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\ihead{Fall 2019 Math 201-003 at Drexel: blackboard notes of 2019-11-13}
\ohead{page \thepage}
\cfoot{}
\begin{document}

\title{Fall 2019 Math 201-003 at Drexel: blackboard notes of 2019-11-13}
\author{Darij Grinberg}
\date{\today}
\maketitle

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

\subsection{Complex numbers}

\subsubsection{Some definitions done last time}

Recall the definition of complex numbers:

\begin{definition}
\textbf{(a)} A \textbf{complex number} is defined as a pair $\left(
a,b\right)  $ of two real numbers.

\textbf{(b)} We let $\mathbb{C}$ be the set of all complex numbers.

\textbf{(c)} For each real number $r$, we denote the complex number $\left(
r,0\right)  $ as $r_{\mathbb{C}}$ (and we will later equate it with $r$).

\textbf{(d)} We let $i$ be the complex number $\left(  0,1\right)  $.

\textbf{(e)} We define three binary operations $+$, $-$ and $\cdot$ on
$\mathbb{C}$ by setting%
\begin{align}
\left(  a,b\right)  +\left(  c,d\right)   &  =\left(  a+c,b+d\right)
,\label{eq.CC.formal.sum}\\
\left(  a,b\right)  -\left(  c,d\right)   &  =\left(  a-c,b-d\right)
,\label{eq.CC.formal.diff}\\
\left(  a,b\right)  \cdot\left(  c,d\right)   &  =\left(  ac-bd,ad+bc\right)
. \label{eq.CC.formal.prod}%
\end{align}
(Thus, the operations $+$ and $-$ are just entrywise addition and subtraction,
just as for row vectors. The operation $\cdot$ is more complicated, and we
will soon see why we have defined it in this particular way.)

\textbf{(f)} As usual, we write $\alpha\beta$ for $\alpha\cdot\beta$ if
$\alpha$ and $\beta$ are complex numbers.

As usual, we write $-\alpha$ for $0_{\mathbb{C}}-\alpha$ if $\alpha$ is a
complex number.
\end{definition}

We equated each real number $r$ with the complex number $r_{\mathbb{C}%
}=\left(  r,0\right)  $.

The complex number $i=\left(  0,1\right)  $ satisfies $i^{2}=-1$.

We represent complex numbers on the \textbf{Argand diagram}: To each complex
number $a+bi=\left(  a,b\right)  $ corresponds the point $\left(  a,b\right)
$.

\begin{definition}
Let $z$ be a complex number. Write $z$ in the form $z=\left(  a,b\right)
=a+bi$ for two real numbers $a,b$.

Let $P_{z}$ be the point corresponding to the complex number $z$ on the Argand
diagram. (As you remember, this is the point with coordinates $\left(
a,b\right)  $.)

\textbf{(a)} The real numbers $a$ and $b$ are called the \textbf{real part}
and the \textbf{imaginary part} of $z$. They are the Cartesian coordinates of
the point $P_{z}$.

\textbf{(b)} The \textbf{absolute value} of $z$ is defined to be the real
number $\sqrt{a^{2}+b^{2}}$. This is the distance between the origin and
$P_{z}$. This is the first (radial) \textbf{polar coordinate} of $P_{z}$. It
is denoted by $\left\vert z\right\vert $.

\textbf{(c)} Assume that $z\neq0$. Then, consider the angle $\varphi$ (with
$-\pi<\varphi\leq\pi$) at which the ray from $0$ to $P_{z}$ stands to the ray
from $0$ to $1$ (i.e., the positive half-axis). This is the second (angular)
\textbf{polar coordinate} of $P_{z}$. It is denoted by $\arg z$, and is called
the \textbf{argument} of $z$.
\end{definition}

\subsubsection{More about angles}

\begin{remark}
Let $z=a+bi$ (with $a,b\in\mathbb{R}$ and $z\neq0$) be a complex number.

How to compute $\arg z$ through $a$ and $b$ ?

