- My updated notes on all topics that we covered:
- My updated notes on Inner Product Spaces.
- My updated notes on Inner Product Spaces.
- My notes on Inner Product Spaces.
- My notes on Eigensystems of Linear Operators.
- Small updates to my notes on Linear Operators.
- My notes on Bases.
- My notes on Vector Spaces.
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Yesterday, I posted Assignment 2. Here, I rewrite the narrative to the assignment that I posted on Canvas.
- I find it essential to highlight the connection between functions and their graphs. Problem 1 reflects that spirit. In some operator theory contexts, subspaces of the direct product vector space \(\mathscr{V}\times\mathscr{W}\) are viewed as multivalued operators.
- Problem 2 not only challenges you to recognize and apply the tools we developed in class but also explores the interplay between writing mathematics in English and Mathish. I think this is a great opportunity to practice our translation skills.
- Problem 3 is standard. I want to emphasise (I)(c) for creating the inverse of a matrix by looking at the meaning of the operator, not doing calculations. This idea might be helpful in Problem 5. Part (II)(a) is an exercise in functions; in (II)(b) and (II)(c) you can use the power of the isomorphism \[ M_{\mathscr{M}}^{\mathscr{M}} : \mathscr{L}\bigl(\mathscr{P}_3\bigr) \to \mathbb{R}^{4\times 4} \] and a special kind of matrices that you can look up on the Internet if you have not encountered them in an undergraduate Linear Algebra class.
- It would be nice if you reviewed and wrote down a proof—the simpler, the better—of the fundamental fact of Matrix Linear Algebra that inspired Problem 4.
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The context of all questions in Problem 5 is the vector space \(\mathscr{L}\bigl(\mathscr{P}_n\bigr)\), the space of all linear operators on the vector space of all polynomials of degree at most \( n \) with coefficients in \( \mathbb{R} \). However, in this context, it is essential to rewrite all statements explicitly in terms of polynomials, including the variable and all relevant quantifiers.
In part (a), one should aim for the simplest possible proof. Applying heavy theorems from Mathematical Analysis to polynomials seems inappropriate. One-line proofs that merely cite a major theorem should, in my opinion, be avoided. If using such a theorem is absolutely unavoidable, one should review and cite it in full detail. Proofs from first principles are always preferable.
- Above, in my narrative about Assignment 2, I share my thoughts on the problems. I hope they are useful. However, the ultimate goal is for you to develop your own thinking about the problems and everything you study. If you ever feel that the existing narrative stifles your own thoughts, disregard it.
- To prove the existence of eigenvalues, we used the Fundamental Theorem of Algebra. In keeping with the principle that we should review major theorems in detail when using them, I am posting my variation of the proof of this theorem.
- I added two examples that we discussed today to my notes on Linear Operators.
- Thank you for your questions today. Based on the discussion in class today, I updated my notes on
- My notes on Eigensystems of Linear Operators.
- I updated all three sections with minor changes.
- My notes on Linear Operators. The same notes with cropped pages.
- My notes on Bases. The same notes with cropped pages.
- My notes on Vector Spaces. The same notes with cropped pages.
- It is often convenient to identify a function with its graph. In my webpage Functions, I illustrate why this perspective is useful. In this post I use this perspective.
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We use the following notation: \(\mathbb{N}\) denotes the set of positive integers, and \(\mathbb{F}\) represents a scalar field, which is either \(\mathbb{R}\) or \(\mathbb{C}\). Let \(m, n \in \mathbb{N}\). We consider finite-dimensional vector spaces \(\mathscr{V}\) and \(\mathscr{W}\) over \(\mathbb{F}\), with \(\dim \mathscr{V} = m\) and \(\dim \mathscr{W} = n\). The vector space of all \(m\)-tuples of elements in \(\mathscr{V}\) is denoted by \(\mathcal{V}^m\). Throughout the post below, all subsets of vector spaces are denoted using calligraphic letters, such as \(\mathscr{A}\), \(\mathscr{B}\), and \(\mathscr{C}\).
The vectors in abstract vector spaces are denoted by lowercase Latin letters, such as \(a, b, v, w\), while vectors in \(\mathbb{F}^m\), which are represented as \(m\)-tuples, are written in bold lowercase Latin letters, such as \(\mathbf{a}, \mathbf{b}, \mathbf{x}\).
In displayed formulas, we will use the following logical symbols: \(\land\) (conjunction) for "and", \(\lor\) (disjunction) for "or", \(\exists\) (existential quantifier) for "there exists", \(\forall\) (universal quantifier) for "for all", \(\Rightarrow\) (implication) for "implies", \(\Leftrightarrow\) (logical equivalence) for "if and only if".
- Objective. In today's post, I will introduce a special class of linear operators between the vector spaces \(\mathscr{V}\) and \(\mathscr{W}\). This perspective on linear operators highlights the fact that a linear operator on a finite-dimensional vector space \(\mathscr{V}\) is uniquely determined by its values on a basis of \(\mathscr{V}\). The definition of this class of linear operators is inspired by the definition of the Coordination Operator expressed in terms of its graph. Therefore, we begin with this definition.
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The following theorem introduces what I call the Coordination Operator using graph notation.
Coordination Theorem. Let \( m \in \mathbb{N} \), and let \( \mathscr{V} \) be a finite-dimensional vector space over \( \mathbb{F} \) with \( \dim \mathscr{V} = m \). Let \[ \mathscr{B} = \bigl( b_1, \ldots, b_m \bigr) \in \mathscr{V}^m \] be a basis for \( \mathscr{V}\). Then, the subset of \( \mathscr{V} \times \mathbb{F}^m \) given by \[ C_{\mathscr{B}} = \left\{ (v, \mathbf{a}) \in \mathscr{V} \times \mathbb{F}^m \,:\, v = \sum_{k=1}^{m} \alpha_k b_k \ \ \land \ \ \mathbf{a} = \begin{bmatrix} \alpha_1 \\[-3pt] \vdots \\[-5pt] \alpha_m \end{bmatrix} \right\} \] is the graph of the linear bijection (isomorphism) \( C_{\mathscr{B}} : \mathscr{V} \to \mathbb{F}^m \). (The linear operator \( C_{\mathscr{B}} \) is called the Coordination Operator.)
