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\begin{document}



\title{Review of Algebra}

\author{}

\date{}

\maketitle




\section{A little abstract algebra}

\subsection{Abelian groups}
Let $S$ be a set. By a binary operation, or simply an operation, we mean a
map $S \times S \to S$.

\begin{definition}
Let $A$ be a set, together with an operation $(x, y) \mapsto x+y$. $A$ is
called an abelian group if the following conditions hold.
\begin{enumerate}
\item The operation is assocative, i.e. $(x+y)+z = x + (y+z)$ for all $x,
    y, z \in A$.

\item The operation is commutative, i.e. $x+y = y+x$ for all $x, y \in A$.

\item There is an element $e \in A$ so that $x+e = x$ for all $x \in A$.
    The element $e$ is usually called the unit element.

\item For any $x \in A$ there is an element $(-x) \in A$ so that $x+(-x) =
    e$.
\end{enumerate}
\end{definition}

\begin{example}
The basic examples of abelian groups are the following.

\begin{itemize}
\item $\Z$, or $\Q$, or $\R$, or $\C$ together with the usual addition. In
    these cases, $e = 0$.

\item $\Q^\rtimes$, or $\R^\times$, or $\C^\times$ with the usual
    multiplication. In these cases $e = 1$ and $(-x)$ is the reciprocal of
    $x$.

\item $\Z/n\Z$. This is the set
    \[
    \{0, \cdots, n-1\}
    \]
    and the additions are taken modulo $n$.
\end{itemize}
\end{example}

If $A_1$ and $A_2$ are abelian groups, there is a natural abelian group
structure on the Cartesian product $A_1 \times A_2$ given by
    \[
    (x_1, y_1)+(x_2, y_2) = (x_1+ x_2, y_1+y_2),
    \]
namely ``adding component-wise''. This is called the direct sum of $A_1$ and
$A_2$ and is denoted by $A_1 \oplus A_2$.

Following usual convention, let us write write $nx$ for $x+x+ \cdots +x$ ($n$
times) in an abelian group $A$. The abelian group $A$ is called finitely
generated if we can find $x_1, \cdots, x_n \in A$ so that for all $g \in A$,
we can find integers $a_1, \cdots, a_n$ so that
    \[
    g = a_1 x_1 + \cdots + a_n x_n.
    \]

The following theorem completely describes the structure of a finitely
abelian group.

\begin{theorem}
Let $A$ be a finitely generated abelian group. Then there is an integer $r$
and finite cylic groups of prime power order $\Z/p_i^{e_i}\Z$, $i = 1,
\cdots, n$, so that
    \[
    A \simeq \Z^{\oplus r} \oplus \left( \Z/p_1^{e_1}\Z \oplus \cdots  \Z/p_n^{e_n}\Z \right).
    \]
Moreover the prime powers $p_1^{e_1}, \cdots, p_n^{e_n}$ are uniquely
determined by $A$.
\end{theorem}

We have not defined what isomorphism ``$\simeq$'' means. Roughly speaking it
means that there is a bijection of two sides which preserves the group
operations.



\subsection{Fields}

\begin{definition}
Let $F$ be a set, together with two operations $(x, y) \mapsto x+y$ and $(x,
y) \mapsto x\cdot y$. Then $F$ is called a field if the following conditions
are satisfied.

\begin{enumerate}
\item $(F, +)$ is an abelian group. Its unit element is denoted by $0$.

\item $(F \bs \{0\}, \cdot)$ is an abelian group. It unit element is
    denoted by $1$.

\item The distributive law holds, i.e. we have
    \[
    x\cdot (y+z) = x\cdot y+ x\cdot z, \quad (x+y) \cdot z = x\cdot z+ y\cdot z
    \]
    for all $x, y, z \in F$.
\end{enumerate}
\end{definition}


\begin{example}
The basic examples of fields are the following.

\begin{enumerate}
\item $\Q$, $\R$ and $\C$ with the usual addition and multiplication.

\item Any subset of $\C$ that is closed under addition, subtraction,
    multiplication and division is a field, e.g.
        \[
        \Q(\sqrt{2}) = \{ a+ b\sqrt{2} \mid a, b \in \Q\}.
        \]

\item The set of all rational function, with usual addition and
    multiplication, i.e.
        \[
        \left\{ \frac{f(x)}{g(x)} \ \Bigg|\
        \text{$f(x)$ and $g(x)$ are polynomials, $g(x) \not=0$} \right\}.
        \]

\item The set of all meromorphic functions, with usual addition and
    multiplication.