Let $P_{z}$ be the point $\left(  a,b\right)  $ as in the previous definition.
Recall that $\arg z$ is the angle at which the ray from $0$ to $P_{z}$ stands
to the ray from $0$ to $1$. The slope of the former ray is $b/a$ (since
$P_{z}=\left(  a,b\right)  $). Thus, the classical relation between the slope
of a line and its angle against the x-axis shows that%
\[
\tan\left(  \arg z\right)  =\dfrac{b}{a}.
\]
Thus, it makes sense to expect $\arg z=\arctan\dfrac{b}{a}$. But this is not
quite the case, since $\dfrac{b}{a}$ only determines the \textbf{line} from
$0$ to $P_{z}$, whereas $\arg z$ depends on the \textbf{ray} from $0$ to
$P_{z}$; thus, $\arg z$ depends on \textquotedblleft what side of the
origin\textquotedblright\ $z$ lies on. So the correct way to determine $\arg
z$ is the following:

\begin{itemize}
\item We have $\tan\left(  \arg z\right)  =\dfrac{b}{a}$. (If $a=0$, then this
must be interpreted as $\tan\left(  \arg z\right)  =\dfrac{b}{0}=\infty$,
which means that $\arg z$ is either $\dfrac{\pi}{2}$ or $\dfrac{-\pi}{2}$.)

\item We have $\arg z\geq0$ if and only if $b\geq0$.
\end{itemize}

\noindent If you know \href{https://en.wikipedia.org/wiki/Atan2}{the
\textquotedblleft two-variable arctangent function\textquotedblright%
\ $\operatorname*{atan2}$}, then you can rewrite this as follows:%
\[
\arg z=\operatorname*{atan2}\left(  b,a\right)  .
\]

\end{remark}

\begin{definition}
For any angle $\varphi$, we define a complex number $\operatorname*{cis}%
\varphi$ by%
\[
\operatorname*{cis}\varphi=\cos\varphi+i\sin\varphi=\left(  \cos\varphi
,\sin\varphi\right)  .
\]
Geometrically (i.e., on the Argand diagram), the point $P_{\operatorname*{cis}%
\varphi}$ corresponding to this complex number $\operatorname*{cis}\varphi$ is
the point obtained by rotating $P_{1}=\left(  1,0\right)  $ through the
angle\footnotemark$\varphi$ around the origin. (Indeed, the latter point
clearly has abscissa $\cos\varphi$ and ordinate $\sin\varphi$, because of how
cosine and sine are defined; but so does the point $P_{\operatorname*{cis}%
\varphi}$.) Thus, in particular, the point $P_{\operatorname*{cis}\varphi}$
lies on the unit circle (i.e., the circle with center at the origin and radius
$1$).
\end{definition}

\footnotetext{Angles are always measured counterclockwise in mathematics. For
example, the point obtained by rotating $P_{1}$ through the angle
$\pi/2=90^{\circ}$ around the origin is the point $P_{i}=\left(  0,1\right)
$.} Note that every angle $\varphi$ satisfies $\operatorname*{cis}\left(
\varphi+2\pi\right)  =\operatorname*{cis}\varphi$ (since $\cos\left(
\varphi+2\pi\right)  =\cos\varphi$ and $\sin\left(  \varphi+2\pi\right)
=\sin\varphi$). More general, two angles $\alpha$ and $\beta$ satisfy
$\operatorname*{cis}\alpha=\operatorname*{cis}\beta$ if and only if
$\alpha-\beta$ is a multiple of $2\pi$. This is often restated as follows: The
complex number $\operatorname*{cis}\varphi$ uniquely determines the angle
$\varphi$ up to a multiple of $2\pi$.

\begin{proposition}
\label{prop.CC.cis.mult}For any two angles $\alpha$ and $\beta$, we have%
\[
\operatorname*{cis}\left(  \alpha+\beta\right)  =\operatorname*{cis}%
\alpha\cdot\operatorname*{cis}\beta.
\]