- For the proof we considered the "reversal" or "flip" of the graph of $C_{\mathscr{B}}$: \[ C_{\mathscr{B}} = \left\{ (v, \mathbf{a}) \in \mathscr{V} \times \mathbb{F}^m \,:\, \mathbf{a} = \begin{bmatrix} \alpha_1 \\[-3pt] \vdots \\[-5pt] \alpha_m \end{bmatrix} \ \ \land \ \ v = \sum_{k=1}^{m} \alpha_k b_k \right\} \] and we proved that both, the given set $C_{\mathcal{B}}$ and its reversal are graphs of functions.
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The definition of the Coordination Operator in the Coordination Theorem, presented in the preceding item, inspired me to define the following linear operator \( L_{\mathscr{C}}^{\mathscr{B}} \).
Definition. Let \( m, n \in \mathbb{N} \) and let \(\mathscr{V}\) and \(\mathscr{W}\) be finite-dimensional vector spaces over \( \mathbb{F} \) with \( \dim \mathscr{V} = m \) and \( \dim \mathscr{W} = n\). Let \[ \mathscr{B} = \bigl( b_1, \ldots, b_m \bigr) \in \mathscr{V}^m \] be a basis for \( \mathscr{V}\), and let \[ \mathscr{C} = \bigl( c_1, \ldots, c_m \bigr) \in \mathscr{W}^m, \] where \( \mathscr{C} \) is an arbitrary \( m \)-tuple in \( \mathscr{W}^m \). We define \begin{equation*} L_{\mathscr{C}}^{\mathscr{B}} = \left\{ (v, w ) \in \mathscr{V} \times \mathscr{W} : \mkern 15mu \begin{array}{l} \mkern 10mu \exists\mkern 0.5mu \begin{bmatrix} \alpha_1 \\[-3pt] \vdots \\[-5pt] \alpha_m \end{bmatrix} \in \mathbb{F}^m \ \ \text{such that} \\[7pt] \displaystyle v = \sum_{k = 1}^{m} \alpha_k b_k \land w = \sum_{k = 1}^{m} \alpha_k c_k \end{array} \right\}. \end{equation*}
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Let \(m \in \mathbb{N}\). Recall that the columns of the $m\times m$ identity matrix $\mathsf{I}_m$ form the standard basis for $\mathbb{F}^m$. This basis I denote by $\mathscr{E}_m$ and its vectors I denote by $\mathbf{e}_1, \ldots, \mathbf{e}_m$:
\[
\mathbf{e}_1 =\begin{bmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{bmatrix} , \quad
\mathbf{e}_2 = \begin{bmatrix} 0 \\ 1 \\ \vdots \\ 0 \end{bmatrix} , \quad \cdots \quad
\mathbf{e}_m = \begin{bmatrix} 0 \\ 0 \\ \vdots \\ 1 \end{bmatrix} .
\]
With this notation, we have the following equivalence
\[
\mathbf{a} = \begin{bmatrix} \alpha_1 \\[-3pt] \vdots \\[-5pt] \alpha_m
\end{bmatrix} \quad \Leftrightarrow \quad \mathbf{a} = \sum_{k = 1}^{m} \alpha_k \mathbf{e}_k.
\]
- Notice that the coordination operator \(C_{\mathscr{E}_m}\) is the identity operator on \(\mathbb{F}^m\).
- Comparing the definitions in the preceding two ■ items, we obtain \[ C_{\mathscr{B}} = L^{\mathscr{B}}_{\mathscr{\mathscr{E}_m}}, \] which served as the inspiration for the definition of \(L_{\mathscr{C}}^{\mathscr{B}}\).
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Here we prove that the set \(L_{\mathscr{C}}^{\mathscr{B}}\) defined above is a graph of a linear operator.
Theorem. Let \( m, n \in \mathbb{N} \) and let \(\mathscr{V}\) and \(\mathscr{W}\) be finite-dimensional vector spaces over \( \mathbb{F} \) with \( \dim \mathscr{V} = m \) and \( \dim \mathscr{W} = n\). Let \[ \mathscr{B} = \bigl( b_1, \ldots, b_m \bigr) \in \mathscr{V}^m \] be a basis for \( \mathscr{V}\), and let \[ \mathscr{C} = \bigl( c_1, \ldots, c_m \bigr) \in \mathscr{W}^m, \] where \( \mathscr{C} \) is an arbitrary \( m \)-tuple in \( \mathscr{W}^m \). Then the set \begin{equation*} L_{\mathscr{C}}^{\mathscr{B}} = \left\{ (v, w ) \in \mathscr{V} \times \mathscr{W} : \mkern 15mu \begin{array}{l} \mkern 10mu \exists\mkern 0.5mu \begin{bmatrix} \alpha_1 \\[-3pt] \vdots \\[-5pt] \alpha_m \end{bmatrix} \in \mathbb{F}^m \ \ \text{such that} \\[7pt] \displaystyle v = \sum_{k = 1}^{m} \alpha_k b_k \land w = \sum_{k = 1}^{m} \alpha_k c_k \end{array} \right\} \end{equation*} is a graph of a linear operator \(L_{\mathscr{C}}^{\mathscr{B}} : \mathscr{V} \to \mathscr{W}\). That is, \[ L_{\mathscr{C}}^{\mathscr{B}} \in \mathscr{L}( \mathscr{V},\mathscr{W}). \]
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Here we prove that
\[
\mathscr{W}^m \ni \mathscr{C} \ \mapsto \ L_{\mathscr{C}}^{\mathscr{B}}
\in \mathscr{L}( \mathscr{V},\mathscr{W})
\]
is an isomorphism between vector spaces.
