark CMY colors,
- Teal (vector $(0,1/2,1/2)$) is the dark Cyan,
- Purple (vector $(1/2,0,1/2)$) is the dark Magenta,
- Olive (vector $(1/2,1/2,0)$) is the dark Yellow.
-
Step Six:
three (out of total of six) colors between neighbouring RGB and CMY. The logic of our choice here is more complicated. R neighbours with M and Y. G neighbours with C and Y. B neighbours with C and M. It turns out that only three out of these six colors have names.
- Orange (vector $(1,1/2,0)$) is the color in the middle between Red and Yellow,
- Chartreuse (vector $(1/2,1,0)$) is the color in the middle between Green and Yellow,
- Violet (vector $(1/2,0,1)$) is the color in the middle between Blue and Magenta.
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Step One:
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I recommend that you look into learning
LaTeX which is a superior typesetting system for writing math papers.
Not only that LaTeX is the best way to write mathematical content on computer, but it is the best way to display math
on the Internet. All the mathematics that you see on my website is written in LaTeX.
All the mathematics that you see on Wikipedia is written in LaTeX. In fact, if you go to a mathematics page,
on the line below the title, next to the Read button, you will find the Edit button.
Clicking this edit button will take you to the code for that webpage. You will see a lot of LaTeX code there.
In fact, if you need exactly that formula, you can copy and paste the formula in your document.
- I created Getting Started with LaTeX page to help you with this.
- Here is a simple LaTeX sample document in which I prove an interesting inequality.
- If you need help starting with LaTeX, feel free to ask me for help. It is as important as learning one nice piece of mathematics. And I want to help you with that, not only because that is my job, but because I deeply believe that both math - through learning rigorous thinking - and professional writing skills - supported by rigorous thinking - will help you in your life more than anything else.
- I have noticed that many students use the online LaTeX editor Overleaf. I do not have any experience with Overleaf, except that I have seen that many students use it with nice results.
- LaTeX is just like a computer program that compiles into your mathematical document. As you probably heard ChatGPT is exceptionally good at writing code in all kind of different languages. So, it is good at writing LaTeX code. If you are confused on how to do a particular task in LaTeX, you can ask ChatGPT for advise. All advice that I obtained was superb. Not only for LaTeX, but for Wolfram Mathematica as well. Sometimes ChatGPT is sloppy, and makes mistakes, parenthesis not matching and stuff like that. It helps to write very detailed user prompts.
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You:
Can you please write a complete LaTeX file with instructions on using basic mathematical operations, like fractions, sums, integrals, basic functions, like cosine, sine, and exponential function, and how to structure a document and similar features? Please explain the difference between the inline and displayed mathematical formulas. Please include examples of different ways of formatting displayed mathematical formulas. Please include what you think would be useful to a mathematics student. Also, can you please include your favorite somewhat complicated mathematical formula as an example of the power of LaTeX? I emphasize I want a complete file that I can copy into a LaTeX compiler and compile into a pdf file. Please ensure that your document contains the code for the formulas you are writing, which displays both as code separately from compiled formulas. Also, please double-check that your code compiles correctly. Remember that I am a beginner and cannot fix the errors. Please act as a concerned teacher would do.
This is the LaTeX document that ChatGPT produced base on the above prompt. Here is the compiled PDF document.
You can ask ChatGPT for specific LaTeX advise. To get a good response, think carefully about your prompt. Also, you can offer to ChatGPT a sample of short mathematical writing from the web or a book as a PNG file and it convert that writing to LaTeX. You can even try with neat handwriting. The results will of course depend on the clarity of the file, ChatGPT makes mistakes, but I found it incredibly useful.
- It is tempting to make LaTeX a mandatory tool for creating submissions in this class. However, I do not appreciate having mandates imposed on me; I prefer the freedom to think my own thoughts. An important part of a teacher's role is to encourage students to think their own thoughts, and imposing a mandate could contradict that spirit.
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A student asked me about the technology that we will use in this class. I am open minded about that. You can use any technology that you are familiar with. I know and use only Wolfram Mathematica. All the calculations and mathematical illustrations on my website are done using Wolfram Mathematica.