\end{proposition}

\begin{proof}
This follows by comparing%
\begin{align*}
\operatorname*{cis}\left(  \alpha+\beta\right)   &  =\underbrace{\cos\left(
\alpha+\beta\right)  }_{=\cos\alpha\cos\beta-\sin\alpha\sin\beta
}+i\underbrace{\sin\left(  \alpha+\beta\right)  }_{=\sin\alpha\cos\beta
+\cos\alpha\sin\beta}\\
&  =\cos\alpha\cos\beta-\sin\alpha\sin\beta+i\left(  \sin\alpha\cos\beta
+\cos\alpha\sin\beta\right)
\end{align*}
with%
\begin{align*}
&  \underbrace{\operatorname*{cis}\alpha}_{=\cos\alpha+i\sin\alpha}%
\cdot\underbrace{\operatorname*{cis}\beta}_{=\cos\beta+i\sin\beta}\\
&  =\left(  \cos\alpha+i\sin\alpha\right)  \cdot\left(  \cos\beta+i\sin
\beta\right) \\
&  =\cos\alpha\cos\beta+\cos\alpha\cdot i\sin\beta+i\sin\alpha\cos
\beta+\underbrace{i^{2}}_{=-1}\sin\alpha\sin\beta\\
&  =\cos\alpha\cos\beta-\sin\alpha\sin\beta+i\left(  \sin\alpha\cos\beta
+\cos\alpha\sin\beta\right)  .
\end{align*}

\end{proof}

Note that Proposition \ref{prop.CC.cis.mult} \textquotedblleft
packages\textquotedblright\ both formulas%
\begin{align*}
\cos\left(  \alpha+\beta\right)   &  =\cos\alpha\cos\beta-\sin\alpha\sin
\beta\ \ \ \ \ \ \ \ \ \ \text{and}\\
\sin\left(  \alpha+\beta\right)   &  =\sin\alpha\cos\beta+\cos\alpha\sin\beta.
\end{align*}


\begin{proposition}
\label{prop.CC.cisargz}Let $z$ be a nonzero complex number. Then,%
\[
z=\left\vert z\right\vert \cdot\operatorname*{cis}\left(  \arg z\right)  .
\]

\end{proposition}

\begin{proof}
[Proof sketch.]Geometrically, this can be seen as follows: The complex numbers
$\operatorname*{cis}\left(  \arg z\right)  $ and $z$ have the same argument
(namely, $\arg z$); thus, the points corresponding to them lie on the same ray
from the origin. Hence, $z=r\cdot\operatorname*{cis}\left(  \arg z\right)  $
for some nonnegative real $r$. To find this $r$, we just compare $\left\vert
z\right\vert $ with $\left\vert \operatorname*{cis}\left(  \arg z\right)
\right\vert =1$. Thus, we get $r=\left\vert z\right\vert $.
\end{proof}

\begin{corollary}
When we multiply two complex numbers, their absolute values get multiplied,
while their arguments get added (modulo $2\pi$).
\end{corollary}

\begin{proof}
[Proof sketch.]We are claiming that any complex numbers $z$ and $w$ satisfy%
\[
\left\vert zw\right\vert =\left\vert z\right\vert \cdot\left\vert w\right\vert
\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \arg\left(  zw\right)
\equiv\arg z+\arg w\operatorname{mod}2\pi,
\]
where \textquotedblleft$\alpha\equiv\beta\operatorname{mod}2\pi$%
\textquotedblright\ means \textquotedblleft$\alpha-\beta$ is an integer
multiple of $2\pi$\textquotedblright.

How do we prove this?

To prove $\left\vert zw\right\vert =\left\vert z\right\vert \cdot\left\vert
w\right\vert $, it suffices to recall the definition of absolute value and
compute both sides. (This is Exercise 10 on
\href{http://www.cip.ifi.lmu.de/~grinberg/t/19fla/mt2t.pdf}{MT2 preparation}.)

