- Notes on Inner Product Spaces.
- Notes on Eigensystems of Linear Mappings. Instead of "linear mapping" in these notes I use the synonym "linear operator."
- Notes on Linear mappings. Instead of "linear mapping" in these notes I use the synonym "linear operator."
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Notes on Bases. In these notes instead of lists I work with finite sets. For the Steinitz exchange lemma see also its Wikipedia page. The Wikipedia's proof is very similar to the proof in the notes.
- Notes on Vector Spaces.
- In Math 304, our second quarter of undergraduate Linear Algebra, I talk more about Fourier Series. It is the most important application of orthogonal systems of vectors in positive definite inner product spaces, in particular Theorem 3.4 in Inner Product Spaces notes. I explain the basics of this application in my post about Fourier Series at Math 304 webpage.
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The cumulative effect of time spent engaged in creative pursuits is an amazing, often neglected aspect of life.
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The school exists for us to engage in creative pursuits and experience the resulting amazing cumulative effect.
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In class I presented a different proof of the Cauchy–Bunyakovsky–Schwarz inequality. The advantage of the proof in the notes is that it does not involve cases. However, it uses a theorem about quadratic polynomials which is commonly known but really proved, see Rigor in the context of quadratic functions.
As a side note here: Could the following statement be an interesting exercise in a college algebra or a precalculus class:
Let $m$ and $b$ be real numbers. Prove the following equivalence \[ \forall x \in \mathbb{R} \quad m x + b \geq 0 \qquad \Leftrightarrow \qquad m = 0 \land b \geq 0. \] -
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.
Theorem. (Cauchy–Bunyakovsky–Schwarz Inequality) Let $\mathcal{V}$ be a vector space over a scalar field $\mathbb{F}$. Let $\langle \mkern 2mu\cdot \mkern 0.5mu, \cdot \mkern 1.5mu \rangle: \mathcal{V}\times\mathcal{V}\to \mathbb{F}$ be a positive semi-definite (nonnegative) inner product. Then \[ \forall \, u, v \in \mathcal{V} \qquad \bigl| \langle u, v \rangle \bigr|^2 \leq \langle u, u \rangle \langle v, v \rangle. \] The equality in the preceding inequality holds if and only if there exist $\alpha, \beta \in \mathbb{F}$ such that \[ |\alpha|^2 + |\beta|^2 \gt 0 \quad \text{and} \quad \bigl\langle \alpha u + \beta v, \alpha u + \beta v \bigr\rangle = 0. \]Proof. Let $u, v \in \mathcal{V}$ be arbitrary.
Case 1. $\langle u, u \rangle \gt 0$ or $\langle v, v \rangle \gt 0$. Case 1a. Assume $\langle u, u \rangle \gt 0$. Since the vector \[ \frac{\langle v, u \rangle}{\langle u, u \rangle} u \] is the orthogonal projection of the vector $v$ onto the subspace $\operatorname{span}\{u\}$, we have that the vectors \[ u \quad \text{and} \quad w = v - \frac{\langle v, u \rangle}{\langle u, u \rangle} u \] are orthogonal. That is \[ \langle w, u \rangle = \left\langle v - \frac{\langle v, u \rangle}{\langle u, u \rangle} u , u\right\rangle = \langle v, u \rangle - \frac{\langle v, u \rangle}{\langle u, u \rangle} \langle u, u \rangle = 0. \] Thus, the sum \[ \langle u, u \rangle v = \langle u, u \rangle w + \langle v, u \rangle u \] is an orthogonal sum. By the Pythagorean Theorem \[ \langle u, u \rangle^2 \langle v, v \rangle = \langle u, u \rangle^2 \langle w, w \rangle + |\langle v, u \rangle|^2 \langle u, u \rangle. \] Consequently, as \[ \langle u, u \rangle^2 \langle w, w \rangle \geq 0, \] we obtain \[ \langle u, u \rangle^2 \langle v, v \rangle \geq |\langle v, u \rangle|^2 \langle u, u \rangle. \] Dividing by $\langle u, u \rangle \gt 0$, we get the Cauchy–Bunyakovsky–Schwarz Inequality \[ \langle u, u \rangle \langle v, v \rangle \geq |\langle v, u \rangle|^2. \] The Case 1b. $\langle v, v \rangle \gt 0$ is proved similarly.
Case 2. Assume $\langle u, u \rangle = 0$ and $\langle v, v \rangle = 0$. Since for every $w \in \mathcal{V}$ we have $\langle w,w\rangle \geq 0$, with $w = u -\langle u, v \rangle v$ we have \begin{align*} 0 &\leq \bigl\langle u -\langle u, v \rangle v , u -\langle u, v \rangle v \bigr\rangle \\ & =\langle u,u \rangle - \langle v, u \rangle \langle u, v \rangle - \langle u, v \rangle \langle v, u \rangle +|\langle u, v \rangle|^2 \langle v,v \rangle \\ & = - 2 |\langle u, v \rangle|^2 \\ & \leq 0. \end{align*} Consequently, \[ \langle u, v \rangle = 0. \] Since, in this case, u \[ \langle u, v \rangle = 0, \quad \langle u, u \rangle = 0, \quad \langle v, v \rangle = 0, \] the Cauchy–Bunyakovsky–Schwarz Inequality holds in this case. This completes the proof of the Cauchy–Bunyakovsky–Schwarz Inequality.