Theorem. Let \( m, n \in \mathbb{N} \) and let \(\mathscr{V}\) and \(\mathscr{W}\) be finite-dimensional vector spaces over \( \mathbb{F} \) with \( \dim \mathscr{V} = m \) and \( \dim \mathscr{W} = n\). Let \[ \mathscr{B} = \bigl( b_1, \ldots, b_m \bigr) \in \mathscr{V}^m \] be a (fixed, arbitrary) basis for \(\mathscr{V}\). The function \[ \Xi : \mathscr{W}^m \to \mathscr{L}( \mathscr{V},\mathscr{W}) \] defined by \[ \forall\mkern 2mu \mathscr{C} \in \mathscr{W}^m \qquad \Xi (\mathscr{C}) = L_{\mathscr{C}}^{\mathscr{B}} \] is an isomorphism between vector spaces \(\mathscr{W}^m\) and \(\mathscr{L}( \mathscr{V},\mathscr{W})\).
- For the proof we define the function \(\Theta: \mathscr{L}( \mathscr{V},\mathscr{W}) \to \mathscr{W}^m\) by setting: \[ \forall\mkern 2mu T \in \mathscr{L}( \mathscr{V},\mathscr{W}) \qquad \Theta(T) = \bigl( Tb_1, \ldots, Tb_m\bigr). \] Then we prove that \(\Theta\) is a function and that it satisfies the following two conditions: \begin{alignat*}{2} \forall\mkern 2mu T &\in \mathscr{L}( \mathscr{V},\mathscr{W}) \qquad & \Xi\bigl(\Theta(T)\bigr) & = T, \\ \forall\mkern 2mu \mathscr{C} &\in \mathscr{W}^m \qquad & \Theta\bigl(\Xi (\mathscr{C})\bigr) & = \mathscr{C}. \end{alignat*} That is, we prove that \(\Theta\) is the inverse of \(\Xi\).
- The symbol \(\Xi\) is the capital Greek letter xi; its lowercase is \(\xi\). I use \(\xi\) when \(x\) might be confusing, as in \(\displaystyle \int_a^x\mkern -5mu f(\xi)\mkern 1mu\mathrm{d}\xi\). The symbol \(\Theta\) is the capital Greek letter theta; its lowercase is \(\theta\), which is often used in polar and spherical coordinates.
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Here we recall theorems done earlier and state a corollary to the theorem in the preceding item.
Theorem 1. Let \(\mathscr{V}\) and \(\mathscr{W}\) be vector spaces over \(\mathbb{F}\).
- If \(\mathscr{V}\) and \(\mathscr{W}\) are isomorphic and \(\mathscr{V}\) is finite-dimensional, then \(\mathscr{W}\) is is finite-dimensional and \(\dim \mathscr{W} = \dim \mathscr{V}.\)
- Assume that \(\mathscr{V}\) and \(\mathscr{W}\) are finite-dimensional. Then \(\mathscr{V}\) and \(\mathscr{W}\) are isomorphic if and only if \(\dim \mathscr{W} = \dim \mathscr{V}.\)
Theorem 2. Let \(\mathscr{V}\) and \(\mathscr{W}\) be finite-dimensional vector spaces over \(\mathbb{F}\). Then \[ \dim \bigl( \mathscr{V}\times\mathscr{W} \bigr) = \dim \mathscr{V} + \dim \mathscr{W}. \]Theorem 3. Let \( m, n \in \mathbb{N} \) and let \(\mathscr{W}\) be finite-dimensional vector space over \( \mathbb{F} \) with \(\dim \mathscr{W} = n\). Then \[ \dim \bigl(\mathscr{W}^m\bigr) = n m. \]- A proof of Theorem 3 uses Mathematical Induction on \(m\), the fact that the vector space \(\mathscr{W}^{m+1}\) is isomorphic to \(\bigl(\mathscr{W}\times\mathscr{W}^{m}\bigr)\), and Theorem 2.
Corollary. Let \( m, n \in \mathbb{N} \) and let \(\mathscr{V}\) and \(\mathscr{W}\) be finite-dimensional vector spaces over \( \mathbb{F} \) with \( \dim \mathscr{V} = m \) and \( \dim \mathscr{W} = n\). Then \[ \dim \bigl(\mathscr{L}( \mathscr{V},\mathscr{W}) \bigr) = m n. \] -
In a preceding ■ item we introduced the isomorphism \[ \mathscr{W}^m \ni \mathscr{C} \ \mapsto \ L_{\mathscr{C}}^{\mathscr{B}} \in \mathscr{L}( \mathscr{V},\mathscr{W}), \] where \(\mathscr{V}\) and \(\mathscr{W}\) are finite-dimensional vector spaces over \(\mathbb{F}\) with \( \dim \mathscr{V} = m\) and \( \dim \mathscr{W} = n\), \(\mathscr{B} \in \mathscr{V}^m\) is a fixed basis of \(\mathscr{V}\) and \(\mathscr{C} \in \mathscr{W}^m\) is a "variable" \(m\)-tuple consisting of vectors in \(\mathscr{W}\).
Let \(m, n \in \mathbb{N}\) and let us specialize the isomorphism in the preceding paragraph for the most popular vector spaces \(\mathscr{V} = \mathbb{F}^m\) and \(\mathscr{W} = \mathbb{F}^n\). For a fixed basis of \(\mathbb{F}^m\), again we choose the most popular basis \(\mathscr{E}_m\) introduced in a different preceding ■ item.
- One special case of the operator \(\ L_{\mathscr{C}}^{\mathscr{B}}\) is worthy of our attention. Consider \begin{alignat*}{2} \mathscr{V} & = \mathbb{F}^m, \quad & \mathscr{W} & = \mathbb{F}^n, \\ \mathscr{B} & = \mathscr{E}_m, \quad & \mathscr{C} & = \bigl(\mathbf{c}_1, \ldots, \mathbf{c}_m \bigr) \in \bigl(\mathbb{F}^n\bigr)^m, \end{alignat*} where \[ \mathbf{c}_1 = \begin{bmatrix} \gamma_{11} \\[-1pt] \gamma_{21} \\[-1pt] \vdots \\[-2pt] \gamma_{n1} \end{bmatrix}, \ \mathbf{c}_2 = \begin{bmatrix} \gamma_{12} \\[-1pt] \gamma_{22} \\[-1pt] \vdots \\[-2pt] \gamma_{n2} \end{bmatrix}, \ \ldots, \ \mathbf{c}_m = \begin{bmatrix} \gamma_{1m} \\[-1pt] \gamma_{2m} \\[-1pt] \vdots \\[-2pt] \gamma_{nm} \end{bmatrix} \in \mathbb{F}^n \] is an \(m\)-tuple of vectors in \(\mathbb{F}^n \). Then, \[ \forall\mkern 1mu \mathbf{x} = \begin{bmatrix} \xi_{1} \\[-1pt] \xi_{2} \\[-1pt] \vdots \\[-2pt] \xi_{m} \end{bmatrix} \in \mathbb{F}^m \qquad L_{\mathscr{C}}^{\mathscr{\mathscr{E}_m}}\mathbf{x} = \sum_{k=1}^m \xi_k \mathbf{c}_k \] The linear operator \[ L_{\mathscr{C}}^{\mathscr{\mathscr{E}_m}} : \mathbb{F}^m \to \mathbb{F}^n \] is completely determined by the \(m\)-tuple \[ \bigl(\mathbf{c}_1, \ldots, \mathbf{c}_m \bigr) \in \bigl(\mathbb{F}^n\bigr)^m. \] A more convenient way to record this \(m\)-tuple is a table of \(m\) columns each containing the entries of the vectors in \(\bigl(\mathbf{c}_1, \ldots, \mathbf{c}_m \bigr)\). Such a table is called an \(n\times m\) matrix: \[ \begin{bmatrix} \gamma_{11} & \gamma_{12} & \cdots & \gamma_{1m} \\[-1pt] \gamma_{21} & \gamma_{22} & \cdots & \gamma_{2m} \\[-1pt] \vdots & \vdots & \ddots & \vdots \\[-2pt] \gamma_{n1} & \gamma_{n2} & \cdots & \gamma_{nm} \end{bmatrix} = \Bigl[ \mathbf{c}_1 \ \ \mathbf{c}_2 \ \cdots \ \mathbf{c}_m \Bigr]. \] The equality \[ L_{\mathscr{C}}^{\mathscr{\mathscr{E}_m}}\mathbf{x} = \sum_{k=1}^m \xi_k \mathbf{c}_k \] is a motivation for the definition of the matrix-vector multiplication: \[ \forall\mkern 1mu \mathbf{x} = \begin{bmatrix} \xi_{1} \\[-1pt] \xi_{2} \\[-1pt] \vdots \\[-2pt] \xi_{m} \end{bmatrix} \in \mathbb{F}^m \qquad \Bigl[ \mathbf{c}_1 \ \ \mathbf{c}_2 \ \cdots \ \mathbf{c}_m \Bigr] \mathbf{x} \overset{\scriptstyle \text{def}}{=} \sum_{k=1}^m \xi_k \mathbf{c}_k. \]
- Thank you for pointing out where my writing needs improvement. With your help, I look forward to making my notes and website error-free.
- My notes on Linear Operators (updated 2025-02-04).
- My notes on Bases. The same notes with cropped pages.
- My notes on Vector Spaces. The same notes with cropped pages.
- The next section of my notes: Linear Operators.
- I like our proof of the Steinitz Exchange Lemma because it is a structured proof by Mathematical Induction. Proofs by Mathematical Induction are often somewhat straightforward, leading authors to abbreviate them and, in doing so, completely sacrifice the sophisticated structure of the Principle of Mathematical Induction.
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Below, I present a remarkable example of a proof using the Principle of Mathematical Induction. This proof appears as Example 4.5 in the book:
Titu Andreescu, Vlad Crsan Mathematical Induction: A Powerful and Elegant Method of Proof. -
First recall the exact statement of the Principle of Mathematical Induction that we are using. By $\mathbb{N}$ we denote the set of positive integers.
Principle of Mathematical Induction. Let $P(n)$ be a propositional function with $n\in \mathbb{N}$. The following implication holds: \[ P(1) \land \Bigl( \forall \mkern 2mu k \in \mathbb{N} \ \ P(k) \Rightarrow P(k+1) \Bigr) \quad \Rightarrow \quad \forall \mkern 2mu n \in \mathbb{N} \ \ \ P(n) \]
- By $\mathbb{R}_+$ we denote the set of positive real numbers. For $n \in \mathbb{N}$ by $\mathbb{R}_+^n$ we denote the set of all $n$-tuples of positive real numbers.
-
Proof.Problem. Prove that the following implication holds for all $n\in\mathbb{N}$: \[ \forall \mkern 2mu (a_1,\ldots,a_n) \in \mathbb{R}_+^n \qquad \prod_{i=1}^n a_i = 1 \quad \Rightarrow \quad n \leq \sum_{i=1}^n a_i . \]
- Let $n\in\mathbb{N}$. Denote the displayed implication in the preceding boxed statement by $P(n)$.
- Base case. The statement $P(1)$ reads: \[ \forall \mkern 1mu a_1 \in \mathbb{R}_+ \qquad a_1 = 1 \quad \Rightarrow \quad a_1 = 1. \] The preceding implication is clearly true.