Computer algebra system Mathematica is very useful in for explorations in mathematics. To get started with Mathematica see my Mathematica page. Please watch the videos that are on my Mathematica page. Watching the movies is very helpful to get started with Mathematica efficiently! Mathematica is available in the computer labs in BH 215 and BH 209.
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I have cherished the interplay between my mind and computer through Mathematica since I began using computers in the early 1990s. Recently, I encountered a beautiful analogy from Steve Jobs that perfectly encapsulates this relationship. In an old interview, Jobs described the power of computers with a striking image: "For me, a computer has always been a bicycle for the mind." This metaphor resonates deeply with me, as it reflects how tools like Mathematica and ChatGPT amplify our cognitive abilities, allowing us to explore mathematical ideas with the efficiency and elegance of a bicycle. And isn't that wonderful!
See the Bicycle for the Mind video. Or, a longer
version.
- The information sheet
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We start with a review. Please review
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To celebrate the concept of reduced row echelon form of a matrix and the full power of its utility I wrote the webpage
- The definition of an abstract vector space in Section 4.1, page 192.
- The definition of a linearly independent set and the definition of a basis in Section 4.3; Examples 3, 4, 5, 6 and 10; Practice Problems 1, 2, 3, Exercises 1-8 and 38.
- Section 4.4: Theorem 7 (the unique representation theorem), the definition of coordinates with respect to a basis, the definition of a change-of-coordinates matrix on page 249 and the definition and the properties of a coordinate mapping; Examples 1, 2, 4, 5, 6; Practice Problems 1, 2; Exercises 3, 4, 5, 7, 9, 10, 11, 13, 18, 21, 32.
- Section 4.5: Theorem 10, the definition of a finite-dimensional vector space and its dimension and the Basis Theorem; Examples 1, 2, 3, 4; Practice Problems 1, 2; Exercises 2, 3, 7, 22, 24, and 34.
- Section 4.7 Change of Basis. Suggested exercises are 2, 3, 4, 6, 8, 9, 11, 12, 15, 16, 19 (you do not need a calculator for this problem).
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To celebrate the concept of reduced row echelon form of a matrix and the full power of its utility I wrote the webpage
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Before briefly reflecting on the history of Linear Algebra, I want to celebrate the history of mathematics with one long sentence:
Throughout history, human civilizations continue to develop and share mathematical knowledge; succeeding civilizations recognize and admire the contributions of the preceding ones; prior mathematical creations serve as a foundation and inspiration for further advancements, all in unity, giving mathematics the true spirit of a collective endeavor of the entirety of humanity through history, the present, and the future.
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What is the oldest linear algebra problem?
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Clay tablet VAT 8389 from the Old Babylonian period, from 2000 to 1600 BC, contains what is believed to be the earliest word problem that can be interpreted as a system of linear equations:
A translation of this word problem into a system of linear equations is as follows: \begin{alignat*}{4} &x_1 & &\ + &x_2 & = 1800 \\ \tfrac{2}{3} &x_1 & &- \tfrac{1}{2} &x_2 & = \phantom{1}500. \end{alignat*} -
Problem 40 of the Rhind papyrus which is dated to 1550 BC is:
Denote by $x_1$ the smallest number and by $x_2$ the common difference. After simplification the above problem translates into the following system of linear equations: \begin{alignat*}{5} 5 &x_1 & & + 10 &x_2 & = 100 \\ \tfrac{11}{7} &x_1 & & - \phantom{1}\tfrac{2}{7} &x_2 & = \phantom{10}0. \end{alignat*} -
Most importantly for us, the oldest known treatment of systems of linear equations from antiquity which resembles the methods that we will use in this class is in Chapter 8 of the Chinese textbook Nine Chapters of the Mathematical Art which is at least 1800 years old.
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Clay tablet VAT 8389 from the Old Babylonian period, from 2000 to 1600 BC, contains what is believed to be the earliest word problem that can be interpreted as a system of linear equations:
- If the history of mathematics might inspire you to study mathematics with more fascination, below I link to some websites with more about the history of Linear Algebra.