Proof of the statement involving the equality is in the notes.
- Since the main topic in this class are linear mappings, that is linear functions, it is appropriate to review the basic facts about Functions.
- I posted my notes on Eigensystems of Linear Mappings. I kept rewriting these notes the whole day today. I hope the notes make sense. If not, I will keep improving them until the notes start making sense.
- We are skipping Section 3E: Products and Quotients of Vector Spaces and Section 3F: Duality. After offering a short overview of what is the core of these topics, I do not feel better about skipping. However, quarters are short.
- We are also skipping Chapter 4: Polynomials. You have covered that kind of stuff in other classes. I will include the basics in the lectures on Chapter 5: Eigenvalues and Eigenvectors.
- You have covered eigenvalues and eigenvectors in your undergraduate linear algebra courses. When I introduce the topic of eigenvalues and eigenvectors in beginning linear algebra classes, I like to start with movies emphasizing that for $2\times 2$ matrices, you can see the eigenvalues and the eigenvectors. Please see the post on Tuesday, February 22, 2022 at my Winter 2022 Math 204 webpage.
- I will post my notes on Eigenvalues and Eigenvectors tomorrow.
- My presentation of the concept of a matrix of a linear mapping differed from the presentation in the book. Here are my notes on Linear mappings. Instead of "linear mapping" in these notes I use the synonym "linear operator."
- I wrote a web-page about finding solutions of the Chebyshev differential equation using linear algebra. This is a good example of the utility of a matrix of a linear transformation. In the linked webpage we study a linear operator $T:\mathcal{P}_5(\mathbb{R}) \to \mathcal{P}_5(\mathbb{R})$ and we use the $6\times 6$ matrix $M^{\mathcal{M}}_{\mathcal{M}}(T),$ where $\mathcal{M}$ is a basis of $\mathcal{P}_5(\mathbb{R})$ which consists of monomials. The notation in the webpage is different, but hopefuly one can see the connection. For example, in class I used $C_{\mathcal{B}}(v)$ to denote the coordinate vector of $v$ relative to a basis $\mathcal{B}$. Below I use $\bigl[v\bigr]_{\mathcal{B}}$ for the coordinate vector.
- In the next item I copy what I posted for my Math 304 (the second quarter of our undergraduate linear algebra sequence) on April 15, 2022. The references are to Lay's Linear Algebra and Applications book. The formulas that I use should be familiar from what we did in class. The example below illustrates the process known as polynomial interpolation by using a matrix of a linear transformation. There are many different ways how polynomial interpolation can be accomplished. I present this one to illustrate the utility of what we did in class.
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Today we considered a 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 \(M\) \[ \left[\!\begin{array}{cccc} 1 & -1 & 1 & -1 \\ 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 \\ 1 & 2 & 3 & 4 \end{array} \!\right]\left[\!\begin{array}{c} a_0 \\ a_1 \\ a_2 \\ a_3 \end{array} \!\right] = \left[\! \begin{array}{l} a_0 - a_1 + a_2 - a_3 \\ a_0 \\ a_0 + a_1 + a_2 + a_3 \\ a_0 + 2 a_1 + 4 a_2 + 8 a_3 \end{array} \!\right]. \]
- 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\.)
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Inspired by a question that I got in the class, 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.
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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;
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Step Two:
the coordinate colors,
- Red (vector $(1,0,0)$),
- Green (vector $(0,1,0)$),
- Blue (vector $(0,0,1)$);
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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;
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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;
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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.
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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:
- The reading homework for Tuesday is Section 3B Null Spaces and Ranges. Also do as many exercises in Exercises 3B. If you find an exercise that attracts your attention, please report it in Discussions on Canvas.
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Today we talked about Problem 1 on Assignment 1. The vectors in this problem are the real numbers in the open interval $(-1,1).$ The vector addition in this vector space is defined as
\[
\forall u,v \in (-1,1) \quad \mathbf{\mathsf{VectorPlus}}(u,v) = \frac{u+v}{1 + u v}.
\]
You pointed out correctly that to justify that this is a valid addition in this vector space we need to prove
\[
\forall u,v \in (-1,1) \quad \text{we have} \quad \frac{u+v}{1 + u v} \in (-1,1).
\]
That is, we need to prove the implication
\[
(-1 \lt u) \land (u \lt 1) \land (-1 \lt v) \land (v \lt 1) \quad \Rightarrow \quad -1 \lt \frac{u+v}{1 + u v} \ \ \land \ \ \frac{u+v}{1 + u v} \lt 1.
\]
Before finding a rigorous proof, it is good idea to look for a very strong conformation that the implication is true. Below I show the graph of the function of two variables
\[
f(u,v) = \frac{u+v}{1 + u v} \quad \text{for} \quad (u,v) \in \bigl([-1,1]\times[-1,1]\bigr)\setminus\bigl\{(-1,1),(1,-1)\bigr\}.