\item $\Z/p\Z$ where $p$ is a prime. This is the set
    \[
    \{0,1, \cdots, p-1\},
    \]
    the additions and multiplications are all modulo $p$. What might be a
    little non-obvious is that why every nonzero element $x \in \Z/p\Z$ has
    an inverse. For that we need to find a $y$ so that $xy \equiv 1 \pmod
    p$. As $p$ is a prime, we have $\gcd (x, p) = 1$. Elementary number
    theory tells us that there are integers $a$ and $z$ so that
        \[
        xz - ap = 1.
        \]
    Of course $p \nmid z$ for otherwise the left hand side would be been
    divisible by $p$. By Euclidean algorithm we have
        \[
        z = pb+y, \quad 1 \leq y \leq p-1.
        \]
    It follows that
        \[
        xy - p (xb- a) = 1.
        \]
    Thus we have found a $y$, $1 \leq y \leq p-1$, with $xy \equiv 1 \pmod
    p$.
\end{enumerate}
\end{example}


A field is called algebraically closed if every polynomial of positive degree
over $F$ has a root in $F$. For all fields $F$, there is an algebraically
closed field containing $F$. A smallest such field is called an algebraic
closure of $F$. The question concerning the existence and uniqueness of
algebraic closure has some subtle issues. We will not discuss them here.

\begin{theorem}
The field of complex numbers $\C$ is algebraically closed.
\end{theorem}

This is an easy consequence of complex analysis. Indeed if $f(z)$ is a
polynomial with no root, then $1/f(z)$ is a bounded holomorphic function on
the whole complex plane. It follows from Liouville's theorem that $1/f(z)$,
and hence $f(z)$ must be a constant.


\section{Vector spaces}

\subsection{Vector spaces}

We make use of the language of fields in this subsection. If you are not
comfortable with this, just replace $F$ everywhere by $\R$ or $\C$.

\begin{definition}
A vector space $(V,+,\cdot)$ over a field $F$ is a non empty set $V$ with a
binary operation $+:V\times V\rightarrow V$ called addition and an operation
$\cdot:F\times V\rightarrow V$ called scalar multiplication. These operations
have the following properties:

\begin{enumerate}
\item Addition is associative: $\forall u,v,w\in V$, $(u+v)+w=u+(v+w);$

\item Addition is commutative: $\forall v,w\in V$, $v+w=w+v;$

\item There is an additive identity: $\exists0\in V$ such that $\forall
    v\in V$, $v+0=0+v=v;$

\item Every element in $V$ has an additive identity: $\forall v\in V$,
    $\exists w\in V$ such that $v+w=w+v=0;$

(We also say that $(V,+)$ forms an abelian group.);

\item For all $v\in V$, $1\cdot v=v$.

\item For all $a,b\in F$, and all $v\in V$, $a\cdot(b\cdot v)=(ab)\cdot
    v$.;

\item For all $a,b\in F$, and all $v\in V$, $(a+b)\cdot v=a\cdot v+b\cdot
    v;$

\item For all $a\in F$, and all $v,w\in V$, $a\cdot(v+w)=a\cdot v+a\cdot
    w.$
\end{enumerate}
\end{definition}

\begin{definition}
Let $V$ be a vector space and $W\subseteq V$. We say $W$ is a subspace of $V$
when $W$ is a vector space using the same addition and multiplication as $V$.
\end{definition}

Let $V$ be a vector space of $F$, and $\cB$ is a set of vectors in $V$. The
most important notions are the following.

\begin{enumerate}
\item We say that $\cB$ is linearly dependent if there are $a_1, \cdots,
    a_r \in F$, not all zero, and $v_1, \cdots, v_n \in \cB$ so that
        \[
        a_1 v_1 + \cdots + a_n v_n = 0.
        \]
    Otherwise we say that $\cB$ is linearly independent.

\item The span of $\cB$ is the subset of $V$ consisting of vectors of the
    form
        \[
        a_1 v_1 + \cdots + a_n v_n, \quad a_1, \cdots, a_n \in F, \quad v_1, \cdots, v_n \in \cB.
        \]
    This is a subspace of $V$.

\item We say that $\cB$ is a basis of $V$ if $\cB$ spans $V$ and $\cB$ is
    linearly independent. If this is the case then we define $\dim V$, the
    dimension of $V$, to be the cardinality of $\cB$. This number is
    independent of the choice of $\cB$.
\end{enumerate}

Using axiom of choice we can show that $V$ has a basis. Assume that $\dim V$
is finite and $\cB = \{v_1, \cdots, v_n\}$ is a basis. Then every vector $v
\in V$ can be uniquely written as
    \[
    v = x_1 v_1 + \cdots + x_n v_n.
    \]
The vector $\tp{(x_1, \cdots, x_n)}$ is called the coordinate of $v$. Thus
every finite dimensional vector space, upon choosing a basis, ``looks like''
the familiar vector space $F^n$.






\subsection{Linear transform}

Let $V$ and $W$ be vector spaces over $F$ and $f: V \to W$ be a map. We say
that $f$ is a linear transform if for all $a_1, a_2 \in V$ and $v_1, v_2 \in
V$ we have
    \[
    f(a_1v_1+a_2v_2) = a_1 f(v_1) + a_2 f(v_2).
    \]
Note that $f(0) = 0$. We say that $f$ is an isomorphism if it is a bijection.

