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\ihead{Fall 2019 Math 201-003 at Drexel: blackboard notes of 2019-11-04}
\ohead{page \thepage}
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\begin{document}

\title{Fall 2019 Math 201-003 at Drexel: blackboard notes of 2019-11-04}
\author{Darij Grinberg}
\date{\today}
\maketitle

\section{Determinants}

\subsection{Laplace expansion (recall)}

Recall from
\href{http://www.cip.ifi.lmu.de/~grinberg/t/19fla/2019-10-30.pdf}{last time}:

\begin{theorem}
[Laplace expansion along the $p$-th row]\label{thm.det.lap-row}Let $A$ be an
$n\times n$-matrix. For each $p,q\in\left[  n\right]  $, we let $M_{p,q}$ be
the $\left(  n-1\right)  \times\left(  n-1\right)  $-matrix obtained from $A$
by removing row $p$ and column $q$. Then, for each $p\in\left[  n\right]  $,
we have%
\[
\det A=\sum_{q=1}^{n}\left(  -1\right)  ^{p+q}A_{p,q}\det\left(
M_{p,q}\right)  .
\]

\end{theorem}

Recall also a basic fact from
\href{http://www.cip.ifi.lmu.de/~grinberg/t/19fla/2019-10-23.pdf}{2019-10-23}:

\begin{theorem}
\label{thm.det.equal-rows}If an $n\times n$-matrix $A$ has two equal rows,
then $\det A=0$.
\end{theorem}

\subsection{Laplace expansion in a column}

Here is the analogue of Theorem \ref{thm.det.lap-row} for columns instead of rows:

\begin{theorem}
[Laplace expansion along the $q$-th column]\label{thm.det.lap-col}Let $A$ be
an $n\times n$-matrix. For each $p,q\in\left[  n\right]  $, we let $M_{p,q}$
be the $\left(  n-1\right)  \times\left(  n-1\right)  $-matrix obtained from
$A$ by removing row $p$ and column $q$. Then, for each $q\in\left[  n\right]
$, we have%
\[
\det A=\sum_{p=1}^{n}\left(  -1\right)  ^{p+q}A_{p,q}\det\left(
M_{p,q}\right)  .
\]

\end{theorem}

\begin{example}
Let's compute $\det A$ for $A=\left(
\begin{array}
[c]{ccc}%
1 & 0 & 2\\
0 & 1 & 0\\
3 & 0 & 4
\end{array}
\right)  $. Theorem \ref{thm.det.lap-col} (for $q=2$) yields%
\begin{align*}
&  \det A\\
&  =-A_{1,2}\det\left(
\begin{array}
[c]{cc}%
A_{2,1} & A_{2,3}\\
A_{3,1} & A_{3,3}%
\end{array}
\right)  +A_{2,2}\det\left(
\begin{array}
[c]{cc}%
A_{1,1} & A_{1,3}\\
A_{3,1} & A_{3,3}%
\end{array}
\right)  -A_{3,2}\det\left(
\begin{array}
[c]{cc}%
A_{1,1} & A_{1,3}\\
A_{2,1} & A_{2,3}%
\end{array}
\right) \\
&  =\underbrace{-0\det\left(
\begin{array}
[c]{cc}%
0 & 0\\
3 & 4
\end{array}
\right)  }_{=0}+1\det\left(
\begin{array}
[c]{cc}%
1 & 2\\
3 & 4
\end{array}
\right)  -\underbrace{0\left(
\begin{array}
[c]{cc}%
1 & 2\\
0 & 0
\end{array}
\right)  }_{=0}\\
&  =\det\left(
\begin{array}
[c]{cc}%
1 & 2\\
3 & 4
\end{array}
\right)  =1\cdot4-2\cdot3=-2.
\end{align*}

\end{example}

Theorem \ref{thm.det.lap-col} is \cite[Proposition B.25]{Strickland}.

\subsection{The adjugate matrix}

Let me define a rather weird matrix:

\begin{definition}
Let $A$ be an $n\times n$-matrix. Let $M_{p,q}$ be as in Theorem
\ref{thm.det.lap-row}. Then, we define the \textbf{adjugate matrix
}$\operatorname*{adj}A$ of $A$ to be the $n\times n$-matrix%
\[
\left(  \left(  -1\right)  ^{p+q}\det\left(  M_{q,p}\right)  \right)  _{1\leq
p\leq n,\ 1\leq q\leq n}.
\]


(Sic! This is really saying $M_{q,p}$, not $M_{p,q}$.)

In other words,%
\[
\operatorname*{adj}A=\left(
\begin{array}
[c]{cccc}%
\det\left(  M_{1,1}\right)   & -\det\left(  M_{2,1}\right)   & \cdots &
\left(  -1\right)  ^{n+1}\det\left(  M_{n,1}\right)  \\
-\det\left(  M_{1,2}\right)   & \det\left(  M_{2,2}\right)   & \cdots &
\left(  -1\right)  ^{n+2}\det\left(  M_{n,2}\right)  \\
\vdots & \vdots & \ddots & \vdots\\
\left(  -1\right)  ^{1+n}\det\left(  M_{1,n}\right)   & \left(  -1\right)
^{2+n}\det\left(  M_{2,n}\right)   & \cdots & \det\left(  M_{n,n}\right)
\end{array}
\right)  .
\]
The signs in this matrix follow the \textquotedblleft chessboard
pattern\textquotedblright: The top-left cell has a $+$ sign; any two adjacent
cells (with a common edge, not just a common corner) always have opposite
signs. (Of course, we are talking about the $\left(  -1\right)  ^{p+q}$ signs
here; the $\det\left(  M_{q,p}\right)  $ factors may include their own signs.)
\end{definition}

\begin{example}
Let $n=2$ and $A=\left(
\begin{array}
[c]{cc}%
a & b\\
c & d
\end{array}
\right)  $. Then,%
\begin{align*}
\operatorname*{adj}A  &  =\left(
\begin{array}
[c]{cc}%
\det\left(  M_{1,1}\right)  & -\det\left(  M_{2,1}\right) \\
-\det\left(  M_{1,2}\right)  & \det\left(  M_{2,2}\right)
\end{array}
\right)  =\left(
\begin{array}
[c]{cc}%
\det\left(
\begin{array}
[c]{c}%
d
\end{array}
\right)  & -\det\left(
\begin{array}
[c]{c}%
b
\end{array}
\right) \\
-\det\left(
\begin{array}
[c]{c}%
c
\end{array}
\right)  & \det\left(
\begin{array}
[c]{c}%
a
\end{array}
\right)
\end{array}
\right) \\
&  =\left(
\begin{array}
[c]{cc}%
d & -b\\
-c & a
\end{array}
\right)
\end{align*}
(since the determinant of a $1\times1$-matrix is just its unique entry).
\end{example}

