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\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}
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