It remains to prove that%
\[
\arg\left(  zw\right)  \equiv\arg z+\arg w\operatorname{mod}2\pi.
\]
In other words, it remains to prove that%
\[
\operatorname*{cis}\left(  \arg\left(  zw\right)  \right)
=\operatorname*{cis}\left(  \arg z+\arg\left(  w\right)  \right)  ,
\]
because $\operatorname*{cis}\varphi$ uniquely determines $\varphi$ up to a
multiple of $2\pi$.

Proposition \ref{prop.CC.cisargz} yields $z=\left\vert z\right\vert
\cdot\operatorname*{cis}\left(  \arg z\right)  $, thus%
\[
\operatorname*{cis}\left(  \arg z\right)  =\dfrac{z}{\left\vert z\right\vert
}.
\]
Similarly,%
\[
\operatorname*{cis}\left(  \arg w\right)  =\dfrac{w}{\left\vert w\right\vert
}\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \operatorname*{cis}\left(
\arg\left(  zw\right)  \right)  =\dfrac{zw}{\left\vert zw\right\vert }.
\]
Hence,%
\begin{align*}
\operatorname*{cis}\left(  \arg\left(  zw\right)  \right)   &  =\dfrac
{zw}{\left\vert zw\right\vert }=\dfrac{zw}{\left\vert z\right\vert
\cdot\left\vert w\right\vert }\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\left\vert zw\right\vert =\left\vert z\right\vert \cdot\left\vert
w\right\vert \right) \\
&  =\underbrace{\dfrac{z}{\left\vert z\right\vert }}_{=\operatorname*{cis}%
\left(  \arg z\right)  }\cdot\underbrace{\dfrac{w}{\left\vert w\right\vert }%
}_{=\operatorname*{cis}\left(  \arg w\right)  }=\operatorname*{cis}\left(
\arg z\right)  \cdot\operatorname*{cis}\left(  \arg w\right) \\
&  =\operatorname*{cis}\left(  \arg z+\arg w\right)
\end{align*}
(since Proposition \ref{prop.CC.cis.mult} yields $\operatorname*{cis}\left(
\arg z+\arg w\right)  =\operatorname*{cis}\left(  \arg z\right)
\cdot\operatorname*{cis}\left(  \arg w\right)  $).
\end{proof}

Thus, for a fixed complex number $z$, the map%
\begin{align*}
\mathbb{C}  &  \rightarrow\mathbb{C},\\
w  &  \mapsto zw
\end{align*}
is a rotation (around the origin, with angle $\arg z$) composed with a
homothety (around the origin, with scaling factor $\left\vert z\right\vert $).

\begin{remark}
\label{rmk.CC.exp.euler}Here is a digression for those who know a bit of
complex analysis (or at least the complex exponential function).

The complex exponential function $\exp:\mathbb{C}\rightarrow\mathbb{C}$ is
defined by%
\begin{equation}
\exp t=1+\dfrac{t}{1!}+\dfrac{t^{2}}{2!}+\dfrac{t^{3}}{3!}+\dfrac{t^{4}}%
{4!}+\cdots. \label{eq.rmk.CC.exp.euler.expt=}%
\end{equation}
The infinite sum on the right hand side of this equation converges (once you
have appropriately defined convergence of sums of complex numbers), so this
function $\exp$ is well-defined. If $t$ is a real number, then $\exp t=e^{t}$
(where $e\approx2.\,\allowbreak718\,3$ is
\href{https://en.wikipedia.org/wiki/E_(mathematical_constant)}{the famous
\textquotedblleft number $e$\textquotedblright}); thus, it is common to write
$e^{t}$ for $\exp t$ even when $t$ is a complex number (although there is no
general well-defined concept of $a^{b}$ for complex numbers $a$ and $b$).

Now, \href{https://en.wikipedia.org/wiki/Euler's_formula}{Euler's formula}
says that $\operatorname*{cis}\varphi=\exp\left(  i\varphi\right)  $ for any
angle $\varphi$. Restated in explicit language, this is saying that if you
plug $i\varphi$ for $t$ in the power series (\ref{eq.rmk.CC.exp.euler.expt=}),
then you get a power series that converges to $\operatorname*{cis}\varphi$. In
other words,%
\[
1+\dfrac{i\varphi}{1!}+\dfrac{\left(  i\varphi\right)  ^{2}}{2!}%
+\dfrac{\left(  i\varphi\right)  ^{3}}{3!}+\dfrac{\left(  i\varphi\right)
^{4}}{4!}+\cdots=\operatorname*{cis}\varphi=\cos\varphi+i\sin\varphi.
\]
Since the powers of $i$ are%
\begin{align*}
i^{0}  &  =1,\ \ \ \ \ \ \ \ \ \ i^{1}=i,\ \ \ \ \ \ \ \ \ \ i^{2}%
=-1,\ \ \ \ \ \ \ \ \ \ i^{3}=-i,\\
i^{4}  &  =1,\ \ \ \ \ \ \ \ \ \ i^{5}=i,\ \ \ \ \ \ \ \ \ \ i^{6}%
=-1,\ \ \ \ \ \ \ \ \ \ i^{7}=-i,\\
i^{8}  &  =1,\ \ \ \ \ \ \ \ \ \ \ldots\ \ \ \ \ \ \ \ \ \ \text{(so }%
i^{n+4}=i^{n}\text{ for each }n\text{),}%
\end{align*}
you can rewrite this as%
\[
1+\dfrac{i\varphi}{1!}-\dfrac{\varphi^{2}}{2!}-\dfrac{i\varphi^{3}}{3!}%
+\dfrac{\varphi^{4}}{4!}+\dfrac{i\varphi^{5}}{5!}-\dfrac{\varphi^{6}}%
{6!}-\dfrac{i\varphi^{7}}{7!}+\cdots=\cos\varphi+i\sin\varphi.
\]
Splitting this equality into its real and imaginary parts, we obtain%
\begin{align*}
1-\dfrac{\varphi^{2}}{2!}+\dfrac{\varphi^{4}}{4!}-\dfrac{\varphi^{6}}%
{6!}+\cdots &  =\cos\varphi\ \ \ \ \ \ \ \ \ \ \text{and}\\
\dfrac{\varphi}{1!}-\dfrac{\varphi^{3}}{3!}+\dfrac{\varphi^{5}}{5!}%
-\dfrac{\varphi^{7}}{7!}+\cdots &  =\sin\varphi.
\end{align*}
These are the classical Maclaurin power series for $\cos$ and $\sin$.

Also, $e^{i\varphi}=\exp\left(  i\varphi\right)  =\operatorname*{cis}\varphi$
yields $e^{i\pi}=\operatorname*{cis}\pi=-1$, the famous Euler identity.
\end{remark}

\subsection{Application of diagonalization to ODEs}

Let us talk a bit about an application of diagonalization of matrices.

This section follows \cite[\S 15]{Strickland}.

Recall one of the simplest forms of an ODE (ordinary differential equation):
We are looking for a function $x$ in a single variable $t$ that satisfies
$x^{\prime}=ax$, where $a$ is a real constant, and where $x^{\prime}$ denotes
the derivative of $x$ in $t$. (People often write $\dot{x}$ for $x^{\prime}$.)

The solutions of this ODE are all functions of the form $ce^{at}$ for
constants $c\in\mathbb{R}$.

Now, imagine we want to solve a system of two ODEs for two functions $x$ and
$y$. For example, let us solve%
\[
\left\{
\begin{array}
[c]{c}%
x^{\prime}=x+y\\
y^{\prime}=x+y
\end{array}
\right.  .
\]


Rewrite this system as%
\[
\left(
\begin{array}
[c]{c}%
x^{\prime}\\
y^{\prime}%
\end{array}
\right)  =A\left(
\begin{array}
[c]{c}%
x\\
y
\end{array}
\right)  ,\ \ \ \ \ \ \ \ \ \ \text{where }A=\left(
\begin{array}
[c]{cc}%
1 & 1\\
1 & 1
\end{array}
\right)  .
\]
Note that the column vectors now have functions as entries (rather than
numbers), but this doesn't change anything.

Now, let us diagonalize $A$:%
\[
A=UDU^{-1},\ \ \ \ \ \ \ \ \ \ U=\left(
\begin{array}
[c]{cc}%
1 & 1\\
1 & -1
\end{array}
\right)  ,\ \ \ \ \ \ \ \ \ \ D=\operatorname*{diag}\left(  2,0\right)
=\left(
\begin{array}
[c]{cc}%
2 & 0\\
0 & 0
\end{array}
\right)  .
\]
Note that%
\[
U^{-1}=\left(
\begin{array}
[c]{cc}%
\dfrac{1}{2} & \dfrac{1}{2}\\
\dfrac{1}{2} & -\dfrac{1}{2}%
\end{array}
\right)  .
\]


Thus, our system rewrites as%
\[
\left(
\begin{array}
[c]{c}%
x^{\prime}\\
y^{\prime}%
\end{array}
\right)  =UDU^{-1}\left(
\begin{array}
[c]{c}%
x\\
y
\end{array}
\right)  .
\]
Rewrite this further by multiplying both sides by $U^{-1}$:%
\[
U^{-1}\left(
\begin{array}
[c]{c}%
x^{\prime}\\
y^{\prime}%
\end{array}
\right)  =DU^{-1}\left(
\begin{array}
[c]{c}%
x\\
y
\end{array}
\right)  .
\]
Note that%
\[
U^{-1}\left(
\begin{array}
[c]{c}%
x\\
y
\end{array}
\right)  =\left(
\begin{array}
[c]{cc}%
\dfrac{1}{2} & \dfrac{1}{2}\\
\dfrac{1}{2} & -\dfrac{1}{2}%
\end{array}
\right)  \left(
\begin{array}
[c]{c}%
x\\
y
\end{array}
\right)  =\left(
\begin{array}
[c]{c}%
\dfrac{x+y}{2}\\
\dfrac{x-y}{2}%
\end{array}
\right)
\]
and%
\[
U^{-1}\left(
\begin{array}
[c]{c}%
x^{\prime}\\
y^{\prime}%
\end{array}
\right)  =\left(
\begin{array}
[c]{cc}%
\dfrac{1}{2} & \dfrac{1}{2}\\
\dfrac{1}{2} & -\dfrac{1}{2}%
\end{array}
\right)  \left(
\begin{array}
[c]{c}%
x^{\prime}\\
y^{\prime}%
\end{array}
\right)  =\left(
\begin{array}
[c]{c}%
\dfrac{x^{\prime}+y^{\prime}}{2}\\
\dfrac{x^{\prime}-y^{\prime}}{2}%
\end{array}
\right)  =\left(
\begin{array}
[c]{c}%
\left(  \dfrac{x+y}{2}\right)  ^{\prime}\\
\left(  \dfrac{x-y}{2}\right)  ^{\prime}%
\end{array}
\right)  .
\]
Setting
\[
z=\dfrac{x+y}{2}\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ w=\dfrac
{x-y}{2},
\]
we can rewrite these as%
\[
U^{-1}\left(
\begin{array}
[c]{c}%
x\\
y
\end{array}
\right)  =\left(
\begin{array}
[c]{c}%
z\\
w
\end{array}
\right)  \ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ U^{-1}\left(
\begin{array}
[c]{c}%
x^{\prime}\\
y^{\prime}%
\end{array}
\right)  =\left(
\begin{array}
[c]{c}%
z^{\prime}\\
w^{\prime}%
\end{array}
\right)  .
\]
So our system becomes%
\[
\left(
\begin{array}
[c]{c}%
z^{\prime}\\
w^{\prime}%
\end{array}
\right)  =D\left(
\begin{array}
[c]{c}%
z\\
w
\end{array}
\right)  .
\]
But
\[
D=\operatorname*{diag}\left(  2,0\right)  ,\ \ \ \ \ \ \ \ \ \ \text{so}%
\ \ \ \ \ \ \ \ \ \ D\left(
\begin{array}
[c]{c}%
z\\
w
\end{array}
\right)  =\left(
\begin{array}
[c]{c}%
2z\\
0w
\end{array}
\right)  .
\]
So our system becomes%
\[
\left(
\begin{array}
[c]{c}%
z^{\prime}\\
w^{\prime}%
\end{array}
\right)  =\left(
\begin{array}
[c]{c}%
2z\\
0w
\end{array}
\right)  .
\]
This is tantamount to $z^{\prime}=2z$ and $w^{\prime}=0w$. The first of these
equations involves $z$ only, while the second involves $w$ only. Solving them,
we obtain $z=ce^{2t}$ and $w=d\underbrace{e^{0t}}_{=1}=d$ (for two real
constants $c$ and $d$).

So we know that
\[
\dfrac{x+y}{2}=z=ce^{2t}\ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ \dfrac{x-y}{2}=w=d.
\]
We can recover $x$ and $y$ from these by treating these equations as a system
of linear equations and solving them by Gaussian elimination. We get%
\[
x=ce^{2t}+d;\ \ \ \ \ \ \ \ \ \ y=ce^{2t}-d.
\]


See \cite[\S 15]{Strickland} for further examples.

See also \cite[\S 16]{Strickland} for another application of diagonalization:
the study of Markov chains. See \cite[\S 17]{Strickland} for an application to
ranking websites for web search (Google's PageRank algorithm).

\section{More about determinants}

Let us state (without proof) a few more facts about determinants.

\subsection{Determinants of block matrices}

\begin{theorem}
Let $\left[
\begin{array}
[c]{cc}%
A & B\\
0_{q\times p} & D
\end{array}
\right]  $ be a block matrix, where $A$ is a $p\times p$-matrix and $D$ is a
$q\times q$-matrix. Then,%
\[
\det\left[
\begin{array}
[c]{cc}%
A & B\\
0_{q\times p} & D
\end{array}
\right]  =\det A\cdot\det D.
\]

\end{theorem}

\begin{example}
Let $p=2$ and $q=2$. Then, this theorem is saying%
\[
\det\left(
\begin{array}
[c]{cccc}%
a & b & c & d\\
a^{\prime} & b^{\prime} & c^{\prime} & d^{\prime}\\
0 & 0 & c^{\prime\prime} & d^{\prime\prime}\\
0 & 0 & c^{\prime\prime\prime} & d^{\prime\prime\prime}%
\end{array}
\right)  =\det\left(
\begin{array}
[c]{cc}%
a & b\\
a^{\prime} & b^{\prime}%
\end{array}
\right)  \cdot\det\left(
\begin{array}
[c]{cc}%
c^{\prime\prime} & d^{\prime\prime}\\
c^{\prime\prime\prime} & d^{\prime\prime\prime}%
\end{array}
\right)  .
\]

\end{example}

\begin{remark}
It is \textbf{not} true that if $A$ is a $p\times p$-matrix and $D$ is a
$q\times q$-matrix, then%
\begin{equation}
\det\left[
\begin{array}
[c]{cc}%
A & B\\
C & D
\end{array}
\right]  =\det A\cdot\det D-\det B\cdot\det C. \label{eq.rmk.det.block.wrong}%
\end{equation}
Indeed, if $p\neq q$, then $\det B$ and $\det C$ make no sense to begin with
(since $B$ and $C$ are not square matrices). But even when $p=q$, you can find
counterexamples to (\ref{eq.rmk.det.block.wrong}).
\end{remark}

However, here is something that is true:

\begin{theorem}
(The Schur complement theorem for determinants.)

Let $\left[
\begin{array}
[c]{cc}%
A & B\\
C & D
\end{array}
\right]  $ be a block matrix, where $A$ is an invertible $p\times p$-matrix
and $D$ is a $q\times q$-matrix. Then,%
\[
\det\left[
\begin{array}
[c]{cc}%
A & B\\
C & D
\end{array}
\right]  =\det A\cdot\det\left(  D-CA^{-1}B\right)  .
\]

\end{theorem}

There is also a way to compute the inverse of a block matrix $\left[
\begin{array}
[c]{cc}%
A & B\\
C & D
\end{array}
\right]  ^{-1}$ using its blocks. See
\href{https://en.wikipedia.org/wiki/Schur_complement}{Wikipedia: Schur
complement}.

\subsection{An application of determinants}

The following exercise is an example of how identifying a polynomial as a
determinant can sometimes help us understand this polynomial better.

\begin{exercise}
Factor the polynomial%
\[
ab^{2}+bc^{2}+ca^{2}-ac^{2}-ba^{2}-cb^{2}%
\]
(in three indeterminates $a,b,c$).
\end{exercise}

\begin{proof}
[Solution.]The polynomial $ab^{2}+bc^{2}+ca^{2}-ac^{2}-ba^{2}-cb^{2}$ looks
very much like the determinant of a $3\times3$-matrix (it is a sum of $6$
terms, $3$ of which have negative signs). A bit of experimentation confirms
that it is indeed the determinant of a very simple $3\times3$-matrix:%
\[
ab^{2}+bc^{2}+ca^{2}-ac^{2}-ba^{2}-cb^{2}=\det\left(
\begin{array}
[c]{ccc}%
1 & 1 & 1\\
a & b & c\\
a^{2} & b^{2} & c^{2}%
\end{array}
\right)  .
\]
We shall now compute this determinant in a different way: Rather than
expanding it, we will use row operations to gradually simplify it:\footnote{As
we have already done many times, we draw boxes around pivot entries.}%
\begin{align*}
&  \det\left(
\begin{array}
[c]{ccc}%
\fbox{$1$} & 1 & 1\\
a & b & c\\
a^{2} & b^{2} & c^{2}%
\end{array}
\right)  \\
&  =\det\left(
\begin{array}
[c]{ccc}%
\fbox{$1$} & 1 & 1\\
0 & \fbox{$b-a$} & c-a\\
a^{2} & b^{2} & c^{2}%
\end{array}
\right)  \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have subtracted }a\cdot
\operatorname{row}1\text{ from }\operatorname{row}2\right)  \\
&  =\det\left(
\begin{array}
[c]{ccc}%
\fbox{$1$} & 1 & 1\\
0 & \fbox{$b-a$} & c-a\\
0 & b^{2}-a^{2} & c^{2}-a^{2}%
\end{array}
\right)  \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have subtracted }a^{2}%
\cdot\operatorname{row}1\text{ from }\operatorname{row}3\right)  \\
&  =\det\left(
\begin{array}
[c]{ccc}%
\fbox{$1$} & 1 & 1\\
0 & \fbox{$b-a$} & c-a\\
0 & 0 & c^{2}-a^{2}-\left(  b+a\right)  \left(  c-a\right)
\end{array}
\right)  \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have subtracted }\left(
b+a\right)  \cdot\operatorname{row}2\text{ from }\operatorname{row}3\right)
\\
&  =1\left(  b-a\right)  \underbrace{\left(  c^{2}-a^{2}-\left(  b+a\right)
\left(  c-a\right)  \right)  }_{\substack{=c^{2}-a^{2}-cb-ca+ab+a^{2}%
\\=c^{2}-cb-ca+ab=\left(  c-a\right)  \left(  c-b\right)  }}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since the determinant of an upper-triangular matrix}\\
\text{is the product of its diagonal entries}%
\end{array}
\right)  \\
&  =\left(  b-a\right)  \left(  c-a\right)  \left(  c-b\right)  .
\end{align*}
Thus,%
\[
ab^{2}+bc^{2}+ca^{2}-ac^{2}-ba^{2}-cb^{2}=\det\left(
\begin{array}
[c]{ccc}%
1 & 1 & 1\\
a & b & c\\
a^{2} & b^{2} & c^{2}%
\end{array}
\right)  =\left(  b-a\right)  \left(  c-a\right)  \left(  c-b\right)  .
\]

\end{proof}

\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}