\begin{example}
Let $n=3$ and $A=\left(
\begin{array}
[c]{ccc}%
a & b & c\\
d & e & f\\
g & h & i
\end{array}
\right)  $. Then,%
\begin{align*}
\operatorname*{adj}A  &  =\left(
\begin{array}
[c]{ccc}%
\det\left(  M_{1,1}\right)  & -\det\left(  M_{2,1}\right)  & \det\left(
M_{3,1}\right) \\
-\det\left(  M_{1,2}\right)  & \det\left(  M_{2,2}\right)  & -\det\left(
M_{3,2}\right) \\
\det\left(  M_{1,3}\right)  & -\det\left(  M_{2,3}\right)  & \det\left(
M_{3,3}\right)
\end{array}
\right) \\
&  =\left(
\begin{array}
[c]{ccc}%
\det\left(
\begin{array}
[c]{cc}%
e & f\\
h & i
\end{array}
\right)  & -\det\left(
\begin{array}
[c]{cc}%
b & c\\
h & i
\end{array}
\right)  & \det\left(
\begin{array}
[c]{cc}%
b & c\\
e & f
\end{array}
\right) \\
-\det\left(
\begin{array}
[c]{cc}%
d & f\\
g & i
\end{array}
\right)  & \det\left(
\begin{array}
[c]{cc}%
a & c\\
g & i
\end{array}
\right)  & -\det\left(
\begin{array}
[c]{cc}%
a & c\\
d & f
\end{array}
\right) \\
\det\left(
\begin{array}
[c]{cc}%
d & e\\
g & h
\end{array}
\right)  & -\det\left(
\begin{array}
[c]{cc}%
a & b\\
g & h
\end{array}
\right)  & \det\left(
\begin{array}
[c]{cc}%
a & b\\
d & e
\end{array}
\right)
\end{array}
\right) \\
&  =\left(
\begin{array}
[c]{ccc}%
ie-fh & ch-ib & bf-ce\\
fg-id & ia-cg & cd-af\\
dh-ge & bg-ah & ae-bd
\end{array}
\right)  .
\end{align*}

\end{example}

Any entry of the matrix $\operatorname*{adj}A$ is a degree-$\left(
n-1\right)  $ polynomial in the entries of $A$.

\begin{theorem}
\label{thm.adj.Aadj}Let $A$ be an $n\times n$-matrix. Then,%
\[
A\cdot\operatorname*{adj}A=\operatorname*{adj}A\cdot A=\det A\cdot I_{n}.
\]

\end{theorem}

\begin{example}
Let $n=2$ and $A=\left(
\begin{array}
[c]{cc}%
a & b\\
c & d
\end{array}
\right)  $. Then, $\operatorname*{adj}A=\left(
\begin{array}
[c]{cc}%
d & -b\\
-c & a
\end{array}
\right)  $. Thus,%
\begin{align*}
A\cdot\operatorname*{adj}A  &  =\left(
\begin{array}
[c]{cc}%
a & b\\
c & d
\end{array}
\right)  \cdot\left(
\begin{array}
[c]{cc}%
d & -b\\
-c & a
\end{array}
\right)  =\left(
\begin{array}
[c]{cc}%
ad-bc & 0\\
0 & ad-bc
\end{array}
\right) \\
&  =\underbrace{\left(  ad-bc\right)  }_{=\det A}I_{2}=\det A\cdot I_{2}%
\end{align*}
and%
\begin{align*}
\operatorname*{adj}A\cdot A  &  =\left(
\begin{array}
[c]{cc}%
d & -b\\
-c & a
\end{array}
\right)  \cdot\left(
\begin{array}
[c]{cc}%
a & b\\
c & d
\end{array}
\right)  =\left(
\begin{array}
[c]{cc}%
ad-bc & 0\\
0 & ad-bc
\end{array}
\right) \\
&  =\underbrace{\left(  ad-bc\right)  }_{=\det A}I_{2}=\det A\cdot I_{2}.
\end{align*}

\end{example}

\begin{example}
Let $n=3$ and $A=\left(
\begin{array}
[c]{ccc}%
a & b & c\\
d & e & f\\
g & h & i
\end{array}
\right)  $, so%
\[
\operatorname*{adj}A=\left(
\begin{array}
[c]{ccc}%
ie-fh & ch-ib & bf-ce\\
fg-id & ia-cg & cd-af\\
dh-ge & bg-ah & ae-bd
\end{array}
\right)  .
\]
Then,%
\[
A\cdot\operatorname*{adj}A=\left(
\begin{array}
[c]{ccc}%
a & b & c\\
d & e & f\\
g & h & i
\end{array}
\right)  \cdot\left(
\begin{array}
[c]{ccc}%
ie-fh & ch-ib & bf-ce\\
fg-id & ia-cg & cd-af\\
dh-ge & bg-ah & ae-bd
\end{array}
\right)  .
\]


Let us compute this product by hand. The $\left(  1,1\right)  $-entry is%
\begin{align*}
&  \left(
\begin{array}
[c]{ccc}%
a & b & c
\end{array}
\right)  \left(
\begin{array}
[c]{c}%
ie-fh\\
fg-id\\
dh-ge
\end{array}
\right)  \\
&  =a\left(  ie-fh\right)  +b\left(  fg-id\right)  +c\left(  dh-ge\right)  \\
&  =\sum_{q=1}^{n}\left(  -1\right)  ^{1+q}A_{1,q}\det\left(  M_{1,q}\right)
=\det A\ \ \ \ \ \ \ \ \ \ \left(  \text{by Laplace expansion along
}1\text{-st row}\right)  .
\end{align*}
Similarly, all other diagonal entries of $A\cdot\operatorname*{adj}A$ are
$\det A$.

The $\left(  1,2\right)  $-entry of $A\cdot\operatorname*{adj}A$ is%
\begin{align*}
&  \left(
\begin{array}
[c]{ccc}%
a & b & c
\end{array}
\right)  \cdot\left(
\begin{array}
[c]{c}%
ch-ib\\
ia-cg\\
bg-ah
\end{array}
\right)  \\
&  =a\left(  ch-ib\right)  +b\left(  ia-cg\right)  +c\left(  bg-ah\right)  \\
&  =\underbrace{\det\left(
\begin{array}
[c]{ccc}%
a & b & c\\
a & b & c\\
g & h & i
\end{array}
\right)  }_{\substack{=0\\\text{(since this matrix has two equal rows)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{because if we compute }\det\left(
\begin{array}
[c]{ccc}%
a & b & c\\
a & b & c\\
g & h & i
\end{array}
\right)  \\
\text{by Laplace expansion along the }2\text{-nd row,}\\
\text{then we get precisely }a\left(  ch-ib\right)  +b\left(  ia-cg\right)
+c\left(  bg-ah\right)
\end{array}
\right)  \\
&  =0.
\end{align*}
Similarly, all off-diagonal entries of $A\cdot\operatorname*{adj}A$ are $0$.

Thus, $A\cdot\operatorname*{adj}A=\det A\cdot I_{3}$.

A similar argument using columns instead of rows shows that
$\operatorname*{adj}A\cdot A=\det A\cdot I_{3}$. This uses Laplace expansion
along a column (i.e., Theorem \ref{thm.det.lap-col}). Thus, $A\cdot
\operatorname*{adj}A=\operatorname*{adj}A\cdot A=\det A\cdot I_{3}$.
\end{example}

In the last example, we have seen why Theorem \ref{thm.adj.Aadj} holds for
$n=3$. The general case can be proved along the same lines. See
\cite[Proposition B.28]{Strickland} for the general proof written out rigorously.

\begin{corollary}
\label{cor.adj.inverse}If an $n\times n$-matrix $A$ is invertible, then
$A^{-1}=\dfrac{1}{\det A}\cdot\operatorname*{adj}A$.
\end{corollary}

\begin{example}
For $n=2$, this leads to%
\[
\left(
\begin{array}
[c]{cc}%
a & b\\
c & d
\end{array}
\right)  ^{-1}=\dfrac{1}{ad-bc}\left(
\begin{array}
[c]{cc}%
d & -b\\
-c & a
\end{array}
\right)  =\left(
\begin{array}
[c]{cc}%
\dfrac{d}{ad-bc} & \dfrac{-b}{ad-bc}\\
\dfrac{-c}{ad-bc} & \dfrac{a}{ad-bc}%
\end{array}
\right)  .
\]
We've seen this before.
\end{example}

\begin{proof}
[Proof of Corollary \ref{cor.adj.inverse}.]Assume that $A$ is invertible.
Then, $\det A\neq0$ (by Theorem 1.4.1 in
\href{http://www.cip.ifi.lmu.de/~grinberg/t/19fla/2019-10-30.pdf}{the
classwork from 2019-10-30}). But Theorem \ref{thm.adj.Aadj} yields
$A\cdot\operatorname*{adj}A=\operatorname*{adj}A\cdot A=\det A\cdot I_{n}$. We
can divide all sides of this equality by $\det A$ (since $\det A\neq0$), and
thus obtain%
\[
\dfrac{1}{\det A}\cdot A\cdot\operatorname*{adj}A=\dfrac{1}{\det A}%
\cdot\operatorname*{adj}A\cdot A=I_{n}.
\]
This rewrites as%
\[
A\cdot\left(  \dfrac{1}{\det A}\operatorname*{adj}A\right)  =\left(  \dfrac
{1}{\det A}\operatorname*{adj}A\right)  \cdot A=I_{n}%
\]
(since $\dfrac{1}{\det A}$ is just a number, and thus can be moved around the
products freely). But this shows that $\dfrac{1}{\det A}\operatorname*{adj}A$
is an inverse of $A$. Hence, $A^{-1}=\dfrac{1}{\det A}\operatorname*{adj}A$.
This proves Corollary \ref{cor.adj.inverse}.
\end{proof}

\begin{remark}
The adjugate of a matrix is also called its \textbf{classical adjoint} or
sometimes just \textbf{adjoint}. But beware of the latter word, as it can mean
many different things. I suggest you use \textquotedblleft
adjugate\textquotedblright\ simply because it only has one meaning.
\end{remark}

Corollary \ref{cor.adj.inverse} is known as \textbf{Cramer's rule}. However,
there is something else also called Cramer's rule (see, e.g.,
\href{http://en.wikipedia.org/wiki/Cramer's_rule}{the eponymous Wikipedia
page}), so beware of confusion.

In order to compute $\operatorname*{adj}A$ according to the definition, you
have to define $n^{2}$ determinants of $\left(  n-1\right)  \times\left(
n-1\right)  $-matrices. Thus, I do not recommend using Corollary
\ref{cor.adj.inverse} for finding $A^{-1}$ (unless $n$ is small), let alone
for solving systems of linear equations (as it only applies when $A$ is invertible).

\section{Eigenvalues and eigenvectors (\textquotedblleft
eigenstuff\textquotedblright)}

We shall now study \textbf{eigenvalues} and \textbf{eigenvectors} of (square)
matrices. This subject, at first, will look like an intellectual game; but
soon, we will see what it is good for, and later will (if time allows...) even
see what it \textquotedblleft really means\textquotedblright.

We follow \cite[\S 13]{Strickland} at first.

\subsection{Definition and examples}

\begin{definition}
Let $A$ be an $n\times n$-matrix. Let $\lambda$ be a scalar (i.e., a real number).

\textbf{(a)} A $\lambda$\textbf{-eigenvector of }$A$ means a nonzero vector
$v\in\mathbb{R}^{n}$ such that $Av=\lambda v$.

\textbf{(b)} We say that $\lambda$ is an \textbf{eigenvalue} of $A$ if and
only if there exists a $\lambda$-eigenvector of $A$.
\end{definition}

\textbf{Caution:} Many authors drop the \textquotedblleft
nonzero\textquotedblright\ in the definition of \textquotedblleft$\lambda
$-eigenvector of $A$\textquotedblright. Then, of course, $0$ (the zero vector)
is \textbf{always} a $\lambda$-eigenvector of $A$, since $A\cdot
0=0=\lambda\cdot0$. Thus, if you do so, you need to modify the definition of
\textquotedblleft eigenvalue\textquotedblright\ so that it requires the
existence of a \textbf{nonzero} $\lambda$-eigenvector.