Define
    \[
    \Ker f = \{v \in V \mid f(v) = 0 \}, \quad
    \im f = \{ w \in W \mid \text{ there is some $v$ such that $f(v) = w$ }\}.
    \]
They are subspaces of $V$ and $W$ respectively. We define the rank of $f$ to
be $\dim \im f$.

The most important result concerning kernel and image is the following.

\begin{theorem}
    \[
    \dim \ker f + \dim \im f = \dim V.
    \]
\end{theorem}

\begin{example}
The most important example of linear transforms are given by matrix
multiplication. Let $A$ be an $m \times n$ matrix. Then the map
    \[
    F^n \to F^m, \quad x \mapsto Ax
    \]
is a linear transform.
\end{example}

In fact all linear transforms between finite dimensional vector spaces are of
this form. Let $f: V \to W$ be a linear transform. Suppose that we choose a
basis $\{v_1, \cdots, v_n\}$ of $V$ and $\{w_1, \cdots, w_m\}$ of $W$. Then
we form a matrix
    \[
    f(v_1, \cdots, v_n) = (w_1, \cdots, w_m) A
    \]
(I guess you need to figure out what this notation means first!) If $v \in V$
with coordinate $x \in F^n$ and $f(v) \in W$ with coordinate $y$, then we
have $y = Ax$.

Of course the matrix $A$ depends on the choice of the basis of $V$ and $W$.
What happens if we choose different bases? The most interesting case is $V =
W$. In this case a linear transform $f: V \to V$ is called an linear operator
on $V$. Suppose that we have a linear operator $f: V \to V$ and $\cB$ is a
basis of $V$. Then we get a matrix $A$ of $f$. What if we switch to a
different basis $\cB'$?

To analyze this question, let us introduce some notation. Suppose
    \[
    \cB =\{v_1, \cdots, v_n\}, \quad \cB' = \{ v_1', \cdots, v_n'\}.
    \]
Then we can find a (unique) invertible matrix $P$ so that
    \[
    (v_1, \cdots, v_n) = ( v_1', \cdots, v_n') P.
    \]
By definition, we have
    \[
    f(v_1, \cdots, v_n) = (v_1, \cdots, v_n) A.
    \]
It follows that
    \[
    f \left( (v_1', \cdots, v_n')P \right)  = \left((v_1', \cdots, v_n')P \right) A.
    \]
Using linearity of $f$, we conclude that
    \[
    f(v_1', \cdots, v_n') = (v_1', \cdots, v_n') PAP^{-1},
    \]
in other words, the matrix of $f$ under the basis $\cB$ is $PAP^{-1}$. This
observation leads to the following definition.

\begin{definition}
We say that two $n \times n$ matrices $A$ and $B$ are similar to each other
if there is an invertible matrix $P$ so that $B = P A P^{-1}$.
\end{definition}



The above discussion indicates the following proposition.

\begin{prop}
Let $f: V \to V$ be a linear operator. The matrices associated to $f$ under
different choices of bases of $V$ are all similar to each other.
\end{prop}

We can define the determinant and trace of a linear operator $f: V \to V$.
Choose a basis of $V$ and let $A$ be the matrix of $f$ under this basis. We
define
    \[
    \det f = \det A, \quad \Tr f = \Tr A.
    \]
It follows from the above proposition that $\det f$ and $\Tr f$ are
independent from the choice of the basis $\cB$. Of course $f$ is invertible
if and only if $A$ is so. Therefore we conclude
    \[
    \text{$f$ is invertible if and only if $\det f = 0$.}
    \]


\subsection{Eigenvalues}
Again we consider a linear operator $f: V \to V$. If $W \subset V$ is a
subspace and $f(W) = W$ then we say that $W$ is an invariant subspace and it
is clear that $f|_W: W \to W$ is a linear operator on $W$.

One dimensional invariant subspaces are of particular interest. Let us
elaborate this point. Let $v \in V$ be a nonzero element which spans a one
dimensional invariant subspace. Then there is a $\lambda \in F$ so that
    \[
    f(v) = \lambda v.
    \]
We call $\lambda$ an eigenvalue of $f$ and $v$ an eigenvector corresponding
to $\lambda$.

The next question is how to find eigenvalues and eigenvectors. Note that the
defining equation of eigenvalues and eigenvectors is equivalent to
    \[
    (\lambda 1_n - f) v = 0.
    \]
(If you are not comfortable with this write down everything in terms of
matrices!) This means that $\lambda 1_n -f$ is not invertible and hence
    \[
    \det (\lambda 1_n -f) = 0.
    \]
Therefore
    \[
    \text{$\lambda$ is an eigenvalue of $f$ if and only if $\det (\lambda 1_n -f) = 0$}.
    \]
A more souped-up version is the following.

\begin{definition}
Let $f: V \to V$ be a linear operator. The characteristic polynomial of $f$
is $\det (t 1_n - f)$.
\end{definition}

The roots of the characteristic polynomial give all eigenvalues of $f$. It
follows that there are precisely $n$ eigenvalues of $f$ (counting
multiplicity) if $F$ is algebraically closed, e.g. $F = \C$.