\]
Here $[-1,1]\times[-1,1]$ is the Cartesian product, that is,
\[
[-1,1]\times[-1,1] = \bigl\{ (u,v)\in \mathbb{R}^2 : u \in (-1,1) \land v \in (-1,1) \bigr\}.
\]
From the square $[-1,1]\times[-1,1]$ we exclude the verteces $(-1,1)$ and $(1,-1)$ since the function $f$ is not defined at those two points. Below is the graph of this function.
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Today, there was a question related to this problem in Discussions on Canvas.
. Below is what I posted in Discussions on Canvas.
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Only late this evening, I realized that a beautiful building could be created by putting together the four graphs we studied above.
Or, if the design fails as architecture, we can try to market it as a serving dish.
The statements $$-1 \lt \frac{u+v}{1 + u v} \ \ \land \ \ \frac{u+v}{1 + u v} \lt 1$$ and $$\left|\frac{u+v}{1 + u v}\right| \lt 1$$ are equivalent. I did not use the absolute value function in class since I do not see how it can be used it in proof. However, using the absolute value might be the simplest way to write what we need to prove $$|u|\lt 1 \ \land \ |v| \lt 1 \quad \Rightarrow \quad \left|\frac{u+v}{1 + u v} \right|\lt 1.$$
I did not use the absolute value function in the proof that I constructed. It does not mean that it cannot be used; just I do not see how. The absolute value function always hides two inequalities. When it comes to proofs involving the absolute value, I prefer to deal with two inequalities separately. Although, it might be more elegant to do one proof using the absolute value only. Sometimes it is possible, but it might require extra effort.
I feel that it is wrong on my part to push you towards proof that I came up with. However, I violated this rule today in class, and I suggested that you prove that $$u\in (-1,1) \ \land \ v\in (-1,1) \quad \Rightarrow \quad 1+uv \gt 0.$$ Then, I think it is useful to observe the following equivalences:
Assume that $$u\in (-1,1) \ \land \ v\in (-1,1).$$ The following two inequalities are equivalent $$-1 \lt \frac{u+v}{1 + u v} \quad \Leftrightarrow \quad -1-uv \lt u+v.$$ Also, the following two inequalities are equivalent $$\frac{u+v}{1 + u v} \lt 1 \quad \Leftrightarrow \quad u+v \lt 1+ uv.$$ Based on the last two equivalences, we need to prove the following two implications $$u\in (-1,1) \ \land \ v\in (-1,1) \quad \Rightarrow \quad 1 + u + v +uv \gt 0$$ and $$u\in (-1,1) \ \land \ v\in (-1,1) \quad \Rightarrow \quad 1 - u - v +uv \gt 0.$$ Sometimes one gets lucky in math, so the expressions $$1 + u + v +uv$$ and $$1 - u - v +uv$$ factor so beautifully conveniently (or, conveniently beautifully, I cannot decide which order of adverbs to use).
- The reading homework for Monday is Section 3A Vector Space of Linear Maps. Also do as many exercises in Exercises 3A. If you find an exercise that attracts your attention, please report it in Discussions on Canvas.
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Yesterday in class, I proved Proposition 2.43 on page 47 using a different strategy from one presented in the textbook. The proposition states: If $\mathcal{U}$ and $\mathcal{W}$ are subspaces of a finite-dimensional vector space, then
\[
\dim(\mathcal{U}+\mathcal{W}) = \dim\mathcal{U} + \dim\mathcal{W} - \dim(\mathcal{W}\cap\mathcal{W}).
\]
- In this proof we use an extended version of Proposition 2.33 on page 42 which states: If $\mathcal{U}$ is a subspace of a finite-dimensional vector space $\mathcal{V}$, then there exists a subspace $\mathcal{W}$ of $\mathcal{V}$ such that $\mathcal{V} = \mathcal{U}\oplus\mathcal{W}.$ In this case \[ \dim(\mathcal{U}\oplus\mathcal{W}) = \dim\mathcal{U} + \dim\mathcal{W}. \] In the proof of Proposition 2.33 in the textbook there is the sentence "Hence this list can be extended to a basis $u_1, \ldots,u_m,w_1,\ldots,w_n$ of $\mathcal{V}$." Here $m = \dim\mathcal{U},$ $n = \dim\mathcal{W}.$ Thus, in the proof of Proposition 2.33, it is proved that \[ \dim(\mathcal{U}\oplus\mathcal{W}) = \dim\mathcal{U} + \dim\mathcal{W}. \]
- Let $\mathcal{U}$ and $\mathcal{W}$ be subspaces of a finite-dimensional vector space $\mathcal{V}$.