\textbf{Remark:} This all only makes sense for square matrices $A$. If $A$ is
not square, then $Av$ and $\lambda v$ will have different sizes, thus never equal.

\begin{example}
\label{exa.eigen.1111.1}Let $A=\left(
\begin{array}
[c]{cc}%
1 & 1\\
1 & 1
\end{array}
\right)  $. Can we find the eigenvalues of $A$ ? Can we find the corresponding eigenvectors?

Let $\lambda$ be a scalar. We want to find the $\lambda$-eigenvectors of $A$.
These are the nonzero vectors $\left(
\begin{array}
[c]{c}%
x\\
y
\end{array}
\right)  \in\mathbb{R}^{2}$ such that $A\left(
\begin{array}
[c]{c}%
x\\
y
\end{array}
\right)  =\lambda\left(
\begin{array}
[c]{c}%
x\\
y
\end{array}
\right)  $ (by definition). Now, let us rewrite this condition:%
\begin{align*}
&  \left(  A\left(
\begin{array}
[c]{c}%
x\\
y
\end{array}
\right)  =\lambda\left(
\begin{array}
[c]{c}%
x\\
y
\end{array}
\right)  \right) \\
&  \Longleftrightarrow\ \left(  \left(
\begin{array}
[c]{cc}%
1 & 1\\
1 & 1
\end{array}
\right)  \left(
\begin{array}
[c]{c}%
x\\
y
\end{array}
\right)  =\left(
\begin{array}
[c]{c}%
\lambda x\\
\lambda y
\end{array}
\right)  \right) \\
&  \Longleftrightarrow\ \left(  \left(
\begin{array}
[c]{c}%
x+y\\
x+y
\end{array}
\right)  =\left(
\begin{array}
[c]{c}%
\lambda x\\
\lambda y
\end{array}
\right)  \right) \\
&  \Longleftrightarrow\ \left(  x+y=\lambda x\text{ and }x+y=\lambda y\right)
.
\end{align*}
If we regard $\lambda$ as fixed, this is a system of linear equations in
$x,y$, and its augmented matrix is $\left(
\begin{array}
[c]{ccc}%
1-\lambda & 1 & 0\\
1 & 1-\lambda & 0
\end{array}
\right)  $. We can solve it by Gaussian elimination, for example:%
\begin{align*}
&  \left(
\begin{array}
[c]{ccc}%
1-\lambda & 1 & 0\\
1 & 1-\lambda & 0
\end{array}
\right)  \overset{\text{swap row }1\text{ with row }2}{\longrightarrow}\left(
%
\begin{array}
[c]{ccc}%
1 & 1-\lambda & 0\\
1-\lambda & 1 & 0
\end{array}
\right) \\
&  \overset{\text{add }\left(  \lambda-1\right)  \cdot\text{row }1\text{ to
row }2}{\longrightarrow}\left(
\begin{array}
[c]{ccc}%
1 & 1-\lambda & 0\\
0 & 2\lambda-\lambda^{2} & 0
\end{array}
\right)  .
\end{align*}
Thus, we see that:

\begin{itemize}
\item If $2\lambda-\lambda^{2}\neq0$, then there is a pivot in each of the
first two columns; thus, the linear system has a unique solution, which of
course must be the zero vector $0$ (because $0$ is surely a solution), and
this means that $A$ has no $\lambda$-eigenvector (since a $\lambda
$-eigenvector must be nonzero).

\item If $2\lambda-\lambda^{2}=0$, then the second column has no pivot; thus,
the linear system has a free variable, and therefore there exists a nonzero
solution; this means that $A$ has a $\lambda$-eigenvector.
\end{itemize}

Thus, $A$ has a $\lambda$-eigenvector if and only if $2\lambda-\lambda^{2}=0$.
But $2\lambda-\lambda^{2}=0$ is equivalent to $\lambda\in\left\{  0,2\right\}
$. Therefore, $0$-eigenvectors and $2$-eigenvectors exist (for our $A$), and
no other eigenvectors do (i.e., if $\lambda\notin\left\{  0,2\right\}  $, then
$\lambda$-eigenvectors don't exist). In other words, the eigenvalues of $A$
are $0$ and $2$.

How do we find the eigenvectors? By solving the respective systems of linear equations:

\begin{itemize}
\item The $0$-eigenvectors of $A$ are the nonzero solutions of $A\left(
\begin{array}
[c]{c}%
x\\
y
\end{array}
\right)  =0\left(
\begin{array}
[c]{c}%
x\\
y
\end{array}
\right)  $; they are all of the form $\left(
\begin{array}
[c]{c}%
-y\\
y
\end{array}
\right)  $. Picking $y$ nonzero gives a $0$-eigenvector.

\item The $2$-eigenvectors of $A$ are the nonzero solutions of $A\left(
\begin{array}
[c]{c}%
x\\
y
\end{array}
\right)  =2\left(
\begin{array}
[c]{c}%
x\\
y
\end{array}
\right)  $; they are all of the form $\left(
\begin{array}
[c]{c}%
y\\
y
\end{array}
\right)  $. Picking $y$ nonzero gives a $2$-eigenvector.
\end{itemize}
\end{example}

This was rather haphazard: We cannot hope that Gaussian elimination will
always proceed this nicely. If $A$ is more complicated, then the choice of
pivot might eventually depend on the value of $\lambda$. Thus, we have the
following problem: How do we compute the eigenvalues and the corresponding
eigenvectors of an $n\times n$-matrix in general?

We will see this soon enough; for now, let us recall the Inverse Matrix
Theorem (Theorem 1.2.1 in
\href{http://www.cip.ifi.lmu.de/~grinberg/t/19fla/2019-10-16.pdf}{the
classwork from 2019-10-16}), but smuggle two more equivalent statements into it:

\begin{theorem}
\label{thm.IMT.more}(The Inverse Matrix Theorem, updated.)

Let $A$ be an $n\times n$-matrix.

Then, the following statements are equivalent (i.e., if any of them holds,
then so do all the others):

\begin{itemize}
\item \textbf{(a)} The matrix $A$ can be row-reduced to $I_{n}$.

\item \textbf{(b)} The columns of $A$ are linearly independent.

\item \textbf{(c)} The columns of $A$ span $\mathbb{R}^{n}$.

\item \textbf{(d)} The columns of $A$ form a basis of $\mathbb{R}^{n}$.

\item \textbf{(e)} The matrix $A^{T}$ can be row-reduced to $I_{n}$.

\item \textbf{(f)} The columns of $A^{T}$ are linearly independent.

\item \textbf{(g)} The columns of $A^{T}$ span $\mathbb{R}^{n}$.

\item \textbf{(h)} The columns of $A^{T}$ form a basis of $\mathbb{R}^{n}$.

\item \textbf{(i)} The matrix $A$ has a left inverse.

\item \textbf{(j)} The matrix $A$ has a right inverse.

\item \textbf{(k)} The matrix $A$ has an inverse (i.e., $A$ is invertible).

\item \textbf{(l)} We have $\det A\neq0$.

\item \textbf{(m)} The only column vector $v\in\mathbb{R}^{n}$ satisfying
$Av=0$ is the zero vector $0$.
\end{itemize}
\end{theorem}

\begin{proof}
[Proof of Theorem \ref{thm.IMT.more}.]The Inverse Matrix Theorem (Theorem
1.2.1 in \href{http://www.cip.ifi.lmu.de/~grinberg/t/19fla/2019-10-16.pdf}{the
classwork from 2019-10-16}) shows that the $11$ statements \textbf{(a)},
\textbf{(b)}, $\ldots$, \textbf{(k)} are equivalent. Thus, we only need to
link \textbf{(l)} and \textbf{(m)} to these $11$ statements.

The equivalence \textbf{(k)} $\Longleftrightarrow$ \textbf{(l)} is precisely
Theorem 1.4.1 in
\href{http://www.cip.ifi.lmu.de/~grinberg/t/19fla/2019-10-30.pdf}{the
classwork from 2019-10-30}. Thus, statement \textbf{(l)} is added to our list
of $11$ equivalent statements. It remains to link \textbf{(m)}.

It is easy to see that \textbf{(k)} $\Longrightarrow$ \textbf{(m)}: Indeed, if
\textbf{(k)} holds, then the matrix $A$ has an inverse $A^{-1}$, and therefore
every column vector $v\in\mathbb{R}^{n}$ satisfying $Av=0$ must be the zero
vector $0$ (because comparing $\underbrace{A^{-1}A}_{=I_{n}}v=I_{n}v=v$ with
$A^{-1}\underbrace{Av}_{=0}=A^{-1}0=0$, we find $v=0$); but this means that
\textbf{(m)} holds.

Let us now prove that \textbf{(m)} $\Longrightarrow$ \textbf{(k)}. Indeed,
assume that \textbf{(m)} holds. We need to prove that \textbf{(k)} holds.
Equivalently, we need to prove that \textbf{(b)} holds (since we already know
that \textbf{(b)} is equivalent to \textbf{(k)}). Let $v_{1},v_{2}%
,\ldots,v_{n}$ be the columns of $A$ (so that $A=\left[  v_{1}\mid v_{2}%
\mid\cdots\mid v_{n}\right]  $). Let $\lambda_{1}v_{1}+\lambda_{2}v_{2}%
+\cdots+\lambda_{n}v_{n}=0$ be any relation between these columns. We shall
prove that this relation is trivial. Indeed, let $v$ denote the column vector
$\left(
\begin{array}
[c]{c}%
\lambda_{1}\\
\lambda_{2}\\
\vdots\\
\lambda_{n}%
\end{array}
\right)  $. Thus, from $A=\left[  v_{1}\mid v_{2}\mid\cdots\mid v_{n}\right]
$ and $v=\left(
\begin{array}
[c]{c}%
\lambda_{1}\\
\lambda_{2}\\
\vdots\\
\lambda_{n}%
\end{array}
\right)  $, we obtain
\begin{align*}
Av  &  =\left[  v_{1}\mid v_{2}\mid\cdots\mid v_{n}\right]  \left(
\begin{array}
[c]{c}%
\lambda_{1}\\
\lambda_{2}\\
\vdots\\
\lambda_{n}%
\end{array}
\right) \\
&  =\lambda_{1}v_{1}+\lambda_{2}v_{2}+\cdots+\lambda_{n}v_{n}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma 1.1.6 in
\href{http://www.cip.ifi.lmu.de/~grinberg/t/19fla/2019-10-07.pdf}{the
classwork from 2019-10-07}}\right) \\
&  =0.
\end{align*}
Hence, according to statement \textbf{(k)}, the vector $v$ must be the zero
vector $0$. Thus, its entries $\lambda_{1},\lambda_{2},\ldots,\lambda_{n}$ are
all $0$. Hence, the relation $\lambda_{1}v_{1}+\lambda_{2}v_{2}+\cdots
+\lambda_{n}v_{n}=0$ must be trivial. Thus, we have proved that every relation
between the columns of $A$ is trivial. In other words, the columns of $A$ are
linearly independent. In other words, statement \textbf{(b)} holds. As
explained above, this proves that \textbf{(m)} $\Longrightarrow$ \textbf{(k)}.

Combining \textbf{(k)} $\Longrightarrow$ \textbf{(m)} with \textbf{(m)}
$\Longrightarrow$ \textbf{(k)}, we obtain the equivalence \textbf{(k)}
$\Longleftrightarrow$ \textbf{(m)}. Thus, statement \textbf{(m)} is added to
our list of equivalent statements. Theorem \ref{thm.IMT.more} is thus proved.
\end{proof}

Let us state the contrapositive of Theorem \ref{thm.IMT.more}:

\begin{theorem}
\label{thm.NIMT.more}(The Non-Inverse Matrix Theorem, updated.)

Let $A$ be an $n\times n$-matrix.

Then, the following statements are equivalent (i.e., if any of them holds,
then so do all the others):

\begin{itemize}
\item \textbf{(a')} The matrix $A$ cannot be row-reduced to $I_{n}$.

\item \textbf{(b')} The columns of $A$ are linearly dependent.

\item \textbf{(c')} The columns of $A$ do not span $\mathbb{R}^{n}$.

\item \textbf{(d')} The columns of $A$ do not form a basis of $\mathbb{R}^{n}$.

\item \textbf{(e')} The matrix $A^{T}$ cannot be row-reduced to $I_{n}$.

\item \textbf{(f')} The columns of $A^{T}$ are linearly dependent.

\item \textbf{(g')} The columns of $A^{T}$ do not span $\mathbb{R}^{n}$.

\item \textbf{(h')} The columns of $A^{T}$ do not form a basis of
$\mathbb{R}^{n}$.

\item \textbf{(i')} The matrix $A$ has no left inverse.

\item \textbf{(j')} The matrix $A$ has no right inverse.

\item \textbf{(k')} The matrix $A$ has no inverse (i.e., $A$ is not invertible).

\item \textbf{(l')} We have $\det A=0$.

\item \textbf{(m')} There exists a nonzero column vector $v\in\mathbb{R}^{n}$
satisfying $Av=0$.
\end{itemize}
\end{theorem}

\begin{proof}
[Proof of Theorem \ref{thm.NIMT.more}.]Each of the statements in Theorem
\ref{thm.NIMT.more} is the negation of the corresponding statement in Theorem
\ref{thm.IMT.more}. (For example, statement \textbf{(a')} in Theorem
\ref{thm.NIMT.more} is the negation of statement \textbf{(a)} in Theorem
\ref{thm.IMT.more}.) Thus, Theorem \ref{thm.NIMT.more} follows from Theorem
\ref{thm.IMT.more} (because if two statements are equivalent, then so are
their negations).
\end{proof}

We now revisit Example \ref{exa.eigen.1111.1}:

\begin{example}
\label{exa.eigen.1111.using-det}Let us return to Example
\ref{exa.eigen.1111.1}, so $A=\left(
\begin{array}
[c]{cc}%
1 & 1\\
1 & 1
\end{array}
\right)  $, but now let us try a different approach (instead of Gaussian elimination):

For any scalar $\lambda$, we have the following equivalence:%
\begin{align*}
&  \ \left(  \lambda\text{ is an eigenvalue of }A\right) \\
&  \Longleftrightarrow\ \left(  \text{there exists a }\lambda
\text{-eigenvector of }A\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of \textquotedblleft
eigenvalue\textquotedblright}\right) \\
&  \Longleftrightarrow\ \left(  \text{there exists a nonzero column vector
}v\in\mathbb{R}^{n}\text{ satisfying }Av=\lambda v\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since a }\lambda\text{-eigenvector of }A\text{ is defined to be}\\
\text{a nonzero column vector }v\in\mathbb{R}^{n}\text{ satisfying }Av=\lambda
v
\end{array}
\right) \\
&  \Longleftrightarrow\ \left(  \text{there exists a nonzero column vector
}v\in\mathbb{R}^{n}\text{ satisfying }\left(  A-\lambda I_{n}\right)
v=0\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }Av=\lambda v\text{ is equivalent to }\left(  A-\lambda
I_{n}\right)  v=0\\
\text{(because }\left(  A-\lambda I_{n}\right)  v=Av-\lambda\underbrace{I_{n}%
v}_{=v}=Av-\lambda v\text{)}%
\end{array}
\right) \\
&  \Longleftrightarrow\ \left(  \det\left(  A-\lambda I_{n}\right)  =0\right)
\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by the equivalence \textbf{(m')} }\Longleftrightarrow\text{
\textbf{(l')} in Theorem \ref{thm.NIMT.more},}\\
\text{applied to the matrix }A-\lambda I_{n}\text{ instead of }A
\end{array}
\right)  .
\end{align*}
Thus, in order to find the eigenvalues of $A$, you just compute $\det\left(
A-\lambda I_{n}\right)  $ for general $\lambda$, and find out what values of
$\lambda$ make it $0$.

In our case, $A=\left(
\begin{array}
[c]{cc}%
1 & 1\\
1 & 1
\end{array}
\right)  $ and $n=2$, so
\[
A-\lambda I_{n}=\left(
\begin{array}
[c]{cc}%
1 & 1\\
1 & 1
\end{array}
\right)  -\lambda\left(
\begin{array}
[c]{cc}%
1 & 0\\
0 & 1
\end{array}
\right)  =\left(
\begin{array}
[c]{cc}%
1-\lambda & 1\\
1 & 1-\lambda
\end{array}
\right)  .
\]
Thus,%
\[
\det\left(  A-\lambda I_{n}\right)  =\det\left(
\begin{array}
[c]{cc}%
1-\lambda & 1\\
1 & 1-\lambda
\end{array}
\right)  =\left(  1-\lambda\right)  \left(  1-\lambda\right)  -1\cdot
1=-2\lambda+\lambda^{2}.
\]
So $\det\left(  A-\lambda I_{n}\right)  =0$ if and only if $-2\lambda
+\lambda^{2}=0$, which is equivalent to $\lambda\in\left\{  0,2\right\}  $.
Thus, the eigenvalues of $A$ are $0$ and $2$.
\end{example}

The method we just showed works in general: For any $n\times n$-matrix $A$,
the eigenvalues of $A$ are the scalars $\lambda$ for which $\det\left(
A-\lambda I_{n}\right)  =0$. Let's give this determinant a name:

\begin{definition}
Let $A$ be an $n\times n$-matrix. We define%
\[
\chi_{A}\left(  t\right)  =\det\left(  A-tI_{n}\right)  .
\]
This is a polynomial in $t$, and is called the \textbf{characteristic
polynomial} of $A$.
\end{definition}

\begin{proposition}
\label{prop.eigen.evals=roots}Let $A$ be an $n\times n$-matrix. Then, the
eigenvalues of $A$ are the roots of the characteristic polynomial $\chi
_{A}\left(  t\right)  $.
\end{proposition}

\begin{proof}
For any scalar $\lambda$, we have the equivalence%
\begin{align*}
&  \ \left(  \lambda\text{ is an eigenvalue of }A\right) \\
&  \Longleftrightarrow\ \left(  \det\left(  A-\lambda I_{n}\right)  =0\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{as we have seen in Example
\ref{exa.eigen.1111.using-det}}\right) \\
&  \Longleftrightarrow\ \left(  \det\left(  A-tI_{n}\right)  \text{ becomes
}0\text{ if we substitute }\lambda\text{ for }t\right) \\
&  \Longleftrightarrow\ \left(  \lambda\text{ is a root of the polynomial
}\det\left(  A-tI_{n}\right)  \right) \\
&  \Longleftrightarrow\ \left(  \lambda\text{ is a root of the polynomial
}\chi_{A}\left(  t\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\det\left(  A-tI_{n}\right)  =\chi_{A}\left(  t\right)  \right)  .
\end{align*}
Thus, the eigenvalues of $A$ are the roots of the characteristic polynomial
$\chi_{A}\left(  t\right)  $.
\end{proof}

\begin{example}
Let $n=2$ and $A=\left(
\begin{array}
[c]{cc}%
a & b\\
c & d
\end{array}
\right)  $. Then, the characteristic polynomial $\chi_{A}\left(  t\right)  $
of $A$ is%
\begin{align*}
\chi_{A}\left(  t\right)   &  =\det\left(  A-tI_{n}\right)  =\det\left(
\left(
\begin{array}
[c]{cc}%
a & b\\
c & d
\end{array}
\right)  -t\left(
\begin{array}
[c]{cc}%
1 & 0\\
0 & 1
\end{array}
\right)  \right) \\
&  =\det\left(
\begin{array}
[c]{cc}%
a-t & b\\
c & d-t
\end{array}
\right)  =\left(  a-t\right)  \left(  d-t\right)  -bc\\
&  =t^{2}-\underbrace{\left(  a+d\right)  }_{\substack{=\operatorname*{Tr}%
A\\\text{(the trace of }A\text{)}}}t+\underbrace{\left(  ad-bc\right)
}_{=\det A}.
\end{align*}
Thus, the eigenvalues of $A$ are the roots of this polynomial; they are%
\[
\dfrac{a+d\pm\sqrt{\left(  a+d\right)  ^{2}-4\left(  ad-bc\right)  }}%
{2}=\dfrac{a+d\pm\sqrt{\left(  a-d\right)  ^{2}+4bc}}{2}.
\]


For example, if $A=\left(
\begin{array}
[c]{cc}%
1 & 2\\
3 & 4
\end{array}
\right)  $, then the eigenvalues of $A$ are%
\[
\dfrac{1+4\pm\sqrt{\left(  1-4\right)  ^{2}+4\cdot2\cdot3}}{2}=\dfrac
{5\pm\sqrt{33}}{2}.
\]

\end{example}

\begin{example}
We can try to play the same game with $n=3$: Let $n=3$ and $A=\left(
\begin{array}
[c]{ccc}%
a & b & c\\
d & e & f\\
g & h & i
\end{array}
\right)  $. Then,%
\begin{align*}
\chi_{A}\left(  t\right)   &  =\det\left(  A-tI_{n}\right)  =\det\left(
\begin{array}
[c]{ccc}%
a-t & b & c\\
d & e-t & f\\
g & h & i-t
\end{array}
\right) \\
&  =-t^{3}+\underbrace{\left(  a+e+i\right)  }_{=\operatorname*{Tr}A}%
t^{2}-\underbrace{\left(  ia+ie+ae-bd-cg-fh\right)  }_{=\det\left(
M_{1,1}\right)  +\det\left(  M_{2,2}\right)  +\det\left(  M_{3,3}\right)  }t\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\left(  iae-ibd-cge-afh+bfg+cdh\right)
}_{=\det A}.
\end{align*}
Unfortunately, this is a polynomial of degree $3$, and there is no good way to
compute the roots of such a polynomial explicitly.

However, if $A$ is \textquotedblleft nice\textquotedblright\ (e.g.,
strategically chosen for a homework problem), then the polynomial may have
neat rational or integer roots.
\end{example}

\begin{example}
Let $n=3$ and $A=\left(
\begin{array}
[c]{ccc}%
a & 0 & 0\\
b & c & 0\\
d & e & f
\end{array}
\right)  $ (an arbitrary lower-triangular matrix). Then,
\[
\chi_{A}\left(  t\right)  =\det\left(
\begin{array}
[c]{ccc}%
a-t & 0 & 0\\
b & c-t & 0\\
d & e & f-t
\end{array}
\right)  =\left(  a-t\right)  \left(  c-t\right)  \left(  f-t\right)  .
\]
The roots of $\chi_{A}\left(  t\right)  $ clearly are $a,c,f$, which are of
course the diagonal entries of $A$.
\end{example}

This holds more generally:

\begin{proposition}
If a matrix $A$ is triangular, then its eigenvalues are its diagonal entries.
\end{proposition}

\begin{proof}
Let $A$ be triangular. Then, the matrix $A-tI_{n}$ is triangular as well, and
thus its determinant $\det\left(  A-tI_{n}\right)  $ equals the product of its
diagonal entries $A_{1,1}-t,A_{2,2}-t,\ldots,A_{n,n}-t$. Hence,%
\[
\chi_{A}\left(  t\right)  =\det\left(  A-tI_{n}\right)  =\left(
A_{1,1}-t\right)  \left(  A_{2,2}-t\right)  \cdots\left(  A_{n,n}-t\right)  .
\]
Thus, the roots of $\chi_{A}\left(  t\right)  $ are $A_{1,1},A_{2,2}%
,\ldots,A_{n,n}$. But these roots are the eigenvalues of $A$ (by Proposition
\ref{prop.eigen.evals=roots}). Hence, the eigenvalues of $A$ are
$A_{1,1},A_{2,2},\ldots,A_{n,n}$. That is, they are the diagonal entries of
$A$. This proves the proposition.
\end{proof}

Now, assuming you have found all eigenvalues of a matrix $A$, how do you find
the eigenvectors?

\textbf{Method for finding eigenvalues and eigenvectors of a matrix:}

Given an $n\times n$-matrix $A$, we can find all eigenvalues and eigenvectors
of $A$ as follows:

\begin{itemize}
\item Calculate the characteristic polynomial $\chi_{A}\left(  t\right)
=\det\left(  A-tI_{n}\right)  $ of $A$.

\item Find all roots $\lambda_{1},\lambda_{2},\ldots,\lambda_{k}$ of $\chi
_{A}\left(  t\right)  $. These are the eigenvalues of $A$.

\item For each eigenvalue $\lambda_{i}$, compute the nonzero solutions $v$ to
$\left(  A-\lambda_{i}I_{n}\right)  v=0$ (for example, using Gaussian
elimination). These are the $\lambda_{i}$-eigenvectors of $A$.
\end{itemize}

\begin{example}
Let%
\[
A=\left(
\begin{array}
[c]{cccc}%
16 & 2 & 1 & 1\\
2 & 16 & 1 & 1\\
1 & 1 & 16 & 2\\
1 & 1 & 2 & 16
\end{array}
\right)  .
\]
What are the eigenvalues and eigenvectors of $A$ ?

The characteristic polynomial of $A$ is%
\begin{align*}
\chi_{A}\left(  t\right)   &  =\det\left(  A-tI_{n}\right)  =\det\left(
\begin{array}
[c]{cccc}%
16-t & 2 & 1 & 1\\
2 & 16-t & 1 & 1\\
1 & 1 & 16-t & 2\\
1 & 1 & 2 & 16-t
\end{array}
\right) \\
&  =\det\left(
\begin{array}
[c]{cccc}%
16-t & 2 & 1 & 1\\
-14+t & 14-t & 0 & 0\\
1 & 1 & 16-t & 2\\
1 & 1 & 2 & 16-t
\end{array}
\right) \\
&  =\det\left(
\begin{array}
[c]{cccc}%
16-t & 2 & 1 & 1\\
-14+t & 14-t & 0 & 0\\
1 & 1 & 16-t & 2\\
0 & 0 & -14+t & 14-t
\end{array}
\right) \\
&  =\det\left(
\begin{array}
[c]{cccc}%
16-t & 2 & 2 & 1\\
-14+t & 14-t & 0 & 0\\
1 & 1 & 18-t & 2\\
0 & 0 & 0 & 14-t
\end{array}
\right) \\
&  =\det\left(
\begin{array}
[c]{cccc}%
18-t & 2 & 2 & 1\\
0 & 14-t & 0 & 0\\
2 & 1 & 18-t & 2\\
0 & 0 & 0 & 14-t
\end{array}
\right) \\
&  =\left(  14-t\right)  \det\left(
\begin{array}
[c]{ccc}%
18-t & 2 & 2\\
0 & 14-t & 0\\
2 & 1 & 18-t
\end{array}
\right) \\
&  =\underbrace{\left(  14-t\right)  \left(  14-t\right)  }_{=\left(
14-t\right)  ^{2}}\underbrace{\det\left(
\begin{array}
[c]{cc}%
18-t & 2\\
2 & 18-t
\end{array}
\right)  }_{\substack{=\left(  18-t\right)  \left(  18-t\right)
-2\cdot2\\=t^{2}-36t+320\\=\left(  t-16\right)  \left(  t-20\right)  }}\\
&  =\left(  14-t\right)  ^{2}\left(  t-16\right)  \left(  t-20\right)  .
\end{align*}
So the roots of $\chi_{A}\left(  t\right)  $ (thus the eigenvalues of $A$) are
$14$, $16$ and $20$.

What about the eigenvectors?

The $14$-eigenvectors of $A$ are the nonzero vectors $v$ such that $\left(
A-14I_{4}\right)  v=0$. Since
\begin{align*}
A-14I_{4}  &  =\left(
\begin{array}
[c]{cccc}%
16-14 & 2 & 1 & 1\\
2 & 16-14 & 1 & 1\\
1 & 1 & 16-14 & 2\\
1 & 1 & 2 & 16-14
\end{array}
\right) \\
&  =\left(
\begin{array}
[c]{cccc}%
2 & 2 & 1 & 1\\
2 & 2 & 1 & 1\\
1 & 1 & 2 & 2\\
1 & 1 & 2 & 2
\end{array}
\right)  ,
\end{align*}
this boils down to solving%
\[
\left(
\begin{array}
[c]{cccc}%
2 & 2 & 1 & 1\\
2 & 2 & 1 & 1\\
1 & 1 & 2 & 2\\
1 & 1 & 2 & 2
\end{array}
\right)  \left(
\begin{array}
[c]{c}%
x\\
y\\
z\\
w
\end{array}
\right)  =\left(
\begin{array}
[c]{c}%
0\\
0\\
0\\
0
\end{array}
\right)  .
\]
You can do this by Gaussian elimination, and obtain%
\[
\left(
\begin{array}
[c]{c}%
x\\
y\\
z\\
w
\end{array}
\right)  =\left(
\begin{array}
[c]{c}%
-y\\
y\\
-w\\
w
\end{array}
\right)  .
\]
So the $14$-eigenvectors of $A$ are the nonzero vectors of the form $\left(
\begin{array}
[c]{c}%
-y\\
y\\
-w\\
w
\end{array}
\right)  $.

We can compute the $16$-eigenvectors and the $20$-eigenvectors as well; we get%
\begin{align*}
A-16I_{4}  &  =\left(
\begin{array}
[c]{cccc}%
16-16 & 2 & 1 & 1\\
2 & 16-16 & 1 & 1\\
1 & 1 & 16-16 & 2\\
1 & 1 & 2 & 16-16
\end{array}
\right) \\
&  =\left(
\begin{array}
[c]{cccc}%
0 & 2 & 1 & 1\\
2 & 0 & 1 & 1\\
1 & 1 & 0 & 2\\
1 & 1 & 2 & 0
\end{array}
\right)  \longrightarrow\left(
\begin{array}
[c]{cccc}%
1 & 0 & 0 & 1\\
0 & 1 & 0 & 1\\
0 & 0 & 1 & -1\\
0 & 0 & 0 & 0
\end{array}
\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{RREF}\right)  ,
\end{align*}
so the $16$-eigenvectors are%
\[
\left(
\begin{array}
[c]{c}%
-w\\
-w\\
w\\
w
\end{array}
\right)  \ \ \ \ \ \ \ \ \ \ \text{with }w\neq0.
\]


You can find the $20$-eigenvectors in the same way.
\end{example}

Let me comment on what can go \textquotedblleft wrong\textquotedblright\ in
the process:

\begin{example}
Let%
\[
A=\left(
\begin{array}
[c]{cc}%
0 & 1\\
-1 & 0
\end{array}
\right)  .
\]
Does it have eigenvalues? Its eigenvalues are the roots of $\chi_{A}\left(
t\right)  $. But%
\[
\chi_{A}\left(  t\right)  =\det\left(
\begin{array}
[c]{cc}%
0-t & 1\\
-1 & 0-t
\end{array}
\right)  =\left(  0-t\right)  \left(  0-t\right)  -1\left(  -1\right)
=t^{2}+1.
\]
What are the roots of this polynomial?

If we stick to real numbers, then this polynomial has no roots, since
$x^{2}+1>0$ for every real $x$. Thus, if we stick to real numbers, then $A$
has no eigenvalues and thus no eigenvectors. However, there is a way to force
$A$ to have eigenvalues: Just extend the \textquotedblleft field of
scalars\textquotedblright\ to the complex numbers, in which case $i$ and $-i$
become eigenvalues. Linear algebra using complex numbers as scalars is
analogous to linear algebra using real numbers as scalars, but with the nice
feature that each non-constant polynomial has at least one root (and, in fact,
can be factored into linear factors), so every $n\times n$-matrix $A$ (with
$n>0$) has at least one eigenvalue.
\end{example}

\begin{thebibliography}{9999999999}                                                                                       %


\bibitem[Strickland]{Strickland}Neil Strickland, \textit{MAS201 Linear
Mathematics for Applications}, lecture notes, 28 September 2013.\newline\url{http://neil-strickland.staff.shef.ac.uk/courses/MAS201/}
\end{thebibliography}


\end{document}