Once we have found all eigenvectors, then finding all eigenvectors are just a
questions of solving linear equations.

If we denote the eigenvalues of $f$ by $\lambda_1, \cdots, \lambda_n$, then
we have
    \[
    \det f = \lambda_1 \cdots \lambda_n, \quad \Tr f = \lambda_1 + \cdots + \lambda_n.
    \]
This can be seen from the fact that
    \[
    \det (t1_n -f) = (t - \lambda_1) \cdots (t - \lambda_n),
    \]
and comparing the coefficient of $t^{n-1}$ (trace) and the constant term
(determinant).


\subsection{Jordan canonical form}
The next question is that given a linear operator $f: V \to V$, what is the
``best'' choice of the basis so that the resulting matrix is the simplest?
This leads to the notion of Jordan canonical forms. We assume in this
subsection that $F$ is algebraically closed. Let us introduce some notation.

By an $n \times n$ Jordan block, we mean an $n \times n$ matrix of the form
    \[
    J_n(\lambda) = \begin{pmatrix}
    \lambda & 1 \\ & \lambda & \ddots \\ && \ddots & 1 \\ &&& \lambda
    \end{pmatrix}.
    \]
A Jordan canonical form is a blocked diagonal matrix so that all diagonal
blocks are Jordan blocks. Note that the eigenvalues of a Jordan canonical
forms are just given by diagonal elements.

One of the most important results in linear algebra is the following.

\begin{theorem}
Let $f: V \to V$ be a linear operator. Then we can choose a basis of $V$ so
that the matrix that represents $f$ is in Jordan canonical form.
\end{theorem}

Let $A$ be an $n \times n$ matrix. If we view it as a linear transform $F^n
\to F^n$ and apply the above theorem, we obtain a variant of the above
result.

\begin{theorem}
Every (square) matrix is similar to a Jordan canonical form. In other words,
given any (square) matrix $A$, we can find an invertible matrix $P$ so that
    \[
    PAP^{-1} = \begin{pmatrix} J_{n_1}(\lambda_1) \\ & \ddots \\ && J_{n_s}(\lambda_s) \end{pmatrix}.
    \]
\end{theorem}


These are very deep results and have a large number of useful consequences.
Some of them will be discussed in the problems.





\section{Euclidean spaces}
In this section without saying the contrary by a vector space we mean a
finite dimensional vector space over $\R$.

\subsection{Inner products}

\begin{definition}
Let $V$ a vector space.
\begin{enumerate}
\item A bilinear form on $V$ is a map
    \[
    f: V \times V \to \R
    \]
    which is linear in both variables.

\item A bilinear form $f$ is symmetric if $f(v, w) = f(w, v)$.

\item A symmetric bilinear form is semi-positive definite if $f(v, v) \geq
    0$, and it is positive definite if it is semi-positive definite and
    $f(v, v) = 0$ if and only if $v = 0$.

\item An inner product on $V$ is a symmetric positive definite bilinear
    form.
\end{enumerate}
\end{definition}

An inner product on $V$ is usually denote by $\langle-, -\rangle$. A vecotr
space with an inner product is called a Euclidean space, or an inner product
space.

\begin{prop}
Let $(V, \langle-,-\rangle)$ be a Euclidean space. Then
    \[
    \langle v, w \rangle^2 \leq \langle v, v\rangle \langle w, w \rangle,
    \]
for all $v, w \in V$.
\end{prop}

This is the so-called Cauchy's inequality and it plays a definitive role in
the following definition.

\begin{definition}
Let $(V, \langle-,-\rangle)$ be a Euclidean space.
\begin{enumerate}
\item The norm of vector $v$, denote by $\aabs{v}$, is
    \[
    \aabs{v} = \sqrt{\langle v, v\rangle}.
    \]
    Of course $\aabs{v} = 0$ if and only if $v = 0$.

\item Let $v, w \in V$. The angle between $v$ and $w$ is defined to be
    \[
    \theta = \arccos \frac{\langle v, w\rangle}{\aabs{v} \aabs{w}}.
    \]
    The vectors $v$ and $w$ are perpendicular if their angle is $\pi/2$, or
    in other words, $\langle v, w\rangle = 0$.

\item Let $\{v_1, \cdots, v_n\}$ be a basis of $V$. It is called an
    orthonormal basis if
        \[
        \langle v_i, v_j \rangle = \left\{ \begin{aligned} & 1, && \text{if $i = j$};\\
        &0, && \text{if $i \not=j$}. \end{aligned} \right.
        \]
\end{enumerate}
\end{definition}

Now you might want to think, where is Cauchy's inequality used in this
definition?


\begin{example}
The space $\R^n$, together with the usual inner product
    \[
    \langle v, w\rangle = x_1y_1+ \cdots + x_n y_n,
    \]
where $v = \tp{(x_1, \cdots, x_n)}$, $w = \tp{(y_1, \cdots, y_n)}$, is an
inner product space. The standard basis of $\R^n$ is an orthonormal basis.
\end{example}

Now consider an arbitrary Euclidean space $(V, \langle-, -\rangle)$. Let
$\cB$ be an orthonormal basis. Let $v, w \in V$ and $\tp{(x_1, \cdots, x_n)}$
and $\tp{(y_1, \cdots, y_n)}$ be their coordinates respectively. Simple
computations lead to that
    \[
    \langle v, w\rangle = x_1y_1 + \cdots + x_n y_n.
    \]
This means that all Euclidean space, upon choosing an orthonormal basis,
``look like'' the familiar Euclidean space $\R^n$ with the standard inner
product.