- By Exercise 10 in 1.C the set $\mathcal{U}\cap \mathcal{W}$ is a subspace of both vector spaces $\mathcal{U}$ and $\mathcal{W}.$ By Proposition 2.33 there exist subspaces $\mathcal{U}_1$ and $\mathcal{W}_1$ such that \[ \mathcal{U} = (\mathcal{U}\cap \mathcal{W}) \oplus \mathcal{U}_1 \quad \text{and} \quad \mathcal{W} = (\mathcal{U}\cap \mathcal{W}) \oplus \mathcal{W}_1 \] and \[ \dim\mathcal{U} = \dim(\mathcal{U}\cap \mathcal{W}) + \dim\mathcal{U}_1 \quad \text{and} \quad \dim\mathcal{W} = \dim(\mathcal{U}\cap \mathcal{W}) + \dim\mathcal{W}_1. \]
-
Next we prove that \[ \mathcal{U}_1\cap \mathcal{W}_1 = \{0_{\mathcal{V}}\}. \] Let $v \in \mathcal{U}_1\cap \mathcal{W}_1$ be arbitrary. Since $\mathcal{U}_1 \subseteq \mathcal{U}$ and $\mathcal{W}_1 \subseteq \mathcal{W},$ we deduce $v \in \mathcal{U}\cap \mathcal{W}.$ Thus \[ v \in (\mathcal{U}\cap \mathcal{W}) \cap \mathcal{U}_1 = \{0_{\mathcal{V}}\}. \] That is $v = 0_{\mathcal{V}}.$
Hence, the sum $\mathcal{U}_1 + \mathcal{W}_1$ is a direct sum.
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Next we prove that (prove this as an exercise) \[ \mathcal{U}+\mathcal{W} = (\mathcal{U}\cap \mathcal{W}) + \bigl( \mathcal{U}_1 \oplus \mathcal{W}_1 \bigr) \] and \[ (\mathcal{U}\cap \mathcal{W}) \cap \bigl( \mathcal{U}_1 \oplus \mathcal{W}_1 \bigr) = \{0_{\mathcal{V}}\}. \]
Let $u_1 \in \mathcal{U}_1$ and $w_1 \in \mathcal{W}_1$ be arbitrary and assume that $u_1+w_1 \in \mathcal{U}\cap \mathcal{W}.$ Set $v=u_1+w_1 \in \mathcal{U}\cap \mathcal{W}.$ The equality \[ u_1 = v - w_1 \] provides a rich reading: \[ v-w_1 \in \mathcal{W} \quad \text{since} \quad v \in \mathcal{U}\cap \mathcal{W} \subseteq \mathcal{W} \quad \text{and} \quad -w_1 \in \mathcal{W}_1 \subseteq \mathcal{W} \] and \[ v-w_1 = u_1 \in \mathcal{U} \quad \text{since} \quad u_1 \in \mathcal{U}_1 \subseteq \mathcal{U}. \] Thus, \[ v-w_1 = u_1 \in \mathcal{U}\cap \mathcal{W} \quad \text{and} \quad v-w_1 = u_1 \in \mathcal{U}_1. \] Since \[ (\mathcal{U}\cap \mathcal{W}) \cap \mathcal{U}_1 = \{0_{\mathcal{V}}\}, \] we deduce $v-w_1 = u_1 = 0_{\mathcal{V}}.$ Consequently, \[ w_1 = v \in (\mathcal{U}\cap \mathcal{W}) \cap \mathcal{W}_1 = \{0_{\mathcal{V}}\}. \] Thus, both $u_1 = 0_{\mathcal{V}}$ and $w_1 = 0_{\mathcal{V}}.$ This proves that \[ (\mathcal{U}\cap \mathcal{W}) \cap \bigl( \mathcal{U}_1 \oplus \mathcal{W}_1 \bigr) = \{0_{\mathcal{V}}\}, \] and consequently \[ \mathcal{U}+\mathcal{W} = (\mathcal{U}\cap \mathcal{W}) \oplus \bigl( \mathcal{U}_1 \oplus \mathcal{W}_1 \bigr). \]
Therefore, \[ \mathcal{U}+\mathcal{W} = (\mathcal{U}\cap \mathcal{W}) \oplus \bigl( \mathcal{U}_1 \oplus \mathcal{W}_1 \bigr) \]
- By the extended version of Proposition 2.33, the last equality and the direct sum in the third item of this proof imply that \begin{align*} \dim(\mathcal{U}+\mathcal{W}) & = \dim(\mathcal{U}\cap \mathcal{W}) + \dim\bigl( \mathcal{U}_1 \oplus \mathcal{W}_1 \bigr) \\ & = \dim(\mathcal{U}\cap \mathcal{W}) + \dim\mathcal{U}_1 + \dim\mathcal{W}_1 \\ & = \dim(\mathcal{U}\cap \mathcal{W}) + \dim\mathcal{U}_1 + \dim(\mathcal{U}\cap \mathcal{W}) + \dim\mathcal{W}_1 - \dim(\mathcal{U}\cap \mathcal{W}) \\ & = \dim\mathcal{U} + \dim\mathcal{W} - \dim(\mathcal{U}\cap \mathcal{W}). \end{align*} This completes the proof.
- In Problem 2 on Assignment 1, I ask you to explore a subspace $\mathcal{S}_{\omega}$ of the vector space $\mathbb{R}^\mathbb{R}.$ A starting point for the explorations related to this problem is to familiarize ourselves with the individual members of $\mathcal{S}_{\omega}$ for a given $\omega$. It is useful to start with $\omega=1$ and plot several functions in $\mathcal{S}_1.$ One can do this by hand and stay with a small number of functions.
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Which functions are in $\mathcal{S}_1?$ For example, with $a=1$ and $b=0$, the function $\sin(x)$ is in the set $\mathcal{S}_1,$ with $a=1$ and $b=\pi$, the function $\sin(x+\pi) = -\sin(x)$ is in the set $\mathcal{S}_1.$ One can continue with specific values with $a$ and $b$ and plot individual functions.
Below I present 180 functions from $\mathcal{S}_1$ with the coefficients \begin{align*} a & \in \left\{\frac{1}{6}, \frac{1}{3}, \frac{1}{2}, \frac{2}{3}, \frac{5}{6}, 1, \frac{7}{6}, \frac{4}{3}, \frac{3}{2}, \frac{5}{3}, \frac{11}{6},2, \frac{13}{6}, \frac{7}{3}, \frac{5}{2} \right\}, \\ b & \in \left\{ 0, \frac{\pi}{6},\frac{\pi}{3},\frac{\pi}{2},\frac{2\pi}{3}, \frac{5\pi}{6}, \pi, \frac{7\pi}{6},\frac{4\pi}{3},\frac{3\pi}{2},\frac{5\pi}{3}, \frac{11\pi}{6} \right\} \end{align*}
Place the cursor over the image to see individaual functions.
- I created the above animation and picture using the Computer algebra system Mathematica. Enjoyment of mathematics is considerable enriched with Mathematica. I would like to encourage you to get familiar with it. 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.
- The reading homework for Thursday is Section 2B Bases and Section 2C Dimension. Also do as many exercises in Exercises 2B and 2C as you can. If you find an exercise that you cannot solve please report them in Discussions on Canvas.
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We discussed Exercise 16 in Section 2A.
This exercise requires us to prove that the vector space of all continuous real-valued functions defined on the interval $[0,1]$ is infinite dimensional. Denote this space by $\mathcal{C}[0,1]$.
I suggested that Exercise 14 in the same section can be used. This exercise states that a vector space $\mathcal{V}$ is infinite-dimensional if and only if there is a sequence \[ v_1,\ldots, v_n,\ldots \] in $\mathcal{V}$ such that for every $m\in\mathbb{N}$ the vectors $v_1,\ldots, v_m$ are linearly independent.
- To apply Exercise 14 to Exercise 16 we need to create a sequence of continuous functions \[ \phi_n : [0,1] \to \mathbb{R} \quad \text{with} \quad n \in \mathbb{N}, \] such that \[ \forall m \in \mathbb{N} \quad \phi_1,\ldots, \phi_m \] are linearly independent.
- We quickly came up with two such sequences that some of you were familiar with from your prior math expedience. One sequence was \[ \forall n \in \mathbb{N} \quad \phi_n(t) = \sin(\pi n t) \quad \text{with} \quad t\in [0,1]. \] The other sequence was \[ \forall n \in \mathbb{N} \quad \phi_n(t) = t^{n-1} \quad \text{with} \quad t\in [0,1]. \]
- To prove that the sequence \[ \forall n \in \mathbb{N} \quad \phi_n(t) = \sin(\pi n t) \quad \text{with} \quad t\in [0,1] \] has the property \[ \forall m \in \mathbb{N} \quad \phi_1,\ldots, \phi_m \] are linearly independent, we need an inner product in $\mathcal{C}[0,1]$ in which the functions are orthogonal. We will do this in Chapter 6. At Western we do this in Math 304. But in Math 504 we cover it in more detail.
- So, to complete the solution of Exercise 16 the second sequence might be more appropriate. Let \[ \forall n \in \mathbb{N} \quad \phi_n(t) = t^{n-1} \quad \text{with} \quad t\in [0,1]. \] We need to prove that this sequence has the property \[ \forall m \in \mathbb{N} \quad \phi_1,\ldots, \phi_m \] are linearly independent. Let us state precisely the meaning of the last statement. We need to prove the following implication for all $m\in\mathbb{N}$ the following implication holds \[ \forall t \in [0,1] \quad \alpha_1 + \alpha_2 t + \cdots + \alpha_m t^{m-1} = 0 \quad \Rightarrow \quad \forall k \in \{1,\ldots,m\} \ \ \alpha_k = 0. \]
- The discussion in the textbook starting at Definition 2.10 and ending Example 2.14 is relevant to this question. But, the most relevant is Proposition 4.6 on page 116. However, there is a little detail here that needs to be addressed. In Proposition 4.6 the domain is $\mathbb{R}$ and here the domain is $[0,1]$.
- Another suggestion was to use the Fundamental Theorem of Algebra. I agree, the Fundamental Theorem of Algebra can be used, but to make such a proof complete, please write down the exact statement of this theorem and explain in detail how you use this theorem. My reluctance to use the Fundamental Theorem of Algebra comes from my desire to come up with a simpler proof. Also, whenever we use a powerful tool it is a good idea to review its proof. Here are my notes with a simple proof of the Fundamental Theorem of Algebra.
- We briefly discussed bases of arbitrary vector spaces. I found a very nice presentation of this topic at the webpage of William Fulton Distinguished University Professor Karen E. Smith at the University of Michigan. Here is the link to her notes.
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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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The following reasoning is inspired by the Ode to RREF. It leads to an alternative construction of a basis starting from a spanning list. If you spend some time reading the Ode, please pay attention to the following dichotomy relationship among the columns, which is present in each list of vectors. I will call a list a tuple since it seems to me that it is more common mathematical terminology. Although, I should get used to using "list" since that is commonly used term in computer science.
- Let $n\in\mathbb{N}.$ Let \[ \bigl(v_1, v_2, \ldots, v_n\bigr) \] be an $n$-tuple of vectors in a vector space $\mathcal{V}$ over $\mathbb{F}$. Assume: \[ \exists k' \in \{1,\ldots,n\} \quad \text{such that} \quad v_{k'} \neq 0_{\mathcal{V}}. \] This assumption is equivalent to \[ \operatorname{span} \{ v_1, v_2, \ldots, v_n \} \neq \{0_{\mathcal{V}}\} \]
- Here we define an $n$-tuple $\delta$ consisting of $0$-s and $1$-s. That is an $n$-tuple in $\{0,1\}^{\{1,\ldots,n\}}$. For every $k\in \{1,\ldots,n\}$ we set \[ \delta_k = \begin{cases} 1 &\text{if} \quad v_k \not\in \operatorname{span}\bigl\{v_j: j \in\{1,\ldots,n\} \land j \lt k \bigr\}, \\[6pt] 0 &\text{if} \quad v_k \in \operatorname{span}\bigl\{v_j: j \in\{1,\ldots,n\} \land j \lt k \bigr\}. \end{cases} \] There might be a little confusion about the definition of $\delta_1$. What is \[ \bigl\{v_j: j \in\{1,\ldots,n\} \land j \lt 1 \bigr\}? \] It seems to me that it is clear that \[ \bigl\{v_j: j \in\{1,\ldots,n\} \land j \lt 1 \bigr\} = \emptyset. \] By definition \[ \operatorname{span} \emptyset = \{0_{\mathcal{V}}\}. \] Thus \[ \delta_1 = \begin{cases} 1 &\text{if} \quad v_1 \neq 0_{\mathcal{V}}, \\[6pt] 0 &\text{if} \quad v_1 = 0_{\mathcal{V}}. \end{cases} \] If you look at the matrix discussed in the Ode, you will notice that $n=5$ and $\delta = (1,1,0,1,0).$
- Set \[ d = \sum_{k=1}^{n} \delta_k. \] Since we assume that at least on vector in the $n$-tuple is nonzero, we deduce that $d \in \mathbb{N}$, that is $d \geq 1$.
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Consider the subset of $\bigl\{v_1, v_2, \ldots, v_n\bigr\}$ defined as \[ \mathcal{P} = \bigl\{ v_k : k \in \{1,\ldots,n\} \land \delta_k = 1 \bigr\}. \] I claim that the set $\mathcal{P}$ has $d$ elements, that $\mathcal{P}$ is linearly independent and the span of $\mathcal{P}$ is identical to the span of $\bigl\{v_1, v_2, \ldots, v_n\bigr\}.$
In the Ode, the set $\mathcal{P}$ is the set of all yellow columns. In the context of matrices these columns are called the pivot columns of a matrix.
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In fact, I would prefer to write the set $\mathcal{P}$ is a $d$-tuple. But how to do that? In plain English, from the $n$-tuple $\bigl(v_1, v_2, \ldots, v_n\bigr)$ I want to select those vectors $v_k$ that received $\delta_k =1$ in exactly the same order as their appear in the original $n$-tuple. Notice that this is done in the Ode just by visual inspection, so to say. Below is one way how to describe this process rigorously that I see now.
First define the function \[ \eta : \{1,\ldots,n\} \to \{0,1,\ldots,d\} \] by \[ \forall k\in \{1,\ldots,n\} \quad \eta_k = \sum_{j=1}^{k} \delta_j. \] Observe the facts: $0$ is in the range of $\eta$ if and only if $v_1 = 0_{\mathcal{V}}$ and $\{1,\ldots,d\}$ is a subset of the range of $\eta.$ Therefore \[ \forall p \in \{1,\ldots,d\} \quad \bigl\{ k \in \{1,\ldots,n\} : \eta_k = p \bigr\} \neq \emptyset. \] Define \[ \mu : \{1,\ldots,d\} \to \{1,\ldots,n\} \] by \[ \forall p \in \{1,\ldots,d\} \quad \mu_p = \min \bigl\{ k \in \{1,\ldots,n\} : \eta_k = p \bigr\}. \] Now we can write the set $\mathcal{P}$ as a $d$-tuple: \[ \bigl( v_{\mu_1}, \ldots, v_{\mu_d} \bigr). \]
- For the matrix in the Ode we have $d=3$ and \[ \eta = (1,2,2,3,3). \] Therefore \begin{align*} \bigl\{ k \in \{1,2,3,4,5\} : \eta_k = 1 \bigr\} & = \{1\} \\ \bigl\{ k \in \{1,2,3,4,5\} : \eta_k = 2 \bigr\} & = \{2,3\} \\ \bigl\{ k \in \{1,2,3,4,5\} : \eta_k = 3 \bigr\} & = \{4,5\}. \end{align*} Therefore \begin{align*} \mu_1 & = \min \bigl\{ k \in \{1,2,3,4,5\} : \eta_k = 1 \bigr\} = \min \{1\} = 1 \\ \mu_2 & = \min \bigl\{ k \in \{1,2,3,4,5\} : \eta_k = 2 \bigr\} = \min \{2,3\} = 2 \\ \mu_3 & = \min \bigl\{ k \in \{1,2,3,4,5\} : \eta_k = 3 \bigr\} = \min \{4,5\} = 4. \end{align*} Thus the pivot columns of the matrix in the Ode are: the first, the second and the fourth column.
- The procedure introduced in for the $n$-tuple of vectors \[ \bigl(v_1, v_2, \ldots, v_n\bigr) \] is analogous to what is performed by row reduction for a matrix. The $n$-tuple $\delta$ selects the special linearly independent vectors from an $n$-tuple of vectors. These vectors could be called the pivot vectors of an $n$-tuple of vectors.
- The reading homework for Thursday is to continue with the reading of Section 2A Span and Linear Independence. Also do as many exercises in Exercises 2A as you can. If you find an exercise that you cannot solve please report them in Discussions on Canvas.
- I promised to present a list of Exercises 2A which catch my interest. Thank you for asking about Exercise 14. This is definitely one of the most interesting exercises in this set. My list is: 5, 9, 10, 11, 14, 15 (I prefer the notation $\mathbb{F}^\mathbb{N}$), 16, 17. Is this list too long? too short? More problems you do, better your problem solving skills will develop. But, never forget to think of your own variations on the problems that you work on. Asking your own questions and pursuing them truly individualizes the learning experience.
- Experiencing mathematics
Understanding a theorem, mathematical concept, or a problem is an individual act of creative sensemaking that engages all our senses and creativity.
- These are my notes on Vector Spaces. Please check Problems 1, 2, 3 at the end of the notes.
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These are my notes on Bases. In these notes instead of lists I work with finite sets. For the Steinitz exchange lemma see its Wikipedia page. The Wikipedia's proof is very similar to the proof in my notes.
For a finite set $S$ the symbol $|S|$ denotes the number of elements in that set. This notation is identical to the notation for the absolute value function. There should be no confusion since the reader should always consider the nature of the object which is enclosed by the vertical bars. If the object enclosed by the vertical bars is a real number, then the symbol is the absolute value of that real number; the object enclosed by the vertical bars is a complex number, then the symbol is the modulus of that complex number; the object enclosed by the vertical bars is a finite set, then the symbol is the number of elements of that set; see the Wikipedia page Cardinality.
- The reading homework for Tuesday is Section 2A Span and Linear Independence. Also as many exercises in Exercises 2A as you can. If you find an exercise that you cannot solve please report them in Discussions on Canvas.
- In this class we present a rigorous approach to Linear Algebra. Since rigorous reasoning in Mathematics is based on Mathematical Logic, I wrote a Brief Review of Mathematical Logic. After reading Brief Review of Mathematical Logic you can read my webpage Mathematical Rigor in the Context of Quadratic Functions where I rigorously prove some statements that you might have seen in precalculus context.
- The reading homework for Monday is Section 1C Subspaces. Also as many exercises in Exercises 1C as you can. If you find an exercise that you cannot solve please report them in Discussions on Canvas.
- Today we discussed Section 1B Definition of Vector Space. I proved Proposition 1.30. Please read and internalize Propositions 1.26, 1.27, 1.31, 1.32. There are more Propositions of this kind in Exercises 1B: 1 and 2. Exercise 5 is really interesting.
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I recall the definition of a vector space as I stated it in class. Here $\mathbb{F}$ is either $\mathbb{R}$ or $\mathbb{C}.$
Definition. A nonempty set $\mathcal{V}$ is said to be a vector space over $\mathbb{F}$ if it satisfies the following ten axioms.
Axiom 1. (AE) There exists a function $+: \mathcal{V}\!\times\!\mathcal{V} \to \mathcal{V}.$That is, for each pair $(u,v) \in \mathcal{V}\!\times\!\mathcal{V}$ there exists a unique $u+v \in \mathcal{V}$ which is called the sum of $u$ and $v.$Axiom 2. (AA) For all $u, v, w \in \mathcal{V}$ we have $u+(v+w) = (u+v)+w.$Axiom 3. (AC) For all $u, v \in \mathcal{V}$ we have $u+v = v+u.$Axiom 4. (AZ) There exists $0_{\mathcal{V}} \in \mathcal{V}$ such that for every $v \in \mathcal{V}$ we have $v+0_{\mathcal{V}} = v.$Axiom 5. (AO) For every $v \in \mathcal{V}$ there exists $w \in \mathcal{V}$ such that $v+w = 0_{\mathcal{V}}.$Axiom 6. (SE) There exists a function $\cdot: \mathbb{F}\!\times\!\mathcal{V} \to \mathcal{V}.$That is, for each real number $\alpha \in \mathbb{F}$ and each $v \in \mathcal{V}$ there exists a unique $\alpha v \in \mathcal{V}$ which is called the scalar product of $\alpha$ and $v.$Axiom 7. (SA) For every $\alpha, \beta \in \mathbb F$ and every $v \in \mathcal{V}$ we have $\alpha (\beta v) = (\alpha\beta) v$Axiom 8. (SD) For every $\alpha, \beta \in \mathbb F$ and every $v \in \mathcal{V}$ we have $(\alpha +\beta) v = \alpha v + \beta v$Axiom 9. (SD) For every $\alpha \in \mathbb F$ and every $u, v \in \mathcal{V}$ we have $\alpha (u + v) = \alpha u + \alpha v$Axiom 10. (S0) For every $v \in \mathcal{V}$ we have $1 v = v$Explanation of the abbreviations: AE--addition exists, AA--addition is associative, AC--addition is commutative, AZ--addition has zero, AO--addition has opposites, SE-- scaling exists, SA--scaling is associative, SD--scaling distributes over addition of real numbers, SD--scaling distributes over addition of vectors, SO--scaling with one.
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In class I made the following daring claim: Each example of a vector space over $\mathbb{F}$ that I know is a subspace of a vector space $\mathbb{F}^D$ for some nonempty set $D.$
While talking to students during the office hour today, I realized that my daring claim is not true. Below I give an example of an exotic vector space for which I do not see how to present it as a subspace of $\mathbb{R}^D$ for some nonempty set $D.$
- Here I give an example of an "exotic" vector space: \[ \require{bbox} \mathcal{V} = \left\{ \bbox[yellow]{\left[\! \begin{array}{c} x_1 \\ x_2 \end{array} \!\right]} \, : \, x_1, x_2 \in \mathbb{R} \ \ \text{and} \ \ x_1, x_2 \gt 0 \right\} \] with the addition defined as \[ \bbox[yellow]{\left[\! \begin{array}{c} x_1 \\ x_2 \end{array} \!\right]} \bbox[yellow]{+} \bbox[yellow]{\left[\! \begin{array}{c} y_1 \\ y_2 \end{array} \!\right]} = \bbox[yellow]{\left[\! \begin{array}{c} x_1 y_1 \\ x_2 y_2 \end{array} \!\right]} \quad \text{for all} \quad \bbox[yellow]{\left[\! \begin{array}{c} x_1 \\ x_2 \end{array} \!\right]}, \bbox[yellow]{\left[\! \begin{array}{c} y_1 \\ y_2 \end{array} \!\right]} \in \mathcal{V} \] and scaling defined as \[ \alpha \bbox[yellow]{\phantom{*}} \bbox[yellow]{\left[\! \begin{array}{c} x_1 \\ x_2 \end{array} \!\right]} = \bbox[yellow]{\left[\! \begin{array}{c} (x_1)^\alpha \\ (x_2)^\alpha \end{array} \!\right]} \quad \text{for all} \quad \bbox[yellow]{\left[\! \begin{array}{c} x_1 \\ x_2 \end{array} \!\right]}\in \mathcal{V} \quad \text{and for all} \quad \alpha \in \mathbb{R}. \] It is a nice exercise to verify that all the axioms of a vector space are satisfied. Interestingly, the zero vector in the vector space $\mathcal{V}$ is the vector $\bbox[yellow]{\left[\! \begin{array}{c} 1 \\ 1 \end{array} \!\right]}.$
- I colored the vectors in $\mathcal{V}$ yellow to distinguish them from the vectors in $\mathbb{R}^2.$ As an exercise you can prove that the above defined set $\mathcal{V}$ with the given addition and scaling satisfies all the axioms of a vector space.
- It is interesting to study subspaces of the vector space $\mathcal{V}.$ For example, what is \[ \operatorname{Span} \left\{ \bbox[yellow]{\left[\! \begin{array}{c} 1 \\ 2 \end{array} \!\right]} \right\}, \] or, what is \[ \operatorname{Span} \left\{ \bbox[yellow]{\left[\! \begin{array}{c} 2 \\ 1 \end{array} \!\right]} \right\}, \] or, \[ \operatorname{Span} \left\{ \bbox[yellow]{\left[\! \begin{array}{c} 2 \\ 2 \end{array} \!\right]} \right\}, \] or \[ \operatorname{Span} \left\{ \bbox[yellow]{\left[\! \begin{array}{c} 4 \\ 2 \end{array} \!\right]} \right\}, \] or, \[ \operatorname{Span} \left\{ \bbox[yellow]{\left[\! \begin{array}{c} 2 \\ 4 \end{array} \!\right]} \right\}. \]
- The information sheet
- Some relevant Wikipedia links:
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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: