\documentclass{amsbook} \usepackage{enumitem} \usepackage{amssymb} \newtheorem{theorem}{Theorem}[chapter] \newtheorem{lemma}[theorem]{Lemma} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{xca}[theorem]{Exercise} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \numberwithin{section}{chapter} \numberwithin{equation}{chapter} \begin{document} \frontmatter \title{Minqi's Solutions to the Exercises of Sheldon Axler (2014), Linear Algebra Done Right (3rd Edition)} \author{Minqi Pan\footnote{Website: http://www.minqi-pan.com}} \date{\today} \maketitle \setcounter{page}{4} \tableofcontents \mainmatter \include{minqi_s_solutions_to_the_exercises_of_sheldon_axler_2014_linear_algebra_done_right_3rd_edition_ch1} \chapter{Finite-Dimensional Vector Space} \section{Exercise 2.A.1} Suppose $v_1,v_2,v_3,v_4$ spans $V$. Prove that the list $v_1-v_2,v_2-v_3,v_3-v_4,v_4$ also spans $V$. \begin{proof} Denote $w_1=v_1-v_2, w_2=v_2-v_3, w_3=v_3-v_4, w_4=v_4$. Let $A=\begin{bmatrix} 1& -1& 0&0\\ 0& 1&-1&0\\ 0& 0&1&-1\\ 0& 0&0&1 \end{bmatrix},$ then $A\begin{bmatrix} v_1\\ v_2\\ v_3\\ v_4\end{bmatrix}=\begin{bmatrix} w_1\\ w_2\\ w_3\\ w_4\end{bmatrix}, \begin{bmatrix} v_1\\ v_2\\ v_3\\ v_4\end{bmatrix}=A^{-1}\begin{bmatrix} w_1\\ w_2\\ w_3\\ w_4\end{bmatrix}.$ Moreover, it can be calculated that $A^{-1}= \begin{bmatrix} 1& 1& 1& 1\\ 0& 1& 1& 1\\ 0& 0& 1& 1\\ 0& 0& 0& 1 \end{bmatrix}.$ \begin{enumerate} \item Pick $x\in\text{span}(v_1,v_2,v_3,v_4)$, then there exists $x_1,x_2,x_3,x_4\in\mathbb{F}$ such that \begin{equation*} \begin{split} v&=\begin{bmatrix} x_1&x_2&x_3&x_4 \end{bmatrix} \begin{bmatrix}v_1\\v_2\\v_3\\v_4\end{bmatrix}\\ &=\begin{bmatrix} x_1&x_2&x_3&x_4 \end{bmatrix} \begin{bmatrix} 1& 1& 1& 1\\ 0& 1& 1& 1\\ 0& 0& 1& 1\\ 0& 0& 0& 1 \end{bmatrix} \begin{bmatrix} w_1\\ w_2\\ w_3\\ w_4\end{bmatrix}\\ &= \begin{bmatrix} x_1& x_1+x_2& x_1+x_2+x_3& x_1+x_2+x_3+x_4 \end{bmatrix} \begin{bmatrix} w_1\\ w_2\\ w_3\\ w_4 \end{bmatrix}. \end{split} \end{equation*} Therefore $v\in\text{span}(w_1,w_2,w_3,w_4)$. \item Pick $w\in\text{span}(w_1,w_2,w_3,w_4)$, then there exists $y_1,y_2,y_3,y_4\in\mathbb{F}$ such that \begin{equation*} \begin{split} w&=\begin{bmatrix} y_1&y_2&y_3&y_4 \end{bmatrix} \begin{bmatrix}w_1\\w_2\\w_3\\w_4\end{bmatrix}\\ &=\begin{bmatrix} y_1&y_2&y_3&y_4 \end{bmatrix} \begin{bmatrix} 1& -1& 0&0\\ 0& 1&-1&0\\ 0& 0&1&-1\\ 0& 0&0&1 \end{bmatrix} \begin{bmatrix} v_1\\ v_2\\ v_3\\ v_4\end{bmatrix}\\ &= \begin{bmatrix} y_1& y_2-y_1& y_3-y_2& y_4-y_3 \end{bmatrix} \begin{bmatrix} v_1\\ v_2\\ v_3\\ v_4 \end{bmatrix}. \end{split} \end{equation*} Therefore $w\in\text{span}(v_1,v_2,v_3,v_4)$. \end{enumerate} The above (1) and (2) imply that $\text{span}(w_1,w_2,w_3,w_4)=\text{span}(v_1,v_2,v_3,v_4)$. \end{proof} \section{Exercise 2.A.2} Verify the assertions in Example~2.18. \begin{enumerate}[label=(\alph*)] \item A list $v$ of one vector $v\in V$ is linearly independent if and only if $v\ne 0$. \item A list of two vectors in $V$ is linearly independent if and only if neither vector is a scalar multiple of the other. \item $(1,0,0,0),(0,1,0,0),(0,0,1,0)$ is linearly independent in $\mathbb{F}^4$. \item The list $1,z,\dots,z^m$ is linearly independent in $\mathcal{P}(\mathbb{F})$ for each nonnegative integer $m$. \end{enumerate} \begin{proof} \begin{enumerate}[label=(\alph*)] \item If the list $v$ is linearly independent but $v=0$, then $1v=0$ is a nontrivial representation of $0$ using $v$ which is a contradiction. If $v\ne 0$ but the list $v$ is not linearly independent, then there exists $a\in\mathbb{F}$ such that $av=0$ and $a\ne 0$ which contradicts 1.30 of \cite{axler2014linear}. \item If a list of two vectors $v_1,v_2$ in $V$ is linearly independent and assume that, without loss of generality, $v_1=av_2$ for some $a\in\mathbb{F}$, then $1v_1-av_2=0$ is a nontrivial representation of $0$ using $v_1,v_2$ which is a contradiction. If a list of two vectors $v_1,v_2$ in $V$ is not linearly independent, then there exist some $a_1,a_2\in\mathbb{F}$ such that $a_1v_1+a_2v_2=0$ and not both of $a_1$ and $a_2$ are zero. If $a_1\ne 0$, then $v_1=-\frac{a_2}{a_1}v_2$; if $a_1=0$, then $v_2=-\frac{a_1}{a_2}v_1$. \item Consider the equation $x(1,0,0,0)+y(0,1,0,0)+z(0,0,1,0)=0$ where $x,y,z\in\mathbb{F}$. In matrix form, $\begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{bmatrix} \begin{bmatrix} x\\ y\\ z \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$ Since the coefficient matrix is nonsingular, the only solution is $x=y=z=0$. \item Consider the function $f(z)=x_0+x_1z+\dots+x_{N-1}z^{N-1}$ where $N-1\geq 0$ and $x_0,x_1,\dots,x_{N-1}\in\mathbb{F}$. Suppose $f(z)=0$ for all $z\in\mathbb{F}$. Then, in particular, $f(1)=0, f(2)=0, \dots, f(N)=0$. In matrix form, $\begin{bmatrix} 1&1&1&1&\cdots &1 \\ 1&2&2^2&2^3&\cdots&2^{N-1} \\ 1&3&3^2&3^3&\cdots&3^{N-1}\\ 1&4&4^2&4^3&\cdots&4^{N-1}\\ \vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\ 1&N&N^{2}&N^{3}&\cdots&N^{N-1} \end{bmatrix} \begin{bmatrix} x_0\\ x_1\\ \vdots\\ x_{N-1} \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ \vdots\\ 0 \end{bmatrix}.$ Note that the above coefficient matrix is a Vandermonde matrix. Since all the numbers $1, 2, \dots, N$ are distinct, its determinant is non-zero. Therefore the only solution is $x_0=x_1=\dots=x_{N-1}=0$. \end{enumerate} \end{proof} \section{Exercise 2.A.3} Find a number $t$ such that $(3,1,4),(2,-3,5),(5,9,t)$ is not linearly independent in $\mathbb{R}^3$. \begin{proof} We claim that $t=2$ makes the list not linearly independent in $\mathbb{R}^3$. This can be shown by \begin{equation*} \begin{split} &-3\cdot (3,1,4)+2\cdot (2,-3,5)+1\cdot (5,9,2)\\ &=((-9)+4+5,(-3)+(-6)+9,(-12)+10+2)\\ &=0. \end{split} \end{equation*} \end{proof} \section{Exercise 2.A.4} Verify the assertion in the second bullet point in Example 2.20: the list $(2,3,1),(1,-1,2),(7,3,c)$ is linearly dependent in $\mathbb{F}^3$ if and only if $c=8$. \begin{proof} Let $A=\begin{bmatrix} 2&1&7\\ 3&-1&3\\ 1&2&c \end{bmatrix}.$ Then the list $(2,3,1),(1,-1,2),(7,3,c)$ is linearly dependent if and only if $\text{det}(A)=0$. Since $\text{det}(A)=40-5c$, it follows that $\text{det}(A)=0$ if and only if $c=8$. \end{proof} \section{Exercise 2.A.5} \begin{enumerate}[label=(\alph*)] \item Show that if we think of $\mathbb{C}$ as a vector space over $\mathbb{R}$, then the list $(1+i,1-i)$ is linearly independent. \item Show that if we think of $\mathbb{C}$ as a vector space over $\mathbb{C}$, then the list $(1+i,1-i)$ is linearly dependent. \end{enumerate} \begin{proof} Suppose that there exists $a+bi, c+di\in\mathbb{C}$ such that $a,b,c,d\in\mathbb{R}$ and $(a+bi)(1+i)+(c+di)(1-i)=0$, then $(a-b+c+d)+(a+b-c+d)i=0,$ Rewrite the above linear system in the matrix form, $\begin{bmatrix} 1&-1&1&1\\ 1&1&-1&1 \end{bmatrix} \begin{bmatrix} a\\ b\\ c\\ d \end{bmatrix} = \begin{bmatrix} 0\\ 0 \end{bmatrix},$ then its solutions are $\left\{\begin{bmatrix}-y\\x\\x\\y\end{bmatrix}:x,y\in\mathbb{R}\right\}.$ \begin{enumerate}[label=(\alph*)] \item If we think of $\mathbb{C}$ as a vector space over $\mathbb{R}$, then the list is linearly independent because $\left\{\begin{bmatrix}-y\\x\\x\\y\end{bmatrix}:x,y\in\mathbb{R}\right\}\cap\left\{\begin{bmatrix}x\\0\\y\\0\end{bmatrix}:x,y\in\mathbb{R}\right\}=\left\{\begin{bmatrix}0\\0\\0\\0\end{bmatrix}\right\}.$ \item If we think of $\mathbb{C}$ as a vector space over $\mathbb{C}$, then the list is linearly dependent because $\begin{bmatrix}-1\\1\\1\\1\end{bmatrix}\in\left\{\begin{bmatrix}-y\\x\\x\\y\end{bmatrix}:x,y\in\mathbb{R}\right\}$. \end{enumerate} \end{proof} \section{Exercise 2.A.6} Suppose $v_1,v_2,v_3,v_4$ is linearly independent in $V$. Prove that the list$v_1-v_2,v_2-v_3,v_3-v_4,v_4$ is also linearly independent. \begin{proof} Use the same notation as the proof of Exercise~2.A.1. If the list$v_1-v_2,v_2-v_3,v_3-v_4,v_4$ is linearly dependent, then there exists $a_1,a_2,a_3,a_4\in\mathbb{F}$, such that$\begin{bmatrix} a_1&a_2&a_3&a_4 \end{bmatrix} \begin{bmatrix} w_1\\w_2\\w_3\\w_4 \end{bmatrix} =0$ and $\begin{bmatrix} a_1&a_2&a_3&a_4 \end{bmatrix}\ne \begin{bmatrix} 0&0&0&0 \end{bmatrix}.$ Therefore,$\begin{bmatrix} a_1&a_2&a_3&a_4 \end{bmatrix} A \begin{bmatrix} v_1\\v_2\\v_3\\v_4 \end{bmatrix} =0.$ Since $A$ is nonsingular, $\begin{bmatrix} a_1&a_2&a_3&a_4 \end{bmatrix} A\ne \begin{bmatrix} 0&0&0&0 \end{bmatrix}.$ Therefore it follows that the list $v_1,v_2,v_3,v_4$ is linearly dependent. \end{proof} \section{Exercise 2.A.7} Prove or give a counterexample: If $v_1,v_2,\dots,v_m$ is a linearly independent list of vectors in $V$, then$5v_1-4v_2,v_2,v_3,\dots,v_m$ is linearly independent. \begin{proof} We claim that the proposition is true. Define $A=\begin{bmatrix} 5&-4&0&\dots &0\\ 0&1&0&\dots &0\\ 0&0&1&\dots &0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&0&\dots &1 \end{bmatrix}.$ Suppose that $5v_1-4v_2,v_2,v_3,\dots,v_m$ is linearly dependent, then there exists $a_1,a_2,a_3,\dots,a_m\in\mathbb{F}$, such that $\begin{bmatrix} a_1&a_2&a_3&\dots &a_m \end{bmatrix} A \begin{bmatrix} v_1\\v_2\\v_3\\\dots\\v_m \end{bmatrix} = 0$ and $\begin{bmatrix} a_1&a_2&a_3&\dots &a_m \end{bmatrix}\ne \begin{bmatrix} 0&0&0&\dots &0 \end{bmatrix}.$ Since $\text{det}(A)=5$, the matrix $A$ is nonsingular. Thus $\begin{bmatrix} a_1&a_2&a_3&\dots &a_m \end{bmatrix}A\ne \begin{bmatrix} 0&0&0&\dots &0 \end{bmatrix}.$ Therefore it follows that the list $v_1,v_2,\dots,v_m$ is linearly dependent. \end{proof} \section{Exercise 2.A.8} Prove or give a counterexample: If $v_1,v_2,\dots,v_m$ is a linearly independent list of vectors in $V$ and $\lambda\in\mathbb{F}$ with $\lambda\ne 0$, then $\lambda v_1,\lambda v_2,\dots,\lambda v_m$ is linearly independent. \begin{proof} We claim that the proposition is true. Suppose that $\lambda v_1,\lambda v_2,\dots,\lambda v_m$ is linearly dependent, then there exists $a_1,a_2,\dots,a_m$, not all $0$, such that$a_1\lambda v_1+a_2\lambda v_2+\dots+a_m\lambda v_m=0.$ Since $\lambda\ne 0$, $\lambda^{-1}$ exists in $\mathbb{F}$. Multiplying the above equation by $\lambda^{-1}$ yields $a_1 v_1+a_2 v_2+\dots+a_m v_m=0.$ Note that not all of $a_1, a_2,\dots,a_m$ are zero. Therefore it follows that $v_1,v_2,\dots,v_m$ is a linearly dependent list. \end{proof} \section{Exercise 2.A.9} Prove or give a counterexample: If $v_1,\dots,v_m$ and $w_1,\dots,w_m$ are linearly independent lists of vectors in $V$, then $v_1+w_1,\dots,v_m+w_m$ is linearly independent. \begin{proof}[Counterexample] Set $V=\mathbb{R}^2, m=2, v_1=(0,1), v_2=(1,0), w_1=(0,-1), w_2=(-1,0)$. Then $v_1, v_2$ is a linearly independent list, and $w_1, w_2$ is also a linearly independent list. But$v_1+w_1=(0,0)$ and $v_2+w_2=(0,0)$ is not a linearly independent list. \end{proof} \section{Exercise 2.A.10} Suppose $v_1,\dots,v_m$ is linearly independent in $V$ and $w\in V$. Prove that if $v_1+w,\dots,v_m+w$ is linearly dependent, then $w\in \text{span}(v_1,\dots,v_m)$. \begin{proof} If $v_1+w,\dots,v_m+w$ is linearly dependent, then there exists $a_1,\dots,a_m$, not all $0$, such that$a_1(v_1+w)+\dots+a_m(v_m+w)=0.$ The above equation implies that$w=-\frac{a_1}{a_1+\dots+a_m}v_1-\dots-\frac{a_m}{a_1+\dots+a_m}v_m,$ thus $w\in\text{span}(v_1,\dots,v_m)$. \end{proof} \backmatter \bibliographystyle{amsalpha} \bibliography{minqi_s_solutions_to_the_exercises_of_sheldon_axler_2014_linear_algebra_done_right_3rd_edition} \end{document}