- Inductive step. Let $k\in\mathbb{N}$ be arbitrary. We need to prove \[ P(k) \quad \Rightarrow \quad P(k+1). \] Assume $P(k)$. That is assume \[ \forall \mkern 2mu (b_1,\ldots,b_k) \in \mathbb{R}_+^k \qquad \prod_{i=1}^k b_i = 1 \quad \Rightarrow \quad k \leq \sum_{i=1}^k b_i. \] The preceding implication is the Inductive hypothesis. We will prove $P(k+1)$. That is we will prove \[ \forall \mkern 2mu (a_1,\ldots,a_{k+1}) \in \mathbb{R}_+^{k+1} \qquad \prod_{i=1}^{k+1} a_i = 1 \quad \Rightarrow \quad k+1 \leq \sum_{i=1}^{k+1} a_i. \] Let $(a_1,\ldots,a_{k+1}) \in \mathbb{R}_+^{k+1}$ be arbitrary and assume \[ \prod_{i=1}^{k+1} a_i = 1. \] Since neither product, nor sum depend on the order of the $(k+1)$-tuple, we can also assume that \[ \forall \mkern 1mu i \in \{1,\ldots,k+1\} \quad a_1 \leq a_i \leq a_{k+1}. \] With this order we have \[ a_{k+1} \lt 1 \quad \Rightarrow \quad \prod_{i=1}^{k+1} a_i \lt 1. \] Therefore, the contrapositive yields that $a_1\cdots a_{k+1} = 1$ implies $a_{k+1} \geq 1$. Similarly, the assumed order yields \[ a_{1} \gt 1 \quad \Rightarrow \quad \prod_{i=1}^{k+1} a_i \gt 1. \] Therefore, the contrapositive yields that $a_1\cdots a_{k+1} = 1$ implies $a_1 \leq 1$. Notice that \begin{align*} a_1 \leq 1 \ \land \ a_{k+1} \geq 1 &\quad \Rightarrow \quad \bigl(1-a_1\bigr) \bigl(1-a_{k+1}\bigr) \leq 0 \\ &\quad \Rightarrow \quad a_1 a_{k+1} \leq a_1 + a_{k+1} - 1. \end{align*} Now we use the Inductive hypothesis. Let $k$-tuple $(b_1,\ldots,b_k)$ consist of the product $b_1 = a_1a_{k+1}$ and, if $k \gt 1$, $b_i = a_i$ for all $i \in \{2,\ldots,k\}$. Since \[ \prod_{i=1}^k b_i = \prod_{i=1}^{k+1} a_i = 1, \] we can apply the Inductive hypothesis and deduce that \[ k \leq \sum_{i=1}^k b_i. \] Since we proved that \[ b_1 = a_1 a_{k+1} \leq a_1 + a_{k+1} - 1, \] we deduce that \[ k \leq \sum_{i=1}^k b_i = a_1a_{k+1} + \sum_{i=2}^k a_i \leq a_1 + a_{k+1} - 1 + \sum_{i=2}^k a_i. \] Thus, \[ k+1 \leq \sum_{i=1}^{k+1} a_i. \]
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For the sake of completeness, in the above problem, we should add the condition for the equality to hold in the inequality.
Problem. Prove that the following implication holds for all $n\in\mathbb{N}$: \[ \forall \mkern 2mu (a_1,\ldots,a_n) \in \mathbb{R}_+^n \qquad \prod_{i=1}^n a_i = 1 \quad \Rightarrow \quad n \leq \sum_{i=1}^n a_i . \] The equality in the above inequality holds if and only if $(a_1,\ldots,a_n) = (1,\ldots,1)$.
-
An application of the above problem provides one of the simplest proofs of the Inequality of Arithmetic and Geometric Means, see the Wikipedia page AM-GM Inequality
Proof.AM-GM Inequality By $\mathbb{R}_+$ we denote the set of positive real numbers. For $n \in \mathbb{N}$ by $\mathbb{R}_+^n$ we denote the set of all $n$-tuples of positive real numbers. Prove that the following implication holds for all $n\in\mathbb{N}$: \[ \forall\mkern 2mu (x_1,\ldots,x_n) \in \mathbb{R}_+^n \qquad \sqrt[n]{x_1 \cdots x_n} \leq \frac{x_1 + \cdots + x_n}{n}. \] The equality holds if and only if $x_1 = \cdots = x_n$.
- Let $n\in\mathbb{N}$ and $(x_1,\ldots,x_n) \in \mathbb{R}_+^n$ be arbitrary. Set $\alpha = \sqrt[n]{x_1 \cdots x_n}$. Then $\alpha \gt 0$ and $\alpha^n = x_1 \cdots x_n$. Set \[ \forall\mkern 1mu i \in \{1,\ldots,n\} \quad a_i = \frac{x_i}{\alpha}. \] Then, \[ (a_1,\ldots,a_n) \in \mathbb{R}_+^n \ \ \land \ \ \prod_{i=1}^n a_i = \frac{x_1 \cdots x_n}{\alpha^n} = 1. \] By the above Problem we deduce that \[ n \leq \sum_{i=1}^n a_i = \frac{1}{\alpha}\bigl( x_1 + \cdots + x_n \bigr). \] Consequently, since $n \gt 0$ and $\alpha \gt 0$, we have \[ \sqrt[n]{x_1 \cdots x_n} = \alpha \leq \frac{1}{n}\bigl( x_1 + \cdots + x_n \bigr). \] This proves the AM-GM inequality.
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We started with a problem which is a special case of the AM-GM Inequality, then we used that problem to prove the AM-GM Inequality. We could have chosen a different special case of the AM-GM Inequality and formulated it as a problem. The proof is probably similar.
Problem. Prove that the following implication holds for all $n\in\mathbb{N}$: \[ \forall \mkern 2mu (a_1,\ldots,a_n) \in \mathbb{R}_+^n \qquad \sum_{i=1}^n a_i = n \quad \Rightarrow \quad \prod_{i=1}^n a_i \leq 1. \] The equality in the above inequality holds if and only if $(a_1,\ldots,a_n) = (1,\ldots,1)$.
- The content of this post has very little to do with this class. On Friday, I presented a proof that a vector space is infinite dimensional if and only if there exists a sequence of linearly independent vectors in it. In this proof I used a recursive definition. That prompted me to look for a formal justification. Below is the best that I found.
- The following theorem is from the book Naive Set Theory by Paul Halmos. I changed notation and some arguments, but the essence and some formulations are from the book. Halmos writes beautifully. The intention of the theorem is to provide a formal justification for recursive definitions of functions on \(\mathbb{N}\) (sequences).
-
Recursion Theorem.
Let \(X\) be a nonempty set, \(a \in X\) and let \(f: X\to X\) be a function. There exists a unique function \(F:\mathbb{N} \to X\) such that
- \(F(1) = a.\)
- For all \(n \in \mathbb{N}\), we have \(F(n+1) = f\bigl(F(n)\bigr)\).
-
Proof.
Recall that a function from $\mathbb{N}$ to $X$ is a certain kind of subset of $\mathbb{N} \times X$; we shall construct $F$ explicitly as a set of ordered pairs.
-
Consider, for this purpose, the family $\mathcal{F}$ of all those subsets $A$ of $\mathbb{N} \times X$ with the following two properties:
- (i) \((0,a) \in A.\)
- (ii) For all \(n \in \mathbb{N}\), if \((n,x) \in A\), then \(\bigl(n+1,f(x)\bigr) \in A\).
- It is straightforward to see that $F$ itself belongs to $\mathcal{F}$. Notice that \(F\) is the minimal member of $\mathcal{F}$. That is, for all sets \(G \in \mathcal{F}\) we have \(F \subseteq G\).
- To prove that \(F\) is a function, we will prove the following implication: \[ \forall \mkern 2mu G \in \mathcal{F} \quad F \subseteq G \quad \Rightarrow \quad F \ \ \text{is a function}. \] The contrapositive is easier to prove: For all \(H \in \mathcal{F}\) the following implication holds: \begin{equation} \label{eq-implminF} H \ \ \text{is not a function} \quad \Rightarrow \quad \exists \mkern 2mu G \in \mathcal{F} \ \ \text{such that} \quad G \subset H, \end{equation} where \(\subset\) denotes strict inclusion.
- Let \(H \in \mathcal{F}\) be arbitrary. Define the following set of positive integers: \[ \mathbb{H} = \Bigl\{ k \in \mathbb{N} : \exists\mkern 2mu x, y \in X \ \ \text{such that} \ \ x\neq y \land (k,x), (k,y) \in H \Bigr\} \] By the definition of a function, it is straightforward to verify that \(H\) is a function if and only if \(\mathbb{H} =\emptyset\).
- To prove the last implication in (III) assume that \(H\) is not a function, that is \(\mathbb{H} \neq \emptyset\). Since \(\mathbb{H} \subseteq \mathbb{N}\), by the Well Ordering Axiom, \(\mathbb{H}\) has a minimum. Set \(m = \min \mathbb{H}\). Since \(m \in \mathbb{N}\), we can proceed by considering two exclusive cases \(m=1\) and \(m \in \mathbb{N}\setminus\{1\}\).
- Case 1. \(m = 1\). Since \(m = 1 \in \mathbb{H}\), by the definition of \(\mathbb{H}\), there exists \(b \in X\) such that \(b\neq a\) and $(1, b) \in H$. Consider, in this case, the set $G = H\setminus \bigl\{(1, b)\bigr\}$. Observe that \(G\) a proper subset of \(H\). Furthermore, \(G \in \mathcal{F}\). Why? First, \(G\) contains $(1, a)$ (since $a \neq b$), and that for all \(n \in \mathbb{N}\) if $(n, x) \in G$, then $\bigl(n+1, f(x)\bigr) \in G$. The reason for the second assertion is that since $n+1 > 1$, the discarded element \((1,b)\) is not equal to $\bigl(n+1, f(x)\bigr)$. Thus, \(G \subset H\), a proper subset, and \(G \in \mathcal{F}\).
- Case 2. \(m \in \mathbb{N}\setminus\{1\}\). In this case \(m-1 \in \mathbb{N}\) and, since \(m = \min \mathbb{H}\), \(m-1 \notin \mathbb{H}\). Therefore, there exists a unique \(x\in X\) such that \((m-1,x) \in H\). Since \(H\in\mathcal{F}\), by item (ii) in the definition of \(\mathcal{F}\), we have \(\bigl(m,f(x)\bigr) \in H\). Since \(m \in \mathbb{H}\), there exists \(y \in X\) such that \(y \neq f(x)\) and \((m,y) \in H\). Consider, in this case, the set $G = H\setminus \bigl\{(m, y)\bigr\}$. Observe that \(G\) contains $(1, a)$ (since $m > 1$), and that if \(G\) contains $(k, z)$, say, then it contains $\bigl(k+1, f(z)\bigr)$ also. Indeed, if $k = m-1$, then $z$ must be $x$, and the reason \(G\) contains $\bigl(m, f(x)\bigr)$ is that $f(x) \neq y$; if, on the other hand, $k \neq m-1$, then the reason \(G\) contains $\bigl(k+1, f(t)\bigr)$ is that $k+1 \neq m$. In other words, $G \in \mathcal{F}$. This again proves the existence of \(G \subset H\), a proper subset, and \(G \in \mathcal{F}\).
- In conclusion, we have proved the last implication in (III), which proves that \(F\) is a function. The proof of uniqueness is a straightforward application of Mathematical Induction.
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Consider, for this purpose, the family $\mathcal{F}$ of all those subsets $A$ of $\mathbb{N} \times X$ with the following two properties:
- The Recursion Theorem has been proven.
- An application of the Recursion Theorem to define a function on \(\mathbb{N}\) is called a recursive definition.
- Example. The most omnipresent recursive definition is the definition of Fibonacci numbers. Classically, it is written as \[ a_0 = 0, \ \ a_1 = 1, \quad \ \forall n \in \mathbb{N} \ \ a_{n+1} = a_{n} + a_{n-1}. \] To cast this definition in the framework of the Recursion Theorem, we set \[ X = \mathbb{Z} \times \mathbb{Z}, \quad f: X \to X, \ \ f(x,y) = (y,x+y), \] and \[ F(1) = (0, 1), \quad \forall n \in \mathbb{N} \quad F(n+1) = f\bigl(F(n)\bigr). \] With this definition and \(n \in \mathbb{N}\), the \(n\)-th Fibonacci number is the second entry in \(F(n)\).
- Example. The recursive definition of the sequence of partial sums of a given sequence \(a:\mathbb{N} \to \mathbb{R}\) is as follows: \[ X = \mathbb{N} \times \mathbb{R}, \quad f: X \to X, \ \ f(n,x) = \bigl(n+1,x+a_{n+1}\bigr), \] and \[ F(1) = (1, a_1), \quad \forall n \in \mathbb{N} \quad F(n+1) = f\bigl(F(n)\bigr). \] With this definition and \(n \in \mathbb{N}\), the \(n\)-th partial sum of \(a:\mathbb{N} \to \mathbb{R}\) is the second entry in \(F(n)\).
- Ideally, every recursive definition found in the literature can be cast as an application of the Recursion Theorem, though doing so might require some effort.
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A student recently stopped by my office to discuss Problem 3 on Assignment 1. In the narrative accompanying this problem, I hint that polar coordinates might be a useful tool.
Polar coordinates often prove invaluable. For example, working with complex numbers, via Euler's identity, see the end of the post today.
Over the years, I have noticed that students often approach polar coordinates with a certain degree of overconfidence. However, this topic warrants careful attention. As is often said in discussions about Large Language Models: Attention is All You Need.
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First I introduce some less common notation.
- We need a fundamental yet often overlooked function, the Unit Step function, for which I will use the patriotic abbreviation \(\operatorname{us}(x)\) with \(x\in \mathbb{R}\). Its definition is as follows: \[ \operatorname{us}(x) = \begin{cases} 1 & \text{if} \quad x \geq 0, \\ 0 & \text{if} \quad x \lt 0. \end{cases} \] This function effectively highlights the dichotomy between nonnegative and negative real numbers.
- By \(\mathbb{R}^+\) we denote the set of positive real numbers; that is \[ \mathbb{R}^+ = \bigl\{ x \in \mathbb{R} : x \gt 0 \bigr\}. \] With this set notation, the Unit Step function is the indicator function of the set \(\mathbb{R}^+\cup\{0\} \subset \mathbb{R}\).
- Polar Coordinates Theorem. The equations \begin{align*} x & = r \cos\theta, \\ y & = r \sin\theta, \end{align*} establish a bijection between all \((r,\theta)\in\mathbb{R}^+\times (-\pi,\pi]\) and all \((x,y) \in \mathbb{R} \setminus\{(0,0)\}\). Moreover, for \((r,\theta)\in\mathbb{R}^+\times (-\pi,\pi]\) and \((x,y) \in \mathbb{R} \setminus\{(0,0)\}\), these equations are equivalent to \begin{align*} r & = \sqrt{x^2+y^2}, \\ \theta & = \bigl(2\operatorname{us}(y) - 1\bigr) \arccos\biggl( \frac{x}{\sqrt{x^2+y^2}} \biggr). \end{align*}
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For the proof of the Polar Coordinates Theorem, let \((x,y) \in \mathbb{R}^2\setminus\{(0,0)\}\) and solve the equations \(x= r \cos\theta \) and \(y = r \sin\theta\) for \(r \gt 0\) and \(\theta\in (-\pi,\pi]\).
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I. First solve the equations \(x= r \cos\theta \) and \(y = r \sin\theta\) for \(r \gt 0\). We have \begin{align*} 0 \lt x^2 + y^2 & = (r \cos\theta)^2 + (r \sin\theta)^2 \\ & = r^2 \bigl( (\cos\theta)^2 + (\sin\theta)^2 \bigr) \\ & = r^2. \end{align*} Thus, since \(r \gt 0\), \[ r = \sqrt{x^2+y^2}. \]
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II. With the result from I, the point \(\Bigl(\dfrac{x}{r},\dfrac{y}{r}\Bigr)\) is a point on the unit circle. By the definitions of the trigonometric functions cosine and sine there exists a unique \(\theta \in (-\pi,\pi]\) such that \[ \cos\theta = \frac{x}{r} \quad \text{and} \quad \sin\theta = \frac{y}{r}. \] Let us calculate \(\theta \in (-\pi,\pi]\) in terms of \(x\) and \(y\). Notice that for all \((x,y) \in \mathbb{R}^2\setminus\{(0,0)\}\) we have \[ \frac{x}{\sqrt{x^2+y^2}} \in [-1,1]; \] that is \( \dfrac{x}{\sqrt{x^2+y^2}}\) is in the domain of the inverse trigonometric function \(\arccos:[-1,1] \to [0,\pi]\). This observation is the key for this proof.
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III. If \(y \geq 0\), then by the definition of the inverse trigonometric function \(\arccos:[-1,1] \to [0,\pi]\) we have \[ \theta = \arccos\biggl( \frac{x}{\sqrt{x^2+y^2}} \biggr) \in [0,\pi]. \] Let us prove that this \(\theta \in [0,\pi]\) is the solution that we seek. With this \(\theta \in [0,\pi]\) we have \(\cos\theta = \frac{x}{r}\). Furthermore, since \((\cos\theta)^2 + (\sin\theta)^2=1\), we have \[ (\sin\theta)^2 = 1 - \frac{x^2}{x^2+y^2} = \frac{y^2}{x^2+y^2}. \] As both \(\sin\theta \geq 0\) and \(y \geq 0\), we deduce \(\sin\theta = \frac{y}{r}\), and thus \[ \cos\theta = \frac{x}{r} \quad \text{and} \quad \sin\theta = \frac{y}{r} \geq 0, \] the desired equalities hold.
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IV. If \(y \lt 0\), then, with the above defined \(\theta\), we have \(\theta \in (0,\pi)\) and, by the calculations in III, \[ \cos\theta = \frac{x}{r} \quad \text{and} \quad \sin\theta = \frac{|y|}{r} \gt 0. \] Therefore, with \(-\theta \in (-\pi,0)\) we have \[ \cos(-\theta) = \cos\theta = \frac{x}{r} \quad \text{and} \quad \sin(-\theta) = -\sin\theta = -\frac{|y|}{r} = \frac{y}{r} \lt 0, \] the desired equalities again hold.
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V. Since \[ 2\operatorname{us}(y) - 1 = \begin{cases} \phantom{-}1 & \text{if} \quad y \geq 0, \\ -1 & \text{if} \quad y \lt 0, \end{cases} \] we can unify the results from III and IV and set \[ \theta = \bigl(2\operatorname{us}(y) - 1\bigr) \arccos\biggl( \frac{x}{\sqrt{x^2+y^2}} \biggr). \] If \(y \geq 0\), then, by III, we have \(x= r \cos\theta \) and \(y = r \sin\theta\). If \(y \lt 0\), then, by IV, we have \(x= r \cos\theta \) and \(y = r \sin\theta\). In conclusion with \(\theta\) given in the last displayed equality we have both desired equalties regardless of the sign of \(y\).
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This completes the proof of the Polar Coordinates Theorem.
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Since I spent all this space on the details of polar coordinates, let me recall that in the context of complex numbers there is standard terminology for all four real numbers \(x\), \(y\), \(r\), \(\theta\) introduced in the Polar Coordinates Theorem.
\(z= x + \mathrm{i}\mkern 2mu y = r e^{\mathrm{i}\mkern 2mu\theta}\) Terminology Notation \(x\) the real part of \(z\) \(\operatorname{Re}(z)\) \(y\) the imaginary part of \(z\) \(\operatorname{Im}(z)\) \(r\) the modulus of \(z\) \(|z|\) \(\theta\) the principal argument of \(z\) \(\operatorname{Arg}(z)\) -
The website Argument (complex analysis) - Wikipedia distinguishes the principal value of the argument, denoted by \(\operatorname{Arg}(z)\), and the multivalued function \(\arg(z)\). I have seen \(\arg(z)\) used as the notation for the principal value.
I am proud to promote the simple formula for the principal value of the argument: \(\theta = \operatorname{Arg}(z) \in (-\pi,\pi]\) using the Unit Step function: For \(\mathbb{C}\setminus\{0\}\) we have \[ \theta = \operatorname{Arg}(z) = \Bigl(2\operatorname{us}\bigl(\operatorname{Im}(z)\bigr) - 1\Bigr) \arccos\biggl( \frac{\operatorname{Re}(z)}{|z|} \biggr). \]
For \((x,y) \in \mathbb{R}^2\setminus\{(0,0)\}\), compare this simple formula to the function \(\operatorname{arctan2}\) at Computing from the real and imaginary part - Wikipedia \[ \operatorname{Arg}(x + iy) = \operatorname{atan2}(y,\, x) = \begin{cases} \arctan\left(\frac y x\right) &\text{if } x \gt 0, \\[5mu] \arctan\left(\frac y x\right) + \pi &\text{if } x \lt 0 \text{ and } y \ge 0, \\[5mu] \arctan\left(\frac y x\right) - \pi &\text{if } x \lt 0 \text{ and } y \lt 0, \\[5mu] +\frac{\pi}{2} &\text{if } x = 0 \text{ and } y \gt 0, \\[5mu] -\frac{\pi}{2} &\text{if } x = 0 \text{ and } y \lt 0. \end{cases} \] I am not sure why they use \(\arctan\) when \(\arccos\) is much simpler: \[ \operatorname{atan2}(y,\, x) = \bigl(2\operatorname{us}(y) - 1\bigr) \arccos\biggl( \frac{x}{\sqrt{x^2+y^2}} \biggr). \] My only explanation is that they are afraid of \(\sqrt{x^2+y^2}\).
- I disclose that I borrowed the code for the long cases formula for \(\operatorname{atan2}\) from the Wikipedia site by clicking the Edit link next to the title. It is all in LaTeX. The table with terminology was generated by ChatGPT. My HTML skills are not that good.
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Again, I updated the Basis section of my notes: Bases.
The same notes with cropped pages.
Since I often rewrite, to celebrate thinking, writing, and teaching, I created my own saying
I came, I thought, I wrote, I taught, I thought more, I rewrote.
- I updated my notes on Vector Spaces. I have added more notes to my abstract version of the even-odd functions problem. Admittedly, this problem is peculiar—some parts are straightforward, while others require you to discover and formulate new statements. I assign such tasks because exploration and discovery are integral to learning. If you fail to arrive at discoveries for which you are confident make mathematical sense, document your explorations and reasoning; partial credit will be awarded for your explorations and reasoning. The same notes with cropped pages.
- I updated my notes on Vector Spaces. I corrected the errors that you pointed out in Discussions on Canvas. The same notes with cropped pages.
- I updated my notes on Vector Spaces. I corrected the errors that you pointed out today and added some more content that I presented in class. The same notes with cropped pages.
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The reading homework for Thursday:
- Section 1A in the textbook: Review of complex numbers, the concept of a list (I prefer to call lists tuples), the space $\mathbb{F}^n$ (Notice that in this book the author uses the old-fashioned bold face capitals to denote the scalar fields. I prefer to use, so called blackboldbold font, which seems to be more common nowdays.)
- Section 1B in the textbook: Definition of a vector space and exercises.
- If you find an exercise that you cannot solve, please report them in Discussions on Canvas.
- (updated) My notes on Vector Spaces. The same notes with cropped pages.
- My notes on Complex Numbers. There are 11 exercises at the end of my notes. Almost entirely the content of my notes is mentioned in the Common Core State Standards for Mathematics. Some of the exercises are very similar to the exercises mentioned in the Common Core State Standards for Mathematics.
- My notes on Mathematical Logic. The same notes with cropped pages.
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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 it 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 advise that I obtained was superb. Not only in LaTeX, but in Wolfram Mathematica as well. Sometimes ChatGPT is sloppy, and makes trivial 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 the 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.
- Professor Sheldon Axler, the author of Linear Algebra Done Right has made PDF file of the fourth edition of this popular linear algebra book available free of charge. We are taking advantage of this generous gesture and we are using this book this quarter.
- The information sheet
- Some relevant Wikipedia links:
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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: