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\begin{document}

\title{Vectors and Matrices}
\author{}
\date{\today}
\maketitle

\tableofcontents

\newpage
\setcounter{section}{-1}
\section{Introduction}

This course covers ``linear algebra'', topics in
\emph{algebra} \& \emph{geometry}

\noindent It involves approaches that are
\begin{center}
    concrete \& abstract

    computational \& conceptual
\end{center}

\noindent The key ideas to develop / build on are:
\begin{itemize}
    \item Elementary geometry (Euclidean):
        points, lines, planes in 2d or 3d; length,
        angles
    \item Points described by coordinates
    \item Points described by vectors; what is a
        vector?
    \item Simple transformations e.g. rotations \&
        reflections $\to$ linear maps.
\end{itemize}

\subsection{Plan}

\begin{enumerate}
    \item Complex Numbers
    \item Vectors in 3 dimensions
    \item Vectors in General, $\RR^n$ \& $\CC^n$
    \item Matrices \& Linear Maps
    \item Determinants \& Inverses
    \item Eigenvalues \& Eigenvectors
    \item Changing Bases, Canonical Forms \&
        Symmetries
\end{enumerate}

\newpage
\section{Complex Numbers}

\subsection{Basic Definitions}

The following terms will not be defined here but
assumed to be understood:
\begin{itemize}
    \item $\CC$, $+$, $\times$
    \item conjugate, modulus, argument
    \item complex plane / Argand diagram
\end{itemize}

\noindent
Construct $\CC$ by adding an element $i$ to real
numbers $\RR$, with
\[ i^2 = -1 .\]
Any complex number $z \in \CC$ has the form
\[ z = x + iy \qquad \text{with $x, y \in \RR$} ;\]
$x = \mathrm{Re}(z)$ is the \emph{real part};
$y = \mathrm{Im}(z)$ is the \emph{imaginary part}.

\noindent $\RR \subset \CC$ consisting of elements
$x = i0 = x$.

\noindent In following, use notation above \&
\[ z_1 = x_1 + iy_1, \quad z_2 = x_2 + iy_2 \quad
\text{etc} .\]

\begin{enumerate}
    \item \underline{Addition} (\& subtraction). Define
        \[ z_1 \pm z_2 = (x_1 \pm x_2)
        + i(y_1 \pm y_2) \]
    \item \underline{Multiplication}. Define
        \[ z_1z_2 = (x_1x_2 - y_1y_2)
        + i(x_1y_2 + x_2y_1) \]
        If $z \neq 0$, note that
        \[ z^{-1} = \frac{x}{x^2 + y^2}
        - \frac{y}{x^2 + y^2} \]
        satisfies $zz^{-1} = 1$.
    \item \underline{Complex conjugate} Define
        \[ \ol{z} = z^* = x - iy \]
        Then:
        \[ \mathrm{Re}(z) = \half(z + \ol{z}) \]
        and
        \[ \mathrm{Im}(z) = \frac{1}{2i}
        (z - \ol{z}) \]
        $\ol{(\ol{z})} = z$ \& further
        \[ \ol{z_1 + z_2} = \ol{z_1} + \ol{z_2} \]
        \[ \ol{z_1z_2} = \ol{z_1}\ol{z_2} \]
    \item \underline{Modulus} is defined by
        $r =|z|$, real \& $\ge 0$, with
        $r^2 = |z|^2 = z\ol{z} = x^2 + y^2$
    \item \underline{Argument} $\theta = \arg(z)$
        real, defined for $z \neq 0$ by
        \[ z = r(\cos \theta + i \sin \theta) \]
        for some real $\theta$ (this is known as
        \emph{polar form})
        \[ \cos \theta = \frac{x}{\sqrt{x^2 + y^2}},
        \quad \sin\theta = \frac{y}{\sqrt{x^2 + y^2}} \]
        \[ \implies \tan \theta = \frac{y}{x} \]
        $\arg(z)$ is determined only
        \emph{$\bmod\ 2\pi$} i.e. can change
        $\theta \to \theta + 2n\pi$ for $n \in
        \ZZ$.

        To make it unique we can restrict the
        range, e.g. the \emph{principal value}
        defined by
        \[ -\pi < \theta \le \pi \]
    \item \underline{Argand diagram \& Complex
        Plane} Plot $\mathrm{Re}(z)$ \&
        $\mathrm{Im}(z)$ on orthogonal axes, then
        $r = |z|$ \& $\theta = \arg(z)$ are length
        \& angle shown
        \begin{center}
        \begin{tsqx}
            (-0.2,0)--(1.5,0) EndArrow
            (0,-0.2)--(0,1.5) EndArrow
            label Re($z$) @ (1.5,-0.15)
            label Im($z$) @ (0.15,1.5)
            z = (1.3,1)
            origin--z
            label $x$ @ (0.65,-0.1)
            label $y$ @ (-0.1,0.5)
            label $r$ @ 0.5*z+(-0.05,0.05)
            z--(0,1) dashed
            z--(1.3,0) dashed
            anglemark4 (1,0) origin z 10
            label $\alpha$ @ 0.4*dir(17)
        \end{tsqx}
        \end{center}
\end{enumerate}

\begin{example*}
    Consider
    \[ z = -1 + i\sqrt{3} = 2\left( -\half +
    i\frac{\sqrt{3}}{2} \right) \]
    here $z = 2$ \& $\arg(z) = \frac{2\pi}{3} +
    2n\pi$. Note $\tan \theta = -\sqrt{3} \implies
    \theta = \frac{2\pi}{3} + 2n\pi = \arg(z)$
    \emph{or} $\theta = -\frac{\pi}{3} + 2n\pi =
    \arg(-z)$.
    \begin{center}
    \begin{tsqx}
        (-1.1,0)--(1.1,0) EndArrow
        (0,-1.1)--(0,1.1) EndArrow
        label Re @ (1.1,-0.15)
        label Im @ (0.15,1.1)
        z = dir(120)
        label $-z$ @ 0.1*dir(-z)-z
        origin--z EndArrow
        origin--origin-z EndArrow
        anglemark4 (1,0) origin z 5
        label $\frac{2\pi}{3}$ @ 0.3*dir(45)
    \end{tsqx}
    \end{center}
\end{example*}

\subsection{Basic Properties \& Consequences}

\noindent\textbf{Aside (motivating the
definitions leading to $\CC$)} \\
Note that $\ZZ$ can be seen as a way to solve some
equations involving $\ZZ$, for example $x + 3 =
0$. Rational numbers can then be used to solve
other equations such as $5x + 1 = 0$, and real
numbers are used to solve some quadratics and
other higher degree polynomials, such as $x^2 -
2$. Finally, the complex numbers are used to allow
us to solve more equations that we couldn't
before, such as $x^2 + 4 = 0$. This leads to the
fundamental theorem of algebra.

\begin{enumerate}[(i)]
    \item $\CC$ with operations $+$, $\times$ is a
        \emph{field}.

        \noindent i.e. $\CC$ with $+$ is an
        abelian group \& distributive laws hold,
        i.e.
        \[ z_1(z_2 + z_3) = z_1z_2 + z_1z_3 .\]
    \item \underline{Fundamental Theorem of
        Algebra} A polynomial of degree $n$ with
        coefficients in $\CC$ can be written as a
        product of $n$ linear factors
        \[ P(z) = c_nz^n + \cdots + c_1z + c_0
        \qquad c_i \in \CC, c_n \neq 0 \]
        \[ = c_n(z - \alpha_1) \cdots (z -
        \alpha_n) \qquad \alpha_i \in \CC .\]
        Hence $P(z) = 0$ has at least one root
        \& $n$ roots counted with multiplicity.
    \item Addition \& Subtraction as parallelogram
        constructions:
        \begin{center}
        \begin{tsqx}
            (-1.5,0)--(1.5,0) EndArrow
            (0,-0.3)--(0,1.5) EndArrow
            label Re @ (1.5,-0.15)
            label Im @ (0.15,1.5)
            z_2 = (0.3,0.8)
            z_1 = (1.1,0.3)
            origin--z_1 EndArrow
            origin--z_2 EndArrow
            origin--z_1+z_2 EndArrow
            origin--z_2-z_1 EndArrow
            z_1--z_2--z_1+z_2--z_1 dashed
            z_2-z_1--z_2 dashed
            labelled z_1 + z_2
            labelled z_2 - z_1
        \end{tsqx}
        \end{center}
        Complex conjugation is reflection in real
        axis
        \begin{center}
        \begin{tsqx}
            (-0.3,0)--(1.5,0) EndArrow
            (0,-1.5)--(0,1.5) EndArrow
            label Re @ (1.5,-0.15)
            label Im @ (0.15,1.5)
            z := (1,0.8)
            label $z = x + iy$ @ z+0.1*dir(z)
            label $\overline{z} = z - iy$ @ (1,-0.8)+0.1*dir((1,-0.8))
            origin--z EndArrow
            origin--(1,-0.8) EndArrow
            z--(1,-0.8) dashed
        \end{tsqx}
        \end{center}
    \item \phantom{.}\\[-3\baselineskip]\noindent
\begin{proposition*}[Composition Property]
   Modulus / length obeys
   \[ |z_1z_2| = |z_1||z_2| \]
\end{proposition*}

\begin{proof}
    This result follows immediately by just
    expanding.
\end{proof}

\begin{proposition*}[Triangle Inquality]
    \[ |z_1 + z_2| \le |z_1| + |z_2| \]
\end{proposition*}

\begin{proof}
    Compare
    \[ LHS^2 = (z_1 + z_2)\ol{(z_1 + z_2)} \]
    \[ RHS^2 = |z_1|^2 + 2|z_1||z_2| + |z_2|^2 \]
    Compare ``cross terms'':
    \begin{align*}
        z_1\ol{z_2} + z_2\ol{z_1} &\le 2|z_1||z_2|
        \\
        \iff \half (z_1\ol{z_2} + \ol{(z_1\ol{z_2})})
        &\le |z_1||\ol{z_2}| \\
        \iff \mathrm{Re}(z_1\ol{z_2}) &\le
            |z_1\ol{z_2}|
    \end{align*}
    as desired.
\end{proof}

\begin{proposition*}[Alternative form of triangle
    inequality]
    Replace $z_1$ by $z_2 - z_1$ and rearrange to
    get
    \begin{align*}
        |z_2 - z_1| &\ge |z_2| - |z_1| \\
        \text{or} &\ge |z_1| - |z_2
    \end{align*}
    so
    \[ |z_2 - z_1| \ge \left| |z_2| - |z_1| \right| \]
\end{proposition*}
    \item \phantom{.}\\[-3\baselineskip]\noindent
        \begin{proposition*}
            $z_1 = r_1(\cos\theta_1 +
            i\sin\theta_1)$ and $z_2 =
            r_2(\cos\theta_2 + i\sin\theta_2)$
            implies that
            \[ z_1z_2 = r_1r_2(\cos(\theta_1 +
            \theta_2) + i\sin(\theta_1 + \theta_2)) \]
        \end{proposition*}
        \begin{proof}
            Just expand and apply trig formulae.
        \end{proof}
        \begin{theorem*}[De Moivre's Theorem]
            \[ (\cos\theta + i\sin\theta)^n
            = \cos n\theta + i \sin n \theta
            \qquad \forall n \in \ZZ \]
            (for $z \neq 0$, $z^0 = 1$ \& $z^{-n} =
            (z^{-1})^n$ for $n > 0$.)
        \end{theorem*}
        \begin{proof}
            Use the proposition above and induct.
        \end{proof}
\end{enumerate}

\subsection{Exponential \& Trigonometric Functions}

Define $\exp$, $\cos$, $\sin$ as functions on
$\CC$ by
\[ \exp(z) = e^z
= \sum_{n = 0}^\infty \frac{z^n}{n!} \]
\begin{align*}
    \cos(z) &= \half (e^{iz} + e^{-iz}) \\
    &= 1 - \frac{1}{2!} z^2 + \frac{1}{4!} z^4 +
    \cdots
\end{align*}
\begin{align*}
    \sin(z) &= \frac{1}{2i} (e^{iz} - e^{-iz}) \\
    &= z - \frac{1}{3!} z^3 + \frac{1}{5!} z^5 +
    \cdots
\end{align*}

\noindent These series converge $\forall z \in \CC$ and such
series can be multiplied, rearranged, and
differentiated. \\
Furthermore
\[ e^z e^w = e^{z + w} \]
From above
\[ e^0 = 1 \qquad \text{and} \qquad (e^z)^n =
e^{nz} \qquad n \in \ZZ \]
\begin{proof}
    Induction for positive integers, and for
    negative integers use
    \[ e^z e^{-z} = 1 \implies e^{-z} = (e^z)^{-1} \]
\end{proof}

\begin{lemma*}
    For $z = x + iy$
    \begin{enumerate}[(i)]
        \item $e^z = e^x(\cos y + i\sin y)$
        \item $\exp$ on $\CC$ takes all complex
            values except 0.
        \item $e^z = 1 \iff z = 2n\pi i$,
            $n \in \ZZ$.
    \end{enumerate}
\end{lemma*}

\begin{proof}
    \begin{enumerate}[(i)]
        \item $e^{x + iy} = e^x e^{iy}$ but
            $e^{iy} = \cos y + i\sin y$.
        \item $|e^z| = e^x$ takes all real values
            $> 0$. $\arg(e^z) = y$ taking all
            possible values.
        \item
            \begin{align*}
                e^z = 1 &\iff e^x = 1, \cos y = 1,
                    \sin y = 0 \\
                &\iff x = 0 \text{ and } y = 2\pi n
            \end{align*}
            as required.
    \end{enumerate}
\end{proof}

\noindent Returning to polar form or $\bmod$ / $\arg$ form
(Subsection 1.1 (v)), this can be written
\[ z = r(\cos\theta + i\sin\theta) = re^{i\theta} \]
for $r = |z|$ and $\theta = \arg(z)$. \\
De Moivre's Theorem now follows from
\[ (e^{i\theta})^n = e^{in\theta} .\]

\subsubsection*{Roots of unity}

$z$ is an $N$-th root of unity if $z^N = 1$. To
find all solutions:
\begin{align*}
    z = re^{i\theta} \text{ satisfies }& z^N = 1 \\
    \iff r^N e^{iN\theta} &= 1 \\
    \iff r^N = 1 \text{ and }& N\theta = 2n\pi
        \qquad n \in \ZZ
\end{align*}

\noindent This gives $N$ distinct solutions:
\begin{align*}
    z &= e^{2\pi n / N}
        \qquad n = 0, 1, \dots, N - 1 \\
    &= \cos \frac{2\pi n}{N} + i \sin \frac{2 \pi
    n}{N} \\
    &= \omega^n
\end{align*}
where $\omega = e^{2\pi / N}$.

\begin{center}
\begin{tsqx}
    \omega^0 = dir 0 R-45
    \omega^1 = dir 60
    \omega^2 = dir 120
    \omega^3 = dir 180 R-135
    \omega^4 = dir 240
    \omega^5 = dir 300
    unitcircle
    (-1.5,0)--(1.5,0) EndArrow
    (0,-1.5)--(0,1.5) EndArrow
    label Re @ (1.5,-0.15)
    label Im @ (0.15,1.5)
\end{tsqx}
\end{center}

\subsection{Transformations; lines \& circles}

Consider the following transformations on $\CC$
(maps $\CC \to \CC$).
\begin{align*}
    z &\mapsto z + a &&\text{translation by $a \in
    \CC$} \\
    z &\mapsto \lambda z &&\text{scaling by
    $\lambda \in \RR$} \\
    z &\mapsto e^{i\alpha}z &&\text{rotation by
    $\alpha \in \RR$} \\
    z &\mapsto \ol{z} &&\text{reflection in real
    axis} \\
    z &\mapsto \frac{1}{z} &&\text{inversion}
\end{align*}
\noindent
Consider general point on a \emph{line} in $\CC$
through $z_0$ and parallel to $w \neq 0$ (fixed
$z_0$, $w \in \CC$):
\begin{center}
\begin{tsqx}
    (-1.5,0)--(1.5,0) EndArrow
    (0,-0.5)--(0,1.5) EndArrow
    label Re @ (1.5,-0.15)
    label Im @ (0.15,1.5)
    z_0 = (-0.5,1)
    w = (0.75,0.3)
    z_0+w--z_0-w
    origin--z_0 EndArrow
    origin--w EndArrow
\end{tsqx}
\end{center}

\[ z = z_0 + \lambda w \]
for any real parameter $\lambda$. \\
To eliminate $\lambda$, take conjugate
\[ \ol{z} = \ol{z_0} + \lambda \ol{w} \]
and then combine
\[ \ol{w}z - w \ol{z} = \ol{w} z_0 - w \ol{z_0} \]
Consider general point on a \emph{circle} with
centre $c \in \CC$ and radius $\rho$:
\begin{center}
\begin{tsqx}
    (-1.5,0)--(1.5,0) EndArrow
    (0,-0.5)--(0,1.5) EndArrow
    label Re @ (1.5,-0.15)
    label Im @ (0.15,1.5)
    c ;= (1,0.5) R-90
    CR c 0.3
    P := c+0.3*dir(70)
    c--P EndArrow
    c-(0.4,0)--c+(0.4,0)
    anglemark4 c+(1,0) c c+dir(70) 3
    label $\alpha$ @ c+0.15*dir(35)
    label $\rho$ @ c+0.15*dir(90)
\end{tsqx}
\end{center}
%
\[ z = c + \rho e^{i\alpha} \qquad \text{for any
real $\alpha$} \]
Equivalently
\[ |z - c| = \rho \]
or $|z^2| - \ol{c}z - c \ol{z} = \rho^2 - |c|^2$.
(squaring sides above).
\smallskip\noindent
M\"obius transformations are generated by
translations, scalings, rotations and inversion.
They can be viewed as acting on
\[ \CC_{\infty} = \CC \cup \{\infty\} \]
which is geometrically a sphere (see IA Groups).

\subsection{Logarithms \& Complex Powers}

Define
\[ w = \log z \qquad z \in \CC, z \neq 0 \]
by
\[ e^w = \exp w = z \]
i.e. $\log$ is inverse of $\exp$ but $\exp$ is
many-to-one ($e^z = e^{z + 2n\pi i}$) and so
$\log$ is \emph{multi-valued}.
\[ z = re^{i\theta} = e^{\log r} e^{i\theta}
= e^{\log r + i\theta} \implies
\log z = \log(r + i\theta) = \log|z| + i\arg|z| \]
Multiple values of $\arg$ and $\log$ are related:
\[ \theta \to \theta + 2n\pi \]
\[ \log z \to \log z + 2n\pi i \]
where $n \in \ZZ$. To make them single valued we
can restrict e.g. $0 \le \theta < 2\pi$ or
$-\pi < \theta \le \pi$ (called the
\emph{principal value}).

\begin{example*}
    \[ z = -3i = 3(-i) = e^{\log 3} e^{-i\pi/2 +
    2n\pi i} = e^{\log 3 - i\pi/2 + 2n\pi i} \]
    Hence
    \[ \log z = \log 3 - \frac{i\pi}{2} + 2n\pi i \]
    \[ \arg z = \begin{cases}
        3 \pi/2 & \text{ if we use $0 \le \theta <
        2\pi$} \\
        -\pi/2 & \text{ if we use $-\pi <
        \theta \le \pi$}
    \end{cases} \]
\end{example*}

We define \emph{complex powers} by
\[ z^\alpha = e^{\alpha \log z} \qquad z \in \CC,
z \neq 0 \& \alpha \in \CC \]
This is multi-valued in general under the change
$\arg z \to \arg z + 2n \pi$
\[ z^\alpha \to z^\alpha e^{2\pi in\alpha} \]
\begin{enumerate}[(i)]
    \item If $\alpha = P \in \ZZ$ then $z^\alpha =
        z^p$ unique.
    \item If $\alpha = \frac{p}{q} \in \QQ$, then
        $z^\alpha = z^{p/q}$ takes finitely many
        values.
\end{enumerate}
but in general we have \emph{infinitely} many
values.

\subsubsection*{Examples}

\begin{itemize}
    \item $(1 + i)^{1/2}$: $1 + i = \sqrt{2}
        e^{i\pi/4} = e^{\half\log 2 + i\pi/4}$
        Hence
        \begin{align*}
            \log(1 + i) &= \half \log 2 +
            \frac{i\pi}{4} + 2n\pi i \\
            \implies (1 + i)^{1/2} &=
            e^{\half\log(1 + i)} \\
            &= e^{\frac{1}{4}\log 2 + i\pi/8
            + n\pi i} \\
            &= 2^{1/4}e^{i\pi/8}(-1)^n
        \end{align*}
        \begin{center}
        \begin{tsqx}
            (-1.5,0)--(1.5,0) EndArrow
            (0,-1.5)--(0,1.5) EndArrow
            label Re @ (1.5,-0.15)
            label Im @ (0.15,1.5)
            origin--(1,1) EndArrow
            label $1 + i$ @ (1.1,1.1)
            X := 1.189*dir(22.5)
            origin--X EndArrow
            origin--origin-X EndArrow
            anglemark4 (1,0) origin X 5
            anglemark4 (-1,0) origin -X 5
            label $(1 + i)^{1/2}$ @ (0.1,0.1)+X
            label $(1 + i)^{1/2}$ @ (-0.1,-0.1)-X
        \end{tsqx}
        \end{center}
    \eqitem \begin{align*}
        (-3i)^i &= e^{i \log(-3i)} \\
        &= e^{i(\log 3 - i\pi/2 + 2n\pi i)} \\
        &= e^{i\log 3} e^{\pi/2 - 2n\pi}
        \qquad n \in \ZZ
    \end{align*}
\end{itemize}

\section{Vectors in 3 Dimensions}

A vector is a quantity with magnitude and
direction (e.g. force, electric and magnetic
fields) - all examples modelled on
\emph{position}. \\\\
Take geometrical approach to position vectors in
3D space based on standard (Euclidean) notions of
points, lines, planes, length, angle etc. Choose
point $O$ as the origin, then points $A$, $B$ have
position vectors
\[ \ul{a} = \vec{OA}, \qquad \ul{b} = \vec{OB} \]
\begin{center}
\begin{tsqx}
    ! size(5cm);
    O = origin R-90
    B = (1,1)
    A = (2,0.3)
    O--A EndArrow
    O--B EndArrow
    label $\underline{a}$ @ 0.5*A-(0.02,0.1)
    label $\underline{b}$ @ 0.5*B+(-0.1,0.1)
\end{tsqx}
\end{center}
lengths denoted by $|\ul{a}| = |\vec{OA}|$. Also,
$\ul{o}$ is the position vector for $O$.

\subsection{Vector Addition and Scalar
Multiplication}

\begin{enumerate}[(i)]
    \item \ul{Scalar Multiplication} Given
        $\ul{a}$, position vector for $A$, and a
        \emph{scalar} $\lambda \in \RR$, $\lambda
        \ul{a}$ is position vector of point $A'$
        on $OA$ with
        \[ |\lambda \ul{a}| = |\vec{OA'}|
        = |\lambda| |ul{a}| \]
        as shown
        \begin{center}
        \begin{tsqx}
            O .= origin
            A = dir 30 R130
            A' = 2.5*A
            label $A'$ @ O-A'+0.15*dir(210)
            label $\underline{a}$ @ 0.5*A+(0.1,-0.1)
            label $\lambda\underline{a}, \lambda > 0$ @ A'+(0.1,-0.3)
            label $\lambda\underline{a}, \lambda < 0$ @ O-A'+(0.5,-0.1)
            O--A EndArrow
            O--O-A' EndArrow
            A--A' EndArrow
        \end{tsqx}
        \end{center}
        Say $\ul{a}$ and $\ul{b}$ are
        \emph{parallel}, $\ul{a} \parallel \ul{b}$
        iff $\ul{a} = \lambda \ul{b}$ or $\ul{b} =
        \lambda \ul{a}$. This definition allows
        $\lambda < 0$, and $\lambda = 0$ so
        $\ul{a} \parallel \ul{o}$ for any
        $\ul{a}$.
    \item Given $\ul{a}$, $\ul{b}$ position
        vectors of $A$, $B$, construct a
        parallelogram $OACB$
        \begin{center}
        \begin{tsqx}
            O = origin R225
            B = (0.5,1)
            A = (2,0)
            C = A+B
            O->B
            O->C
            O->A
            B--C--A dashed
            label $\underline{b}$ @ 0.5*B+0.1*dir(170)
            label $\underline{c}$ @ 0.5*C+0.1*dir(120)
            label $\underline{a}$ @ 0.5*A+0.1*dir(-90)
        \end{tsqx}
        \end{center}
        and define $\ul{a} + \ul{b} = \ul{c}$,
        position vector of point $C$ provided
        $\ul{a} \not\,\parallel \ul{b}$; if $\ul{a}
        \parallel \ul{b}$ then we can write
        $\ul{a} = \alpha \ul{u}$, $\ul{b} = \beta
        \ul{u}$ for some $\ul{u}$, and then
        \[ \ul{a} + \ul{b} = (\alpha + \beta)
        \ul{u} \]
    \item \ul{Properties} For any vectors $\ul{a},
        \ul{b}, \ul{c}$
        \[ \ul{a} + \ul{o} = \ul{o} + \ul{a}
        = \ul{a} \]
        so $\ul{o}$ is the identity for $+$.
        We also have that there exists some
        $-\ul{a}$ such that
        \[ \ul{a} = (-\ul{a}) = (-\ul{a}) + \ul{a}
        = \ul{o} \]
        so there exists an inverse of every
        vector. We also have
        \[ \ul{a} + \ul{b} = \ul{b}+ \ul{a} \]
        so $+$ is commutative. It is also
        associative, i.e.
        \[ \ul{a} + (\ul{b} + \ul{c})
        = (\ul{a} + \ul{b}) + \ul{c} \]
        We also have the following properties
        \[ \lambda(\ul{a} + \ul{b}) =
        \lambda\ul{a} + \lambda\ul{b} \]
        \[ (\lambda + \mu)\ul{a} = \lambda\ul{a} +
        \mu\ul{a} \]
        \[ \lambda(\mu\ul{a}) = (\lambda\mu)\ul{a} \]
        All can be checked geometrically i.e.
        associativity from parallelepiped.
    \item \ul{Linear Combinations and Span}
        A \emph{linear combination} of vectors
        $\ul{a}, \ul{b}, \dots, \ul{c}$ is an
        expression
        \[ \alpha\ul{a} + \beta\ul{b} + \cdots
        + \gamma\ul{c} \]
        for some $\alpha, \beta, \dots, \gamma \in
        \RR$.
        The \emph{span} of a set of vectors is
        \begin{align*}
            \mathrm{span}\{\ul{a}, \ul{b}, \dots, \ul{c}\}
            \\
            = \{\alpha \ul{a} + \beta\ul{b} +
            \cdots + \gamma\ul{c} : \alpha, \beta,
            \dots, \gamma \in \RR \}
        \end{align*}
        If $\ul{a} \neq \ul{a}$ then
        $\mathrm{span}\{\ul{a}\} =
        \{\lambda\ul{a}\}$, i.e. the \emph{line}
        through $O$ and $A$.
        If $\ul{a} \not\,\parallel \ul{b}$ then
        \[ \mathrm{span}\{\ul{a}, \ul{b}\}
        = \{\alpha \ul{a} + \beta\ul{b} : \alpha,
        \beta \in \RR\} \]
        i.e. the \emph{plane} through $O$, $A$ and
        $B$.
\end{enumerate}

\subsection{Scalar or Dot Product}

\begin{enumerate}[(i)]
    \item \ul{Definition}: Given $\ul{a}$ and
        $\ul{b}$ let $\theta$ be the angle between
        them; then
        \[ \ul{a} \cdot \ul{b} = |\ul{a}||\ul{b}|
        \cos\theta \]
        \begin{center}
        \begin{tsqx}
            ! size(5cm);
            O = origin R180
            A := dir -20
            B := 0.8*dir(30)
            label $\underline{a}$ @ A+0.05*dir(-20)
            label $\underline{b}$ @ B+0.05*dir(30)
            O->B
            O->A
            anglemark4 A O B 5
            label $\theta$ @ 0.2*dir(5)
        \end{tsqx}
        \end{center}
        \emph{scalar} or \emph{dot product} or
        \emph{inner product}
        ($\theta$ defined unless $|\ul{a}|$ or
        $|\ul{b}| = 0$ and then $\ul{a} \cdot
        \ul{b} = 0$.)
    \item \ul{Properties}
        \[ \ul{a} \cdot \ul{b} = \ul{b} \cdot
        \ul{a} \]
        \[ \ul{a} \cdot \ul{a} = |\ul{a}|^2 \ge 0
        \& = 0 \text{ iff } \ul{a} = 0 \]
        \[ (\lambda \ul{a})\cdot\ul{b} =
        \lambda(\ul{a}\cdot\ul{b}) =
        \ul{a}\cdot(\lambda\ul{b}) \]
        \[ \ul{a} \cdot(\ul{b} + \ul{c}) =
        \ul{a}\cdot\ul{b} + \ul{a}\cdot\ul{c} \]
    \item \ul{Interpretation}
        For $\ul{a} \neq 0$, consider $\ul{u} =
        \frac{\ul{a}}{|\ul{a}|}$
        \[ \ul{u} \cdot \ul{b} =
        \frac{1}{\ul{a}}\ul{a}\cdot\ul{b} =
        |\ul{b}| \cos\theta \]
        is \emph{component} of $\ul{b}$ along
        $\ul{a}$.
        \begin{center}
        \begin{tsqx}
            O := origin
            (-0.2,0)--(0,0)
            (0,-0.2)--(0,0)
            B := (1.3,1.3)
            label $\underline{b}$ @ B+0.1*dir(45)
            A := (0,0.7)
            label $\underline{a}$ @ A+(-0.1,0)
            O->A
            A->(0,1.3)
            O->(1.3,0)
            (1.3,0)->(1.5,0)
            (0,1.3)->(0,1.5)
            label $\underline{b}_\perp$ @ (1.3,-0.1)
            label $\underline{b}_\parallel$ @ (-0.1,1.3)
            O->B
            (0,1.3)--B--(1.3,0) dashed
            anglemark4 B O A 5
            label $\theta$ @ 0.2*dir(67)
        \end{tsqx}
        \end{center}
        We can \emph{resolve} $\ul{b} =
        \underbrace{\ul{b}_\parallel}_{\parallel
        \ul{a}} + \underbrace{\ul{b}_\perp}_{
        \perp \ul{a}}$
        where $\ul{a} \perp \ul{b}$ iff $\ul{a}
        \cdot \ul{b} = 0$. Note $\ul{a} \cdot
        \ul{b} = \ul{a} \cdot \ul{b}_\parallel$.
        (The expressions can be computed as
        $\ul{b}_\parallel = (\ul{b} \cdot \ul{u})
        \ul{u}$, $\ul{b}_\perp = \ul{b} - (\ul{b}
        \cdot \ul{u}) \ul{u}$.
\end{enumerate}

In general, vectors $\ul{a}$ and $\ul{b}$ are
\emph{orthogonal} or \emph{perpendicular}, written
\[ \ul{a} \perp \ul{b} \iff \ul{a} \cdot \ul{b}
= 0 \]
definition allows $\ul{a}$ or $\ul{b}$ = $\ul{o}$;
$\ul{o} \perp \text{any vector}$.

\subsection{Orthonormal Bases and Components}

Choose vectors $\ul{e_1}$, $\ul{e_2}$, $\ul{e_3}$
that are \emph{orthonormal} i.e. each of unit
length and mutually perpendicular.
\[ \ul{e_j} \cdot \ul{j} = \begin{cases}
    1 & \text{if $i = j$} \\
    0 & \text{if $i \neq j$}
\end{cases} \]
Equivalent to choosing Cartesian axes along these
directions, $\{\ul{e_i}\}$ is a \emph{basis}: any
vector can be expressed
\[ \ul{a} = \sum_i a_i\ul{e_i} = a_1\ul{e_1} +
a_2\ul{e_2} = a_3\ul{e_3} \]
and each component $a_i$ is \emph{uniquely}
determined.
\[ a_i = \ul{e_i} \cdot \ul{a} \]
\begin{center}
\begin{tsqx}
    ! size(5cm);
    O := origin
    O->>(0,1)
    O->>(1,0)
    O->>dir(-30)
    (1,0)--(2.5,0)
    (0,1)--(0,2.5)
    dir(-30)--2.5*dir(-30)
    label $\underline{e_3}$ @ (-0.2,1)
    label $\underline{e_2}$ @ (1,0.2)
    label $\underline{e_1}$ @ rotate(-30)*(1,-0.2)
    (-0.5,0)--O dashed
    A := (1.1,2.2)
    label $\underline{a}$ @ A+(0.2,0)
    O->A
    A--(-0.5,2.2) dashed
    label $a_3$ @ (-1,1.1)
    brace (-0.5,0) (-0.5,2.2)
\end{tsqx}
\end{center}

Each $\ul{a}$ can now be identified with set of
components in
\[ \ub{(a_1, a_2, a_3)}_{\text{row vector}} \qquad
\text{or} \qquad
\ub{\begin{pmatrix}
    a_1 \\ a_2 \\ a_3
\end{pmatrix}}_{\text{column vector}} \]
Note
\begin{align*}
    \ul{a} \cdot \ul{b} 
    &= asdf \left( \sum_i
    a_i\ul{e_i} \right) \cdot \left( \sum_j b_j
    \ul{e_j} \right) \\
    &= a_1b_1 + a_2b_2 + a_3b_3 \\
    \text{and} |\ul{a}|^2 &= a_1^2 + a_2^2 + a_3^2
    \qquad \text{Pythagoras}
\end{align*}
$\ul{e_1}$, $\ul{e_2}$, $\ul{e_3}$ are also often
written $\ul{i}$, $\ul{j}$, $\ul{k}$.

\subsection{Vector or Cross Product}

\begin{definition*}
    Given $\ul{a}$ and $\ul{b}$, let $\theta$ be
    angle between them measured in sense shown
    relative to a unit normal $\ul{n}$ to the plan
    they span
    \begin{center}
    \begin{tsqx}
        ! size(5cm);
        B := 0.9*dir(15)
        A := dir(-20)
        N := 0.3*dir(90)
        O := origin
        O->>N
        O->>A
        O->>B
        label $\underline{b}$ @ B+0.05*dir(B)
        label $\underline{a}$ @ A+0.05*dir(A)
        label $\underline{n}$ @ N+0.05*dir(N)
        arc3 O 0.15*dir(-20) 0.15*dir(15) EndArcArrow(SimpleHead)
        label $\theta$ @ 0.2*dir(-2.5)
    \end{tsqx}
    \end{center}
    ``right-handed sense''. (unit normal $\equiv$
    unit vector $\perp$ plane); \\
    then
    \[ \ul{a} \times \ul{b} = |\ul{a}||\ul{b}|
    \sin\theta \ul{n} \]
    (sometimes $\wedge$ is used instead of
    $\times$) is \emph{vector} or \emph{cross}
    product. \\
    Note $\emph{n}$ is defined up to a choice of
    sign if $\ul{a} \not\,\parallel \ul{b}$, but
    changing sign of $\ul{n}$ means changing
    $\theta$ to $2\pi - \theta$ so definition is
    undefined; $\ul{n}$ is not defined it $\ul{a}
    \parallel \ul{b}$, and $\theta$ is not defined
    it $|\ul{a}|$ or $|\ul{b}| = 0$, but $\ul{a}
    \times \ul{b} = \ul{o}$ in these cases.
\end{definition*}

\subsubsection*{Properties}

\[ \ul{a} \times \ul{b} = -\ul{b} \times \ul{a} \]
\[ (\lambda \ul{a}) \times \ul{b} = \lambda
(\ul{a} \times \ul{b}) = \ul{a} \times (\lambda
\ul{b}) \]
\[ \ul{a} \times (\ul{b} + \ul{c}) = \ul{a} \times
\ul{b} + \ul{a} \times \ul{c} \]
\[ \ul{a} \times \ul{b} = \ul{o} \iff \ul{a}
\parallel \ul{b} \]
\[ \ul{a} \times \ul{b} \perp \ul{a} \& \ul{b} \]
\[ \ul{a} \cdot (\ul{a} \times \ul{b}) = \ul{b}
\cdot (\ul{a} \times \ul{b}) = 0 .\]

\subsubsection*{Interpretations}

\begin{itemize}
    \item $\ul{a} \times \ul{b}$ is the \emph{vector} area
    of the parallelogram shown
    \begin{center}
    \begin{tsqx}
        O := origin
        A := (2,0)
        B := (0.5,0.8)
        label $\underline{a}$ @ A-(0,0.1)
        label $\underline{b}$ @ B+(-0.1,0.1)
        anglemark4 A O B 4
        O->>B
        O->>A
        X := foot B O A
        B--A+B--A dashed
        B--X dashed
        rightanglemark4 A X B 4
        label $\theta$ @ 0.2*dir(27)
    \end{tsqx}
    \end{center}
    \[ \ul{a} \times \ul{b} =
    |\ul{a}||\ul{b}|\sin\theta \qquad\text{for
    $\sin\theta \ge 0$} = \text{``base''} \times
    \text{``$\perp$ height''} \]
    \emph{scalar} area \\
    Direction of normal $\ul{n}$ gives orientation of
    parallelogram in space.
\item Fix $\ul{a}$ and consider $\ul{x} \perp
    \ul{a}$; then $\ul{x} \mapsto \ul{a} \times
    \ul{x}$ scales $|\ul{x}|$ by a factor of
    $|\ul{a}|$ and rotates $\ul{x}$ by $\pi/2$ in
    plane $\perp \ul{a}$ as shown.
    \begin{center}
    \begin{tsqx}
        ! size(5cm);
        A := (-0.8,0.8)
        B := (-1.2,-0.8)
        C := (0.8,-0.8)
        D := (1.2,0.8)
        A--B--C--D--A
        X := (-0.6,-0.48)
        AX := (0.8,-0.324)
        A' := (0,0.96)
        O := origin
        O->>X
        O->>AX
        O->>A'
        label $\underline{x}$ @ X+0.1*dir(X)
        label $\underline{a}$ @ A'+0.1*dir(A')
        label $\underline{a} \times \underline{x}$ @ AX+(0,0.1)
        arc3 O 0.2*dir(X) 0.2*dir(AX) EndArcArrow(SimpleHead)
        label $\frac{\pi}{2}$ @ 0.3*dir(-85)
    \end{tsqx}
    \end{center}
\end{itemize}

\subsubsection*{Component Expressions}

Consider $\ul{e_1}$, $\ul{e_2}$, $\ul{e_3}$
orthonormal basis as in section 2.3 but assume in
addition that
\[ \ul{e_1} \times \ul{e_2} = \ul{e_3} = -\ul{e_2}
\times \ul{e_1} \]
\[ \ul{e_1} \times \ul{e_3} = \ul{e_1} = -\ul{e_3}
\times \ul{e_2} \]
\[ \ul{e_3} \times \ul{e_1} = \ul{e_2} = -\ul{e_1}
\times \ul{e_3} \]
(all equalities from any one)
This is called a \emph{right-handed} orthonormal
basis. Now for
\[ \ul{a} = \sum_i a_i\ul{e_i} = (a_1 \ul{e_1} +
a_2\ul{e_2} + a_3\ul{e_3}) \]
and
\[ \ul{b} = \sum_j b_j\ul{e_j} = (b_1\ul{e_1} +
b_2\ul{e_2} + b_3\ul{e_3}) \]
we get
\begin{align*}
    \ul{a} \times \ul{b}
    &= (a_2b_3 - a_3b_2)\ul{e_1} \\
    &\,\,\,\, + (a_3b_1 - a_1b_3) \ul{e_2} \\
    &\,\,\,\, + (a_1b_2 - a_2b_1) \ul{e_3}
\end{align*}

\subsection{Triple Products}

\subsubsection*{Scalar Triple Product}

\begin{notation*}
    Define
    \begin{align*}
        &\,\,\,\,\,\,\,\,\ul{a} \cdot (\ul{b} \times \ul{c})
        = \ul{b} \cdot (\ul{c} \times \ul{a})
        = \ul{c} \cdot (\ul{a} \times \ul{b}) \\
        &= -\ul{a} \cdot (\ul{c} \times \ul{b})
        = -\ul{b} \cdot (\ul{a} \times \ul{c})
        = -\ul{c} \cdot (\ul{b} \times \ul{a}) \\
        &= [\ul{a}, \ul{b}, \ul{c}]
    \end{align*}
\end{notation*}

Interpretation: $|\ul{c} \cdot (\ul{a} \times
\ul{b})|$ is volume of parallelepiped shown $ =
(\text{area of parallelogram base}) \times (\perp
\text{ height}) = |\ul{a} \times
\ul{b}||c||\cos\phi|$
\begin{center}
\begin{tsqx}
    A := (1.5,0)
    B := (0.5,0.5)
    C := (0.35,0.8)
    AB := (0,1)
    label $\underline{a}$ @ A+(0,-0.1)
    label $\underline{b}$ @ B+(0.1,-0.1)
    label $\underline{c}$ @ C+(0.1,-0.1)
    label $\underline{a} \times \underline{b}$ @ AB+(-0.25,0)
    O := origin
    O->>AB
    O->>A
    O->>B
    O->>C
    A--A+B--B--B+C--C--A+C--A dashed
    A+C--A+B+C--B+C dashed
    A+B--A+B+C dashed
\end{tsqx}
\end{center}
$\ul{c} \cdot \ul{a} \times \ul{b}$ ``signed
volume''; if $\ul{c} \cdot \ul{a} \times \ul{b} >
0$ say $\ul{a}$, $\ul{b}$, $\ul{c}$ right-handed
set.

\begin{remark*}
    $\ul{a} \cdot \ul{b} \times \ul{c} = 0$ if and
    only if $\ul{a}$, $\ul{b}$ and $\ul{c}$ are
    \emph{co-planar} meaning one of them lies in
    plane spanned by other two.
    For example $\ul{c} = \alpha \ul{a} + \beta
    \ul{b}$ belonging to $\mathrm{span}\{\ul{a},
    \ul{b}\}$.
\end{remark*}

\begin{example*}
    \[ \ul{a} = (2,0,-1) \qquad \ul{b} = (7,-3, 5) \]
    \begin{align*}
        \implies \ul{a} \times \ul{b}
        &= (0.5-(-1)(-3))\ul{e_1} \\
        &\,\,\,\, + ((-1)\cdot 7 - 2.5) \ul{e_2}
        \\
        &\,\,\,\, + (2 \cdot (-3) - 0.7) \ul{e_3}
        \\
        &= (-3,-17, -6)
    \end{align*}
    Test whether $\ul{a}$, $\ul{b}$, $\ul{c}$
    coplanar with $\ul{c} = (3, -3, 7)$
    \[ \ul{c} \cdot \ul{a} \times \ul{b} = 3(-3) +
    (-3)(-17) + 7(-6) = 0 ;\]
    consistent with $\ul{c} = \ul{b} - 2\ul{a}$.
\end{example*}

\subsubsection*{Vector Triple Product}

\[ \ul{a} \times (\ul{b} \times \ul{c}) = (\ul{a}
\cdot \ul{c})\ul{b} - (\ul{a} \cdot \ul{b})\ul{c} \]
\[ (\ul{a} \times \ul{b}) \times \ul{c} = (\ul{a}
\cdot \ul{c})\ul{b} - (\ul{b} \cdot \ul{c})\ul{a} \]
Form of RHS is constrained by definitions above,
or could check explicitly. Return to these
formulas using index notation and summation
convention.

\subsection{Lines, Planes and Other Vector Equations}

\begin{enumerate}[(a)]
    \item \ul{Lines} \\
        General point on a line through
        $\ul{a}$ with direction $\ul{u} (\neq
        \ul{o})$ has position vector
        \[ \ul{r} = \ul{a} = \lambda \ul{u}
        \qquad \lambda \in \RR \]
        parametric form
        \begin{center}
        \begin{tsqx}
            O := origin
            label $\underline{o}$ @ O+0.1*dir(-100)
            A := (-0.5,1)
            R := (1,1.4)
            O->>A
            O->>R
            A->>0.7*A+0.3*R
            A--1.3*A-0.3*R
            0.7*A+0.3*R->>R
            R--1.3*R-0.3*A
            label $\underline{a}$ @ A+0.1*dir(-135)
            label $\underline{r}$ @ R+0.1*dir(-50)
            label $\underline{u}$ @ 0.7*A+0.3*R+0.1dir(100)
            label $\lambda\underline{u}$ @ R+0.1*dir(100)
        \end{tsqx}
        \end{center}
        Alternative form without parameter
        $\lambda$ obtained by crossing with
        $\ul{u}$:
        \[ \ul{u} \times \ul{r} = \ul{u} \times
        \ul{a} \]
        Conversely
        \[ \ul{u} \times (\ul{r} - \ul{a}) =
        \ul{o} \]
        and this holds if and only if
        \[ \ul{r} - \ul{a} = \lambda \ul{u} \]
        for some real $\lambda$.
        Now consider
        \[ \ul{u} \times \ul{r} = \ul{C} \]
        where $\ul{u}$, $\ul{c}$ are given vectors
        with $\ul{u} \neq \ul{o}$. Note that 
        \[ \ul{u} \cdot (\ul{u} \times \ul{r}) =
        \ul{u} \cdot \ul{c} = 0 \]
        If $\ul{u} \cdot \ul{c} \neq 0$ then
        we hae a contradiction i.e. no solutions.
        If $\ul{u} \cdot \ul{c} = 0$, try a
        particular solution by considering
        \[  \ul{u} \times (\ul{u} \times \ul{c}) =
        (\ul{u} \cdot \ul{c})\ul{u} - (\ul{u}
        \cdot \ul{u})\ul{c}
        = -|\ul{u}|^2\ul{c} \]
        Hence
        \[ \ul{a} = -\frac{1}{|\ul{u}|^2}(\ul{u}
        \times \ul{c}) \]
        is a solution. General solution (arguing
        as before) is
        \[ \ul{r} = \ul{a} + \lambda \ul{u} \]
    \item \ul{Planes} \\
        General point on a plane through $\ul{a}$
        with directions $\ul{u}$, $\ul{v}$ in
        plane ($\ul{u} \not\,\parallel \ul{v}$)
        has position vector
        \[ \ul{r} = \ul{a} + \lambda \ul{u} +
        \mu\ul{v} \qquad \lambda, \mu \in \RR \]
        parametric form
        \begin{center}
        \begin{tsqx}
            W := (-1.6,1)
            X := (0.6,1)
            Y := (1.6,2.5)
            Z := (-0.6,2.5)
            W--X--Y--Z--W
            O .= origin
            A := (-0.5,1.5)
            U := A+(0.3,0)
            V := A+(0.1,0.35)
            O->>A
            R := A+(1,0)+(0.2,0.7)
            O->>R
            A->>U
            U->>A+(1,0)
            A->>V
            V->>A+(0.2,0.7)
            A+(0.2,0.7)--R--A+(1,0) dashed
            label $\underline{a}$ @ A+(-0.1,0.1)
            label $\underline{v}$ @ V+(-0.05,0.1)
            label $\underline{u}$ @ U+(0,-0.1)
        \end{tsqx}
        \end{center}
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/c1df80022e6a11ec.png}
        \end{center}
        Alternative form without parameters
        obtained by dotting with normal
        \[ \ul{n} = \ul{u} \times \ul{v} \neq
        (\ul{o} \text{ since $\ul{u}
        \not\,\parallel \ul{v}$ but not necessarily a
        unit vector}) \]
        This gives
        \[ \ul{n} \cdot \ul{r} = \ul{n} \cdot \ul{a} =
        k \]
        where $k$ is a constant. Note component of
        $\ul{r}$ along $\ul{n}$ is
        \[ \frac{\ul{n} \cdot \ul{r}}{|\ul{n}|}
        = \frac{k}{|\ul{n}|} \qquad \text{(constant)} \]
        is clearly a plane and moreover
        $\frac{|k|}{|\ul{n}|}$ is perpendicular
        distance of plane from $\ul{o}$.
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/51981eca2e6b11ec.png}
        \end{center}
    \item \ul{Other Vector Equations} \\
        Consider equations for $\ul{r}$ (unknown)
        written in vector notation with given
        (constant) vectors. Possible approaches:
        \begin{itemize}
            \item Can re-write and convert to some
                standard form, e.g.
                \[ |\ul{r}|^2 + \ul{r} \cdot
                \ul{a} = k, \qquad \text{constant} \]
                Then we can complete the square:
                \[ |\ul{r} + \half\ul{a}|^2 =
                (\ul{r} + \half\ul{a}) \cdot
                (\ul{r} + \half \ul{a}) = k +
                \frac{1}{4} |\ul{a}|^2 \]
                Equation of a sphere, centre
                $-\half \ul{a}$ and radius $\sqrt{k +
                \frac{1}{4} |\ul{a}|^2}$, provided
                $k + \frac{1}{4} |\ul{a}|^2 > 0$.
                For equations linear in $\ul{r}$.
            \item Try dotting and crossing with
                constant vectors to learn more
                (see examples).
            \item Can try expressing
                \[ \ul{r} = \alpha\ul{a} +
                \beta\ul{b} + \gamma\ul{c} \]
                for some non-$\omega$-planar
                $\ul{a}$, $\ul{b}$, $\ul{c}$ and
                solve for $\alpha$, $\beta$,
                $\gamma$.
            \item Can choose basis and use index /
                matrix notation.
        \end{itemize}
\end{enumerate}

\subsection{Index (suffix) Notation and the Summation Convention}

\begin{enumerate}[(a)]
    \item \ul{Components; $\delta$ and
        $\varepsilon$} \\
        Write vectors $\ul{a}$, $\ul{b}$, \dots in
        terms of components. $a_i$, $b_i$, \dots,
        with respect to an orthonormal,
        right-handed basis
        \[ \ul{e_1}, \ul{e_2}, \ul{e_3} \]
        Indices or suffices $i, j, k, l, p, q,
        \dots$ take values 1, 2, 3. Then
        \[ \ul{c} = \alpha \alpha \ul{a} +
        \beta\ul{b} \]
        \[ \iff c_i = [\alpha \ul{a} +
        \beta\ul{b}] = \alpha a_i \beta b_i \]
        for $i = 1, 2, 3$ (\emph{free index})
        \[ \ul{a} \cdot \ul{b} = \sum_i a_ib_i =
        \sum_j a_jb_j \]
        \[ \ul{x} = \ul{a} = (\ul{b} \cdot \ul{c})
        \ul{d}\]
        for $j = 1, 2, 3$ free index.
        \[ \iff x_j = a_j + \left(\sum_k b_k
        c_k\right)d_j \]
        \begin{definition*}[Kronecker Delta]
            \[ \delta_{ij} = \begin{cases}
                1 & \text{if $i = j$} \\
                0 & \text{if $i \neq j$}
            \end{cases} \]
            \[ \delta{ij} = \delta{ji} \qquad
            \text{(symmetric)} \]
            As an asdfasdf matrix
            \[ 
            \begin{pmatrix}
                \delta_{11} & \delta{12} & \delta_{13} \\
                \delta_{21} & \delta_{22} & \delta_{23} \\
                \delta_{31} & \delta_{32} & \delta_{33}
            \end{pmatrix}
            =
            \begin{pmatrix}
                1 & 0 & 0 \\
                0 & 1 & 0 \\
                0 & 0 & 1
            \end{pmatrix}
            \]
            Then
            \[ \ul{e_i} \cdot \ul{j} = \delta_{ij} \]
            and
            \begin{align*}
                \ul{a} \cdot \ul{b}
                &= \left(\sum_i a_i
                \ul{e_i} \right) \cdot \left( \sum_j
                b_j \ul{e_j} \right) \\
                &= \sum_{ij} a_ib_j \ul{e_i} \cdot
                \ul{e_j} \\
                &= \sum_{ij} a_ib_j \delta_{ij} \\
                &= \sum_i a_ib_i
            \end{align*}
        \end{definition*}
        \begin{definition*}[Levi-Civita Epsilon]
            \[ \varepsilon_{ijk} = \begin{cases}
                +1 & \text{if $(i, j, k)$ even
                permutation of $(1, 2, 3)$} \\
                -1 & \text{if $(i, j, k)$ odd
                permutation of $(1, 2, 3)$} \\
                0 & \text{else}
            \end{cases} \]
            i.e.
            \[ \varepsilon_{123} =
            \varepsilon{231} = \varepsilon_{312} =
            +1 \]
            \[ \varepsilon_{321} =
            \varepsilon_{213} = \varepsilon_{132}
            = -1 \]
            \[ \varepsilon_{ijk} = 0 \qquad
            \text{if any two index values match} \]
            Note that $\varepsilon_{ijk}$ is
            \emph{totally} antisymmetric:
            exchanging any pair of indices
            produces a change in sign. Then
            \[ \ul{e_i} \times \ul{e_j} = \sum_k
            \varepsilon_{ijk} \ul{e_k} \]
            e.g.
            \[ \ul{e_2} \times \ul{e_1} = \sum_k
            \varepsilon_{21k} \ul{e_k}
            = \varepsilon_{213} \ul{e_3} \]
            And
            \begin{align*}
                \ul{a} \times \ul{b} = \left(
                \sum_i a_i\ul{e_i} \right)  \times
                \left( \sum_j b_j \ul{e_j} \right)
                \\
                &= \sum_{ij} a_ib_j \ul{e_i}
                \times \ul{e_j} \\
                &= \sum_{ij} a_ib_j \sum_k
                \varepsilon_{ijk} \ul{e_k} \\
                &= \sum_k \left( \sum_{ij}
                \varepsilon_{ijk} a_ib_j \right)
                \ul{e_k}
            \end{align*}
            Hence
            \[ (\ul{a} \times \ul{b})_k =
            \sum_{ij} \varepsilon_{ijk} a_ib_j \]
            e.g.
            \begin{align*}
                (\ul{a} \times \ul{b})_3
                &= \sum_{ij} \varepsilon_{ij3}
                a_ib_j \\
                &= \varepsilon_{123} a_1b_2 +
                \varepsilon_{213} a_2b_1 \\
                &= a_1b_2 - a_2b_1
            \end{align*}
        \end{definition*}
    \item \ul{Summation Convention} \\
        With component / index notation, we
        observe that indices that appear
        \emph{twice} in a given term are (usually)
        summed over. In the summation convention
        we \emph{omit} $\sum$ signs for repeated
        indices: the sum is understood. \\
        \ul{Examples}
        \begin{enumerate}[(i)]
            \eqitem \begin{align*}
                &a_i \delta_{ij} \qquad
                \text{$\sum_i$ understood} \\
                =&a_1\delta_{1j} + a_2\delta_{2j}
                + a_3\delta_{3j} \\
                =&\begin{cases}
                    a_1 &\text{if $j = 1$} \\
                    a_2 &\text{if $j = 2$} \\
                    a_3 &\text{if $j = 3$}
                \end{cases}
            \end{align*}
                \emph{or}
                \[ a_i\delta_{ij} = a_j \]
                true for $j = 1, 2, 3$.
            \item Here on the first line we have
                $\sum_{i, j}$ is understood, and
                on the second line we have the
                $\sum_i$ is understood
            \begin{align*}
                \ul{a} \cdot \ul{b}
                &= \delta_{ij} a_ib_j \\
                &= a_ib_i
            \end{align*}
            \item Here $\sum_{j, k}$ is understood
                \[ (\ul{a} \times \ul{b})_i =
                \varepsilon_{ijk} a_j b_k \]
            \item Here $\sum_{ijk}$ is understood
                \[ \ul{a} \cdot \ul{b} \times \ul{c}
                = \varepsilon_{ijk} a_i b_j c_k \]
            \item Here $\sum_i$ is understood
                \[ \delta_{ii} = \delta_{11} +
                \delta{22} + \delta_{33} = 3 \]
            \item On the last line we have that
                $\sum_j$ is understood
                \begin{align*}
                    [(\ul{a} \cdot \ul{c}) \ul{b}
                    - (\ul{a} \cdot \ul{b})\ul{c}]_i
                    &= (\ul{a} \cdot \ul{c})
                    b_i - (\ul{a} \cdot \ul{b})c_i
                    \\
                    &= a_j c_j b_i - a_j b_j c_i
                \end{align*}
        \end{enumerate}
        \ul{Summation Convention Rules}
        \begin{enumerate}[(i)]
            \item An index occurring exactly
                \emph{once} in any term must
                appear once in \emph{every} term
                and it can take any value - a
                \emph{free} index.
            \item An index occurring exactly
                \emph{twice} in a given term is
                summed over - a \emph{repeated} or
                \emph{contracted} or \emph{dummy}
                index.
            \item No index can occur more than
                twice.
        \end{enumerate}
        \ul{Application}: proof of the vector
        triple product identity. Consider
        \begin{align*}
            [\ul{a} \times (\ul{b} \times
            \ul{c})]_i
            &= \varepsilon_{ijk} a_j (\ul{b}
            \times \ul{c})_k \\
            &= \varepsilon_{ijk} a_j
            \varepsilon_{kpq} b_p c_q \\
            &= \varepsilon_{ijk} \varepsilon_{pqk}
            a_j b_p c_q
        \end{align*}
        Now
        \[ \varepsilon_{ijk} \varepsilon_{pqk} =
        \delta_{ip} \delta_{jq} - \delta_{iq}
        \delta_{jp} \]
        (see section (c) velow). Then
        \begin{align*}
            [\ul{a} \times (\ul{b} \times \ul{c})]_i
            &= (\delta_{ip} \delta_{jq} -
            \delta_{iq} \delta_{jp}) a_j b_p c_q
            \\
            &= a_j \delta_{ip} b_p \delta_{jq} c_q
            - a_j \delta_{jp} b_p \delta_{iq} c_q
            \\
            &= a_j b_i c_j - a_j b_j c_i \\
            &= (a_j c_j) b_i - (a_j b_j) c_i \\
            &= (\ul{a} \cdot \ul{c}) b_i - (\ul{a}
            \cdot \ul{b}) c_i \\
            &= [(\ul{a} \cdot \ul{c}) \ul{b} -
            (\ul{a} \cdot \ul{b}) \ul{c}]_i
        \end{align*}
        True for $i = 1, 2, 3$ hence
        \[ \ul{a} \times (\ul{b} \times \ul{c})
        = (\ul{a} \cdot \ul{c}) \ul{b} - (\ul{a}
        \cdot \ul{b}) \ul{c} \]
    \item \ul{$\varepsilon$ $\varepsilon$
        identities} \\
        \begin{itemize}
            \item Expected to know this and quote
                it:
                \[ \varepsilon_{ijk}
                \varepsilon_{pqk}
                = \delta_{ip} \delta_{jq}
                - \delta_{iq} \delta_{jp} 
                = \varepsilon_{kij}
                \varepsilon_{kpq} \]
                Check: RHs and LHS are both
                antisymmetric (change sign) under
                \[ i \leftrightarrow j \qquad
                \text{or} \qquad p \leftrightarrow
                q \]
                SO both sides vanish if $i$ and
                $j$ or $p$ and $q$ take same
                values. Now suffices to check
                \[ i = p = 1
                \qquad\text{and}\qquad
                j = q = 2 \]
                \[ LHS = \varepsilon_{123}
                \varepsilon_{123} = +1 \]
                \[ RHS = \delta_{11} \delta_{22}
                - \delta_{12} \delta_{21} = +1 \]
                or $i = q = 1$ and $j = p = 2$
                \[ LHS = \varepsilon_{123}
                \varepsilon_{213}
                = (+1)(-1) = -1 \]
                \[ RHS = \delta_{12} \delta{21} -
                \delta_{11} \delta_{22} = -1 \]
                All other index choices work
                similarly.
            \item $\varepsilon_{ijk}
                \varepsilon_{pjk} = 2\delta_{ip}$
                \ul{contract} result aove
                \begin{align*}
                    \varepsilon_{ijk}
                    \varepsilon_{pjk}
                    &= \delta_{ip} \delta_{jj} -
                    \delta_{ij} \delta_{jp} \\
                    &= 3 \delta_{ip} - \delta_{ip}
                    \\
                    &= 2 \delta_{ip}
                \end{align*}
            \item $\varepsilon_{ijk}
                \varepsilon_{ijk} = 6$.
        \end{itemize}
\end{enumerate}

\newpage
\section{Vectors in General; $\RR^n$ and $\CC^n$}

\subsection{Vectors in $\RR^n$}

\subsubsection*{(a) Definitions}

If we regard vectors as sets of components, it is
easy to generalise from 3 to $n$ dimensions.

\begin{itemize}
    \item Let $\RR^n = \{ \ul{x} = (x_1, \dots,
        x_n : x_i \in \RR\}$ and define
        \begin{enumerate}[(i)]
            \item \ul{addition} \\
               \[ \ul{x} + \ul{y} = (x_1 + y_1,
                \dots, x_n + y_n) \]
            \item \ul{scalar multiplication}
                \[ \lambda \ul{x} = (\lambda x_1,
                \dots, \lambda x_n) \]
        \end{enumerate}
        for any $\ul{x}, \ul{y} \in \RR^n$ and
        $\lambda \in \RR$.
    \item \emph{Inner product} or \emph{scalar
        product} on $\RR^n$ is defined by
        \[ \ul{x} \cdot \ul{y} = \sum_i x_i y_i =
        x_1 y_1 + \cdots x_n y_n \]
        \ul{Properties}
        \begin{enumerate}[(i)]
            \item Symmetric $\ul{c} \cdot \ul{y} =
                \ul{y} \cdot \ul{x}$
            \item Bilinear  (linear in each
                vector)
                \[ (\lambda \ul{X} + \lambda'
                \ul{x}') \cdot \ul{y} = \lambda
                (\ul{x} \cdot \ul{y}) + \lambda'
                (\ul{x}' \cdot \ul{y}) \]
                and
                \[ \ul{x} \cdot (\mu \ul{y} + \mu'
                \ul{y}' = \mu(\ul{x} \cdot \ul{y})
                + \mu' (\ul{x} \cdot \ul{y}') \]
            \item \ul{Positive definite}
                \[ \ul{x} \cdot \ul{x} = \sum_{i}
                x_i^2 \ge 0 \]
                and is equal to 0 if and only if
                $\ul{x} = \ul{o}$. The
                \emph{length} or \emph{norm}  of
                vector $\ul{x}$ is $|\ul{x}| (\ge
                0)$ defined by $|\ul{z}|^2 =
                \ul{x} \cdot \ul{x}$.
            \item For $\ul{x} \in \RR^n$ we can
                write
                \[ \ul{x} = \sum_i x_i \ul{e_i} \]
                where
                \begin{align*}
                    \ul{e_1}
                    &= (1, 0, \dots, 0) \\
                    \ul{e_1}
                    &= (0, 1, \dots, 0) \\
                    &\vdots \\
                    \ul{e_n}
                    &= (0, 0, \dots, 1)
                \end{align*}
                call $\{\ul{e_i}\}$ the
                \emph{standard basis} for $\RR^n$.
                Note that it is orthonormal:
                \[ \ul{e_i} \cdot \ul{e_j} =
                \delta_{ij} = \begin{cases}
                    1 & \text{if $i = j$} \\
                    0 & \text{if $i \neq j$}
                \end{cases} \]
        \end{enumerate}
\end{itemize}

\subsubsection*{(b) Cauchy-Schwarz and $\triangle$ Inequalities Proposition}

\begin{proposition*}[Cauchy-Schwarz]
    \[ |\ul{x} \cdot \ul{y}| \le |\ul{x}||\ul{y}| \]
    for $\ul{x}, \ul{y} \in \RR^n$ and equality
    holds if and only if $\ul{x} = \lambda \ul{y}$
    or $\ul{y} = \lambda \ul{x}$ ($\ul{x}
    \parallel \ul{y}$) for some $\lambda \in \RR$.
\end{proposition*}

Deductions reveal geometrical aspects of inner
product:
\begin{enumerate}[(i)]
    \item Set 
        \[ \ul{x} \cdot \ul{y} = |\ul{x}||\ul{y}|
        \cos\theta \]
        to define angle $\theta$ between $\ul{x}$
        and $\ul{y}$
    \item $\triangle$ inequality holds
        \[ |\ul{x} + \ul{y}| \le |\ul{x}| +
        |\ul{y}| \]
\end{enumerate}

Now we present a proof of the Cauchy-Schwarz
inequality

\begin{proof}
If $\ul{y} = \ul{o}$, result is immediate. \\
If $\ul{y} \neq \ul{o}$, consider
\begin{align*}
    |\ul{x} - \lambda \ul{y}|^2
    &= (\ul{x} - \lambda\ul{y}) \cdot (\ul{x} -
    \lambda \ul{y}) \\
    &= |\ul{x}|^2 - 2\lambda \ul{x} \cdot \ul{y} +
    \lambda^2|\ul{y}|^2 \ge 0
\end{align*}
This is a quadratic in real $\lambda$ with at most
one real root, so discriminant satisfies
\[ (-2\ul{x} \cdot \ul{y})^2 -
4|\ul{x}|^2|\ul{y}|^2 \le 0 \]
Equality holds if and only if $disc = 0$ which holds
if and only if $\lambda \ul{y} = \ul{x}$ for some
$\lambda \in \RR$.
\end{proof}
\noindent
Now we present a proof of the $\triangle$
inequality.

\begin{proof}
    \[ LHS^2 = |\ul{x} + \ul{y}|^2 = |\ul{x}|^2 +
    2 \ul{x} \cdot \ul{y} + |\ul{y}|^2 \]
    \[ RHS^2 = (|\ul{x}| + |\ul{y}|)^2
    = |\ul{x}|^2 + 2|\ul{x}||\ul{y}| + |\ul{y}|^2 \]
    and compare using Cauchy-Schwarz.
\end{proof}

\subsubsection*{(c) Comments}

Inner product on $\RR^n$.
\[ \ul{a} \cdot \ul{b} = \delta_{ij} a_i b_i \]
Component definition matches geometrical
definition for $n = 3$ (section 2.2). \\
In $\RR^3$ also have a cross product with
component definition
\[ (\ul{a} \times \ul{b})_i = \varepsilon_{ijk}
a_j b_k \]
(geometrical definition given in section 2.4) \\
In $\RR^n$ we have $\varepsilon_{ij\dots l}$
totally antisymmetric. (see chapter 5). Cannot use
this to define vector-valued product except in $n
= 3$. But in $\RR^2$ have $\varepsilon_{ij}$ with
\[ \varepsilon_{12} = -\varepsilon_{21} = 1 \]
and can use this to define an additional scalar
cross product in 2D.
\begin{align*}
    [\ul{a}, \ul{b}]
    &= \varepsilon_{ij} a_i b_j \\
    &= a_1 b_2 - a_2 b_1 \qquad
    \text{for $\ul{a}, \ul{b} \in \RR^2$}
\end{align*}
Geometrically, this gives (signed) area of
parallelogram
\[ [\ul{a}, \ul{b}] = |\ul{a}||\ul{b}| \sin\theta \]
\begin{center}
\begin{tsqx}
    ! size(5cm);
    O := origin
    A := 1.5*dir(-5)
    B := dir(45)
    O->A
    O->B
    A--A+B--B dashed
    label $\underline{a}$ @ A+(0,-0.1)
    label $\underline{b}$ @ B+(0,0.1)
    anglemark4 A O B 4
    label $\theta$ @ 0.3*dir(20)
\end{tsqx}
\end{center}
Compare with
\[ [\ul{a}, \ul{b}, \ul{c}] = \ul{a} \times \ul{b}
\times \ul{c} = \varepsilon_{ijk} a_i b_j c_k \]
(signed) volume of parallelepiped.

\subsection{Vector Spaces}

\subsubsection*{(a) Axioms; span; subspaces}

Let $V$ be a set of objects called \emph{vectors}
with operations
\begin{enumerate}[(i)]
    \item $\ul{v} + \ul{w} \in V$
    \item $\lambda \ul{v} \in V$
\end{enumerate}
(the above expressions are defined $\forall
\ul{v}, \ul{w} \in V$ and $\forall \lambda \in
\RR$) \\
Then $V$ is a \emph{real vector space} if $V$ is
an abelian group under $+$ and
\begin{align*}
    \lambda (\ul{v} + \ul{w})
    &= \lambda \ul{v} + \lambda \ul{w} \\
    (\lambda + \mu) \ul{v}
    &= \lambda \ul{v} + \mu \ul{v} \\
    \lambda (\mu \ul{v})
    &= (\lambda\mu) \ul{v} \\
    1 \ul{v}
    &= \ul{v}
\end{align*}
These axioms or key properties apply to
geometrical vectors with $V$ 3D space or to
vectors in $V = \RR^n$, as above, as well as other
examples. \\
For vectors $\ul{v_1}, \ul{v_2}, \dots, \ul{v_r}
\in V$ we can form a \emph{linear combination}
\[ \lambda_1 \ul{v_1} + \lambda_2 \ul{v_2} +
\cdots + \lambda_r \ul{v_r} \in V \]
for any $\lambda_i \in \RR$; the span is defined
\[ \mathrm{span}\{\ul{v_1}, \ul{v_2}, \dots,
\ul{v_r}\} = \{\sum_i \lambda_i \ul{v_i} :
\lambda_i \in \RR\} \]
A \emph{subspace} of $V$ is a subset that is
itself a vector space. \\
Note $V$ and $\{\ul{o}\}$ are subspaces.
\[ \mathrm{span}\{\ul{v_1}, \ul{v_2}, \dots,
\ul{v_r}\} \]
is a subspace for any vectors $\ul{v_1}, \dots,
\ul{v_r}$. Note: a non-empty subset $U \subseteq
V$ is a subspace if and only if
\[ \ul{v}, \ul{w} \in U \implies \lambda \ul{v} +
\mu\ul{w} \in U \,\,\,\forall \lambda, \mu \in \RR \]

\begin{example*}
    In 3D or $\RR^3$ a line or plane through
    $\ul{o}$ is a subspace but a line or plane
    that doesn't contain $\ul{o}$ is not a
    subspace. For example
    \[ \ul{v_1} =
    \begin{pmatrix}
        1 \\
        0 \\
        -1
    \end{pmatrix}
    , \ul{v_2} =
    \begin{pmatrix}
        1 \\
        1 \\
        -2
    \end{pmatrix}
    , \ul{n} =
    \begin{pmatrix}
        1 \\
        1 \\
        1
    \end{pmatrix}
    \]
    \[ \mathrm{span}\{\ul{v_1}, \ul{v_2}\} = \{\ul{r} :
    \ul{n} \cdot \ul{r} = 0\} \]
    which is a plane and subspace. But
    \[ \{ul{r} : \ul{n} \cdot \ul{r} = 1\} \]
    is a plane but not a subspace ($\ul{r}$,
    $\ul{r'}$ on plane then $(\ul{r} + \ul{r'})
    \cdot \ul{n} = 2$)
\end{example*}

\subsubsection*{(b) Linear Dependence and Independence}

For vectors $\ul{v_1}, \ul{v_2}, \dots, \ul{v_r}
\in V$, with $V$ a real vector space, consider the
\emph{linear relation}
\[ \lambda_1 \ul{v_1} + \lambda_2 \ul{v_2} + \cdots
+ \lambda_r \ul{v_r} = \ul{o} \tag{$*$} \]
If $(*) \implies \lambda_i = 0$ for every $i$ then
the vectors form a \emph{linearly independent} set
(they obey only the \emph{trivial} linear relation
with $\lambda_i = 0$). \\
If $(*)$ holds with at least one $\lambda_i \neq
0$ then the vectors form a \emph{linearly
dependent} set (they obey a \emph{non-trivial}
linear relation.)

\textbf{Examples}

\begin{itemize}
    \item \[
        \begin{pmatrix}
            1 \\
            0
        \end{pmatrix}
        ,
        \begin{pmatrix}
            0 \\
            1
        \end{pmatrix}
        ,
        \begin{pmatrix}
            0 \\
            2
        \end{pmatrix}
        \]
        is linearly dependent because
        \[ 0
        \begin{pmatrix}
            1 \\
            0
        \end{pmatrix}
        + 2
        \begin{pmatrix}
            0 \\
            1
        \end{pmatrix}
        + (-1)
        \begin{pmatrix}
            0 \\
            2
        \end{pmatrix}
        = 0 \]
        Note that we cannot express
        $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ in
        terms of the others, but it is still
        linearly dependent.
    \item Any set containing $\ul{o}$ is linearly
        dependent. For example
        \[ \{
        \begin{pmatrix}
            1 \\
            0
        \end{pmatrix}
        ,
        \begin{pmatrix}
            0 \\
            0
        \end{pmatrix}
        \} \]
        we have
        \[ 0
        \begin{pmatrix}
            1 \\
            0
        \end{pmatrix}
        + 412
        \begin{pmatrix}
            0 \\
            0
        \end{pmatrix}
        = \ul{o} \]
        non-trivial linear relation.
    \item $\{\ul{a}, \ul{b}, \ul{c}\}$ in $\RR^3$
        linearly independent if $\ul{a}\cdot
        \ul{b} \times \ul{c} \neq 0$. Consider
        \[ \alpha \ul{a} + \beta\ul{b} + \gamma
        \ul{c} = \ul{o} \]
        Take dot with $\ul{b} \times \ul{c}$ to
        get
        \[ \alpha \ul{a} \cdot \ul{b} \times
        \ul{c} = 0 \implies \alpha = 0 \]
        and we can get $\beta = \gamma = 0$ with a
        similar argument.
\end{itemize}

\subsubsection*{(c) Inner Product}

This is an additional structure on a real vector
space $V$, also characterised by axioms. For
$\ul{v}, \ul{w} \in V$ write inner product $\ul{v}
\cdot \ul{w}$ or $(\ul{v}, \ul{w}) \in \RR$. This
satisfies axioms corresponding to the properties
in section 3.1(a)
\begin{enumerate}[(i)]
    \item symmetric
    \item bilinear
    \item positive definite
\end{enumerate}

\begin{lemma*}
    In a real vector space $V$ with inner product,
    if $\bf{v}_1, \dots, \bf{v}_r$ are non-zero
    and orthogonal:
    \[ \ub{(\bf{v}_i, \bf{v}_i) \neq 0}_{\text{fixed $i$}}
    \qquad \text{and} \qquad
    \ub{(\bf{v}_i, \bf{v}_j) = 0}_{i \neq j} \]
    then $\bf{v}_1, \dots, \bf{v}_r$ are linearly
    independent.
\end{lemma*}

\begin{proof}
    \begin{align*}
        \sum_i \alpha_i \bf{v}_i
        &= \bf{0} \\
        (\bf{v}_j, \sum_i \alpha_i \bf{v}_i)
        &= \sum_i \alpha_i (\bf{v}_j, \bf{v}_i) \\
        &= \alpha_j (\bf{v}_j, \bf{v}_j) \\
        &= 0 \\
        \implies \alpha_j &= 0
    \end{align*}
    as claimed.
\end{proof}

\subsection{Bases and Dimension}

For a vector space $V$, a basis is a set
\[ \mathfrak{B} = \{\bf{e}_1, \dots, \bf{e}_n\} \]
such that
\begin{enumerate}[(i)]
    \item $\mathfrak{B}$ spans $V$, i.e. any
        $\bf{v} \in V$ can be written
        \[ \bf{v} = \sum_{i = 1}^n v_i \bf{e}_i \]
    \item $\mathfrak{B}$ is linearly independent.
\end{enumerate}

\noindent
Given (ii), the coefficients $v_i$ in (i) are
unique since
\[ \sum_i v_i \bf{e}_i = \sum_i v_i' \bf{e}_i \]
\[ \implies \sum_i (v_i - v_i') \bf{e}_i = \bf{0} \]
\[ \implies v_i = v_i' \]
$v_i$ are \emph{components} of $\bf{v}$ with
respect to $\mathfrak{B}$.

\noindent
\ul{Examples}

Standard basis for $\RR^n$ consisting of
\[ \bf{e}_1 =
\begin{pmatrix}
    1 \\
    0 \\
    \vdots \\
    0
\end{pmatrix}
, \bf{e}_2 =
\begin{pmatrix}
    0 \\
    1 \\
    \vdots \\
    0
\end{pmatrix}
, \dots, \bf{e}_n =
\begin{pmatrix}
    0 \\
    \vdots \\
    0 \\
    1
\end{pmatrix}
\]
is a basis according to general definition.
\begin{enumerate}[(i)]
    \eqitem \begin{align*}
        \bf{x} =
        \begin{pmatrix}
            x_1 \\
            \vdots \\
            x_n
        \end{pmatrix}
        = x_1 \bf{e}_1 + \cdots + x_n\bf{e}_n
    \end{align*}
    \item $\bf{x} = \bf{0}$ if and only if $x_1 =
        x_2 = \cdots = x_n = 0$.
\end{enumerate}
Many other bases can be chosen, for example in
$\RR^2$ we have bases
\[ \left\{
\begin{pmatrix}
    1 \\
    0
\end{pmatrix}
,
\begin{pmatrix}
    1 \\
    1
\end{pmatrix}
\right\} ,
\left\{ 
\begin{pmatrix}
    1 \\
    1
\end{pmatrix}
,
\begin{pmatrix}
    1 \\
    -1
\end{pmatrix}
\right\} ,\]
or $\{\bf{a}, \bf{b}\}$ for any $\bf{a}, \bf{b}
\in \RR^2$ with $\bf{a} \not\,\,\parallel \bf{b}$.

In $\RR^3$, $\{\bf{a}, \bf{b}, \bf{c}\}$ is a
basis if and only if
\[ \bf{a} \cdot \bf{b} \times \bf{c} \neq 0 .\]
Consider previous example of plane through
$\bf{0}$, subspace in $\RR^3$
\[ \bf{n} \cdot \bf{r} = 0 \qquad \text{with}
\qquad \bf{n} =
\begin{pmatrix}
    1 \\
    1 \\
    1
\end{pmatrix}
;\]
we have $\{\bf{v}_1, \bf{v}_2\}$ basis with
\[ \bf{v}_1 =
\begin{pmatrix}
    1 \\
    0 \\
    -1
\end{pmatrix}
, \bf{v}_2 =
\begin{pmatrix}
    1 \\
    1 \\
    -2
\end{pmatrix}
\]
not normalised or $\perp$ but could choose
orthonormal basis
\[ \{\bf{u}_1, \bf{u}_2\} \qquad \text{with}
\qquad \bf{u}_1 = \frac{1}{\sqrt{2}}
\begin{pmatrix}
    1 \\
    -1 \\
    0
\end{pmatrix}
, \qquad \text{and}\qquad
\bf{u}_2 = \frac{1}{\sqrt{6}}
\begin{pmatrix}
    1 \\
    1 \\
    -2
\end{pmatrix}
.\]
\begin{theorem*}
    If $\{\bf{e}_1, \dots, \bf{e}_n\}$ and
    $\{\bf{f}_1, \dots, \bf{f}_m\}$ are bases for
    a real vector space $V$, then
    \[ n = m .\]
\end{theorem*}

\begin{definition*}
    The number of vectors in any basis is the
    \emph{dimension} of $V$, $\mathrm{dim}\,\,V$.
\end{definition*}

\begin{note*}
    $\RR^n$ has dimension $n$ (!)
\end{note*}

\begin{proof}
    \[ \bf{f}_a = \sum_i A_{ai} \bf{e}_i \]
    and
    \[ \bf{e}_i = \sum_a B_{ia} \bf{f}_a \]
    for constants $A_{ai}$ and $B_{ia}$ and we use
    ranges of indices $i, j = 1, \dots, n$ and $a,
    b = 1, \dots, m$ [since $\{\bf{e}_i\}$ and
    $\{\bf{f}_a\}$ are bases]
    \begin{align*}
        \implies \bf{f}_a
        &= \sum_i A_{ai} \left(
        \sum_b B_{ib} \bf{f}_b \right) \\
        &= \sum_b \left( \sum_i A_{ai} B_{ib}
        \bf{f}_b \right)
    \end{align*}
    But coefficients with respect to a basis are
    unique so
    \[ \sum_i A_{ai} B_{ib} = \delta_{ab} \]
    Similarly
    \[ \bf{e}_i = \sum_j \left( \sum_a B_{ia}
    A_{aj} \right) \bf{e}_j \]
    and hence
    \[ \sum_a B_{ia} A_{aj} = \delta_{ij} \]
    Now
    \begin{align*}
        \sum_{ia} A_{ai} B_{ia}
        &= \sum_a \delta_{aa} = m \\
        = \sum_{ia} B_{ia} A_{ai}
        &= \sum_i \delta_{ii} = n \\
        \implies m = n, &\text{ as required.}
    \end{align*}
\end{proof}

The steps in the proof above are within the scope
of the course; but the proof without prompts is
\emph{non-examinable}.

\begin{note*}
    By convention the vector space $\{\bf{0}\}$
    has dimension $\bf{0}$. Not every vector space
    is finite dimensional!
\end{note*}

\begin{proposition*}
    Let $V$ be a vector space of dimension $n$
    (for example $\RR^n$).
    \begin{enumerate}[(i)]
        \item If $Y = \{\bf{w}_1, \dots,
            \bf{w}_m\}$ spans $V$, then $m \ge n$
            and in the case where $m > n$, we can
            remove vectors can be removed from $Y$
            to get a basis.
        \item If $X = \{\bf{u}_1, \dots,
            \bf{u}_k\}$ are linearly independent
            then $k \le n$ and in the case $k < n$
            we can add vectors to $X$ to get a
            basis.
    \end{enumerate}
\end{proposition*}

\subsection{Vectors in $\CC^n$}

\subsubsection*{(a) Definitions}

Let $\CC^n = \{\bf{z} = (z_1, \dots, z_n) : z_j
\in \ZZ\}$ and define:
\begin{itemize}
    \item \ul{addition} $\bf{z} + \bf{w} = (z_1 +
        w_1, \dots, z_n + w_n)$
    \item \ul{scalar multiplication} $\lambda
        \bf{z} = (\lambda z_1, \dots, \lambda
        z_n)$ for any $\bf{z}, \bf{w} \in \CC^n$.
\end{itemize}
Taking \emph{real} scalars $\lambda, \mu \in \RR$,
$\CC^n$ is a real vector space obeying axioms or
key properties in section 3.2(a). \\
Taking \emph{complex} scalars $\lambda, \mu \in
\CC$, $\CC^n$ is a \emph{complex} vector space -
same axioms or key properties hold, and
definitions of linear combinations, linear
(in)dependence, span, bases, dimension all
generalise to complex scalars.

\bigskip

\noindent
The distinction matters, for example
\[ \bf{z} = (z_1, \dots, z_n) \in \CC^n \]
with $z_j = x_j + iy_j$, $x_j, y_j \in \RR$ then
\[ \bf{z} = \sum_j x_j \bf{e}_j + \sum_j y_j
\bf{f}_j \]
(real linear combination) where
\[ \bf{e}_j = \ub{(0, \dots, 1, \dots,
0)}_{\text{position $j$}} \]
\[ \bf{f}_j = \ub{(0, \dots, i, \dots,
0)}_{\text{position $j$}} \]
therefore $\{\bf{e}_1, \dots, \bf{e}_n, \bf{f}_1,
\dots, \bf{f}_n\}$ basis for $\CC^n$ as a
\emph{real} vector space. So real dimension is
$2n$. But
\[ \bf{z} = \sum_j = z_j \bf{e}_j \qquad \text{and}
\qquad \{\bf{e}_1, \dots, \bf{e}_n\} \]
is a basis for $\CC^n$ as a complex vector space,
dimension $n$ (over $\CC$).

\subsubsection*{(b) Inner Product}

Inner product or scalar product on $\CC^n$ is
defined by
\[ (\bf{z}, \bf{w}) = \sum_j \ol{z_j} w_j
= \ol{z_1} w_1 + \cdots + \ol{z_n} w_n \]

\ul{Properties}

\begin{enumerate}[(i)]
    \item \ul{hermitian} $(\bf{w}, \bf{z}) =
        \ol{(\bf{z}, \bf{w})}$.
    \item \ul{Linear / anti-linear}
        \[ (\bf{z}, \lambda \bf{w} + \lambda'
        \bf{w}') = \lambda (\bf{z}, \bf{w}) +
        \lambda' (\bf{z}, \bf{w}') \]
        \[ (\mu\bf{z} + \mu' \bf{z}', \bf{w})
        = \ol{\mu} (\bf{z}, \bf{w}) + \ol{\mu'}
        (\bf{z}', \bf{w}) \]
    \item \ul{positive definite} \\
        $(\bf{z}, \bf{z}) = \sum_i |z_i|^2$ is
        real and $\ge 0$, and $0$ if and only if
        $\bf{z} = \bf{0}$.
\end{enumerate}
Defined \emph{length} or \emph{norm} of $\bf{z}$
to be $|\bf{z}| \ge 0$ with $|\bf{z}|^2 =
(\bf{z}, \bf{z})$. \\
Define $\bf{z}, \bf{w} \in \CC^n$ to be
\emph{orthogonal} if $(\bf{z}, \bf{w}) = 0$. \\
Note: the standard basis $\{\bf{e_j}\}$ for
$\CC^n$ (see part (a)) is orthonormal
\[ (\bf{e_i}, \bf{e_j}) = \delta_{ij} \]
Also, if $\bf{z_1}, \bf{z_2}, \dots, \bf{z_k}$ are
non-zero and orthogonal in sense above, then they
are linearly independent over $\CC$ (same argument
as in real case).

\begin{example*}
    Complex inner product on $\CC$ $(n = 1)$ is
    \[ (z, w) = \ol{z}w \]
    Let $z = a_1 + ia_2$ (real and imaginary part)
    and $w = b_1 + ib_2$. Then
    \[ \bf{a} = (a_1, a_2), \bf{b} = (b_1, b_2)
    \in \RR^2 \qquad \text{corresponding vectors.} \]
    \begin{align*}
        \ol{z}w
        &= (a_1 b_1 + a_2 b_2) + i(a_1 b_2 -
        a_2 b_1) \\
        &= \bf{a} \cdot \bf{b} + i [\bf{a}, \bf{b}]
    \end{align*}
    recover scalar dot and cross product in
    $\RR^2$.
\end{example*}

\newpage
\section{Matrices and Linear Maps}

\subsection{Introduction}

\subsubsection*{(a) Definitions}

A \emph{linear map} or \emph{linear
transformation} is a function
\[ T : V \to W \]
between vector spaces $V$ (dim $n$) and
$W$ (dim $m$) such that
\[ T(\lambda \bf{x} + \mu \bf{y}) = \lambda T(
\bf{x}) + \mu T(\bf{y}) \]
for all $\bf{x}, \bf{y} \in V$ and $\lambda, \mu
\in \RR$ or $\CC$ for $V$, $W$ both real or
complex vector spaces. [mostly concerned with $V =
\RR^n$ or $\CC^n$, $W = \RR^m$ or $\CC^m$]

\begin{note*}
    A linear map is completely determined by its
    action on a basis $\{\bf{e}_n, \dots,
    \bf{e}_n\}$ for $V$, since
    \[ T\left(\sum_i x_i \bf{e}_i\right)
    = \sum_i x_i T(\bf{e}_i) \]
    ---
    \[ \bf{x}' = T(\bf{x}) \in W \]
    is the \emph{image} of $\bf{x} \in V$
    \[ \mathrm{Im}(T) = \{\bf{x}' \in W : \bf{x}'
    = T(\bf{x}) \text{ for some $\bf{x} \in V$}\} \]
    is the \emph{image} of $T$.
\end{note*}

\begin{lemma*}
    $\mathrm{Ker}(T)$ is a subspace of $V$ and
    $\mathrm{Im}(T)$ is a subspace of $W$.
\end{lemma*}

\begin{customproof}{Check.}
    $\bf{x}, \bf{y} \in \mathrm{Ker}(T) \implies
    T(\lambda \bf{x} + \mu \bf{y}) = \lambda
    T(\bf{x}) + \mu T(\bf{y}) = \bf{0}$ and
    $\bf{0} \in \mathrm{Ker}(T)$ so the result
    follows. \\
    Also $\bf{0} \in \mathrm{Im}(T)$ and $\bf{x}',
    \bf{y}' \in \mathrm{Im}(T)$ then
    \[ T(\lambda \bf{x} + \mu \bf{y}) = \lambda
    T(\bf{x}) + \mu T(\bf{y}) = \lambda \bf{x}' +
    \mu \bf{y}' \in \mathrm{Im}(T) \]
    for some $\bf{x}, \bf{y} \in V$.
\end{customproof}

\bigskip

\noindent
\ul{Examples}
\begin{enumerate}[(i)]
    \item zero linear map $T : V \to W$ is given
        by $T(\bf{x}) = \bf{0}\,\,\forall \bf{x}
        \in V$. Then $\mathrm{Im}\,\,T =
        \{\bf{0}\}$, $\mathrm{Ker}\,\,T = V$.
    \item For $V = W$, the identity linear map $T
        : V \to V$ is given by
        \[ T(\bf{x}) = \bf{x} \qquad \forall
        \bf{x} \in V \]
        then $\mathrm{Im}\,\,T = V$,
        $\mathrm{Ker}\,\,T
        = \{\bf{0}\}$.
    \item $V = W = \RR^3$, $\bf{x}' = T(\bf{x})$
        given by
        \[ \bf{x}_1' = 3x_1 + x_2 + 5x_3 \]
        \[ \bf{x}_2' = -x_1 - 2x_3 \]
        \[ \bf{x}_3' = 2x_1 + x_2 + 3x_3 \]
        \[ \mathrm{Ker}(T) = \left\{\lambda
        \begin{pmatrix}
            2 \\
            -1 \\
            -1
        \end{pmatrix}
        \right\} \qquad \text{(dim 1)} \]
        \[ \mathrm{Im}(T) = \left\{\lambda
        \begin{pmatrix}
            3 \\
            -1 \\
            2
        \end{pmatrix}
        + \mu
        \begin{pmatrix}
            1 \\
            0 \\
            1
        \end{pmatrix}
        \right\} \qquad \text{(dim 2)} \]
\end{enumerate}

\subsubsection*{(b) Rank and Nullity}

$\mathrm{dim}\,\,\mathrm{Im}(T)$ is the
\emph{rank} of $T$ ($\le m$) and
$\mathrm{dim}\,\,\mathrm{Ker}(T)$ is the
\emph{nullity} of $T$ ($\le n$)

\begin{theorem*}[Rank-nullity]
    For
    \[ T : V \to W \]
    a linear map (as in (a) above)
    \[ \mathrm{rank}(T) + \mathrm{null}(T) = n =
    \mathrm{dim}\,\,V \]
\end{theorem*}

\bigskip

\noindent
\ul{Examples} - refer to part (a) above
\begin{enumerate}[(i)]
    \item $\mathrm{null}(T) + \mathrm{rank}(T) = n
        + 0 = n$
    \item $\mathrm{null}(T) + \mathrm{rank}(T) = 0
        + n = n$
    \item $\mathrm{null}(T) + \mathrm{rank}(T) = 1
        + 2 = 3$
\end{enumerate}
\noindent
Note that the following proof is
\emph{non-examinable}.

\begin{proof}
    Let $\bf{e}_1, \dots, \bf{e}_k$ be a basis for
    $\mathrm{Ker}(T)$ so $T(\bf{e}_i) = \bf{0}$
    for $i = 1, \dots, k$. Extend by $\bf{e}_{k +
    1}, \dots, \bf{e}_n$ to get a basis for $V$.
    Claim
    \[ \mathfrak{B} = \{T(\bf{e}_{k + 1}, \dots,
    T(\bf{e}_n)\} \]
    is a basis for $\mathrm{Im}(T)$. The result
    then follows since $\mathrm{null}(T) = k$ and
    $\mathrm{rank}(T) = n - k$, implying
    $\mathrm{null}(T) + \mathrm{rank}(T) = n$. \\
    To check the claim:
    \begin{itemize}
        \item $\mathfrak{B}$ spans
            $\mathrm{Im}(T)$ since
            \begin{align*}
                \bf{x}
                &= \sum_1^n x_i \bf{e}_i \\
                \implies T(\bf{x})
                &= \sum_{i = k + 1}^n x_i
                T(\bf{e}_i)
            \end{align*}
        \item $\mathfrak{B}$ is linearly
            independent since
            \begin{align*}
                \sum_{i = k + 1}^n \lambda_i
                T(\bf{e}_i)
                &= \bf{0} \\
                \implies T\left( \sum_{i = k =
                1}^n \lambda_i \bf{e}_i \right)
                &= \bf{0} \\
                \implies \sum_{i = k + 1}^n
                \lambda_i \bf{e}_i
                &\in \mathrm{Ker}(T) \\
                \implies \sum_{i = k + 1}^n
                \lambda_i \bf{e}_i
                &= \sum_{i = 1}^k \mu_i \bf{e}_i
            \end{align*}
            But $\bf{e}_1, \dots, \bf{e}_n$ are
            linearly independent in $V$
            \[ \implies \lambda_i = 0 \qquad
            (i = k + 1, \dots, n) \]
            \[ \implies \mu_i = 0 \qquad (i = 1,
            \dots, k) \]
            hence $\mathfrak{B}$ is linearly
            independent.
    \end{itemize}
    Therefore $\mathfrak{B}$ is a basis.
\end{proof}

\subsection{Geometrical Examples}

\subsubsection*{(a) Rotations}

In $\RR^2$, rotation about $\bf{0}$ through angle
$\theta$ is defined by
\[ \bf{e}_1 \mapsto \bf{e}_1' = (\cos\theta)
\bf{e}_1 + (\sin\theta)\bf{e}_2 \]
\[ \bf{e}_2 \mapsto \bf{e}_2' = -(\sin\theta)
\bf{e}_1 + (\cos\theta) \bf{e}_2 \]
\begin{center}
\begin{tsqx}
    ! size(5cm);
    (0.6,0)->>(1,0)
    (0,0.6)->>(0,1)
    O := origin
    e_1 := (0.6,0)
    e_2 := (0,0.6)
    e_1' := rotate(30)*e_1
    e_2' := rotate(30)*e_2
    anglemark4 e_1 O e_1' 5
    anglemark4 e_2 O e_2' 5
    label $\mathbf{e}_1$ @ e_1+(0,-0.1)
    label $\mathbf{e}_2$ @ e_2+(0.1,0)
    label $\mathbf{e}_1'$ @ e_1'+0.1*dir(30)
    label $\mathbf{e}_2'$ @ e_2'+0.1*dir(120)
    label $\theta$ @ 0.2*dir(15)
    label $\theta$ @ 0.2*dir(105)
    (-0.5,0)->>e_1
    (0,-0.2)->>e_2
    O->>e_1'
    O->>e_2'
\end{tsqx}
\end{center}
In $\RR^3$, rotation about axis given by
$\bf{e}_3$ is defined as above, with
\[ \bf{e}_3 \mapsto \bf{e}_3' = \bf{e}_3 \]
Now consider rotation about axis $\bf{n}$ (unit
vector). \\
Given $\bf{x}$, resolve $\parallel$ and $\perp$ to
$\bf{n}$:
\[ \bf{x} + \bf{x}_\parallel + \bf{x}_\perp \]
with $\bf{x}_\parallel = (\bf{x} \cdot \bf{n})
\bf{n}$, and hence $\bf{n} \cdot \bf{x}_\perp =
0$. Under rotation
\begin{align*}
    \bf{x}_\parallel
    &\mapsto \bf{x}_\parallel' = \bf{x}_\parallel \\
    \bf{x}_\perp
    &\mapsto \bf{x}_\perp' = (\cos\theta)
    \bf{x}_\perp + (\sin\theta) \bf{n} \times
    \bf{x}
\end{align*}
by considering plane $\perp \bf{n}$, comparing to
rotation in $\RR^2$ and noting that
\[ |\bf{x}_\perp = |\bf{n} \times \bf{n} \times
\bf{x}| \]
\begin{center}
\begin{tsqx}
    ! size(5cm);
    (-0.2,0)->>(1.5,0)
    (0,-0.2)->>(0,1.5)
    label $\mathbf{x}_\perp$ @ (1.6,0)
    label $\mathbf{n} \times \mathbf{x}$ @ (0,1.6)
    X := (1.2,0.8)
    label $\mathbf{x}_\perp'$ @ X+0.1*dir(X)
    anglemark4 (1,0) origin X 5
    label $\theta$ @ 0.2*dir(18)
    origin->>X
\end{tsqx}
\end{center}
Re-assemble:
\[ \bf{x} \mapsto \bf{x}' = \bf{x}_\parallel' +
\bf{x}_\perp' = (\cos\theta) \bf{X} + (1 -
\cos\theta)(\bf{n} \cdot \bf{x}) \bf{n} +
\sin\theta \bf{n} \times \bf{x} .\]

\subsubsection*{(b) Reflections}

Consider \emph{reflection} in plane in $\RR^3$ (or
line in $\RR^2$ through $\bf{0}$ with unit normal
$\bf{n}$. \\
Given $\bf{x}$, resolve $\parallel$ and $\perp$ to
$\bf{n}$:
\begin{align*}
    \bf{x}_\parallel
    &\mapsto \bf{x}_\parallel' = -\bf{x}_\parallel
    \\
    \bf{x}_\perp
    &\mapsto \bf{x}_\perp' = \bf{x}_\perp
\end{align*}
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/60a1704437d811ec.png}
\end{center}
\[ \bf{x} \mapsto \bf{x}' = \bf{x} - 2(\bf{x}
\cdot \bf{n})\bf{n} \]

\subsubsection*{(c) Dilations}

A dilation by scale factors $\alpha$, $\beta$,
$\gamma$ (real, $> 0$) along axes $\bf{e}_1$,
$\bf{e}_2$, $\bf{e}_3$ in $\RR^3$ is defined by
\[ \bf{x} = x_1 \bf{e}_1 + x_2 \bf{e}_2 + x_3
\bf{e}_3 \mapsto \bf{x}' = \alpha x_1 \bf{e}_1 +
\beta x_2 \bf{e}_2 + \gamma x_3 \bf{e}_3 \]
[unit cube $\to$ cuboid]

\subsubsection*{(d) Shears}

Given $\bf{a}$, $\bf{b}$ orthogonal unit vectors
($|\bf{a}| = |\bf{b}| = 1$ and $\bf{a} \cdot
\bf{b} = 0$) define a shear with parameter
$\lambda$ by
\[ \bf{x} \mapsto \bf{x}' = \bf{x} + \lambda
(\bf{x} \cdot \bf{b}) \bf{a} \]
Definition applies in $\RR^n$ and $\bf{u}' =
\bf{u}$ for any vector $\bf{u} \perp \bf{b}$.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/3397cec637d911ec.png}
\end{center}

\subsection{Matrices as Linear Maps $\RR^n \to \RR^n$}

\subsubsection*{(a) Definitions}

Consider a linear map $T : \RR^n \to \RR^m$ and
standard bases $\{\bf{e}_i\}$ and $\{\bf{f}_a\}$.
Let $\bf{x}' = T(\bf{x})$ with
\[ \bf{x} = \sum_i x_i \bf{e}_i =
\begin{pmatrix}
    x_1 \\
    \vdots \\
    x_n
\end{pmatrix}
, \bf{x}' = \sum_a x_a' \bf{f}_a =
\begin{pmatrix}
    x_1' \\
    \vdots \\
    x_m'
\end{pmatrix}
\]
Linearity implies $T$ is determined by
\[ T(\bf{e}_i) = \bf{e}_i' = \bf{C}_i \in \RR^m (i
= 1, \dots, n) ;\]
take these as \emph{columns} of an $m \times n$
\emph{array} or \emph{matrix} with \emph{rows}
\[ \bf{R}_a \in \RR^n (a = 1, \dots, m) .\]
$M$ has entries $M_{ai} \in \RR$ where $a$ labels
rows and $i$ labels columns.
\[
\begin{pmatrix}
    \uparrow & & \uparrow \\
    \bf{C}_1 & \cdots & \bf{C}_n \\
    \downarrow & & \downarrow
\end{pmatrix}
= M =
\begin{pmatrix}
    \leftarrow & \bf{R}_1 & \rightarrow \\
    & \vdots & \\
    \leftarrow & \bf{R}_m & \rightarrow
\end{pmatrix}
\]
\[ (\bf{C}_i)_a = M_{ai} = (\bf{R}_a)_i \]
Action of $T$ is given by matrix $M$ multiplying
vector $\bf{x}$
\begin{center}
    \fbox{$\bf{x}' = M\bf{x}$ defined by
    $\bf{x}_a' = M_{ai} x_i$ ($\sum$ convention)}
\end{center}
This follows from definitions above since
\[ \bf{x}' = T\left( \sum_i x_i \bf{e}_i \right) =
\sum_i x_i \bf{C}_i \]
\begin{align*}
    \implies (\bf{x}')_a
    &= \sum_i x_i (\bf{C}_i)_a \\
    &= \sum_i M_{ai} x_i \\
    &= \sum_i (\bf{R}_a)_i x_i \\
    &= \bf{R}_a \cdot \bf{x}
\end{align*}
Now regard properties of $T$ as properties of $M$.
\[ \mathrm{Im}(T) = \mathrm{Im}(M) =
\mathrm{span}\{\bf{C}_1, \dots, \bf{C}_n\} \]
\emph{image of $M$} (or $T$) is span of column $S$.
\[ \mathrm{Ker}(T) = \mathrm{Ker}(M) = \{\bf{x} :
\bf{R}_a \cdot \bf{x} = 0 \,\,\forall a\} \]
\emph{kernel of $M$} is subspace $\perp$ all rows

\subsubsection*{(b) Examples}

(Refer to sections 4.1 and 4.2)

\begin{enumerate}[(i)]
    \item Zero map $\RR^n \to \RR^m$ corresponds
        to \emph{zero matrix} $M = 0$ with $M_{ai}
        = 0$.
    \item Identity map $\RR^n \to \RR^n$
        corresponds to \emph{identity matrix}
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/9c9d4c0a37db11ec.png}
        \end{center}
        with $I_{ij} = \delta_{ij}$.
    \item $\RR^3 \to \RR^3$, $\bf{x}' = T(\bf{x})
        = M \bf{x}$ with
        \[ M =
        \begin{pmatrix}
            3 & 1 & 5 \\
            -1 & 0 & -2 \\
            2 & 1 & 3
        \end{pmatrix}
        , \bf{C}_1 =
        \begin{pmatrix}
            3 \\
            -1 \\
            2
        \end{pmatrix}
        , \bf{C}_2 =
        \begin{pmatrix}
            1 \\
            0 \\
            1
        \end{pmatrix}
        , \bf{C}_3 =
        \begin{pmatrix}
            5 \\
            -2 \\
            3
        \end{pmatrix}
        \]
        \begin{align*}
            \mathrm{Im}(T)
            &= \mathrm{Im}(M) \\
            &= \mathrm{span}\{\bf{C}_1, \bf{C}_2,
            \bf{C}_3\} \\
            &= \mathrm{span}\{\bf{C}_1, \bf{C}_2\}
            \qquad \text{since $\bf{C}_3 =
            2\bf{C}_1 - \bf{C}_2$}
        \end{align*}
        \begin{align*}
            \bf{R}_1
            &=
            \begin{pmatrix}
                3 & 1 & 5
            \end{pmatrix} \\
            \bf{R}_2
            &=
            \begin{pmatrix}
                -1 & 0 & -2
            \end{pmatrix} \\
            \bf{R}_3
            &=
            \begin{pmatrix}
                2 & 1 & 3
            \end{pmatrix} \\
            \bf{R}_2 \times \bf{R}_3
            &=
            \begin{pmatrix}
                2 & -1 & -1
            \end{pmatrix}
            = \bf{u}
        \end{align*}
        and we can notice that $\bf{u}$ is
        perpendicular to all rows. In fact
        \[ \mathrm{Ker}(T) = \mathrm{Ker}(M) =
        \{\lambda\bf{u}\} .\]
    \item Rotation through $\theta$ about $\bf{0}$
        in $\RR^2$
        \begin{align*}
            \bf{e}_1 =
            \begin{pmatrix}
                1 \\
                0
            \end{pmatrix}
            &\mapsto
            \begin{pmatrix}
                \cos\theta \\
                \sin\theta
            \end{pmatrix}
            = \bf{C}_1 \\
            \bf{e}_2 =
            \begin{pmatrix}
                0 \\
                1
            \end{pmatrix}
            &\mapsto
            \begin{pmatrix}
                -\sin\theta \\
                \cos\theta
            \end{pmatrix}
            = \bf{C}_2 \\
            \implies M
            &=
            \begin{pmatrix}
                \cos\theta & -\sin\theta \\
                \sin\theta & \cos\theta
            \end{pmatrix}
            .
        \end{align*}
    \item Dilation $\bf{x}' = M\bf{x}$ with scale
        factors $\alpha$, $\beta$, $\gamma$ along
        axes in $\RR^3$:
        \[ M =
        \begin{pmatrix}
            \alpha & 0 & 0 \\
            0 & \beta & 0 \\
            0 & 0 & \gamma
        \end{pmatrix}
        \]
    \item Reflection in plane $\perp \bf{n}$ (unit
        vector) matrix $H$:
        \[ \bf{x}' = H\bf{x} = \bf{x} - 2(\bf{x}
        \cdot \bf{n}) \bf{n} \]
        \begin{align*}
            \bf{x}_i'
            &= x_i - 2x_j n_j n_i \\
            &= (\delta_{ij} - 2n_i n_j) x_j \\
            H_{ij}
            &= \delta_{ij} - 2n_i n_j
        \end{align*}
        For example
        \[ \bf{n} = \frac{1}{\sqrt{3}}
        \begin{pmatrix}
            1 \\
            1 \\
            1
        \end{pmatrix}
        \qquad\qquad n_i n_j = \frac{1}{3}
        \,\,\forall i, j \]
        \[ H = \frac{1}{3}
        \begin{pmatrix}
            1 & -2 & -2 \\
            -2 & 1 & -2 \\
            -2 & -2 & 1
        \end{pmatrix}
        \]
    \item Shear $\bf{x}' = S \bf{x} = \bf{x} +
        \lambda (\bf{b} \cdot \bf{x}) \bf{a}$
        \[ \bf{x}_i' = S_{ij} x_j \]
        with
        \[ S_{ij} = \delta_{ij} + \lambda a_i b_j \]
        for example in $\RR^2$ with $\bf{a} =
        \begin{pmatrix}
            1 \\
            0
        \end{pmatrix}
        $ and $\bf{b} =
        \begin{pmatrix}
            0 \\
            1
        \end{pmatrix}
        $
        \[ S =
        \begin{pmatrix}
            1 & \lambda \\
            0 & 1
        \end{pmatrix}
        \]
    \item Rotation win $\RR^3$ with axis $\bf{n}$
        and angle $\theta$,
        \[ \bf{x}' = R\bf{x} \qquad x_i' = R_{ij}
        x_j \]
        where $R_{ij} = \delta_{ij} \cos\theta +
        (1 - \cos\theta) n_i n_j - (\sin\theta)
        \eps{ijk} n_k$ (see Example Sheet 2).
\end{enumerate}

\subsubsection*{(c) Isometries, area and determinant in $\RR^2$}

Consider linear map $\RR^2 \to \RR^2$ given by a
$2 \times 2$ matrix $M$:
\[ \bf{x} \mapsto \bf{x}' = M\bf{x} \]
\begin{enumerate}[(i)]
    \item When is $M$ an \emph{isometry}
        preserving lengths $|\bf{x}'| = |\bf{x}|$.
        This is equivalent to preserving inner
        products
        \[ \bf{x}' \cdot \bf{y}' = \bf{x} \cdot
        \bf{y} \]
        (since $\bf{x} \cdot \bf{y} =
        \half(|\bf{x} + \bf{y}|^2 - |\bf{x}|^2 -
        |\bf{y}|^2)$). Necessary conditions are
        \[ M
        \begin{pmatrix}
            1 \\
            0
        \end{pmatrix}
        =
        \begin{pmatrix}
            \cos\theta \\
            \sin\theta
        \end{pmatrix}
        \qquad \text{for some $\theta$; most general
        unit vector in $\RR^2$} \]
        \[ M
        \begin{pmatrix}
            0 \\
            1
        \end{pmatrix}
        = \pm
        \begin{pmatrix}
            -\sin\theta \\
            \cos\theta
        \end{pmatrix}
        \qquad \text{general unit vector
        perpendicular to other} \]
        Simple to check that these conditions are
        also sufficient and have two cases:
        \[ M = R =
        \begin{pmatrix}
            \cos\theta & -\sin\theta \\
            \sin\theta & \cos\theta
        \end{pmatrix}
        \qquad \text{\emph{rotation}} \]
        or
        \[ M = H =
        \begin{pmatrix}
            \cos\theta & \sin\theta \\
            \sin\theta & -\cos\theta
        \end{pmatrix}
        \qquad \text{\emph{reflection}} \]
        Compare with expression for reflection in
        Section 4.3(b)(vi)
        \[ H_{ij} = \delta_{ij} - 2n_i n_j \]
        and note for
        \[ \bf{n} =
        \begin{pmatrix}
            n_1 \\
            n_2
        \end{pmatrix}
        =
        \begin{pmatrix}
            -\sin\theta/2 \\
            \cos\theta/2
        \end{pmatrix}
        \]
        we get
        \[ H =
        \begin{pmatrix}
            1 - 2\sin^2 \theta/2 & 2\sin\theta/2 \cos\theta/2 \\
            2\sin\theta/2 \cos\theta/2 & 1 - 2\cos^2\theta/2
        \end{pmatrix}
        \]
        agreeing with $H$ above. This is
        reflection in a line in $\RR^2$ as shown
        \begin{center}
        \begin{tsqx}
            ! size(5cm);
            (-1.5,0)->>(1.5,0)
            (0,-1)->>(0,1)
            label $x_1$ @ (1.6,0)
            label $x_2$ @ (0,1.1)
            -1.5*dir(30)--1.5*dir(30)
            origin->>0.5*dir(120)
            label $\mathbf{n}$ @ 0.6*dir(120)
            anglemark4 (1.5,0) origin dir(20) 5
            label $\theta/2$ @ 0.5*dir(15)
        \end{tsqx}
        \end{center}
    \item How does $M$ change \emph{areas} in
        $\RR^2$ (in general)? Consider unit square
        in $\RR^2$, mapped to parallelogram as
        shown, with area
        \[ [M\bf{e}_1, M\bf{e}_2] \]
        \begin{center}
        \begin{tsqx}
            ! size(5cm);
            (0,0)->>(1,0)
            (0,0)->>(0,1)
            label $\mathbf{e}_1$ @ (1.1,0)
            label $\mathbf{e}_2$ @ (0,1.1)
            (1,0)--(1,1)--(0,1) dashed
            Me_1 := (2,1)
            Me_2 := (0.7,1.5)
            label $M\mathbf{e}_1$ @ 1.1*Me_1
            label $M\mathbf{e}_2$ @ 1.1*Me_2
            origin->>Me_1
            origin->>Me_2
            Me_1--Me_1+Me_2--Me_2 dashed
        \end{tsqx}
        \end{center}
        ``scalar cross product''
        \[ \left[
        \begin{pmatrix}
            M_{11} \\
            M_{21}
        \end{pmatrix}
        ,
        \begin{pmatrix}
            M_{12} \\
            M_{22}
        \end{pmatrix}
        \right] = M_{11}M_{22} - M_{12}M_{21} =
        \det M \]
        where $\det M$ is the \emph{determinant}
        of $2 \times 2$ matrix
        \[ M =
        \begin{pmatrix}
            M_{11} & M_{12} \\
            M_{21} & M_{22}
        \end{pmatrix}
        \]
        This is factor (with sign) by areas are
        scaled under $M$. \\
        Now compare with (i):
        \[ \det R = +1, \qquad \det H = -1 \]
        In either case $|\det M| = +1$. Consider
        shear
        \[ S =
        \begin{pmatrix}
            1 & \lambda \\
            0 & 1
        \end{pmatrix}
        ; \]
        this has $\det S = +1$ but it does not
        preserve lengths.
\end{enumerate}

\subsection{Matrices for Linear Maps in General}

Consider a linear map
\[ T : V \to W \]
between real or complex vector spaces of dimension
$n$, $m$, respectively and choose bases
$\{\bf{e}_i\}$ with $i = 1, \dots, n$ for $V$ and
$\{\bf{f}_a\}$ with $a = 1, \dots, m$ for $W$. The
matrix $M$ for $T$ with respect to these bases is
an $m \times n$ array with entries $M_{ai} \in
\RR \text{ or } \CC$ . It is defined by
\[ T(\bf{e}_i) = \sum_a \bf{f}_a M_{ai} \]
note index positions. This is chosen to ensure
that $T(\bf{x}) = \bf{x}'$ where
\[ \bf{x} = \sum_i x_i \bf{e}_i \]
and
\[ \bf{x}' = \sum_a x_a' \bf{f}_a \]
if and only if
\[ x_a' = \sum_i M_{ai} x_i \]
i.e.
\[
\begin{pmatrix}
    x_1' \\
    \vdots \\
    x_m'
\end{pmatrix}
=
\begin{pmatrix}
    M_{11} & \cdots & M_{1n} \\
    \vdots & \ddots & \vdots \\
    M_{m1} & \cdots & M_{mn}
\end{pmatrix}
\begin{pmatrix}
    x_1 \\
    \vdots \\
    x_n
\end{pmatrix}
\]

\begin{moral*}
    Given choice of bases $\{\bf{e}_i\}$ and
    $\{\bf{f}\}$
    \begin{itemize}
        \item $V$ is identified with $\RR^n$ (or
            $\CC^n$)
        \item $W$ is identified with $\RR^m$ (or
            $\CC^m$)
        \item $T$ is identified with $m \times n$
            matrix $M$
    \end{itemize}
\end{moral*}

\begin{note*}
    There are natural ways to combine linear maps.
    \\
    If $S : V \to W$ is also linear, then so is
    \[ \alpha T + \beta S : V \to W \]
    defined by
    \[ (\alpha T + \beta S)(\bf{x}) = \alpha
    T(\bf{x}) + \beta S(\bf{x}) \]
    Or if $S : U \to V$ is also linear, then so is
    \[ T \circ S : U \to W \]
    composition of maps.
\end{note*}

\subsection{Matrix Algebra}

\subsubsection*{(a) Linear Combinations}

If $M$ and $N$ are $m \times n$ matrices, then
$\alpha M + \beta N$ is an $m \times n$ matrix
defined by
\[ (\alpha M + \beta N)_{ai} = \alpha M_{ai} +
\beta N_{ai} \]
($a = 1, \dots, m$; $i = 1, \dots, n$) [If $M$,
$N$ represent linear maps $T, S : V \to W$, then
$\alpha M + \beta N$ represents $\alpha T + \beta
S$, all with respect to same choice of bases.]

\subsubsection*{(b) Matrix Multiplication}

If $A$ is an $m \times n$ matrix, entries $A_{ai}$
($\in \RR \text{ or } \CC$) and $B$ is an $n
\times p$ matrix, entries $B_{ir}$, then $AB$ is
an $m \times p$ matrix defined by
\[ (AB)_{ar} = A_{ai} B_{ir} \]
The product $AB$ is not defined unless
\[ \text{\# cols of $A$} = \text{\# rows of $B$} \]
\[ a = 1, \dots, m \]
\[ i = 1, \dots, n \]
\[ r = 1, \dots, p .\]
Matrix multiplication corresponds to composition
of linear maps
\[ [(AB)\bf{x}]_a = (AB_{ar})x_r \]
and compare
\begin{align*}
    [A(B\bf{x})]_a
    &= A_{ai} (B\bf{x})_i \\
    &= A_{ai}(B_{ir}x_r) \\
    &= (A_{ai}B_{ir}) x_r
\end{align*}

\begin{example*}
    \[ A =
    \begin{pmatrix}
        1 & 3 \\
        -5 & 0 \\
        2 & 1
    \end{pmatrix}
    ,\qquad B =
    \begin{pmatrix}
        1 & 0 & -1 \\
        2 & -1 & 3
    \end{pmatrix}
    \]
    \[ AB =
    \begin{pmatrix}
        7 & -3 & 8 \\
        -5 & 0 & 5 \\
        4 & -1 & 1
    \end{pmatrix}
    \]
    \[ BA =
    \begin{pmatrix}
        -1 & 2 \\
        13 & 9
    \end{pmatrix}
    \]
\end{example*}

\noindent \textbf{Helpful points of view}

\begin{enumerate}[(i)]
    \item Regarding $\bf{x} \in \RR^n$ as a column
        vector or $n \times 1$ matrix, definition
        of matrix multiplying a matrix as vector
        agree.
    \item For product $AB$ ($A$ is $m \times n$,
        $B$ is $n \times p$) have columns
        \[ \bf{C}_r(B) \in \RR^n \]
        \[ \bf{C}_r(AB) \in \RR^m \]
        related by
        \[ \bf{C}_r(AB) = A\bf{C}_r(B) \]
    \item \eqnoskip
        \[ AB =
        \begin{pmatrix}
            & \vdots & \\
            \leftarrow & \bf{R}_a(A) & \rightarrow \\
            & \vdots &
        \end{pmatrix}
        \begin{pmatrix}
            & \uparrow & \\
            \cdots & \bf{C}_r(B) & \cdots \\
            & \downarrow &
        \end{pmatrix}
        \]
        \begin{align*}
            A(B)_{ar}
            &= [\bf{R}_a(A)]_i [\bf{C}_r(B)]_i \\
            &= \bf{R}_a(A) \cdot \bf{C}_r(B)
        \end{align*}
        dot product in $\RR^n$ for real matrices.
\end{enumerate}

\bigskip

\noindent
\ul{Properties} of matrix products
\begin{align*}
    (\lambda M + \mu N)P
    &= \lambda (MP) + \mu (NP) \\
    P(\lambda M + \mu N)
    &= \lambda (PM) + \mu(PN) \\
    (MN)P
    &= M(NP)
\end{align*}

\subsubsection*{(c) Matrix Inverses}

Consider a $m \times n$ matrix and $B$, $C$ $n
\times m$, $B$ is a \emph{left} inverse for $A$ if
\[ BA = I \qquad (n \times n) ;\]
$C$ is a \emph{right} inverse for $A$ if
\[ AC = I \qquad (m \times m) .\]
If $m = n$, and $A$ is \emph{square}, one of these
implies the other and $B = C = A^{-1}$ \emph{the}
inverse.
\[ AA^{-1} = A^{-1}A = I .\]
Not every matrix has an inverse; if it does it is
called \emph{invertible} or \emph{non-singular}.
\\
Consider map $\RR^N \to \RR^n$ given by real
matrix $M$. If $\bf{x}' = M\bf{x}$ and $M^{-1}$
exists then $\bf{x} = M^{-1} \bf{x}'$. \\
For $n = 2$,
\begin{align*}
    x_1'
    &= M_{11}x_1 + M_{12}x_2 \\
    x_2'
    &= M_{21}x_1 + M_{22}x_2 \\
    \implies M_{22} x_1' - M_{12} x_2'
    &= (\det M) x_1 \\
    \text{and }-M_{21} x_1' + M_{11} x_2'
    &= (\det M) x_2
\end{align*}
So, if $\det M = M_{11} M_{22} - M_{12}M_{21} \neq
0$ then
\[ M^{-1} = \frac{1}{\det M}
\begin{pmatrix}
    M_{22} & -M_{12} \\
    -M_{21} & M_{11}
\end{pmatrix}
\]

\ul{Examples}
\begin{align*}
    R(\theta)
    &=
    \begin{pmatrix}
        \cos\theta & -\sin\theta \\
        \sin\theta & \cos\theta
    \end{pmatrix}
    \\
    R(\theta)^{-1}
    &= R(-\theta) \\
    H(\theta)
    &=
    \begin{pmatrix}
        \cos\theta & \sin\theta \\
        \sin\theta & -\cos\theta
    \end{pmatrix}
    \\
    H(\theta)^{-1}
    &= H(\theta) \\
    S(\lambda)
    &=
    \begin{pmatrix}
        1 & \lambda \\
        0 & 1
    \end{pmatrix}
    \\
    S(\lambda)^{-1}
    &= S(-\lambda)
\end{align*}

\subsubsection*{(d) Transpose and Hermitian Conjugate}

\begin{enumerate}[(i)]
    \item If $M$ is an $m \times n$ matrix, then
        transpose $M^\top$ is an $n \times m$ matrix
        defined by
        \[ (M^\top)_{ia} = M_{ai} \]
        ``exchange rows and columns''
        \[ a = 1, \dots, m; i = 1, \dots, n \]
        \ul{Properties}
        \[ (\alpha + \beta B)^\top = \alpha A^\top + \beta
        B^\top (A, B\,\,\,m \times n) \]
        \[ (AB)^\top = B^\top A^\top \]
        Check:
        \begin{align*}
            [(AB)^\top]_{ra}
            &= (AB)_{ar} \\
            &= A_{ai} B_{ir} \\
            &= (A^\top)_{ia} (B^\top)_{ri} \\
            &= (B^\top)_{ri} (A^\top)_{ia} \\
            &= (B^\top A^\top)_{ra} \qquad \text{as required.}
        \end{align*}
        \begin{note*}
            \[ \bf{x} =
            \begin{pmatrix}
                x_1 \\
                \vdots \\
                x_n
            \end{pmatrix}
            \qquad\text{column vector, $n \times 1$
            matrix} \]
            \[ \implies \bf{x}^\top = (x_1, \dots,
            x_n) \qquad \text{row vector, $1
            \times n$ matrix} \]
            Inner product on $\RR^n$ is
            \[ \bf{x} \cdot \bf{y} = \bf{x}^\top
            \bf{y} \qquad\text{scalar $1 \times 1$
            matrix} \]
            but $\bf{y} \bf{x}^\top = M$, $n \times
            n$ matrix with $M_{ij} = y_i x_j$.
        \end{note*}
    \item If $M$ is square, $n \times n$, then $M$
        is \emph{symmetric} if and only if $M^\top =
        M$ or $M_{ij} = M_{ji}$ and
        \emph{antisymmetric} if and only if
        $M^\top =
        -M$ or $M_{ij} = -M_{ji}$. Any square can
        be written as a sum of symmetric and
        antisymmetric parts:
        \[ M = S + A \]
        where $S = \half (M + M^\top)$ and $A =
        \half(M - M^\top)$.
        \begin{example*}
            If $A$ is $3 \times 3$ antisymmetric,
            then it can be re-written in terms of
            vector $\bf{a}$
            \[ A =
            \begin{pmatrix}
                0 & a_2 & -a_2 \\
                -a_3 & 0 & a_1 \\
                a_2 & -a_1 & 0
            \end{pmatrix}
            \]
            \[ A_{ij} = \varepsilon_{ijk} a_k
            \qquad \text{and}\qquad a_k = \half
            \varepsilon_{kij} A_{ij} \]
            Then
            \begin{align*}
                (A\bf{x})_i
                &= A_{ij} x_j \\
                &= \varepsilon_{ijk} a_k x_j \\
                &= (\bf{x} \times \bf{a})_i
            \end{align*}
        \end{example*}
    \item If $M$ is $m \times n$ matrix the
        \emph{hermitian conjugate} $M^\dag$ is
        defined by
        \[ (M^\dag)_{ia} = \ol{M}_{ai} \]
        or
        \[ M^\dag = \ol{M}^\top = \ol{(M^\top)} \]
        \ul{Properties}
        \[ (\alpha A + \beta B)^\dag = \ol{\alpha}
        A^\dag + \ol{\beta} B^\dag \]
        \[ (AB)^\dag) = B^\dag A^\dag \]
        \begin{note*}
            \[ \bf{z} =
            \begin{pmatrix}
                z_1 \\
                \vdots \\
                z_n
            \end{pmatrix}
            \qquad\text{column vector, $n \times 1$
            matrix} \]
            \[ \implies \bf{z}^\dag = (\ol{z_1},
            \dots, \ol{z_n}) \qquad \text{row
            vector, $1 \times n$ matrix} \]
            Inner product on $\CC^n$ is
            \[ (\bf{z}, \bf{w}) = \bf{z}^\dag
            \bf{w} \qquad \text{scalar $1\times 1$
            matrix} \]
        \end{note*}
    \item If $M$ is \emph{square} $n \times n$
        then $M$ is \emph{hermitian} if $M^\dag =
        M$ or $M_{ij} = \ol{M}_{ji}$ and
        \emph{anti-hermitian} if $M^\dag = -M$ or
        $M_{ij} = -\ol{M}_{ji}$.
\end{enumerate}

\subsubsection*{(e) Trace}

For any square $n \times n$ matrix $M$, the
\emph{trace} is defined by
\[ \mathrm{Tr}(M) = M_{ii} = M_{11} + \cdots +
M{nn} \qquad \text{(sum of diagonal entries)} \]

\ul{Properties}
\[ \mathrm{tr}(\alpha M + \beta N) =
\alpha \mathrm{tr}(M) + \beta \mathrm{tr}(N) \]
\[ \mathrm{tr}(MN) = \mathrm{tr}(NM) \]
check:
\begin{align*}
    (MN)_ii
    &= M_{ia}N_{ai} \\
    &= N_{ai} M_{ia} \\
    &= (NM)_{aa}
\end{align*}
\[ \mathrm{tr}(M) = \mathrm{tr}(M^\top) \]
\[ \mathrm{tr}(I) = n \qquad\text{for $I$ $n
\times n$.} \]
\[ I_{ij} = \delta_{ij} \qquad\text{and}\qquad
I_{ii} = \delta_{ii} = n \]
Previously decomposed
\[ M = S + A \qquad\text{symmetric / antisymmetric
parts} \]
Let $T = S - \frac{1}{n} (\mathrm{tr}(S)) I$ or
$T_{ij} = S_{ij} - \frac{1}{n} \mathrm{tr}(S)
S_{ij}$, then $T_{ii} = \mathrm{tr}(T) = 0$; and
note $\mathrm{tr}(M) = \mathrm{tr}(S)$ and
$\mathrm{tr}(A) = 0$. So
\[ M = \ub{T}_{\text{symm and traceless}} +
\ub{A}_{\text{antisymm part}} + \frac{1}{n}
\ub{\mathrm{tr}(M) I}_{\text{pure trace}} \]
\begin{example*}
    \[ M =
    \begin{pmatrix}
        1 & 2 & 3 \\
        4 & 5 & 6 \\
        1 & 2 & 3
    \end{pmatrix}
    , \qquad S =
    \begin{pmatrix}
        1 & 3 & 2 \\
        3 & 5 & 4 \\
        2 & 4 & 3
    \end{pmatrix}
    , \qquad A =
    \begin{pmatrix}
        0 & -1 & 1 \\
        1 & 0 & 2 \\
        -1 & -2 & 0
    \end{pmatrix}
    \]
    \[ \mathrm{tr}(S) = \mathrm{tr}(M) = 9 \]
    \[ T =
    \begin{pmatrix}
        -2 & 3 & 2 \\
        3 & 2 & 4 \\
        2 & 4 & 0
    \end{pmatrix}
    \]
    \[ M = T + A + 3I \]
    Furthermore $A\bf{x} = \bf{x} \times \bf{a}$
    where $\bf{a} = (2, -1, -1)$.
\end{example*}

\subsubsection*{Orthogonal and Unitary Matrices}

A real $n \times n$ matrix $U$ is orthogonal if
and only if
\[ U^\top U = UU^\top = I \]
i.e.
\[ U^\top = U^{-1} \]
These conditions can be written
\[ U_{ki}U_{kj} = U_{ik}U_{jk} = \delta_{ij} \]
(the left implies the columns are orthonormal, and
the middle implies that the rows are orthonormal).
[recall $[\bf{C}_i(U)]_k = U_{ki} = [R_k(U)]_i$]
\[ \ub{
\begin{pmatrix}
    & \vdots & \\
    \leftarrow & \bf{C}_i & \rightarrow \\
    & \vdots &
\end{pmatrix}
}_{U^\top}
\ub{
\begin{pmatrix}
    & \uparrow & \\
    \cdots & \bf{C}_j & \cdots \\
    & \downarrow &
\end{pmatrix}
}_{U} = I
\]
\[ \bf{C}_i \cdot \bf{C}_j = \delta_{ij} \]
\emph{Equivalent definition} $U$ is orthogonal if
and only if it preserves the inner product on
$\RR^n$
\[ (U\bf{x}) \cdot (U\bf{y}) = \bf{x} \cdot \bf{y}
\,\,\forall \bf{x}, \bf{y} \in \RR^n \]
To check equivalence, write this as
\[ (U\bf{x})^\top(U\bf{y}) = \bf{x}^\top \bf{y} \]
\begin{align*}
    LHS
    &= (\bf{x}^\top U^\top)(U\bf{y}) \\
    &= \bf{x}^\top (U^\top U) \bf{y} \\
    &= RHS \,\,\forall \bf{x}, \bf{y}
\end{align*}
if and only if $U^\top U = I$. Note, since
$\bf{C}_i = U\bf{e}_i$, columns are orthonormal is
equivalent to
\[ (U\bf{e}_i) \cdot (\bf{U} \bf{e}_j) = \bf{e}_i
\cdot \bf{e}_j = \delta_{ij} \]

\bigskip

\noindent
\textbf{Examples} \\
In $\RR^2$ we found all orthogonal matrices
(section 4.3(c)):
\[ \text{rotations } R(\theta) =
\begin{pmatrix}
    \cos\theta & -\sin\theta \\
    \sin\theta & \cos\theta
\end{pmatrix}
\]
\[ \text{and reflections } H(\theta) =
\begin{pmatrix}
    \cos\theta & \sin\theta \\
    \sin\theta & -\cos\theta
\end{pmatrix}
\]
Clearly
\[ R(\theta)^\top = R(-\theta) = R(\theta)^{-1} \]
\[ H(\theta)^\top = H(\theta) = H(\theta)^{-1} \]
In $\RR^3$ found matrix $R(\theta)$ for rotation
through $\theta$ about axis $\bf{n}$
\[ R(\theta)^\top = R(-\theta) \]
since
\[ R(\theta)_{ij} = R(-\theta)_{ji} \]
and can check explicitly
\[ R(\theta)^\top R(\theta) = R(-\theta) R(\theta)
= I \]
or
\[ R(\theta)_{ki} R(\theta)_{kj} = \delta_{ij} \]
A complex $n \times n$ matrix $U$ is
\emph{unitary} if and only if
\[ U^\dag U = UU^\dag = I \]
i.e.
\[ U^\dag = U^{-1} \]
\emph{Equivalent definition}: $U$ is unitary if
and only if it preserves the inner product on
$\CC^n$
\[ (U\bf{z}, U\bf{w}) = (\bf{z}, \bf{w})
\,\,\forall \bf{z}, \bf{w} \in \CC^n \]
To check equivalence write this as
\[ (U\bf{z})^\dag (U\bf{w}) = \bf{z}^\dag \bf{w} \]
\begin{align*}
    LHS
    &= (\bf{z}^\dag U^\dag) (U \bf{w}) \\
    &= \bf{z}^\dag (U^\dag U)\bf{w} \\
    &= RHS \,\,\forall \bf{z}, \bf{w}
\end{align*}
if and only if $U^\dag U = I$.

\newpage
\section{Determinants and Inverses}

\subsection{Introduction}

Consider a linear map
\[ T : \RR^n \to \RR^n \]
If $T$ is invertible then 
\[ \ub{\mathrm{Ker} \,\,T =
\{\bf{0}\}}_{\substack{\text{because $T$}
\\\text{one-to-one}}}
\qquad \text{and} \qquad \ub{\mathrm{Im}\,\,T =
\RR^n}_{\text{$T$ is onto}} \]
These conditions are equivalent by rank-nullity.
Conversely, if these conditions hold, then
\[ \bf{e}_1' = T(\bf{e}_1), \dots, \bf{e}_n' =
T(\bf{e}_n) \]
is a basis (where $\{\bf{e}_i\}$ standard basis)
and we can define a linear map $T^{-1}$ by
\[ T^{-1}(\bf{e}_1') = \bf{e}_1, \dots,
T^{-1}(\bf{e}_n') = \bf{e}_n \]
How can we test whether the conditions hold from
matrix $M$ representing $T$:
\[ T(\bf{x}) = M\bf{x} \]
and how can we find $M^{-1}$ when they do hold? \\
For any $M$ ($n \times n$) we will define a
related matrix $\widetilde{M}$ ($n \times n$) and a
scalar, the \emph{determinant} $\det(M)$ or $|M|$
such that
\[ \widetilde{M}M = (\det M) I \tag{$*$} \]
Then if $\det M \neq 0$, $M$ is invertible with
\[ M^{-1} = \frac{1}{\det M} \widetilde{M} \]
\ul{For $n = 2$} we found in section 4.4(c) that
($*$) holds with
\[ M =
\begin{pmatrix}
    M_{11} & M_{12} \\
    M_{21} & M_{22}
\end{pmatrix}
\qquad \text{ and } \qquad
\widetilde{M} =
\begin{pmatrix}
    M_{22} & -M_{12} \\
    -M_{21} & M_{11}
\end{pmatrix}
\]
and
\begin{align*}
    \det M
    &= \left|
    \begin{matrix}
        M_{11} & M_{12} \\
        M_{21} & M_{22}
    \end{matrix}
    \right| \\
    &= M_{11}M_{22} - M_{12}M_{21} \\
    &= [M\bf{e}_1, M\bf{e}_2] \\
    &= [\bf{C}_1(M), \bf{C}_2(M)] \\
    &= \eps_{ij} M_{i1}M_{j2}
\end{align*}
Factor by which areas are scaled under $M$
\[ \det M \neq 0 \iff \{M\bf{e}_1, M\bf{e}_2\}
\text{ linearly independent } \iff \mathrm{Im}(M) =
\RR^2 \]
\ul{For $n = 3$} consider similarly
\begin{align*}
    [M\bf{e}_1, M\bf{e}_2, M\bf{e}_3]
    &= [\bf{C}_1(M), \bf{C}_2(M), \bf{C}_3(M)]
    &\text{scalar triple product} \\
    &= \eps_{ijk} M_{i1} M_{j2} M_{k3} \\
    &= \det M, \qquad \text{definition for $n = 3$}
\end{align*}
This is factor by which volumes are scaled under
$M$ and
\[ \det M \neq 0 \iff \{M\bf{e}_1, M\bf{e}_2,
M\bf{e}_3\} \text{ linearly independent } \iff
\mathrm{Im}(M) = \RR^3 \]
Now define $\widetilde{M}$ from $M$ using rows /
column notation:
\[ \bf{R}_1(\widetilde{M}) = \bf{C}_2(M) \times
\bf{C}_3(M) \]
\[ \bf{R}_2(\widetilde{M}) = \bf{C}_3(M) \times
\bf{C}_1(M) \]
\[ \bf{R}_3(\widetilde{M}) = \bf{C}_1(M) \times
\bf{C}_2(M) \]
and note that
\begin{align*}
    (\widetilde{M}M)_{ij}
    &= \bf{R}_i(\widetilde{M}) \cdot \bf{C}_j(M)
    \\
    &= \ub{(\bf{C}_1(M) \cdot \bf{C}_2(M) \times
    \bf{C}_3(M))}_{\det M} \delta_{ij}
\end{align*}
as required.

\begin{example*}
    \[ M =
    \begin{pmatrix}
        1 & 3 & 0 \\
        0 & -1 & 2 \\
        4 & 1 & -1
    \end{pmatrix}
    \]
    \[ \bf{C}_2 \times \bf{C}_3 =
    \begin{pmatrix}
        3 \\
        -1 \\
        1
    \end{pmatrix}
    \times
    \begin{pmatrix}
        0 \\
        2 \\
        -1
    \end{pmatrix}
    =
    \begin{pmatrix}
        -1 \\
        3 \\
        6
    \end{pmatrix}
    \]
    \[ \bf{C}_3 \times \bf{C}_1 =
    \begin{pmatrix}
        0 \\
        2 \\
        -1
    \end{pmatrix}
    \times
    \begin{pmatrix}
        1 \\
        0 \\
        4
    \end{pmatrix}
    =
    \begin{pmatrix}
        8 \\
        -1 \\
        -2
    \end{pmatrix}
    \]
    \[ \bf{C}_1 \times \bf{C}_2 =
    \begin{pmatrix}
        1 \\
        0 \\
        4
    \end{pmatrix}
    \times
    \begin{pmatrix}
        3 \\
        -1 \\
        1
    \end{pmatrix}
    =
    \begin{pmatrix}
        4 \\
        11 \\
        -1
    \end{pmatrix}
    \]
    \[ \widetilde{M} =
    \begin{pmatrix}
        -1 & 3 & 6 \\
        8 & -1 & -2 \\
        4 & 1 & -1
    \end{pmatrix}
    \]
    and $\widetilde{M}M = (\det M)I$ where
    \[ \det M = \bf{C}_1 \cdot \bf{C}_2 \times
    \bf{C}_3 = 23 .\]
\end{example*}

\subsection{$\eps$ and Alternating Forms}

\subsubsection*{(a) $\eps$ and Permutations}

Recall: a permutation $\sigma$ on the set $\{1, 2,
\dots, n\}$ is a bijection from this set to
itself, specified by list
\[ \sigma(1), \sigma(2), \dots, \sigma(n) \]
Permutation $\sigma$ form a group, the symmetric
group $S_n$ of order $n!$. The \emph{sign} or
\emph{signature} $\eps(\sigma) = (-1)^k$ where $K$
is the number of transpositions (this is
well-defined). The \emph{alternating} or $\eps$
symbol in $\RR^n$ or $\CC^n$ is defined by
\[ \eps_{\ub{ij\dots }_{\text{$n$ indices}}} =
\begin{cases}
    +1 & \text{if $i, j, \dots, l$ is an even
    permutation} \\
    -1 & \text{if $i, j, \dots, l$ is an odd
    permutaiton} \\
    0 & \text{else}
\end{cases} \]
If $\sigma$ any permutation of $1, 2, \dots, n$
then
\[ \eps_{\sigma(1)\sigma(2)\cdots\sigma(n)} =
\eps(\sigma) \]

\begin{lemma}
    \[ \sigma_{\sigma(i)\sigma(j)\cdots\sigma(l)}
    = \eps(\sigma) \eps_{ij\cdots l}\]
    ($\eps$ totally antisymmetric is a corollary)
\end{lemma}

\begin{proof}
    If $i, j, \dots, l$ is \emph{not} a
    permutation of $1, 2, \dots, n$ then $RHS =
    LHS = 0$. If $i = \rho(1)$, $j = \rho(2)$,
    \dots, $l = \rho(n)$ for some permutation
    $\rho$ then
    \[ RHS = \eps(\sigma) \eps(\rho) =
    \eps(\sigma\rho) = LHS \]
    as required.
\end{proof}

\subsubsection*{(b) Alternating Forms and Linear (In)dependence}

Given $\bf{v}_1, \dots, \bf{v}_n \in \RR^n
\text{ or }\CC^n$ the \emph{alternating form}
combines them to produce scalar, defined by
\begin{align*}
    [\bf{v}_1, \bf{v}_2, \dots, \bf{v}_n]
    &= \eps_{ij\cdots l} (\bf{v}_1)_i (\bf{v}_2)_j
    \cdots (\bf{v}_n)_l \\
    &= \sum_\sigma \eps(\sigma)
    (\bf{v}_1)_{\sigma(1)} (\bf{v}_2)_{\sigma(2)}
    \vdots (\bf{v}_n)_{\sigma(n)}
\end{align*}
($\sum_\sigma$ means sum over all $\sigma \in S_n$)

\bigskip

\noindent
\textbf{Properties} \\
\begin{enumerate}[(i)]
    \item \emph{Multilinear}
        \begin{align*}
            [\bf{v}_1, \dots, \bf{v}_{p - 1},
            \alpha \bf{u} + \beta \bf{w}, \bf{v}_{p
            + 1}, \dots, \bf{v}_n] =
            &\alpha [\bf{v}_1,
            \dots, \bf{v}_{p - 1}, \bf{u}, \bf{v}_{p +
            1}, \dots, \bf{v}_n] \\
            &+ \beta [\bf{v}_1,
            \dots, \bf{v}_{p - 1}, \bf{w}, \bf{w}_{p +
            1}, \dots, \bf{v}_n]
        \end{align*}
    \item \emph{Totally antisymmetric}
        \[ [\bf{v}_{\sigma(1)}, \dots,
        \bf{v}_{\sigma(n)}] = \eps(\sigma)
        [\bf{v}_1, \dots, \bf{v}_n] \]
    \item $[\bf{e}_1, \bf{e}_2, \dots, \bf{e}_n] =
        1$ for $\bf{e}_i$ standard basis vectors.
        Properties (i), (ii), (iii) fix the
        alternating form, and they also imply
    \item If $\bf{v}_p = \bf{v}_q$ for some $p
        \neq q$ then
        \[ [\bf{v}_1, \dots, \bf{v}_p, \dots,
        \bf{v}_q, \dots, \bf{v}_n] = 0 \]
        (from (ii), exchanging $\bf{v}_p
        \leftrightarrow \bf{v}_q$ changes sign of
        alternating form).
    \item If $\bf{v}_p = \sum_{i \neq p} \lambda_i
        \bf{v}_i$ then
        \[ [\bf{v}_1, \dots, \bf{v}_p, \dots,
        \bf{v}_n] = 0 \]
        (sub in and use (i) and (iv)).
\end{enumerate}

\begin{example*}
    In $\CC^4$,
    \[ \bf{v}_1 =
    \begin{pmatrix}
        i \\
        0 \\
        0 \\
        2
    \end{pmatrix}
    , \qquad \bf{v}_2 =
    \begin{pmatrix}
        0 \\
        0 \\
        5i \\
        0
    \end{pmatrix}
    ,\]
    \[ \bf{v}_3 =
    \begin{pmatrix}
        3 \\
        2i \\
        0 \\
        0
    \end{pmatrix}
    , \qquad \bf{v}_4 =
    \begin{pmatrix}
        0 \\
        0 \\
        -i \\
        1
    \end{pmatrix}
    \]
    \begin{align*}
        \implies [\bf{v}_1, \bf{v}_2, \bf{v}_3,
        \bf{v}_4]
        &= 5i [\bf{v}_1, \bf{e}_3, \bf{v}_3,
        \bf{v}_4] \\
        &= 5i[i\bf{e}_1 + 2\bf{e}_4, \bf{e}_3,
        3\bf{e}_1 + 2i\bf{e}_2,
        \cancel{-i\bf{e}_3} +
        \bf{e}_4] \\
        &= 5i[i\bf{e}_1 + \cancel{2\bf{e}_4},
        \bf{e}_3, 3\bf{e}_1 + 2i\bf{e}_2,
        \bf{e}_4] \\
        &= 5i[i\bf{e}_1, \bf{e}_3,
        \cancel{3\bf{e}_1} + 2i\bf{e}_2, \bf{e}_4]
        \\
        &= (5i \cdot i \cdot 2i)
        [\bf{e}_1, \bf{e}_3, \bf{e}_2, \bf{e}_4]
        \\
        &= -10i(-1) \\
        &= 10i
    \end{align*}
\end{example*}

\begin{note*}
    Properties (i) and (iii) immediate from
    definition.
\end{note*}

\begin{proof}[of property (ii)]
    \begin{align*}
        [\bf{v}_{\sigma(1)}, \dots,
        \bf{v}_{\sigma(n)}]
        &= \sum_\rho \eps(\rho)
        \ub{[\bf{v}_{\sigma(1)}]_{\rho(1)} \cdots
        [\bf{v}_{\sigma(n)}]_{\rho(n)}}_{\text{each
        term can be rewritten}} \\
        &= \sum_\rho \eps(\rho)
        [\bf{v}_1]_{\rho\sigma^{-1}(1)} \cdots
        [\bf{v}_n]_{\rho\sigma^{-1}(n)} \\
        &= \sum_\rho \eps(\sigma) \eps(\rho')
        [\bf{v}_1]_{\rho'(1)} \cdots
        [\bf{v}_n]_{\rho'(n)} \\
        &= \eps(\sigma) \sum_{\rho'} \sigma(\rho')
        [\bf{v}_1]_{\rho'(1)} \cdots
        [\bf{v}_n]_{\rho'(n)} \\
        &= \eps(\sigma) [\bf{v}_1, \dots, \bf{v}_n]
    \end{align*}
    as claimed.
\end{proof}

\begin{proposition*}
    \[ [\bf{v}_1, \dots, \bf{v}_n] \neq 0 \iff
    \bf{v}_1, \dots, \bf{v}_n \text{ linearly
    indendent} \]
\end{proposition*}

\begin{proof}
    To show ``$\Rightarrow$'' use property (v). If
    $\bf{v}_1, \dots, \bf{v}_n$ are linearly
    dependent then $\sum \alpha_i \bf{v}_i =
    \bf{0}$ where not all coefficients are zero.
    Suppose without loss of generality that
    $\alpha_p \neq 0$, then express $\bf{v}_p$ as
    a linear combination of $\bf{v}_i$ ($i \neq
    p$) and
    \[ [\bf{v}_1, \dots, \bf{v}_n] = 0 .\]
    To show ``$\Leftarrow$'' note that
    $\bf{v}_1, \dots, \bf{v}_n$ linearly
    independent means they also span (in $\RR^n$
    or $\CC^n$) so we can write standard basis
    vectors as
    \[ \bf{e}_i = A_{ai}\bf{v}_a \]
    for some $A_{ai} \in \RR \text{ or } \CC$.
    But then
    \begin{align*}
        [\bf{e}_1, \dots, \bf{e}_n]
        &= [A_{a1} \bf{v}_a, A_{b2} \bf{v}_b,
        \dots, A_{cn} \bf{v}_c] \\
        &= A_{a1}A_{b2} \cdots A_{cn} [\bf{v}_a,
        \bf{v}_b, \dots, \bf{v}_c] \\
        &= A_{a1} A_{b2} \cdots A_{cn} \eps_{ab
        \cdots c} [\bf{v}_1, \bf{v}_2, \dots,
        \bf{v}_n]
    \end{align*}
    and $LHS = 1$, so $[\bf{v}_1, \bf{v}_2, \dots,
    \bf{v}_n] \neq 0$. Example in $\CC^4$ above:
    $\bf{v}_1$, $\bf{v}_2$, $\bf{v}_3$, $\bf{v}_4$
    linearly independent.
\end{proof}

\subsection{Determinants in $\RR^n$ and $\CC^n$}

\subsubsection*{(a) Definition}

For an $n \times n$ matrix $M$ with columns
\[ \bf{C}_a = M \bf{e}_a \]
the \emph{determinant} $\det M \text{ or } |M| \in \RR
\text{ or } \CC$ is defined by
\begin{align*}
    \det M
    &= [\bf{C}_1, \bf{C}_2, \dots, \bf{C}_n] \\
    &= [M\bf{e}_1, M\bf{e}_1, \dots, M\bf{e}_n] \\
    &= \eps_{ij \cdots l} M_{i1} M_{j2} \cdots
    M_{ln} \\
    \sum_\sigma \eps(\sigma) M_{\sigma(1)1}
    M_{\sigma(2)2} \cdots M_{\sigma(n)n}
\end{align*}

\begin{proposition*}[Tanspose Property]
    \[ \det M = \det M^\top \]
\end{proposition*}

So
\begin{align*}
    \det(M)
    &= [\bf{R}_1, \bf{R}_2, \dots, \bf{R}_n] \\
    &= \eps_{ij \cdots l} M_{1i} M_{2j} \cdots
    M_{nl} \\
    &= \sum_\sigma \eps(\sigma) M_{1\sigma(1)}
    M_{2\sigma(2)} \cdots M_{n \sigma(n)}
\end{align*}

\begin{example*}
    In $\RR^3$ or $\CC^3$
    \begin{align*}
        \det M
        &= \eps_{ijk} M_{i1} M_{j2} M_{k3} \\
        &= M_{11} \left|
        \begin{matrix}
            M_{22} & M_{23} \\
            M_{32} & M_{33}
        \end{matrix}
        \right| - M_{21} \left|
        \begin{matrix}
            M_{12} & M_{13} \\
            M_{32} & M_{33}
        \end{matrix}
        \right| + M_{31} \left|
        \begin{matrix}
            M_{12} & M_{13} \\
            M_{22} & M_{23}
        \end{matrix}
        \right|
    \end{align*}
\end{example*}

\subsubsection*{Properties}

$\det M$ is a function of rows or columns of $M$
that is
\begin{enumerate}[(i)]
    \item multilinear
    \item totally antisymmetric (or alternating)
    \item $\det I = 1$
\end{enumerate}

\begin{theorem*}
    \begin{align*}
        \det M \neq 0
        &\iff \text{cols of $M$ are linearly
        independent} \\
        &\iff \text{rows of $M$ are linearly
        independent} \\
        &\iff \mathrm{rank}\,\,M = n \qquad (M
        \,\,\,n \times n) \\
        &\iff \mathrm{Ker}\,\,M = \{\bf{0}\} \\
        &\iff M^{-1} \text{ exists}
    \end{align*}
\end{theorem*}

\begin{proof}
    All equivalences follow immediately from
    earlier results including discussion in
    section 5.1.
\end{proof}

\bigskip

\begin{proof}[of Transpose Property]
    Suffices to show
    \[ \sum_\sigma \eps(\sigma) M_{\sigma(1)1}
    \cdots M_{\sigma(n)n} = \sum_\sigma
    \eps(\sigma) M_{1\sigma(1)} \cdots M_{2
    \sigma(2)} \]
    But in a given term on the left hand side,
    \[ M_{\sigma(1)1} \cdots M_{\sigma(n)n} =
    M_{1\rho(1)} \cdots M_{n\rho(n)} \]
    by re-ordering factors, where $\rho =
    \sigma^{-1}$. Then $\eps(\sigma) = \eps(\rho)$
    and $\sum_\sigma$ equivalent to $\sum_\rho$,
    so result follows.
\end{proof}

\subsubsection*{(b) Evaluating Determinants: Expanding by Rows or Columns}

For $M$ $n \times n$, for each entry $M_{ia}$
define the \emph{minor} $M^{ia}$ to be the
determinant of $(n - 1) \times (n - 1)$ matrix
obtained by deleting row $i$ and column $a$ from
$M$.

\begin{proposition*}
    \begin{align*}
        \det M
        &= \sum_i (-1)^{i + a} M_{ia} M^{ia}
        &\text{$a$ fixed} \\
        &= \sum_a (-1)^{i + a} M_{ia} M^{ia}
        &\text{$i$ fixed}
    \end{align*}
    called expanding by (or about) column $a$ or
    row $i$ respectively.
\end{proposition*}

\begin{proof}
    See section 5.4.
\end{proof}

\begin{example*}
    \[ M =
    \begin{pmatrix}
        i & 0 & 3 & 0 \\
        0 & 0 & 2i & 0 \\
        0 & 5i & 0 & -i \\
        2 & 0 & 0 & 1
    \end{pmatrix}
    \]
    Expand by row 3 to find
    \[ \det M = \sum_a (-1)^{3 + a} M_{3a} M^{3a} \]
    \[ M_{31} = M_{33} = 0 ;\]
    \[ M_{32} = 5i, \quad M^{32} = \left|
    \begin{matrix}
        i & 3 & 0 \\
        0 & 2i & 0 \\
        2 & 0 & 1
    \end{matrix}
    \right|
    \]
    \[ M_{34} = -i, \quad M^{34} = \left|
    \begin{matrix}
        i & 0 & 3 \\
        0 & 0 & 2i \\
        2 & 0 & 0
    \end{matrix}
    \right|
    \]
    \[ M^{32} = i \left|
    \begin{matrix}
        2i & 0 \\
        0 & 1
    \end{matrix}
    \right| - 3 \left|
    \begin{matrix}
        0 & 0 \\
        2 & 1
    \end{matrix}
    \right| = i(2i) = -2 \tag{row 1}
    \]
    \[ M^{34} = i \left|
    \begin{matrix}
        0 & 2i \\
        0 & 0
    \end{matrix}
    \right| + 3 \left|
    \begin{matrix}
        0 & 0 \\
        2 & 0
    \end{matrix}
    \right| = 0 \tag{row 1}
    \]
    \[ \det M = (-1)^{3 + 2} 5i(-2) = 10i \]
    Alternatively we can expand by column 2:
    \begin{align*}
        \det M
        &= \sum_i (-1)^{2 + i} M_{i2} M^{i2} \\
        &= (-1)^{2 + 3} M_{32} M^{32} \\
        &= 10i
    \end{align*}
    (Calculated this previously as example of
    alternating form in $\CC^n$)
\end{example*}

\begin{lemma*}
    If
    \[ M = \left(
    \begin{tabular}{c|c}
        $A$ & $O$ \\
        \hline
        $O$ & $I$
    \end{tabular}
    \right) \]
    block form with $A$ an $r \times r$ matrix;
    $I$ an $(n - r) \times (n - r)$ identity, then
    $\det M = \det A$.
\end{lemma*}

\begin{proof}
    For $r = n - 1$, result follows by expanding about
    column $n$ or row $n$, and for $r < n - 1$,
    continue process.
\end{proof}

\subsubsection*{(c) Simplifying Determinants: Rows and Column Operations}

From the definitions of $\det M$ in terms of
columns ($a$) or rows ($i$) and the properties
above (including section 5.2(b)) we note the
following

\begin{itemize}
    \item \ul{Row or Column Scalings} \\
        If $\bf{R}_i \mapsto \lambda \bf{R}_i$ for
        some (fixed) $i$ or $\bf{C}_a \mapsto
        \lambda \bf{C}_i$ for some (fixed) $a$
        then $\det M \mapsto \lambda \det M$. If
        \emph{all} rows or columns are scaled, so
        $M \mapsto \lambda M$, then $\det M
        \mapsto \lambda^n \det M$.
    \item \ul{Row or Column Operations} \\
        If $\bf{R}_i \mapsto \bf{R}_i + \lambda
        \bf{R}_j$ for $i \neq j$ or $\bf{C}_a
        \mapsto \bf{C}_a + \lambda \bf{C}_b$ for
        $a \neq b$, then $\det M \mapsto \det M$.
    \item \ul{Row or Column Exchanges} \\
        If $\bf{R}_i \leftrightarrow \bf{R}_j$ for
        $i \neq j$ or $\bf{C}_a \leftrightarrow
        \bf{C}_b$ for $a \neq b$ then $\det M
        \mapsto -\det M$.
\end{itemize}

\begin{example*}
    \[ A =
    \begin{pmatrix}
        1 & 1 & a \\
        a & 1 & 1 \\
        1 & a & 1
    \end{pmatrix}
    \qquad a \in \CC \]
    Considering $\bf{C}_1 \mapsto \bf{C}_1 -
    \bf{C}_3$, which keeps the determinant
    invariant, we get:
    \begin{align*}
        \det A
        &= \det
        \begin{pmatrix}
            1 - a & 1 & a \\
            a - 1 & 1 & 1 \\
            0 & a & 1
        \end{pmatrix} \\
        &= (1 - a) \det
        \begin{pmatrix}
            1 & 1 & a \\
            -1 & 1 & 1 \\
            0 & a & 1
        \end{pmatrix}
    \end{align*}
    Now we consider $\bf{C}_2 \to \bf{C}_2 -
    \bf{C}_3$:
    \begin{align*}
        \det A
        &= (1 - a) \det
        \begin{pmatrix}
            1 & 1 - a & a \\
            -1 & 0 & 1 \\
            0 & a - 1 & 1
        \end{pmatrix} \\
        &= (1 - a)^2 \det
        \begin{pmatrix}
            1 & 1 & a \\
            -1 & 0 & 1 \\
            0 & -1 & 1
        \end{pmatrix}
    \end{align*}
    And finally $\bf{R}_1 \to \bf{R}_1 + \bf{R}_2
    + \bf{R}_3$:
    \begin{align*}
        \det A
        &= (1 - a)^2 \det
        \begin{pmatrix}
            0 & 0 & a + 2 \\
            -1 & 0 & 1 \\
            0 & -1 & 1
        \end{pmatrix} \\
        &= (1 - a)^2(a + 2) \left|
        \begin{matrix}
            -1 & 0 \\
            0 & -1
        \end{matrix}
        \right| \\
        &= (1 - a)^2(a + 2)
    \end{align*}
\end{example*}

\subsubsection*{(d) Multiplicative Property}

\begin{theorem*}
    For $n \times n$ matrices $M$ and $N$,
    \[ \det (MN) = \det(M) \det(N) .\]
\end{theorem*}

\noindent
This is based on the following lemma.

\begin{lemma*}
    \[ \eps_{i_1\dots i_n} M_{i_1a_1} \cdots
    M_{i_na_n} = (\det M) \eps_{a_1\dots a_n} \]
\end{lemma*}

\begin{proof}[of Theorem]
    \begin{align*}
        \det(MN)
        &= \eps_{i_1\dots i_n} (MN)_{i_11} \cdots
        (MN)_{i_nn} \\
        &= \eps_{i_1\dots i_n} M_{i_1k_1} N_{k_11}
        \cdots M_{i_nk_n} N_{k_nn} \\
        &= \eps_{i_1\dots i_n} M_{i_1k_1} \cdots
        M_{i_nk_n} N_{k_11} N_{k_nn} \\
        &= (\det M) \eps_{k_1\dots k_n} N_{k_11}
        \cdots N_{k_nn} \\
        &= (\det M)(\det N)
    \end{align*}
    as required.
\end{proof}

\bigskip

\begin{proof}[of Lemma]
    Use total antisymmetry of left hand side and
    right hand side and then check by taking $a_1
    = 1, \dots, a_n = n$.
\end{proof}

\subsubsection*{Examples}

\begin{enumerate}[(i)]
    \item If
        \[ M = \left(
        \begin{tabular}{c|c}
            $A$ & $O$ \\
            \hline
            $O$ & $B$
        \end{tabular}
        \right) \]
        (block form) with $A$ an $r \times r$ and
        $B$ an $(n - r) \times (n - r)$, then
        \[ \det M = \det A \cdot \det B \]
        Since
        \[ \left(
        \begin{tabular}{c|c}
            $A$ & $O$ \\
            \hline
            $O$ & $B$
        \end{tabular}
        \right) = \left(
        \begin{tabular}{c|c}
            $A$ & $O$ \\
            \hline
            $O$ & $I$
        \end{tabular}
        \right) \left(
        \begin{tabular}{c|c}
            $I$ & $O$ \\
            \hline
            $O$ & $B$
        \end{tabular}
        \right) \]
        and we can use Lemma above.
    \item $M^{-1}M = I \implies \det(M^{-1})
        \det(M) = \det(I) = 1$ so $\det(M^{-1}) =
        (\det M)^{-1}$.
    \item For $R$ real and orthogonal,
        \[ R^\top R = I \implies \det(R^\top)
        \det(R) = (\det R^2) = 1 \]
        \[ \implies \det R = \pm 1 \]
    \item For $U$ complex and unitary
        \[ U^\dag U = UI \implies \det (U^\dag)
        \det(U) = \ol{\det(U)}\det(U) =
        |\det(U)|^2 = 1 \]
        \[ \implies |\det U| = 1 \]
\end{enumerate}

\subsection{Minors, Cofactors and Inverses}

\subsubsection*{(a) Cofactors and Determinants}

Consider column $\bf{C}_a$ of matrix $M$ ($a$
fixed) and write $\bf{C}_a = \sum_i M_{ia}
\bf{e}_i$ in definition of determinant:
\begin{align*}
    \det M
    &= [\bf{C}_1, \dots, \bf{C}_{a - 1}, \bf{C}_a,
    \bf{C}_{a + 1}, \dots, \bf{C}_n] \\
    &= [\bf{C}_1, \dots, \bf{C}_{a - 1}, \sum_i
    M_{ia} \bf{e}_i, \bf{C}_{a + 1}, \dots,
    \bf{C}_n] \\
    &= \sum_i M_ia \Delta_{ia} \qquad \text{no sum
    over $a$}
\end{align*}
where the \emph{cofactor} $\Delta_{ia}$ is defined
by
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/05b7a15a4faa11ec.png}
\end{center}
introduced earlier. We have deduced
\begin{align*}
    \det M
    &= \sum_i M_{ia} \Delta_{ia} \\
    &= \sum_i M_{ia}(-1)^{i + a} M^{ia}
\end{align*}
proving proposition in section 5.3(b). [Similarly,
considering row $i$, find other expression].

\subsubsection*{(b) Adjugates and Inverses}

Reasoning as in (a) with $\bf{C}_b = \sum_i M_{ib}
\bf{e}_i$
\begin{align*}
    [\bf{C}_1, \dots, \bf{C}_{a - 1}, \bf{C}_b,
    \dots, \bf{C}_{a + 1}, \dots, \bf{C}_n]
    &= \sum_i M_{ib} \Delta_{ia} \\
    &= \begin{cases}
        \det M & \text{if $a = b$} \\
        0 & \text{if $a \neq b$}
    \end{cases}
\end{align*}
Hence
\[ \sum_i M_{ib} \Delta_{ia} = (\det M)
\delta_{ab} \]
And similarly
\[ \sum_a M_{ja} \Delta_{ia} = (\det M)
\delta_{ij} \]
Let $\Delta$ be the \emph{matrix} of cofactors
with entries $\Delta_{ia}$, and define
\emph{adjugate} $\widetilde{M} = \mathrm{adj}(M) =
\Delta^\top$. Then relations above because
\begin{align*}
    \Delta_{ia} M_{ib}
    &= (\Delta^\top)_{ai} M_{ib} \\
    &= (\Delta^\top M)_{ab} \\
    &= (\widetilde{M} M)_{ab} \\
    &= (\det M) \delta_{ab}
\end{align*}
and
\[ M_{ja} \Delta_{ia} = (M\widetilde{M})_{ji} =
(\det M) \delta_{ij} \]
This justifies ($*$) in section 5.1 with
\[ \widetilde(M) = \Delta^\top \]
and
\[ \Delta_{ia} = (-1)^{i + a} M^{ia} \]
we have
\[ \widetilde{M}M = M\widetilde{M} = (\det M) I \]
Hence if $\det M \neq 0$ then it is invertible and
\[ M^{-1} = \frac{1}{\det M} \widetilde{M} \]

\begin{example*}
    Consider
    \[ A =
    \begin{pmatrix}
        1 & 1 & a \\
        a & 1 & 1 \\
        1 & a & 1
    \end{pmatrix}
    \]
    previously found $\det A = (a - 1)^2(a + 2)$.
    Hence $A^{-1}$ exists if $a \neq 1$, $a \neq
    -2$. Matrix of cofactors is
    \[ \Delta =
    \begin{pmatrix}
        1 - a & 1 - a & a^2 - 1 \\
        a^2 - 1 & 1 - a & 1 - a \\
        1 - a & a^2 - 1 & 1 - a
    \end{pmatrix}
    \]
    e.g.
    \[ A^{12} = \left|
    \begin{matrix}
        a & 1 \\
        1 & 1
    \end{matrix}
    \right| = a - 1 \]
    \[ \Delta_{12} = (-1)^{1 + 2} A^{12} = 1 - a \]
    Adjugate $\widetilde{A} = \Delta^\top$ and
    \begin{align*}
        A^{-1}
        &= \frac{1}{\det A} \widetilde{A} \\
        &= \frac{1}{(1 - a)(a + 2)}
        \begin{pmatrix}
            1 & -(1 + a) & 1 \\
            1 & 1 & -(1 + a) \\
            -(1 + a) & 1 & 1
        \end{pmatrix}
    \end{align*}
    if $a \neq 1$, $a \neq -2$.
\end{example*}

\subsection{Systems of Linear Equations}

\subsubsection*{(a) Introduction and Nature of Solutions}

Consider a system of $n$ linear equations in $n$
unknowns $x_i$ written in vector / matrix form
\[ A\bf{x} = \bf{b} \qquad \bf{x}, \bf{b} \in
\RR^n \]
and $A$ an $n \times n$ matrix, i.e.
\[ A_{11}x_1 + \cdots + A_{1n} x_n = b_1 \]
\[ \vdots \]
\[ A_{n1}x_1 + \cdots + A_{nn} x_n = b_n \]
There are three possibilities:
\begin{enumerate}[(i)]
    \item $\det A \neq 0 \implies \text{$A^{-1}$
        exists} \implies \text{unique solution }
        \bf{x} = A^{-1}\bf{b}$
    \item $\det A = 0$ and $b \not\in
        \mathrm{Im}\,\,A \implies$ no solution.
    \item $\det A = 0$ and $b \in \mathrm{Im}\,\,A
        \implies$ infinitely many solutions.
\end{enumerate}
Elaboration: a solution exists if and only if
\[ A\bf{x}_0 = \bf{b} \text{ for some $\bf{x}_0$}
\iff \bf{b} \in \mathrm{Im}\,\,A \]
Then $\bf{x}$ is also a solution if and only
if $\bf{u} = \bf{x} - \bf{x}_0$ satisfies
\[ A\bf{u} = \bf{0} \]
\emph{homogeneous problem}. Now
\begin{align*}
    \det A \neq 0
    &\iff \mathrm{Im}\,\,A = \RR^n \\
    &\iff \mathrm{Ker}\,\,A = \{\bf{0}\}
\end{align*}
So in (i) there is a unique solution and it can be
found using $A^{-1}$. But
\begin{align*}
    \det A = 0
    &\iff \mathrm{rank}(A) < n \\
    &\iff \mathrm{null}(A) > 0
\end{align*}
and then either $\bf{b} \not\in \mathrm{Im}\,\,A$
as in case (ii) or $\bf{b} \in \mathrm{Im}\,\,A$
as in case (iii). If $\bf{u}_1, \dots, \bf{u}_k$
is a basis for $\mathrm{Ker}\,\,A$ then general
solution of homogeneous problem is
\[ \bf{u} = \sum_{i = 1}^k \lambda_i \bf{u}_i \]

\bigskip

\noindent
\textbf{Example} \\
$A\bf{x} = \bf{b}$ with $A$ as in section 5.4
and
\[ \bf{b} =
\begin{pmatrix}
    1 \\
    c \\
    1
\end{pmatrix}
\]
with $a, c \in \RR$.
\begin{itemize}
    \item \ul{$a \neq 1, -2$} \\
        Then $A^{-1}$ exists and we have a
        solution for any $c$:
        \[ \bf{x} = A^{-1}\bf{b} = \frac{1}{(1
        - a)(a + 2)}
        \begin{pmatrix}
            2 - c - ca \\
            c - a \\
            c - a
        \end{pmatrix}
        \]
    \item \ul{$a = 1$} \\
        \[ A =
        \begin{pmatrix}
            1 & 1 & 1 \\
            1 & 1 & 1 \\
            1 & 1 & 1
        \end{pmatrix}
        \]
        \[ \mathrm{Im}\,\,A = \left\{\lambda
        \begin{pmatrix}
            1 \\
            1 \\
            1
        \end{pmatrix}
        \right\} \qquad \mathrm{Ker}\,\,A =
        \mathrm{span} \left\{
        \begin{pmatrix}
            -1 \\
            1 \\
            0
        \end{pmatrix}
        ,
        \begin{pmatrix}
            -1 \\
            0 \\
            1
        \end{pmatrix}
        \right\} \]
        $b \in \mathrm{Im}\,\,A$ if and only
        if $c = 1$, particular solution
        \[ \bf{x}_0 =
        \begin{pmatrix}
            1 \\
            0 \\
            0
        \end{pmatrix}
        \]
        general solution
        \[ \bf{x} = \bf{x}_0 + \bf{u} =
        \begin{pmatrix}
            1 - \lambda - \mu \\
            \lambda \\
            \mu
        \end{pmatrix}
        \]
        case (ii). For $a = 1$ and $c \neq 1$
        have no solutions: case (iii).
    \item \ul{$a = -2$}
        \[ A =
        \begin{pmatrix}
            1 & 1 & -2 \\
            -2 & 1 & 1 \\
            1 & -2 & 1
        \end{pmatrix}
        \]
        \[ \mathrm{Im}\,\,A = \mathrm{span}
        \left\{
        \begin{pmatrix}
            1 \\
            -2 \\
            1
        \end{pmatrix}
        \right\} , \left\{
        \begin{pmatrix}
            1 \\
            1 \\
            -2
        \end{pmatrix}
        \right\} \qquad \mathrm{Ker}\,\,A =
        \left\{\lambda
        \begin{pmatrix}
            1 \\
            1 \\
            1
        \end{pmatrix}
        \right\} \]
        $\bf{b} \in \mathrm{Im}\,\,A$ if and only
        if $c = -2$, particular solution
        \[ \bf{x}_0 =
        \begin{pmatrix}
            1 \\
            0 \\
            0
        \end{pmatrix}
        \]
        general solution
        \[ \bf{x} = \bf{x}_0 + \bf{u} =
        \begin{pmatrix}
            1 + \lambda \\
            \lambda \\
            \lambda
        \end{pmatrix}
        \]
        For $c \neq -2$ no solutions.
\end{itemize}

\subsubsection*{(b) Geometrical Intepretation in
$\RR^3$}

Let $\bf{R}_1, \bf{R}_2, \bf{R}_3$ be rows of $A$
($3 \times 3$).
\[ A\bf{u} = \bf{0} \iff \begin{cases}
    \bf{R}_1 \cdot \bf{u} = \bf{0} \\
    \bf{R}_2 \cdot \bf{u} = bf{0} \\
    \bf{R}_3 \cdot \bf{u} = \bf{0}
\end{cases} \]
(these are 3 equations of planes through $\bf{0}$,
normals $\bf{R}_i$, assuming $\neq \bf{0}$). So
solutions of homogeneous problem (finding
$\mathrm{Ker}\,\,A$) given by intersection of
these planes.
\[ \mathrm{rank}(A) = 3 \implies \text{normals
linearly independent and planes intersect in
$\bf{0}$} \]
\[ \mathrm{rank}(A) = 2 \implies \text{normals
span a plane and planes intersect in a line} \]
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/22f689ca4faf11ec.png}
\end{center}
\[ \mathrm{dim}\,\,\mathrm{Ker}\,\,A = 1 .\]
\[ \mathrm{rank}(A) = 1 \implies \text{normals are
parallel and planes coincide} \]
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/48be0eb24faf11ec.png}
\end{center}
Now consider instead
\[ A\bf{x} = \bf{b} \iff
\begin{cases}
    \bf{R}_1 \cdot \bf{x} = b_1 \\
    \bf{R}_2 \cdot \bf{x} = b_2 \\
    \bf{R}_3 \cdot \bf{x} = b_3
\end{cases} \]
planes with normals $\bf{R}_i$ but not passing
through $\bf{0}$ unless $b_i = 0$.
\[ \mathrm{rank}(A) = 3 \iff \det A \neq 0 ,\]
normals linearly independent; planes intersect in
a point and get unique solution for any $\bf{b}$.
\[ \mathrm{rank}(A) = 2 \implies \text{planes may
intersect in a line (as in homogeneous case)} \]
\emph{but} they may not, e.g.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/7139791a4fd911ec.png}
\end{center}
\[ \mathrm{rank}(A) = 1 \implies \text{planes may
coincide (as in homogeneous case)} \]
\emph{but} they may not, e.g.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/85e8a0a24fd911ec.png}
\end{center}

\subsubsection*{Gaussian Elimination and Echelon Form}

Consider $A\bf{x} = \bf{b}$ with $\bf{x} \in
\RR^n$ and $\bf{b} \in \RR^m$ and $A$ an $m \times
n$ matrix. Gaussian elimination is a direct
approach to solving system of equations:
\[ A_{11}x_1 + \cdots + A_{1n} x_n = b_1 \]
\[ \vdots \]
\[ A_{m1}x_1 + \cdots + A_{mn}x_n = b_m \]

\begin{example*}
    \[ 3x_1 + 2x_2 + x_3 = b_1 \tag{$1$} \]
    \[ 6x_1 + 3x_2 + 3x_3 = b_2 \tag{$2$} \]
    \[ 6x_1 + 2x_2 + 4x_3 = b_3 \tag{$3$} \]
    Step (1): subtract multiples of ($1$) from
    ($2$) and ($3$) to eliminate $x_1$:
    \[ 0 - x_2 + x_3 = b_2 - 2b_1 \tag{$2'$} \]
    \[ 0 - 2x_2 + 2x_3 = b_3 - 2b_1 \tag{$3'$} \]
    Step (2): repeat this using ($2'$) to eliminate
    $x_2$:
    \[ 0 + 0 + 0 = b_3 - 2b_2 + 2b_1 \tag{$3''$} \]
    Now consider new system ($1$), ($2'$), ($3'$)
    \[ b_3 - 2b_2 + 2b_1 \neq 0 \implies \text{no
    solution} \]
    \[ b_3 - 2b_2 + 2b_1 = 0 \text{ then
    infinitely many solutions} \]
    $x_3$ is arbitrary and then $x_2$ and $x_1$
    determined from ($2'$) and ($1$). In general
    case we aim to carry out steps as in example
    until we obtain equivalent system
    \[ M\bf{x} = \bf{d}
    \text{ with }
    M = \left(
    \begin{tabular}{c|c}
        $\hat{M}$ & numbers \\
        0 & 0
    \end{tabular}
    \right) \]
    with $M$ an $m \times n$ (block form), with
    \[ \hat{M} =
    \begin{pmatrix}
        M_{11} & & \text{numbers} \\
        & \cdots & \\
        0 & & M_{rr}
    \end{pmatrix}
    \]
    $M_{jj} \neq 0$ for each $j$. $M$ obtained
    from $A$ by row operations including row
    exchanges and column exchanges which relabel
    variables $x_i$. Note $x_{r + 1}, \dots, x_n$
    undetermined, $d_{r + 1}, \dots, d_m = 0$ else
    no solution. And if this is satisfied then
    $x_1, \dots, x_r$ determined successively.
    \[ r = \mathrm{rank}\,\,M = \mathrm{rank}\,\,A \]
    If $n = m$ then $\det A = \pm \det M$ and if
    $r = n = m$ then
    \[ \det M = M_{11} \cdots M_{rr} \neq 0 \]
    \[ \implies A \text{ and } M \text{ invertible} \]
    $M$ as above is an example of \emph{echelon
    form}.
\end{example*}

\newpage
\section{Eigenvalues and Eigenvectors}

\subsection{Introduction}

\subsubsection*{(a) Definitions}

For a linear map $T : V \to V$ ($V$ a real or
complex vector space) a vector $\bf{v} \in V$ with
$\bf{v} \neq \bf{0}$ is an \emph{eigenvector} of
$T$ with \emph{eigenvalue} $\lambda$ if
\[ T(\bf{v}) = \lambda \bf{v} \]
If $V = \RR^n \text{ or } \CC^n$ and $T$ given by
an $n \times n$ matrix $A$, then
\[ A\bf{v} = \lambda \bf{v} \iff (A - \lambda
I)\bf{v} = \bf{0} \]
and for given $\lambda$ this holds for some
$\bf{v} \neq \bf{0}$ if and only if $\det(A -
\lambda I) = 0$ \emph{characteristic equation}
i.e. $\lambda$ is an eigenvalue if and only if it
is a root of $\chi_A(t) = \det(A - tI)$
\emph{characteristic polynomial}. $\chi_A(t)$
polynomial of degree $n$ for $A$ $n \times n$. We
find eigenvalues as roots of characteristic
equation and then find corresponding eigenvectors.

\subsubsection*{(b) Examples}

\begin{enumerate}[(i)]
    \item $V = \CC^2$ and
        \[ A =
        \begin{pmatrix}
            2 & i \\
            -i & 2
        \end{pmatrix}
        \]
        then
        \[ \det(A - \lambda I) = \left|
        \begin{matrix}
            2 - \lambda & i \\
            -i & 2 - \lambda
        \end{matrix}
        \right| = (2 - \lambda)^2 - 1 = 0 \]
        if and only if $\lambda = 1 \text{ or } 3$. To
        find eigenvectors $\bf{v} =
        \begin{pmatrix}v_1\\v_2\end{pmatrix}$:
        \ul{$\lambda = 1$}:
        \[ (A - I)\bf{v} =
        \begin{pmatrix}
            1 & i \\
            -i & 1
        \end{pmatrix}
        \begin{pmatrix}
            v_1 \\
            v_2
        \end{pmatrix}
        = \bf{0}
        \]
        \[ \implies \bf{v} = \alpha
        \begin{pmatrix}
            1 \\
            i
        \end{pmatrix}
        \qquad \text{any $\alpha \neq 0$.} \]
        \ul{$\lambda = 3$}:
        \[ (A - 3I)\bf{v} =
        \begin{pmatrix}
            -1 & i \\
            -i & -1
        \end{pmatrix}
        \begin{pmatrix}
            v_1 \\
            v_2
        \end{pmatrix}
        = \bf{0} \]
        \[ \implies \bf{v} = \beta
        \begin{pmatrix}
            1 \\
            -i
        \end{pmatrix}
        \qquad \text{ any $\beta \neq 0$.} \]
    \item $V = \RR^2$
        \[ A =
        \begin{pmatrix}
            1 & 1 \\
            0 & 1
        \end{pmatrix}
        \]
        \[ \det(A - \lambda I) = \left|
        \begin{matrix}
            1 - \lambda & 1 \\
            0 & 1 - \lambda
        \end{matrix}
        \right| = (1 - \lambda)^2 = 0 \]
        \[ \implies \lambda = 1 \]
        Eigenvector:
        \[ (A - I)\bf{v} =
        \begin{pmatrix}
            0 & 1 \\
            0 & 0
        \end{pmatrix}
        \begin{pmatrix}
            v_1 \\
            v_2
        \end{pmatrix}
        = \bf{0} \]
        \[ \implies
        \begin{pmatrix}
            v_1 \\
            v_2
        \end{pmatrix}
        = \alpha
        \begin{pmatrix}
            1 \\
            0
        \end{pmatrix}
        \qquad \text{for any $\alpha \neq 0$.} \]
    \item $V = \RR^2 \text{ or } \CC^2$
        \[ U =
        \begin{pmatrix}
            \cos\theta & -\sin\theta \\
            \sin\theta & \cos\theta
        \end{pmatrix}
        \]
        \[ \chi_U(t) = \det(U - tI) = t^2 -
        2t\cos\theta + 1 \]
        Eigenvalues $\lambda = e^{\pm i\theta}$
        and eigenvectors
        \[ \bf{v} = \alpha
        \begin{pmatrix}
            1 \\
            \mp i
        \end{pmatrix}
        \qquad (\alpha \neq 0) \]
\end{enumerate}

\subsubsection*{(c) Deductions involving $\chi_A(t)$}

For $A$ an $n \times n$ matrix, characteristic
polynomial has degree $n$:
\begin{align*}
    \chi_A(t)
    &= \det
    \begin{pmatrix}
        A_{11} - t & A_{12} & \cdots & A_{1n} \\
        A_{21} & A_{22 - t} & \cdots & A_{2n} \\
        \vdots & \vdots & \ddots & \vdots \\
        A_{n1} & A_{n2} & \cdots & A_{nn - t}
    \end{pmatrix} \\
    &= \sum_{j = 0}^n c_j t^j \\
    &= (-1)^n(t - \lambda_1) \cdots (t -
    \lambda_n)
\end{align*}

\begin{enumerate}[(i)]
    \item There exists at least one eval (one root of
        $\chi_A$); in fact there exists $n$ roots
        counted with multiplicity (Fundamental
        Theorem of Algebra)
    \item $\mathrm{tr}(A) = A_{ii} = \sum_i
        \lambda_i$ sum of reals by comparing terms
        of order $n - 1$ in $t$.
    \item $\det(A) = \chi_A(0) = \prod_i
        \lambda_i$ (product of eigenvalues)
    \item If $A$ is diagonal:
        \[ A =
        \begin{pmatrix}
            \lambda_1 & \cdots & 0 \\
            \vdots & \ddots & \vdots \\
            0 & \cdots & \lambda_n
        \end{pmatrix}
        \]
        with diagonal entries eigenvalues; (ii) and
        (iii) are then immediate.
    \item If $A$ is real, coefficients $c_i$ are
        real and $\chi_A(\lambda) = 0 \iff
        \chi_A(\ol{\lambda}) = 0$: non-real
        roots occur in conjugate pairs.
\end{enumerate}

\subsection{Eigenspaces and Multiplicities}

\subsubsection*{(a) Definitions}

For an eigenvalue $\lambda$ of matrix $A$, define
the \emph{eigenspace}
\[ E_\lambda = \{\bf{v} : A\bf{v} =
\lambda\bf{v}\} = \mathrm{Ker}(A - \lambda I); \]
the \emph{geometric multiplicity}
\[ m_\lambda = \dim E_\lambda = \mathrm{null}
(A - \lambda I) .\]
(\# linearly independent eigenvalues eigenvectors with
eval $\lambda$); \\
the \emph{algebraic multiplicity}
\[ M_\lambda, \text{ multiplicity of $\lambda$ as
a root of $\chi_A$} \]
i.e. $\chi_A(t) = (t - \lambda)^{M_\lambda} f(t)$
with $f(\lambda) \neq 0$.

\begin{proposition*}
    \eqnoskip
    \[ M_\lambda \ge m_\lambda \]
\end{proposition*}

\noindent
[Further discussion in section 6.3]

\subsubsection*{(b) Examples}

\begin{enumerate}[(i)]
    \item Define:
        \[ A =
        \begin{pmatrix}
            -2 & 2 & -3 \\
            2 & 1 & -6 \\
            -1 & -2 & 0
        \end{pmatrix}
        \]
        \[ \chi_A(t) + \det(A - tI) = (5 - t)(t +
        3)^2 \]
        so we have roots $5$ and $-3$, with $M_5 =
        1$ and $M_{-3} = 2$.
        \begin{itemize}
            \item For $\lambda = 5$ we
                have:
                \[ (A - 5I)\bf{x} =
                \begin{pmatrix}
                    -7 & 2 & -3 \\
                    2 & -4 & -6 \\
                    -1 & -2 & -5
                \end{pmatrix}
                \begin{pmatrix}
                    x_1 \\
                    x_2 \\
                    x_3
                \end{pmatrix}
                = \bf{0} \]
                \[ \implies E_5 = \left\{\alpha
                \begin{pmatrix}
                    1 \\
                    2 \\
                    -1
                \end{pmatrix}
                \right\} \]
            \item For $\lambda = -3$ we have
                \[ (A + 3I)\bf{x} =
                \begin{pmatrix}
                    1 & 2 & -3 \\
                    2 & 4 & -6 \\
                    -1 & -2 & 3
                \end{pmatrix}
                \begin{pmatrix}
                    x_1 \\
                    x_2 \\
                    x_3
                \end{pmatrix}
                = \bf{0} \]
                Solve to find:
                \[ \bf{x} =
                \begin{pmatrix}
                    -2x_2 + 3x_3 \\
                    x_2 \\
                    x_3
                \end{pmatrix}
                \]
                or
                \[ E_{-3} = \left\{\alpha
                \begin{pmatrix}
                    -2 \\
                    1 \\
                    0
                \end{pmatrix}
                + \beta
                \begin{pmatrix}
                    3 \\
                    0 \\
                    1
                \end{pmatrix}
                \right\} \]
        \end{itemize}
        So
        \[ \dim E_5 = m_5 = 1 = M_5 \]
        \[ \dim E_{-3} = m_{-3} = 2 = M_{-3} \]
    \item Consider
        \[ A =
        \begin{pmatrix}
            -3 & -1 & 1 \\
            -1 & -3 & 1 \\
            -2 & -2 & 0
        \end{pmatrix}
        \]
        Then
        \[ \chi_A(t) = \det(A - tI) = -(t + 2)^3 \]
        roots are $\lambda = -2$, with $M_{-2} =
        3$. To find eigenvectors:
        \[ (A + 2I)\bf{x} =
        \begin{pmatrix}
            -1 & -1 & 1 \\
            -1 & -1 & 1 \\
            -2 & -2 & 2
        \end{pmatrix}
        \begin{pmatrix}
            x_1 \\
            x_2 \\
            x_3
        \end{pmatrix}
        = \bf{0} \]
        \[ \implies \bf{x} =
        \begin{pmatrix}
            -x_2 + x_3 \\
            x_2 \\
            x_3
        \end{pmatrix}
        \]
        \[ \implies E_{-2} = \left\{\alpha
        \begin{pmatrix}
            -1 \\
            1 \\
            0
        \end{pmatrix}
        + \beta
        \begin{pmatrix}
            1 \\
            0 \\
            1
        \end{pmatrix}
        \right\} \]
        so $\dim E_{-2} = m_{-2} = 2$ but $M_{-2}
        = 3$. (So we do have $M_{-2} \ge m_{-2}$.)
\end{enumerate}

\subsubsection*{(c) Linear Independence of Eigenvectors}

\begin{proposition*}
    \begin{enumerate}[(i)]
        \item Let $\bf{v}_1, \dots, \bf{v}_r$ be
            eigenvectors of matrix $A$ ($n \times
            n$) with eigenvalues $\lambda_1,
            \dots, \lambda_r$. If the eigenvalues are
            distinct, $\lambda_i \neq \lambda_j$
            for $i \neq j$, then the eigenvectors
            are linearly independent.
        \item With conditions as in (i), let
            $\mathcal{B}_{\lambda_i}$ be a basis
            for $E_{\lambda_i}$, then
            \[ \mathcal{B}_{\lambda_1} \cup
            \mathcal{B}_{\lambda_2} \cup \cdots
            \cup \mathcal{B}_{\lambda_r} \]
            is linearly independent.
    \end{enumerate}
\end{proposition*}

\begin{proof}
    \begin{enumerate}[(i)]
        \item Note
            \begin{align*}
                \bf{w}
                &= \sum_{j = 1}^r \alpha_j
                \bf{v}_j \\
                \implies (A - \lambda I)\bf{w}
                &= \sum_{j = 1}^r \alpha_j
                (\lambda_j - \lambda) \bf{v}_j
            \end{align*}
            First, suppose eigenvectors are
            linearly dependent, so there exists
            linear relations $\bf{w} = \bf{0}$
            with number of non-zero coefficients
            $p \ge 2$. Pick a $\bf{w}$ for which
            $p$ is least and assume (without loss
            of generality) that $\alpha_1 \neq 0$.
            Then
            \[ (A - \lambda_1 I)\bf{w} = \sum_{j >
            1} \alpha_j (\lambda_j - \lambda_1)
            \bf{v}_j = \bf{0} ,\]
            a linear relation with $p - 1$
            non-zero coefficients,
            $\contradiction$ ($p$ was least). \\
            Alternative second proof,
            \[ \bf{w} = \bf{0} \]
            \[ \implies \prod{_j \neq k} (A - \lambda_j I)
            \bf{w} = \alpha_k \left(\prod_{j \neq k}
            (\lambda_k - \lambda_j)\right) \bf{v}_k
            = \bf{0} \]
            (for some chosen $k$).
            \[ \implies \alpha_k = 0 \]
            so the eigenvectors are linearly
            independent.
        \item It suffices to show that if
            \[ \bf{w} = \bf{w}_1 + \bf{w}_2 +
            \cdots + \bf{w}_r = \bf{0} \]
            with $\bf{w}_i \in E_{\lambda_i}$ then
            \[ \implies \bf{w}_i = \bf{0} .\]
            This follows by same arguments as in
            (i).
    \end{enumerate}
\end{proof}

\subsection{Diagonalisability and Similarity}

\subsubsection*{(a) Introduction}

\begin{proposition*}
    For an $n \times n$ matrix $A$ acting on $V =
    \RR^n$ or $\CC^n$, the following conditions
    are equivalent
    \begin{enumerate}[(i)]
        \item There exists a basis of eigenvectors for
            $V$, $\bf{v}_1, \bf{v}_2, \dots, \bf{v}_n$
            with
            \[ A \bf{v}_i = \lambda_i \bf{v}_i \]
            (no summation convention here!)
        \item There exists an $n \times n$ invertible
            matrix $P$ with
            \[ P^{-1}AP = D =
            \begin{pmatrix}
                \lambda_1 & \cdots & 0 \\
                \vdots & \ddots & \vdots \\
                0 & \cdots & \lambda_n
            \end{pmatrix}
            \]
    \end{enumerate}
    If either of these conditions holds, $A$
    is \emph{diagonalisable}.
\end{proposition*}

\begin{proof}
    Note that for any matrix $P$, $AP$ has columns
    $A\bf{C}_i(P)$ and $PD$ has columns
    $\lambda_i\bf{C}_i(P)$ for each $i$. Then (i)
    and (ii) are related by
    \[ \bf{v}_i = \bf{c}_i(P) : P^{-1}AP = D \iff
    AP = PD \iff A\bf{v}_i = \lambda_i \bf{v}_i .\]
\end{proof}

\bigskip

\noindent
\textbf{Example} \\
Refer to section 6.1(b):
\[ U =
\begin{pmatrix}
    \cos \theta & -\sin \theta \\
    \sin \theta & \cos \theta
\end{pmatrix}
\]
eigenvalues $e^{\pm i \theta}$ and eigenvectors
$\begin{pmatrix}1\\\mp1\end{pmatrix}$. Linearly
independent over $\CC$ so
\[ P =
\begin{pmatrix}
    1 & 1 \\
    -i & i
\end{pmatrix}
\implies P^{-1} = \half
\begin{pmatrix}
    1 & i \\
    1 & -i
\end{pmatrix}
\]
and
\[ P^{-1}UP =
\begin{pmatrix}
    e^{i\theta} & 0 \\
    0 & e^{-i\theta}
\end{pmatrix}
\]
$U$ diagonalisable over $\CC$ but \emph{not} over
$\RR$.

\subsubsection*{(b) Criteria for Diagonalisability}

\begin{theorem*}
    Let $A$ be an $n \times n$ matrix and
    $\lambda_1, \dots, \lambda_r$ all its distinct
    eigenvalues.
    \begin{enumerate}[(i)]
        \item A necessary and sufficient
            condition: $A$ is diagonalisable if
            and only if
            \[ M_{\lambda_i} = m_{\lambda_i}
            \qquad \text{for $i = 1, \dots, r$} \]
        \item A sufficient condition: $A$ is
            diagonalisable if there are $n$
            distinct eigenvalues, i.e. $r = n$.
    \end{enumerate}
\end{theorem*}

\begin{proof}
    Use Proposition in section 6.2(c) \\
    For (ii) if $r = n$ we have $n$ distinct
    eigenvalues and hence $n$ linearly independent
    eigenvalues, which form a basis (for $\RR^n$
    or $\CC^n$). \\
    For (i), choosing bases
    $\mathcal{B}_{\lambda_i}$ for each eigenspace,
    \[ \mathcal{B}_{\lambda_i} \cup
    \mathcal{B}_{\lambda_2} \cup \cdots \cup
    \mathcal{B}_{\lambda_r} \]
    is a linearly independent set of
    \[ m_{\lambda_1} + m_{\lambda_2} + \cdots +
    m_{\lambda_r} \]
    vectors. It is a basis (for $\RR^n$ or
    $\CC^n$) if and only if we have $n$ vectors.
    But
    \[ m_{\lambda_i} \le M_{\lambda_i} \]
    and
    \[ M_{\lambda_1} + M_{\lambda_2} + \cdots +
    M_{\lambda_r} = n .\]
    Hence we have a basis if and only if
    \[ M_{\lambda_i} = m_{\lambda_i} \qquad
    \text{for each $i$} \]
\end{proof}

\bigskip

\noindent
\textbf{Examples} \\
Refer to section 6.2(b)
\begin{enumerate}[(i)]
    \item \eqnoskip
        \[ A =
        \begin{pmatrix}
            -2 & 2 & -3 \\
            2 & 1 & -6 \\
            -1 & -2 & 0
        \end{pmatrix}
        \]
        \[ \lambda = 5, -3, -3 \qquad M_5 = m_5 = 1
        \qquad M_{-3} = m_{-3} =
        2 \]
        hence $A$ diagonalisable.
        \[ P =
        \begin{pmatrix}
            1 & -2 & 3 \\
            2 & 1 & 0 \\
            -1 & 0 & 1
        \end{pmatrix}
        , \qquad P^{-1} = \frac{1}{8}
        \begin{pmatrix}
            1 & 2 & -3 \\
            -2 & 4 & 6 \\
            1 & 2 & 5
        \end{pmatrix}
        \]
        \[ P^{-1}AP =
        \begin{pmatrix}
            5 & 0 & 0 \\
            0 & -3 & 0 \\
            0 & 0 & -3
        \end{pmatrix}
        \]
        as expected.
    \item \eqnoskip
        \[ A =
        \begin{pmatrix}
            -3 & -1 & 1 \\
            -1 & -3 & 1 \\
            -2 & -2 & 0
        \end{pmatrix}
        \]
        \[ \lambda = -2,-2,-2 \qquad M_{-2} = 3 >
        m_{-2} = 2 \]
        hence $A$ is not diagonalisable. Check: if
        it was then
        \[ P^{-1}AP = -2I \]
        \[ \implies A = P(-2I)P^{-1} = -2I
        \contradiction .\]
\end{enumerate}

\subsubsection*{(c) Similarity}

Matrices $A$ and $B$ ($n \times n$) are
\emph{similar} if
\[ B = P^{-1}AP \]
for some invertible $P$ ($n \times n$). This is an
equivalence relation.

\begin{proposition*}
    If $A$ and $B$ are similar, then
    \begin{enumerate}[(i)]
        \item $B^r = P^{-1}A^rP$ for $r \ge 0$.
        \item $B^{-1} = P^{-1}A^{-1}P$ (if either
            $A$ or $B$ invertible, so is the
            other).
        \item $\mathrm{tr}(B) = \mathrm{tr}(A)$.
        \item $\det(B) = \det(A)$.
        \item $\chi_B(t) = \chi_A(t)$.
    \end{enumerate}
\end{proposition*}

\begin{proof}
    (i) and (ii) immediate. (iii):
    \begin{align*}
        \mathrm{tr}(B)
        &= \mathrm{tr}(P^{-1}AP) \\
        &= \mathrm{tr}(APP^{-1}) \\
        &= \mathrm{tr}(A)
    \end{align*}
    For (iv):
    \begin{align*}
        \det(B)
        &= \det(P^{-1}AP) \\
        &= \det(P^{-1})\det(A)\det(P) \\
        &= \det(A)
    \end{align*}
    For (v):
    \begin{align*}
        \det(B - tI)
        &= \det(P^{-1}AP - tI) \\
        &= \det(P^{-1}(A - tI)P) \\
        &= \det(A - tI)
    \end{align*}
    as in (iv).
\end{proof}

\subsection{Hermitian and Symmetric Matrices}

\subsubsection*{(a) Real Eigenvalues and Orthogonal Eigenvectors}

Recall: matrix $A$ ($n \times n$) is hermitian if
\[ A^\dag = \ol{A}^\top = A \qquad \text{or}
\qquad A_{ij} = \ol{A_{ji}} \]
special case: $A$ is real and symmetric
\[ \ol{A} = A \qquad A^\top = A \qquad \text{or}
\qquad
\begin{cases}
    A_{ij} = \ol{A_{ij}} \\
    A_{ij} = A_{ji}
\end{cases}
\]
Recall: complex inner-product for $\bf{v}, \bf{w}
\in \CC^n$ IS
\[ \bf{v}^\dag \bf{w} = \sum_i \ol{v_i} w_i \]
and for $\bf{v}, \bf{w} \in \RR^n$ this reduces to
\[ \bf{v}^\top \bf{w} = \bf{v} \cdot \bf{w} =
\sum_i v_i w_i \]
Observation: if $A$ is hermitian then
\[ (A\bf{v})^\dag \bf{w} = \bf{v}^\dag (A\bf{w})
\,\,\,\forall \bf{v}, \bf{w} \in \CC^n \]
[since $LHS = (\bf{v}^\dag A^\dag)\bf{w} =
\bf{v}^\dag A^\dag \bf{w} = \bf{v}^\dag A \bf{w} =
RHS$]

\begin{theorem*}
    For a matrix $A$ ($n \times n$) that is
    hermitian
    \begin{enumerate}[(i)]
        \item Every eigenvalue $\lambda$ is real
        \item Eigenvectors $\bf{v}$, $\bf{w}$ with
            distinct eigenvalues $\lambda$, $\mu$
            respectively ($\lambda \neq \mu$) are
            orthogonal
            \[ \bf{v}^\dag \bf{v} = 0 \]
        \item If $A$ is real and symmetric then
            for each $\lambda$ in (i) we can
            choose a real eigenvector $\bf{v}$ and
            (ii) becomes
            \[ \bf{v}^\top \bf{w} = \bf{v} \cdot
            \bf{w} = 0 \]
    \end{enumerate}
\end{theorem*}

\begin{proof}
    \begin{enumerate}[(i)]
        \eqitem \begin{align*}
            \bf{w}^\dag (A\bf{v})
            &= (A\bf{v})^\dag \bf{v} \\
            \implies \bf{v}^\dag (\lambda \bf{v})
            &= (\lambda \bf{v})^\dag \bf{v} \\
            \implies \lambda \bf{v}^\dag \bf{v}
            &= \ol{\lambda} \bf{v}^\dag \bf{v}
        \end{align*}
        for $\bf{v}$ an eigenvector with
        eigenvalue $\lambda$. But $\bf{v} \neq
        \bf{0}$ so $\bf{v}^\dag \bf{v} \neq 0$ and
        $\lambda = \ol{\lambda}$.
        \eqitem \begin{align*}
            \bf{v}^\dag (A\bf{w})
            &= (A\bf{v})^\dag \bf{w} \\
            \implies \bf{v}^\dag (\mu \bf{w})
            &= (\lambda \bf{v})^\dag) \bf{w} \\
            \implies \mu \bf{v}^\dag \bf{w}
            &= \ol{\lambda} \bf{v}^\dag \bf{w} \\
            &= \lambda \bf{v}^\dag \bf{w}
        \end{align*}
        from (i). But $\lambda \neq \mu$ so
        $\bf{v}^\dag \bf{w} = 0$.
        \item Given $A\bf{v} = \lambda \bf{v}$
            with $\bf{v} \in \CC^n$ and $A$,
            $\lambda$ real, let
            \[ \bf{w} = \bf{u} + i\bf{u}' \]
            with $\bf{u}, \bf{u}' \in \RR^n$. Then
            $A\bf{u} = \lambda \bf{u}$ and
            $A\bf{u}' = \lambda \bf{u}'$ but
            $\bf{v} \neq 0$ implies one of
            $\bf{u}$ or $\bf{u}'$ is nonzero, so
            there is at least one real
            eigenvector.
    \end{enumerate}
\end{proof}

\subsubsection*{Unitary and Orthogonal Diagonalisation}

\begin{theorem*}
    Any $n \times n$ hermitian matrix $A$ is
    diagonalisable (as in section 6.3(a))
    \begin{enumerate}[(i)]
        \item There exists a basis of eigenvectors
            \[ \bf{u}_1, \dots, \bf{u}_n \in \CC^n \]
            with
            \[ A\bf{u}_i = \lambda_i \bf{u}_i \]
            ; or equivalently
        \item There exists $n \times n$ invertible
            matrix $P$ with
            \[ P^{-1}AP = D =
            \begin{pmatrix}
                \lambda_1 & \cdots & 0 \\
                \vdots & \ddots & \vdots \\
                0 & \cdots & \lambda_n
            \end{pmatrix}
            ;\]
            columns of $P$ are eigenvectors
            $\bf{u}_i$.
    \end{enumerate}
    In addition: the eigenvectors $\bf{u}_i$ can
    be chosen to be orthonormal
    \[ \bf{u}_i^\dag \bf{u}_j = \delta_{ij} .\]
    or equivalently the matrix $P$ can be chosen
    to be unitary
    \[ P^\dag = P^{-1} \implies P^\dag AP = D \]
    Special case: for $n \times n$ real symmetric $A$,
    can choose eigenvectors $\bf{u}_1, \dots, \bf{u}_n
    \in \RR^n$ with
    \[ \bf{u}_i^\top \bf{u}_j = \bf{u}_i \cdot
    \bf{u}_j = \delta_{ij} \]
    equivalently, the matrix $P$ can be chosen to be
    orthogonal
    \[ P^\top = P^{-1} \implies P^\top AP = D \]
\end{theorem*}
\noindent
Proof of diagonalisability is \emph{not
examinable} and remaining statements follow by
combining results of section 6.2, 6.3 and choosing
\emph{orthonormal} basis for each eigenspace.

\bigskip

\noindent
\textbf{Examples} \\
\begin{enumerate}[(i)]
    \item Consider hermitian ($A^\dag = A$) as in
        section 6.1(b):
        \[ A =
        \begin{pmatrix}
            2 & i \\
            -i & 2
        \end{pmatrix}
        \]
        then $\lambda_1 = 1$ and $\lambda_2 = 3$
        and choose
        \[ \bf{u}_1 = \frac{1}{\sqrt{2}}
        \begin{pmatrix}
            1 \\
            i
        \end{pmatrix}
        , \qquad \bf{u}_2 = \frac{1}{\sqrt{2}}
        \begin{pmatrix}
            1 \\
            -i
        \end{pmatrix}
        \]
        to ensure $\bf{u}_1^\dag \bf{u}_1 =
        \bf{u}_2^\dag \bf{u}_2 = 1$ and note
        \[ \bf{u}_1^\dag \bf{u}_2 = \half (1 - i)
        \begin{pmatrix}
            1 \\
            -i
        \end{pmatrix}
        = 0 .\]
        Let
        \[ P = \frac{1}{\sqrt{2}}
        \begin{pmatrix}
            1 & 1 \\
            i & -i
        \end{pmatrix}
        \]
        then $P^\dag = P^{-1}$ unitary and
        \[ P^\dag AP =
        \begin{pmatrix}
            1 & 0 \\
            0 & 3
        \end{pmatrix}
        \]
    \item Consider symmetric matrix
        \[ A =
        \begin{pmatrix}
            0 & 1 & 1 \\
            1 & 0 & 1 \\
            1 & 1 & 0
        \end{pmatrix}
        \]
        then $\lambda_1 = \lambda_2 = -1$ and
        $\lambda_3 = 2$ and can choose
        \[ \bf{u}_1 = \frac{1}{\sqrt{2}}
        \begin{pmatrix}
            1 \\
            -1 \\
            0
        \end{pmatrix}
        , \qquad \bf{u}_2 = \frac{1}{\sqrt{6}}
        \begin{pmatrix}
            1 \\
            1 \\
            -2
        \end{pmatrix}
        , \qquad \bf{u}_3 = \frac{1}{\sqrt{3}}
        \begin{pmatrix}
            1 \\
            1 \\
            1
        \end{pmatrix}
        \]
        Let $P$ be matrix with columns $\bf{u}_1$,
        $\bf{u}_2$, $\bf{u}_3$ then $P^\top =
        P^{-1}$ orthogonal
        \[ P^\top AP =
        \begin{pmatrix}
            -1 & 0 & 0 \\
            0 & -1 & 0 \\
            0 & 0 & 2
        \end{pmatrix}
        \]
\end{enumerate}

\subsection{Quadratic Forms}

Consider $\mathcal{F} : \RR^2 \to \RR^2$ defined
by
\[ \mathcal{F}(\bf{x}) = 2x_1^2 - 4x_1x_2 + 5x_2^2 \]
This can be expressed
\[ \mathbf{F}(\bf{x}) = x_1'^2 + 6x_2'^2 \]
where
\[ x_1' = \frac{1}{\sqrt{5}} (2x_1 + x_2) \]
\[ x_2' = \frac{1}{\sqrt{5}}(-x_1 + 2x_2) \]
with $x_1'^2 + x_2'^2 = x_1^2 + x_2^2$. To
understand this better, note
\[ \mathcal{F}(\bf{x}) = \bf{x}^\top A\bf{x} \]
where
\[ A =
\begin{pmatrix}
    2 & -2 \\
    -2 & 5
\end{pmatrix}
\]
and we can diagonalise $A$ because $\lambda_1 =
1$, $\lambda_2 = 6$, and then we can compute
\[ \bf{u}_1 = \frac{1}{\sqrt{5}}
\begin{pmatrix}
    2 \\
    1
\end{pmatrix}
, \qquad \bf{u}_2 = \frac{1}{\sqrt{5}}
\begin{pmatrix}
    -1 \\
    2
\end{pmatrix}
\]
Then
\[ x_1' = \bf{u}_1 \cdot \bf{x} \]
\[ x_2' = \bf{u}_2 \cdot \bf{x} \]
give the simplified form for $\mathcal{F}$. In
general, a \emph{quadratic form} is a function
$\mathcal{F} : \RR^n \to \RR^2$ given by
\[ \mathcal{F}(\bf{x}) = \bf{x}^\top A \bf{x} =
x_i A_{ij} x_j \]
where $A$ is an $n \times n$ real symmetric
matrix. From section 6.4,
\[ P^\top AP = D =
\begin{pmatrix}
    \lambda_1 & \cdots & 0 \\
    \vdots & \ddots & \vdots \\
    0 & \cdots & \lambda_n
\end{pmatrix}
\]
where $\lambda_i$ are eigenvalues of $A$ and $P$
orthogonal with columns $\bf{u}_i$ orthonormal
eigenvectors. Let $\bf{x}' = P^\top \bf{x}$ or
$\bf{x} = P\bf{x}'$. Then
\begin{align*}
    \mathcal{F}(\bf{x}
    &= \bf{x}^\top A\bf{x} \\
    &= (P\bf{x}')^\top A (P\bf{x}') \\
    &= (\bf{x}')^\top (P^\top AP) \bf{x}' \\
    &= (\bf{x}')^\top D\bf{x}'
\end{align*}
$\mathcal{F}$ has been \emph{diagonalised}. Now
\[ \bf{x}' = x_1' \bf{e}_1 + \cdots +
x_n' \bf{e}_n \]
and
\begin{align*}
    \bf{x}
    &= x_1 \bf{e}_1 + \cdots + x_n \bf{e}_n \\
    &= x_1' \bf{u} + \cdots + x_n' \bf{u}_n
\end{align*}
since $x_i' = \bf{u}_i \cdot \bf{x} \iff \bf{x}'
P^\top \bf{x}$. Thus, $x_i'$ are coordinates with
respect to new axes given by orthonormal basis
vector $\bf{u}_i$ and these called \emph{principal
axes} of $\mathcal{F}$. Relation to original axes
along standard basis vectors $\bf{e}_i$ and
coordinates $x_i$ is given by an orthogonal
transformation
\[ |\bf{x}|^2 = x_ix_i = x_i'x_i' \]

\subsubsection*{(b) Examples in $\RR^2$ and $\RR^3$}

\textbf{In $\RR^2$} \\
\[ \mathcal{F}(\bf{x}) = \bf{x}^\top A\bf{x} \]
with
\[ A =
\begin{pmatrix}
    \alpha & \beta \\
    \beta & \alpha
\end{pmatrix}
\]
eigenvalues $\lambda_1 = \alpha + \beta$,
$\lambda_2 = \alpha - \beta$. Eigenvectors:
\[ \bf{u}_1 = \frac{1}{\sqrt{2}}
\begin{pmatrix}
    1 \\
    1
\end{pmatrix}
, \qquad \bf{u}_2 = \frac{1}{\sqrt{2}}
\begin{pmatrix}
    -1 \\
    1
\end{pmatrix}
\]
\begin{align*}
    \mathcal{F}(\bf{x})
    &= \alpha x_1^2 + 2 \beta x_1 x_2 + \alpha
    x_2^2 \\
    &= (\alpha + \beta) x_1'^2 + (\alpha -
    \beta)x_2'^2
\end{align*}
with
\[ x_1' = \frac{1}{\sqrt{2}}(x_1 + x_2 \]
\[ x_2' = \frac{1}{\sqrt{2}}(-x_1 + x_2 \]
\begin{enumerate}[(i)]
    \item $\alpha = \frac{3}{2}$, $\beta =
        -\half$. Then $\lambda_1 = 1$, $\lambda_2
        = 2$.
        \[ \mathcal{F}(\bf{x}) = x_1'^2 + 2x_2'^2
        = 1 \]
        defines an ellipse.
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/5b27d5d851f811ec.png}
        \end{center}
    \item $\alpha = -\half$, $\beta =
        \frac{3}{2}$. Then $\lambda_1 = 1$ and
        $\lambda_2 = -2$
        \[ \mathcal{F}(\bf{x}) = x_1'^2 - 2x_2'^2
        = 1 \]
        defines a hyperbola.
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/91182fee51f811ec.png}
        \end{center}
\end{enumerate}

\bigskip

\noindent
\textbf{In $\RR^3$}
\[ \mathcal{F}(\bf{x}) = \bf{x}^\top A\bf{x} =
\lambda_1 x_1'^2 + \lambda_2 x_2'^2 + \lambda_3
x_3'^2 \]
after diagonalisation.
\begin{enumerate}[(i)]
    \item If $A$ has eigenvalues $\lambda_1,
        \lambda_2, \lambda_3 > 0$ then
        $\mathcal{F} = 1$ defines an ellipsoid.
    \item From section 6.4,
        \[ A =
        \begin{pmatrix}
            0 & 1 & 1 \\
            1 & 0 & 1 \\
            1 & 1 & 0
        \end{pmatrix}
        \]
        has eigenvalues $\lambda_1 = \lambda_2 =
        -1$, $\lambda_3 = 2$. Hence
        \begin{align*}
            \mathcal{F} = 2x_1x_2 + 2x_2x_3 +
            2x_3x_1 \\
            &= -x_1'^2 - x_2'^2 + 2x_3'^2
        \end{align*}
        \[ \mathcal{F} = 1 \iff 2x_3'^2 = 1 +
        x_1'^2 + x_2'^2 \]
        hyperboloid:
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/3b44f2b851f911ec.png}
        \end{center}
        \[ \mathcal{F} = -1 \iff x_1'^2 + x_2'^2 =
        1 + 2x_3'^2 \]
        2 sheeted hyperboloid:
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/5724349e51f911ec.png}
        \end{center}
\end{enumerate}

\subsection{Cayley-Hamilton Theorem}

If $A$ is an $n \times n$ complex matrix and
\[ f(t) = c_0 + c_1t + \cdots + c_k t^k \]
polynomial of degree $k$, then
\[ f(A) = c_0I + c_1A + \cdots + c_k A^k \]
We can also define power series of matrices
subject to convergence, for example
\[ \exp A = I + A + \cdots + \frac{1}{r!} A^r +
\cdots \]
converges for any $A$. Note
\begin{enumerate}[(i)]
    \item If
        \[ D =
        \begin{pmatrix}
            \lambda_1 & \cdots & 0 \\
            \vdots & \ddots & \vdots \\
            0 & \cdots & \lambda_n
        \end{pmatrix}
        \]
        is some diagonal matrix, then
        \[ D^r =
        \begin{pmatrix}
            \lambda_1^r & \cdots & 0 \\
            \vdots & \ddots & \vdots \\
            0 & \cdots & \lambda_n^r
        \end{pmatrix}
        \]
        and
        \[ f(D) =
        \begin{pmatrix}
            f(\lambda_1) & \cdots & 0 \\
            \vdots & \ddots & \vdots \\
            0 & \cdots & f(\lambda_n)
        \end{pmatrix}
        \]
    \item If $B = P^{-1}AP$ for invertible $P$,
        i.e. $A$ and $B$ are similar then
        \[ B^r = P^{-1}AP \qquad \text{and} \qquad
        f(B) = f(P^{-1}AP) = P^{-1}f(A)P \]
        Recall, the characteristic polynomial is
        \[ \chi_A(t) = \det(A - tI) = c_0 + c_1t +
        \cdots c_n t^n \]
        where $c_0 = \det A$ and $c_n = (-1)^n$.
\end{enumerate}

\begin{theorem*}[Cayley-Hamilton]
    \[ \chi_A(A) = c_0 I + c_1 A + \cdots + c_n A^n
    = 0 \]
    ``a matrix satisfies its own characteristic
    equation''
\end{theorem*}

\begin{note*}
    Cayley-Hamilton implies
    \[ c_0 I = -A(c_1 I + \cdots + c_n A^{n - 1}) \]
    and if $c_0 = \det A \neq 0$ then
    \[ A^{-1} = - \frac{1}{c_0} (c_1 I + \cdots
    c_n A^{n - 1}) .\]
\end{note*}

\begin{proof}
    \begin{enumerate}[(i)]
        \item General $2 \times 2$ matrix
            \[ A =
            \begin{pmatrix}
                a & b \\
                c & d
            \end{pmatrix}
            \implies \chi_A(t) = t^2 - (a + d)t +
            (ad - bc) \]
            then check by substitution that
            $\chi_A = 0$ (on example sheet 4).
        \item Diagonalisable $n \times n$ matrix:
            \\
            consider $A$ with eigenvalues
            $\lambda_i$ and invertible $P$ such
            that
            \[ P^{-1}AP = D =
            \begin{pmatrix}
                \lambda_1 & \cdots & 0 \\
                \vdots & \ddots & \vdots \\
                0 & \cdots & \lambda_n
            \end{pmatrix}
            \]
            and hence
            \[ \chi_A(D) =
            \begin{pmatrix}
                \chi_A(\lambda_1) & \cdots & 0 \\
                \vdots & \ddots & \vdots \\
                0 & \cdots & \chi_A(\lambda_n)
            \end{pmatrix}
            = 0 \]
            since $\lambda_i$ are eigenvalues.
            Then
            \begin{align*}
                \chi_A(A)
                &= \chi_A(P^{-1}DP) \\
                &= P^{-1}\chi_A(D)P \\
                &= 0
            \end{align*}
            as required.
        \item The non diagonalisable case is
            beyond the scope of this course, but
            one can use an analytical argument to
            extend the diagonalisable case.
    \end{enumerate}
\end{proof}

\newpage
\section{Changing Bases, Canonical Forms and Symmetries}

\subsection{Changing Bases in General}

\subsubsection*{(a) Definitions and Proposition}

Recall Section 4.4: given linear map $T : V \to W$
(real or complex vector spaces) and choice of
bases
\[ \{\bf{e}_i\} \quad i = 1, \dots, n \quad
\text{for $V$} \]
\[ \{\bf{f}_a\} \quad a = 1, \dots, m \quad
\text{for $W$} \]
the matrix $A$ ($m \times n$) with respect to
these bases is defined by
\[ T(\bf{e}_i) = \sum_a \bf{f}_a A_{ai} \]
This definition is chosen to ensure
\[ \bf{y} = T(\bf{x}) \iff y_A = \sum_i A_{ai}
x_i = A_{ai} x_i \]
where
\[ \bf{x} = \sum_i x_i \bf{e}_i, \qquad \bf{y} =
\sum_a y_a \bf{f}_a ,\]
which holds since
\begin{align*}
    T(\sum_i x_i \bf{e}_i)
    &= \sum_i x_i T(\bf{e}_i) \\
    &= \sum_i x_i(\sum_a \bf{f}_a A_{ai}) \\
    &= \sum_a\ub{\left( \sum_i A_{ai}x_i \right)}
    _{\text{$= y_a$ as required}} \bf{f}_a
\end{align*}
Same linear map $T$ has matrix $A'$ with respect
to bases
\[ \{\bf{e}_i'\} \qquad i = 1, \dots, n \quad
\text{for $V$} \]
\[ \{\bf{f}_a'\} \qquad a = 1, \dots, m \quad
\text{for $W$} \]
defined by
\[ T(\bf{e}_i') = \sum_a \bf{f}_a' A_{ai}' \]
To relate $A$ and $A'$ we need to say how bases
are related, and \emph{change} of \emph{base}
matrices $P$ ($n \times n$) and $Q$ ($m \times n$)
are defined by
\[ \bf{e}_i' = \sum_j \bf{e}_j P_{ji}, \qquad
\bf{f}_a' = \sum_b \bf{f}_b Q_{ba} \]

\begin{note*}
    $P$ and $Q$ invertible; in relation above we
    can exchange $\{\bf{e}_i\}$ and
    $\{\bf{e}_i'\}$ with $P \to P^{-1}$ and
    similarly for $Q$.
\end{note*}

\begin{proposition*}
    With definitions as above
    \[ A' = Q^{-1}AP \]
    change of basis formula for matrix of a linear
    map.
\end{proposition*}

\begin{example*}
    $n = 2$, $m = 3$
    \[ T(\bf{e}_1) = \bf{f}_1 + 2 \bf{f}_2 -
    \bf{f}_3 = \sum_a \bf{f}_a A_{a1} \]
    \[ T(\bf{e}_2) = - \bf{f}_1 + 2 \bf{2}_2 +
    \bf{f}_3 = \sum_a \bf{f}_a A_{a2} \]
    \[ \implies A =
    \begin{pmatrix}
        1 & -1 \\
        2 & 2 \\
        -1 & 1
    \end{pmatrix}
    \]
    New basis for $V$
    \[ \bf{e}_1' = \bf{e}_1 - \bf{e}_2 = \sum_i
    \bf{e}_i P_{i1} \qquad
    \bf{e}_2' = \bf{e}_1 + \bf{e}_2 = \sum_i
    \bf{e}_i P_{i2} \]
    \[ \implies P =
    \begin{pmatrix}
        1 & 1 \\
        -1 & 1
    \end{pmatrix}
    \]
    New basis for $W$
    \[ \bf{f}_1' = \bf{f}_1 - \bf{f}_3
    \qquad \bf{f}_2' = \bf{f}_2
    \qquad \bf{f}_3' = \bf{f}_1 + \bf{f}_3 \]
    \[ \implies Q =
    \begin{pmatrix}
        1 & 0 & 1 \\
        0 & 1 & 0 \\
        -1 & 0 & 1
    \end{pmatrix}
    \]
    Change of basis formula:
    \[ A' = Q^{-1}AP =
    \begin{pmatrix}
        \half & 0 & -\half \\
        0 & 1 & 0 \\
        \half & 0 & \half
    \end{pmatrix}
    \begin{pmatrix}
        1 & -1 \\
        2 & 2 \\
        -1 & 1
    \end{pmatrix}
    \begin{pmatrix}
        1 & 1 \\
        -1 & 1
    \end{pmatrix}
    =
    \begin{pmatrix}
        2 & 0 \\
        0 & 4 \\
        0 & 0
    \end{pmatrix}
    \]
    Direct check
    \[ T(\bf{e}_1') = 2\bf{f}_1' \qquad
    T(\bf{e}_2') = 4 \bf{f}_2' \]
    which agrees.
\end{example*}

\subsubsection*{(b) Proof of Proposition}

\begin{align*}
    T(\bf{e}_i')
    &= T(\sum_j \bf{e}_j P_{ji})
    &&\text{definition of $P$} \\
    &= \sum_j T(\bf{e}_j) P_{ji}
    &&\text{$T$ linear} \\
    &= \sum_j \sum_a \bf{f}_a A_{aj} P_{ji}
    &&\text{definition of $A$}
\end{align*}
\begin{align*}
    T(\bf{e}_i')
    &= \sum_b \bf{f}_b' A_{bi}'
    &&\text{definition of $A'$} \\
    &= \sum_b \sum_a \bf{f}_a Q_{ab} A_{bi}'
    &&\text{definition of $Q$}
\end{align*}
Comparing coefficients of $\bf{f}_a$ (since it's a
basis):
\[ \sum_j A_{aj} P_{ji} = \sum_b Q_{ab} A_{bi}' \]
or
\[ AP = QA' \]
as required.

\subsubsection*{(c) Approach using vector components}

Consider
\begin{align*}
    \bf{x}
    &= \sum_j x_j \bf{e}_j \\
    &= \sum_i x_i' \bf{e}_i' \\
    &= \sum_j \left( \sum_i P_{ji} x_i' \right)
    \bf{e}_j \\
    \implies x_j
    &= P_{ji} x_i'
\end{align*}
Write
\[ X =
\begin{pmatrix}
    x_1 \\
    \vdots \\
    x_n
\end{pmatrix}
\qquad \text{and} \qquad X' =
\begin{pmatrix}
    x_1' \\
    \vdots \\
    x_n'
\end{pmatrix}
\]
then
\[ X = PX' \qquad \text{or} \qquad X' = P^{-1}X \]
Note: some care needed if $V = \RR^n$, e.g. $n =
2$ with
\[ \bf{e}_1 =
\begin{pmatrix}
    1 \\
    1
\end{pmatrix}
, \qquad \bf{e}_2 =
\begin{pmatrix}
    1 \\
    -1
\end{pmatrix}
\]
\[ \bf{x} =
\begin{pmatrix}
    5 \\
    1
\end{pmatrix}
\in \RR^2 \]
has $\bf{x} = 3\bf{e}_1 + 2\bf{e}_2$ so
\[ X =
\begin{pmatrix}
    3 \\
    2
\end{pmatrix}
\]
Similarly
\[ \bf{y} = \sum_b y_b \bf{f}_b = \sum_a y_a'
\bf{f}_a' \]
\[ \implies y_b = Q_{ba} y_a' \]
Then
\[ Y = QY' \qquad \text{or} \qquad Y' = Q^{-1}Y \]
where
\[ Y =
\begin{pmatrix}
    y_1 \\
    \vdots \\
    y_m
\end{pmatrix}
\qquad \text{and} \qquad Y' =
\begin{pmatrix}
    y_1 \\
    \vdots y_m'
\end{pmatrix}
\]
Now, marices $A$, $A'$ are defined to ensure
\[ Y = AX \qquad \text{and} \qquad Y' = A'X' \]
But
\begin{align*}
    Y'
    &= Q^{-1}Y \\
    &= Q^{-1}AX \\
    &= (Q^{-1}AP)X' \\
    &= A'X'
\end{align*}
and true $\forall \bf{x}$ so
\[ A' = Q^{-1}AP .\]

\subsubsection*{Comments}

\begin{enumerate}[(i)]
    \item Definition of matrix $A$ for $T : V \to
        W$ with respect to bases $\{\bf{e}_i\}$
        and $\{\bf{f}_a\}$ can be expressed;
        column $i$ of $A$ consists of components
        of $T(\bf{e}_i)$ with respect to basis
        $\{\bf{f}_a\}$. [For $T : \RR^n \to \RR^m$
        with standard bases, columns of $A$ are
        images of standard basis vectors.]
        Similarly, definitions of $P$ and $Q$ say:
        columns consist of complements of new basis
        vectors with respect to old.
    \item With $V = W$ and same bases and
        $\bf{e}_i = \bf{f}_i$, $\bf{e}_i' =
        \bf{f}_i'$ we have
        \[ P = Q \qquad \text{and} \qquad A' =
        P^{-1}AP \]
        Matrices representing the same linear map
        with respect to different bases are
        similar; conversely if $A$ and $A'$ are
        similar then we can regard them as
        representing same linear map with $P$
        defining change of basis. In section 6.3,
        we observed
        \[ \mathrm{tr}(A') = \mathrm{tr}(A) ,\]
        \[ \det(A') = \det(A) ,\]
        \[ \chi_{A'}(t) = \chi_A(t) \]
        so these are properties of linear map.
    \item $V = W = \RR^n \text{ or } \CC^n$, with
        $\bf{e}_i$ standard basis - matrix $A$ is
        diagonalisable if and only if there exists
        basis of eigenvectors
        \[ \bf{e}_i' = \bf{v}_i \]
        with
        \[ A\bf{v}_i = \lambda_i \bf{v}_i \qquad
        \text{no summation convention!} \]
        and then
        \[ A' = P^{-1}AP = D =
        \begin{pmatrix}
            \lambda_1 & \cdots & 0 \\
            \vdots & \ddots & \vdots \\
            0 & \cdots & \lambda_n
        \end{pmatrix}
        \]
        and
        \[ \bf{v}_i = \sum_j \bf{e}_j P_{ji} \]
        eigenvectors are columns of $P$.
        Specialising further $A^\dag = A$ implies
        exists basis of orthonormal eigenvectors
        \[ \bf{e}_i' = \bf{u}_i \qquad \text{and}
        \qquad P^\dag = P^{-1} \]
\end{enumerate}

\subsection{Jordan Canonical / Normal Form}

This result classifies $n \times n$ complex
matrices up to similarity.

\begin{proposition*}
    Any $2 \times 2$ complex matrix $A$ is similar
    to one of the following:
    \begin{enumerate}[(i)]
        \item For some $\lambda_1 \neq \lambda_2$
            \[ A' =
            \begin{pmatrix}
                \lambda_1 & 0 \\
                0 & \lambda_2
            \end{pmatrix}
            \]
            so
            \[ \chi_A(t) = (t - \lambda_1)(t -
            \lambda_2) \]
        \item For some $\lambda$,
            \[ A =
            \begin{pmatrix}
                \lambda & 0 \\
                0 & \lambda
            \end{pmatrix}
            \]
            so
            \[ \chi_A(t) = (t - \lambda)^2 \]
        \item For some $\lambda$,
            \[ A =
            \begin{pmatrix}
                \lambda & 1 \\
                0 & \lambda
            \end{pmatrix}
            \]
            so
            \[ \chi_A(t) = (t - \lambda)^2 \]
    \end{enumerate}
\end{proposition*}

\begin{proof}
    $\chi_A(t)$ has 2 roots over $\CC$.
    \begin{enumerate}[(i)]
        \item For distinct roots or eigenvalues,
            $\lambda_1$, $\lambda_2$, we have $M_1
            = m_1 = M_2 = m_2 = 1$ and
            eigenvectors $\bf{v}_1$, $\bf{v}_2$
            provide a basis.
        \item For repeated root / eigenvalue
            $\lambda$, if $M_\lambda = m_\lambda =
            2$, then same argument applies.
        \item For repeated root / eigenvalue
            $\lambda$, with $M_\lambda = 2$ and
            $m_\lambda = 1$, let $\bf{v}$ be
            eigenvector for $\lambda$ and $\bf{w}$
            any linearly independent vector. Then
            \[ A\bf{v} = \lambda \bf{v} \]
            \[ A\bf{w} = \alpha \bf{v} + \beta
            \bf{w} \]
            say. Matrix of map with respect to
            basis $\{\bf{v}, \bf{w}\}$ is
            \[
            \begin{pmatrix}
                \lambda & \alpha \\
                0 & \beta
            \end{pmatrix}
            \]
            But $\beta = \lambda$ (otherwise case
            (i)) and $\alpha \neq 0$ (otherwise
            case (ii)). Now set $\bf{u} = \alpha
            \bf{v}$ and note
            \[ A\bf{u} = \lambda \bf{u} \]
            \[ A\bf{w} = \bf{u} + \lambda \bf{w} \]
            so with respect to basis $\{\bf{u},
            \bf{w}\}$ matrix is
            \[ A' =
            \begin{pmatrix}
                \lambda & 1 \\
                0 & \lambda
            \end{pmatrix}
            \]
            as claimed.
    \end{enumerate}
\end{proof}

\begin{example*}[using a slightly different approach]
    \[ A =
    \begin{pmatrix}
        1 & 4 \\
        -1 & 5
    \end{pmatrix}
    \]
    \[ \implies \chi_A(t) = (t - 3)^2 \]
    and
    \[ A - 3I =
    \begin{pmatrix}
        -2 & 4 \\
        -1 & 2
    \end{pmatrix}
    \]
    Choose
    \[ \bf{w} =
    \begin{pmatrix}
        1 \\
        0
    \end{pmatrix}
    \]
    not an eigenvector and then
    \[ \bf{u} = (A - 3I) \bf{w} =
    \begin{pmatrix}
        -2 \\
        -1
    \end{pmatrix}
    \]
    But $(A - 3I)^2 = 0$, and
    \[ A\bf{u} = 3\bf{u} \]
    \[ A\bf{w} = \bf{u} + 3\bf{w} \]
    so basis $\{\bf{u}, \bf{w}\}$ gives JCF.
    Check:
    \[ P =
    \begin{pmatrix}
        -2 & 1 \\
        -1 & 0
    \end{pmatrix}
    \implies P^{-1} =
    \begin{pmatrix}
        0 & -1 \\
        1 & -2
    \end{pmatrix}
    \]
    and
    \[ P^{-1}AP =
    \begin{pmatrix}
        3 & 1 \\
        0 & 3
    \end{pmatrix}
    \]
    Generalisation to larger matrices can be
    consired, starting with
    \[ N =
    \begin{pmatrix}
        0 & 1 & 0 & \cdots & 0 \\
        0 & 0 & 1 & \cdots & 0 \\
        0 & 0 & 0 & \cdots & 0 \\
        \vdots & \vdots & \vdots & \ddots & \vdots \\
        0 & 0 & 0 & \cdots & 0
    \end{pmatrix}
    \]
    $n \times n$. When applied to standard basis
    vectors get
    \[ \bf{w}_n \mapsto \bf{e}_{n - 1} \mapsto
    \cdots \mapsto \bf{e}_1 \mapsto \bf{0} \]
    Note
    \[ J = \lambda I + N \]
    then
    \[ \chi_J(t) = (\lambda - t)^n \]
    but $m_\lambda = 1$ ($M_\lambda = n$).
\end{example*}

\begin{theorem*}
    Any $n \times n$ complex matrix $A$ is
    similar to a matrix $A'$ with block form
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/66db862251fd11ec.png}
    \end{center}
    where each diagonal block is a \emph{Jordan
    block} with form
    \[ J_p(\lambda) =
    \ub{
    \begin{pmatrix}
        \lambda & 1 & 0 & \cdots & 0 \\
        0 & \lambda & 1 & \cdots & 0 \\
        0 & 0 & \lambda & \cdots & 0 \\
        \vdots & \vdots & \vdots & \ddots & \vdots \\
        0 & 0 & 0 & \cdots & \lambda
    \end{pmatrix}
    }_{p \times p}
    \]
    with $n_1 + \cdots n_3 = n$ and $\lambda_1,
    \dots, \lambda_r$ are eigenvectors of $A$ and
    $A'$ (same eigenvalue may appear in more
    than one block). $A$ is diagonalisable if and
    only if $A'$ consists of $1 \times 1$ Jordan
    blocks only.
\end{theorem*}

\begin{proof}
    See Linear Algebra and GRM in Part IB.
\end{proof}

\subsection{Quadrics and Conics}

\subsubsection*{(a) Quadrics in General}

A quadric in $\RR^n$ is a hypersurface defined
\[ Q(\bf{x}) = \bf{x}^\top A\bf{x} + \bf{x}^\top
\bf{x} + x = 0 \]
for some $A$, $n \times n$ real symmetric,
non-zero matrix, $\bf{b} \in \RR^n$, $c \in \RR$.
So
\[ Q(\bf{x}) = A_{ij} x_i x_j + b_ix_i + c = 0 \]
Consider classifying solutions up to geometrical
equivalence: no distinction between solutions
related by \emph{isometries} (length preserving
maps) in $\RR^n$, i.e. related by
\begin{enumerate}[(i)]
    \item translation - change in origin
    \item orthogonal transformation about origin -
        change in axes.
\end{enumerate}
If $A$ is invertible (no zero eigenvalues) then by
setting $\bf{y} = \bf{x} + \half A^{-1}\bf{b}$ we
have
\begin{align*}
    \bf{y}^\top A\bf{y}
    &= (\bf{x} + \half A^{-1}\bf{b})^\top A
    (\bf{x} + \half A^{-1} \bf{b}) \\
    &= \bf{x}^\top A\bf{x} + \bf{b}^\top \bf{x} +
    \frac{1}{4} \bf{b}^\top A^{-1} \bf{b}
\end{align*}
[since $(A^{-1}\bf{b})^\top = \bf{b}^\top
(A^{-1})^\top$ and $(A^{-1})^\top = (A^\top)^{-1}
= A_{-1}$ in this case.]
Then $Q(\bf{x}) = 0 \iff \mathcal{F}(\bf{y}) = k$
with $\mathcal{F}(\bf{y}) = \bf{y}^\top
A\bf{y}$. (quadratic form with respect to new
origin $\bf{y} = \bf{0}$) and $k = \frac{1}{4}
\bf{b}^\top A^{-1}\bf{b} - c$. Diagonalise
$\mathcal{F}$ as in section 6.5: orthonormal
eigenvectors give principal axes, eigenvalues of
$A$ and value of $k$ determine nature of quadric.
Example in $\RR^3$ given in section 6.5(b)
\begin{enumerate}[(i)]
    \item eigenvalues $> 0$ and $k > 0$ get
        ellipsoid
    \item eigenvalues of different sign and $k
        \neq 0$ get hyperboloid
\end{enumerate}
If $A$ has one or more zero eigenvalues then
analysis changes and simplest standard form may
have both linear and quadratic terms.

\subsubsection*{(b) Conics}

Quadrics in $\RR^2$ are curves, \emph{conics}.

\bigskip

\noindent
\ul{$\det A \neq 0$.} \\
By completing square and diagonalising we get a
standard form
\[ \lambda_1 x_1'^2 + \lambda_2 x_2'^2 = k \]
\[ \lambda_1, \lambda_2 > 0 \implies
\begin{cases}
    \text{ellipse for $k > 0$} \\
    \text{point for $k = 0$} \\
    \text{no solution for $k < 0$}
\end{cases} \]
\[ \lambda_1 > 0, \lambda_2 < 0 \implies
\begin{cases}
    \text{hyperbola for $k > 0$ or $k < 0$} \\
    \text{pair of lines for $k = 0$}
\end{cases} \]
e.g.
\[ x_1'^2 - x_2'^2 = (x_1 - x_2)(x_1 + x_2) = 0 \]

\bigskip

\noindent
\ul{$\det A = 0$}
Suppose $\lambda_1 > 0$ and $\lambda_2 = 0$;
diagonalise $A$ in original formula to get
\[ \lambda_1 x_1'^2 + b_1'x_1' + b_2'x_2' + c = 0 \]
\[ \iff \lambda_1 x_1''^2 + b_2'x_2' + c' = 0 \]
where
\[ x_1'' = x_1' + \frac{1}{2\lambda_1} b_1' \qquad
\text{and} \qquad c' = c -
\frac{b_1'^2}{4\lambda_1^2} \]
If $b_2' = 0$ then we get a pair of lines for $c'
< 0$, single line for $c' = 0$ and no solutions
for $c' > 0$. If $b_2' \neq 0$ then equation
becomes
\[ \lambda_1 x_1''^2 + b_2'x_2'' = 0 \]
parabola where
\[ x_2'' = x_2' + \frac{1}{b_2'}c' \]

\subsection{Symmetries and Transformation Groups}

\subsubsection*{(a) Orthogonal Transformation and Rotations in $\RR^n$}

\begin{align*}
    \text{$R$ orthogonal}
    &\iff R^\top R = RR^\top = I \\
    &\iff (R\bf{x}) \cdot (R\bf{y}) = \bf{x} \cdot
    \bf{y} \,\,\forall \bf{x}, \bf{y} \\
    &\iff \text{columns or rows of $R$ orthonormal
    vectors}
\end{align*}
The set of such matrices forms the
\emph{orthogonal group} $\mathrm{O}(n)$.
\[ R \in O(n) \implies \det R = \pm 1 \]
[$\det(R^\top)\det(R) = [\det(R)]^2 = 1$]
\[ \mathrm{SO}(n) = \{R \in \mathrm{O}(n) : \det R
= +1\} \]
is a subgroup, the \emph{special orthogonal
group}.
\[ R \in \mathrm{O}(n) \implies \text{$R$
preserves lengths and $|n\text{-dim vol}|$} \]
\[ R \in \mathrm{SO}(n) \implies \text{$R$ also
preserves orientation} \]
$\mathrm{SO}(n)$ consists of all rotations in
$\RR^n$.

\bigskip

\noindent
Reflections belong to $\mathrm{O}(n) \setminus
\mathrm{SO}(n)$, any element of $\mathrm{O}(n)$ is
of the form
\[ R \text{ or } RH \text{ with } R \in
\mathrm{SO}(n) \]
e.g. if $n$ is odd, we can choose $H = -I$.

\noindent\textbf{Active and Passive Points of
View} \\
For a rotation $R$ (matrix), the transformation
\[ x_i' = R_{ij}x_j \]
can be viewed in two ways.

\bigskip

\noindent
\emph{Active view point}: rotation transforms
vectors
\[ x_i' \text{ components of new vector} \]
\[ \bf{x}' = R\bf{x} \text{ with respect to
standard basis $\{e_i\}$} \]
e.g. $\RR^2$
\begin{center}
\begin{tsqx}
    (-1.5,0)->>(0.5,0)
    (0,-1.5)->>(0,0.5)
    (0.5,0)->>(1.5,0)
    (0,0.5)->>(0,1.5)
    label $\mathbf{e}_2$ @ (-0.1,0.5)
    label $\mathbf{e}_1$ @ (0.5,-0.1)
    unitcircle
    X := dir(30)
    X' := dir(50)
    label $\mathbf{x}$ @ 1.1*X
    label $\mathbf{x}' = R\mathbf{x}$ @ 1.2*X'
    origin->>X
    origin->>X'
\end{tsqx}
\end{center}
\[ |\bf{x}'|^2 = |\bf{x}|^2 \]

\bigskip

\noindent
\emph{Passive view point}: rotation changes basis
\[ x_i' \text{ components of same vector
$\bf{x}$ but with respect to new basis
$\{u_i\}$} \]
e.g. $\RR^2$
\begin{center}
\begin{tsqx}
    (-1.5,0)->>(0.5,0)
    (0,-1.5)->>(0,0.5)
    (0.5,0)->>(1.5,0)
    (0,0.5)->>(0,1.5)
    label $x_1$ @ (1.6,0)
    label $x_2$ @ (0,1.6)
    -1.5*dir(70)->>0.5*dir(70)
    -1.5*dir(-20)->>0.5*dir(-20)
    0.5*dir(70)->>1.5*dir(70)
    0.5*dir(-20)->>1.5*dir(-20)
    label $x_2'$ @ 1.6*dir(70)
    label $x_1'$ @ 1.6*dir(-20)
    label $\mathbf{u}_2$ @ 0.6*dir(78)
    label $\mathbf{u}_1$ @ 0.6*dir(-28)
    label $\mathbf{e}_2$ @ (-0.1,0.5)
    label $\mathbf{e}_1$ @ (0.5,-0.1)
    X := dir(30)
    label $\mathbf{x}$ @ 1.1*X
    origin->>X
\end{tsqx}
\end{center}
\begin{align*}
    \bf{u}_1
    &= \sum_j R_{ij} \bf{e}_j \\
    &= \sum_j \bf{e}_j(R^{-1})_{ji}
\end{align*}
(compare to section 6.5: $P = R^{-1}$)

\subsubsection*{(b) 2D Minkowski Space and Lorentz Transformations}

Define a new ``inner product'' on $\RR^2$ by
\[ (\bf{x}, \bf{y} = \bf{x}^\top J
\bf{y} = x_0y_0 - x_1y_1 \]
where
\[ J =
\begin{pmatrix}
    1 & 0 \\
    0 & -1
\end{pmatrix}
\]
where we now label components
\[ \bf{x} =
\begin{pmatrix}
    x_0 \\
    x_1
\end{pmatrix}
\qquad \text{and} \qquad \bf{y} =
\begin{pmatrix}
    y_0 \\
    y_1
\end{pmatrix}
\]
This is not positive definite, since
\[ (\bf{x}, \bf{x}) = \bf{x}^\top J
\bf{x} = x_0^2 - x_1^2 \]
but still bilinear and symmetric. Standard basis
vectors are ``orthonormal'' in generalised sence:
\[ \bf{e}_0 =
\begin{pmatrix}
    1 \\
    0
\end{pmatrix}
\qquad \text{and} \qquad \bf{e}_1 =
\begin{pmatrix}
    0 \\
    1
\end{pmatrix}
\]
obey
\begin{align*}
    (\bf{e}_0, \bf{e}_0)
    &= 1 \\
    (\bf{e}_1, \bf{e}_1)
    &= -1 \\
    (\bf{e}_0, \bf{e}_1)
    &= 0
\end{align*}
New inner product is called the \emph{Minkowski}
metric and $\RR^2$ equipped with it is called
\emph{Minkowski space}. Consider
\[ M =
\begin{pmatrix}
    M_{00} & M_{01} \\
    M_{10} & M_{11}
\end{pmatrix}
\]
giving a linear map $\RR^2 \to \RR^2$. This
preserves the Minkowski metric if and only if
\begin{align*}
    (M\bf{x}, M\bf{y})
    &= (\bf{x}, \bf{y}
    &&\forall \bf{x}, \,athbf{y} \in \RR^2 \\
    \iff (M\bf{x})^\top J(M\bf{y})
    &= \bf{x}^\top(M^\top JM) \bf{y} \\
    &= \bf{x}^\top J \bf{y}
    &&\forall \bf{x}, \bf{y} \in \RR^2 \\
    \iff M^\top JM
    &= J
\end{align*}
The set of such matrices forms a group. Now
\begin{align*}
    \det(M^\top JM)
    &= \det M^\top \det J \det M \\
    &= \det J
    \implies (\det M)^2
    &= 1 \\
    \implies \det M
    &= \pm 1
\end{align*}
Furthermore, $|M_{00}| \ge 1$, so
\[ M_{00} \ge 1 \qquad \text{or} \qquad M_{00} \le
-1 .\]
The subgroup with
\[ \det M = +1 \qquad \text{and} \qquad M_{00} \ge
1 \]
is the \emph{Lorentz group} in 2D. \\
\emph{General form for $M$}: require columns
$M\bf{e}_0$ and $M\bf{e}_1$ to be
orthonormal, like $\bf{e}_0$, $\bf{e}_0$
(with respect to new inner product). This implies
\[ M(\theta) =
\begin{pmatrix}
    \cosh \theta & \sinh \theta \\
    \sinh \theta & \cosh \theta
\end{pmatrix}
\]
First column fixed by requiring $(M\bf{e}_0,
M\bf{e}_1) = 1$ or $M_{00}^1 - M_{10}^2 = 1$ and
$M_{00} \ge 1$. The second column is then fixed by
$M\bf{e}_0, M\bf{e}_1) = 0$, $(M\bf{e}_1,
M\bf{e}_1) = -1$ and $\det M = +1$ (fixes overall
sign). For such matrices
\[ M(\theta_1)M(\theta_2) = M(\theta_1 + \theta_2) \]
\begin{center}
\begin{tsqx}
    (-1.5,0)->>(1.5,0)
    (0,-1.5)->>(0,1.5)
    label $x_1$ @ (1.6,0)
    label $x_0$ @ (0,1.6)
    ! real f(real x) {return sqrt(x**2 - 1);}
    ! draw(graph(f,1,1.5,operator ..));
    ! real g(real x) {return -sqrt(x**2 - 1);}
    ! draw(graph(g,1,1.5,operator ..));
    ! real h(real x) {return sqrt(x**2 + 1);}
    ! draw(graph(h,-f(1.5),f(1.5),operator ..));
    ! real j(real x) {return -sqrt(x**2 + 1);}
    ! draw(graph(j,-f(1.5),f(1.5),operator ..), dashed);
    x := (0.5,h(0.5))
    Mx := (-0.5,h(-0.5))
    origin->>x
    origin->>Mx
    label $\mathbf{x}$ @ 1.1*x
    label $\mathbf{x}' = M\mathbf{x}$ @ (-0.25,1.3)
    (-1.3,-1.3)--(1.3,1.3)
    (-1.3,1.3)--(1.3,-1.3)
    label $k < 0$ @ (1.2,0.1)
    label $k > 0$ @ (0.9,1.5)
    label $k = 0$ @ (1.4,1.4)
\end{tsqx}
\end{center}
curves with $(\bf{x},\bf{x}) = k$, constant, as
shown.

\bigskip

\noindent
\textbf{Physical application} \\
Set
\[ M(\theta) = \gamma(v)
\begin{pmatrix}
    1 & v \\
    v & 1
\end{pmatrix}
\]
\[ v = \tanh \theta \]
\[ \gamma(v) = (1 - v^2)^{-1/2}, \qquad |v| < 1 .\]
Rename $x_0 \to t$ time coordinate and $x_1 \to x$
space coordinate.
\[ \bf{x}' = M\bf{x} \iff
\begin{cases}
    t' = \gamma(t + vx) \\
    x' = \gamma(x + vt)
\end{cases}
\]
\emph{Lorentz transformation} or \emph{boost}
relating observes moving with relative velocity
$b$ according to Special Relativity (units with $c
= 1$). Factor $\gamma(v) = (1 - v^2)^{-1/2}$ gives
rise to effects such as time dilation and length
contraction.

\end{document}