\subsection{The Gram-Schmidt process}
Let $(V, \langle -, -\rangle)$ be a Euclidean space. We know from the
discussion in the previous subsection that an orthonormal basis is something
nice. However the question is: is there an orthonormal basis to begin with?

The answer is affirmative and in fact there is an algorithm which produces an
orthonormal basis out of any basis of $V$. Let us describe it now. Let $v_1,
\cdots, v_n$ be a basis of $V$. We inductively define
    \[
	\begin{aligned}
		u_1 &:=  v_1 \\
		u_2 &:= v_2 - \frac{\langle v_2, u_1\rangle}
        {\langle u_1, u_1 \rangle} u_1 \\
		\vdots &   \\
		u_n &:= v_n - \frac{\langle v_n, u_1\rangle}
        {\langle u_1,u_1 \rangle} u_1
		 - \frac{\langle v_n, u_2\rangle}
        {\langle u_2,u_2 \rangle} u_2
		 -\cdots - \frac{\langle v_n, u_{n-1}\rangle}
        {\langle u_{n-1},u_{n-1} \rangle} u_{n-1}.
	\end{aligned}
    \]
As $v_i$'s may be recovered from the $u_i$'s, we see that $u_1, \cdots, u_n$
form a basis of $V$, and one easily checks that it is an orthogonal basis.

The matrix version of this process gives an important decomposition of
matrices. Consider $\R^n$ with the standard inner product and let $\{v_1,
\cdots, v_n\}$ be a basis. Consider the matrix $A$ whose columns are $v_1,
\cdots, v_n$.

\begin{lemma}
The basis $\{v_1, \cdots, v_n\}$ is an orthonormal basis if and only if
$\tp{A}A = 1_n$.
\end{lemma}

A matrix $A$ with the property that $\tp{A} A = 1_n$ is called an orthogonal
matrix. Note that if $A$ is orthogonal, then $A$ is invertible and $A^{-1} =
\tp{A}$.

Now if we run the Gram-Schmidt algorithm to the basis $\{v_1, \cdots, v_n\}$,
we obtain an orthonormal basis $\{u_1, \cdots, u_n\}$. Let $O$ be the matrix
whose columns are $u_1, \cdots, u_n$ respectively. Then $O$ is an orthogonal
matrix. We observe that $v_i$ is a linear combination of $u_1, \cdots, u_i$.
We thus conclude the following important result.

\begin{theorem}
Let $A$ be an invertible matrix. There is an orthogonal matrix $O$ and an
upper triangular matrix $B$ so that
    \[
    A = OB.
    \]
Similarly (why?) we can find another orthogonal matrix $O'$ and an upper
triangular matrix $B'$ so that $A = B'O'$.
\end{theorem}

This factorization of a matrix $A$ is usually referred to as the QR
factorization.


\subsection{Self-adjoint operators}
The starting point is the following result.

\begin{prop}
Let $(V, \langle-, -\rangle)$ be a Euclidean space and $f: V \to V$ be a
linear operator. Then there is a unique linear operator $f^*: V \to V$ with
the property that
    \[
    \langle f(v), w \rangle = \langle v, f^*(w) \rangle
    \]
for all $v, w \in V$.
\end{prop}

The $f^*$ in the proposition is called the adjoint operator of $f$. The
description of $f^*$ is the simplest if we fix an orthonormal basis of $V$.
Say $\cB$ is such a basis under which $f$ is represented by a matrix $A$.
Then $f^*$ is represented by the matrix $\tp{A}$ under $\cB$. The equality in
the proposition boils down to the obvious identity
    \[
    \tp{(Ax)} y = \tp{x} (\tp{A}y).
    \]

\begin{definition}
Let $(V, \langle-, -\rangle)$ be a Euclidean space and $f: V \to V$ be a
linear operator.
\begin{enumerate}
\item $f$ is a normal operator if $f f^* = f^* f$;

\item $f$ is a self-adjoint operator if $f = f^*$;

\item $f$ is an orthogonal operator if $f f^* = 1_n$.
\end{enumerate}
\end{definition}

Of course self-dual operators and orthogonal operators are normal.

Let us study self-adjoint operators in greater detail.

\begin{prop}
Let $f: V \to V$ be a linear operator on a Euclidean space $V$. Let $\cB$ be
an orthonormal basis and $A$ the matrix representing $f$. Then $f$ is
self-adjoint if and only if $A$ is symmetric, i.e. $\tp{A} = A$.
\end{prop}


The following result is called the spectral theorem.

\begin{theorem}
Let $f: V \to V$ be a linear operator on a Euclidean space $V$. There is an
orthonormal basis of $V$ so that the matrix representing $f$ is diagonal. In
particular every element in such a basis is an eigenvector, and the
eigenvalues of $f$ are all real.
\end{theorem}

Again let $A$ be a symmetric matrix and view it as a self-adjoint operator on
$\R^n$ (with the standard inner product). Apply the spectral theorem to $A$
we obtain the following variant of the spectral theorem for symmetric
matrices.

\begin{theorem}
Let $A$ be a symmetric matrix. It eigenvalues are all real. There is an
orthogonal matrix $O$ and a diagonal matrix $D$ so that
    \[
    A = O D \tp{O}.
    \]
\end{theorem}

Of course the diagonal entries of $O$ are the eigenvalues of $A$.



\subsection{Positive definite symmetric matrices}
Let $A$ be an $n \times n$ symmetric matrix. We say that $A$ is positive
definite if for all $x \in \R^n$ we have $\tp{x} Ax \geq 0$ and the equality
holds if and only if $x =0$. This type of matrices have wide variety of
applications in all branches of mathematics.

\begin{theorem}
Let $A$ be a symmetric matrix. The following are equivalent.
\begin{itemize}
\item $A$ is positive definite.

\item The eigenvalues of $A$ are all positive.

\item There is an invertible matrix $P$ so that $A = \tp{P} P$.

\item All principal minors of $A$ are positive. Here an $i$-th principal
    minor is the determinant of an upper left $i \times i$ block of $A$.
\end{itemize}
\end{theorem}


Similarly a symmetric matrix $A$ is called semi-positive-definite if $\tp{x}
Ax \geq 0$.

\begin{theorem}
Let $A$ be a symmetric matrix. The following are equivalent.
\begin{itemize}
\item $A$ is semi-positive-definite.

\item The eigenvalues of $A$ are all nonnegative.

\item There is a matrix $P$ so that $A = \tp{P} P$ (note that $P$ does not
    have to be a square matrix).
\end{itemize}
\end{theorem}


\subsection{Singular values}
Let $A$ be an $m \times n$ matrices. Then $\tp{A} A$ is an $n \times n$
semi-positive-definite symmetric matrix. The eigenvalues of $\tp{A} A$ are
all nonnegative, and let us denote them by $\mu_1^2, \cdots, \mu_r^2, 0,
\cdots, 0$ where $\mu_1, \cdots, \mu_r$ are positive real numbers. These real
numbers are called singular values of $A$. Let $D$ be the $m \times n$ matrix
of the form
    \[
    D = \begin{pmatrix} \mu_1 \\& \ddots \\ && \mu_r \\ &&& 0 & \cdots & 0 \\
    && &\vdots  & \vdots & \vdots \\ &&&0 & \cdots & 0 \end{pmatrix}.
    \]
Then by the spectral theorem, there is an $n \times n$ orthogonal matrix $Q$
so that
    \[
    \tp{A} A = \tp{Q} \tp{D} D Q.
    \]
It then follows that there is another $m \times m$ orthogonal matrix $Q$ so
that
    \[
    A = PDQ.
    \]
This is usually refer to as the singular value decomposition of $A$. It plays
a definitive role in principal component analysis in statistics.



\section{Some multilinear algebra}

Without saying to the contrary by a vector space we mean a finite dimensional
vector space.


\subsection{Duality}
Let $V$ and $W$ be vector spaces over $F$. A bilinear form on $V \times W$ is
a map
    \[
    f: V \times W \to F,
    \]
linear in both variables. The bilinear form is nondegenerate if for all $v
\in V$ we can find some $w \in W$ so that $f(v, w) \not=0$ and for all $w \in
W$ we can find some $v \in V$ so that $f(v, w) \not=0$. A nondegenerate
bilinear form is usually refer to as a perfect pairing between $V$ and $W$.

\begin{prop}
If there is a perfect pairing between $V$ and $W$ then $\dim V = \dim W$.
\end{prop}

Let $f$ be a perfect pairing. If $\{v_1, \cdots, v_n\}$ is a basis of $V$,
then there is basis $\{ w_1, \cdots, w_n\}$ of $W$ so that
    \[
    f(v_i, w_j) = \left\{ \begin{aligned} & 1, && i = j;\\
    & 0, && i \not=j \end{aligned} \right.
    \]
Then $\{w_1, \cdots, w_n\}$ is called the dual basis of $\{v_1, \cdots,
v_n\}$.

We denote by $V^*$ the space of all linear functionals on $V$. It is called
the dual space of $V$. We observe that there is a bilinear form between $V$
and $V^*$ given by
    \[
    f: V \times V^* \to F, \quad f(v, l) = l(v).
    \]
This pairing is perfect (we make use of the assumption that $V$ is finite
dimensional here!). It follow that
    \[
    \dim V = \dim V^*.
    \]
It should be warn here that even though one can find an isomorphism from $V$
to $V^*$, these two spaces are usually very different and should not be
treated as the same. For example, in the context of differential geometry,
the dual space of the tangent space is the cotangent space, one being the
space of vector fields and the other the space of differential forms.

On the other hand, any element $v$ can be viewed as a linear functional on
$V^*$. If $v \in V$ and $l \in V^*$ we simply define
    \[
    v(l) = l(v).
    \]
This gives an injective linear transform $V \to V^{**}$. But $\dim V = \dim
V^* = \dim V^{**}$ we conclude that this is indeed an isomorphism.

With this result in mind, the upshot is that $V$ and $V^*$ are dual spaces of
each other.


\subsection{Symmetric bilinear forms}
Let $V$ be an $n$-dimensional vector space and
    \[
    f: V \times V \to F
    \]
be a bilinear form. It is called symmetric if $f(v, w) = f(w, v)$. There is a
close relation between symmetric bilinear forms and symmetric matrices. If $V
= F^n$ and $A$ an $n \times n$ symmetric matrix, then
    \[
    f(x, y) = \tp{x} A y
    \]
is a symmetric bilinear form. Conversely let $V$ be an $n$-dimensional vector
space and $\{v_1, \cdots, v_n\}$ a basis, put
    \[
    A = (a_{ij})_{1 \leq i, j\leq n}, \quad a_{ij} = f(v_i, v_j).
    \]
Then $A$ is an $n \times n$ symmetric matrix. Moreover if $v, w \in V$ and
their coordinates under the basis $\{v_1, \cdots, v_n\}$ is $x, y\in F^n$
respectively, then
    \[
    f(v, w) = \tp{x} A y.
    \]

Similar to the situation with the linear operators, we may ask what if we
change to a different basis. Suppose that
    \[
    (v_1', \cdots, v_n') = (v_1, \cdots, v_n) P
    \]
is another basis of $V$ where $P$ is an invertible $n \times n$ matrix. Then
it is not hard to check that the new matrix
    \[
    A' = (a_{ij}')_{1 \leq i, j\leq n}, \quad a_{ij}' = f(v_i', v_j')
    \]
is given by
    \[
    A' = \tp{P} A P.
    \]

\begin{definition}
We say that two symmetric matrices $A$ and $A'$ are congruent if there is an
invertible matrix $P$ so that $A' = \tp{P} A P$.
\end{definition}

The upshot is the following. Once we fix a basis of $V$, a symmetric bilinear
forms bilinear form on $V$ gives rise to a symmetric matrix. If we choose
different bases, the symmetric matrices we obtain are all congruent to each
other.

It is then an interesting question to ask for ``canonical form'' of a
symmetric bilinear form. If $F = \R, \C$, the situation is quite simple. If
$F = \Q$ (or more generally a number field), this is a classical number
theoretic problem. It was solved by Hilbert and Hasse and the result is
usually referred to as the ``local--global'' principle. We refer the readers
to Serre's book {\it A Course in Arithmetic} for more details.

We now describe the results when $F = \R$ or $\C$.

\begin{theorem}
If $F = \R$, we can choose a basis of $V$ so that the resulting symmetric
matrix is of the form
    \[
    \begin{pmatrix} 1_p \\ & - 1_q\\ && 0_{n-r} \end{pmatrix}
    \]
where $p + q = r$.

If $F = \C$, then we can choose a basis of $V$ so that the resulting
symmetric matrix is of the form
    \[
    \begin{pmatrix} 1_r \\ & 0_{n-r} \end{pmatrix}
    \]
\end{theorem}

The integer $r$ is called the rank of $f$. The bilinear form is nondegenerate
if $r = 0$. In this case if $F = \R$, the pair $(p, q)$ is called the
signature of $f$.

We finally mention the relation between the symmetric bilinear forms and
quadratic form. Let $V = F^n$ and assume that the characteristic of $F$ is
not two. By a quadratic form on $V$ we mean a function $q: V \to F$ so that
    \[
    f(x, y) = \frac{1}{2} q(x+y) - q(x) - q(y)
    \]
is a bilinear form on $V$. The quadratic form $q$ can be recovered from the
bilinear form $f$ via
    \[
    q(x) = f(x, x).
    \]
From now on we assume that $F = \R$. The quadratic form $q$ is called
positive definitive if $q(x) \geq 0$ and $q(x) = 0$ only if $x = 0$. This in
particular implies that the bilinear form $f$ obtained from $q$ is
nondegenerate. We have the following characterizations of positive definite
quadratic forms.

\begin{theorem}
The following are equivalent.
\begin{enumerate}
\item $q$ is a positive definite quadratic form.

\item $f$ is an inner product on $V$.

\item The signature of $q$, namely that of $f$, is $(n, 0)$.

\item The matrix representing $q$, namely the matrix representing $f$, is
    positive definite.
\end{enumerate}
\end{theorem}



\subsection{Tensor product of vector spaces}
Let $V$ and $W$ be vector spaces over $F$.

\begin{definition}
The tensor product $V^* \otimes W^*$ is defined to be the space of all
bilinear forms on $V \times W$.
\end{definition}

Let $l_1 \in V^*$ and $l_2 \in W^*$, then we define a bilinear form on $V
\times W$ by
    \[
    (l_1 \otimes l_1')(v, w) = l_1(v) l_1'(w).
    \]
This then gives rise to a natural map
    \[
    V^* \times W^* \to V^* \otimes W^*, \quad (l_1, l_1') \mapsto l_1 \otimes l_1'.
    \]
linear in both $V^*$ and $W^*$. It can be checked that if we choose bases $\{
l_1, \cdots, l_n\}$ of $V^*$ and $\{l_1', \cdots, l'_m\}$ of $W^*$, then
    \[
    \{ l_i \otimes l_j' \mid 1 \leq i \leq n,\ 1 \leq j \leq m\}
    \]
is a basis of $V^* \otimes W^*$. It follows that
    \[
    \dim V^* \otimes W^* = \dim V^* \dim W^* = \dim V \dim W.
    \]

What if we want to define the tensor product $V \otimes W$? We only need to
make use of duality. Note that $V$ is the dual space of $V^*$ and $W$ is the
dual space of $W^*$. It is then natural to define
    \[
    V \otimes W  = (V^* \otimes W^*)^*.
    \]
We then obtain another natural map
    \[
    V \times W \to V \otimes W,
    \]
linear in both $V$ and $W$. By the discussion above, if $\{v_1, \cdots,
v_n\}$ and $\{w_1, \cdots w_m\}$ are bases of $V$ and $W$ respectively, then
    \[
    \{ v_i \otimes w_j \mid 1 \leq i \leq n,\ 1 \leq j \leq m\}
    \]
is a basis of $V \otimes W$, and
    \[
    \dim V \otimes W = \dim V \dim W.
    \]

The most important property of the tensor product $V \otimes W$ is the
following. This is usually referred to as the ``universal property'' of the
tensor product.

\begin{theorem}
Let $U$ be a vector space and $V \times W \to U$ a map, linear in both $V$
and $W$. Then there is a unique linear transform $V \otimes W \to U$ so that
the diagram
    \[
    \xymatrix{
    V \times W \ar[r] \ar[rd] & V \otimes W \ar[d]\\
    & U
    }
    \]
is commutative.
\end{theorem}

Using this universal property one can define the tensor product
    \[
    V_1 \otimes V_2 \otimes \cdots \otimes V_r
    \]
where $V_1, \cdots, V_r$ are vector spaces over $F$.

\begin{definition}
Let $V_1, \cdots, V_r$ be vector spaces over $F$. The tensor product
    \[
    V_1 \otimes V_2 \otimes \cdots \otimes V_r
    \]
is a vector space over $F$ together with a map
    \[
    V_1 \times \cdots \times V_r \to V_1 \otimes V_2 \otimes \cdots \otimes V_r
    \]
with the property that for all vector spaces $U$ and all maps
    \[
    V_1 \times \cdots \times V_r \to U
    \]
which are linear all variables, there is a unique linear transform
    \[
    V_1 \otimes V_2 \otimes \cdots \otimes V_r \to U
    \]
so that the diagram
    \[
    \xymatrix{
    V_1 \times \cdots \times V_r \ar[rr] \ar[rrd]
    && V_1 \otimes \cdots \otimes V_r \ar[d]\\
    && U
    }
    \]
is commutative.
\end{definition}

The tensor product exists and is unique (up to isomorphism). Indeed one first
define
    \[
    V_1^* \otimes \cdots \otimes V_r^*
    \]
as the space of maps $V_1 \times \cdots \times V_r \to F$ that are linear in
all variables, and put
    \[
    V_1 \otimes \cdots \otimes V_r =
    (V_1^* \otimes \cdots \otimes V_r^*)^*.
    \]
The uniqueness can be checked directly.




Moreover one can prove the following associativity property.

\begin{prop}
Let $V_1, V_2$ and $V_3$ be vector spaces. Then
    \[
    (V_1 \otimes V_2) \otimes V_3 = V_1 \otimes (V_2 \otimes V_3) =
    V_1 \otimes V_2 \otimes V_3.
    \]
\end{prop}



In differential geometry and many other branches of mathematics, tensor
products of the form
    \[
    V \otimes \cdots \otimes V \otimes V^* \otimes \cdots V^*
    \]
are usually considered. If $V$ appears $r$ times and $V^*$ appears $s$ times,
elements in this space is usually referred to as $(r, s)$ type tensor. In
particular tensors of type $(r,0)$ are called contravariant and of type $(0,
s)$ are called covariant.





\begin{comment}
\section{Some group theory}


\subsection{Generalities}

\begin{definition}
A set $G$, together with a binary operation $G$, i.e. a map $G \times G \to
G$, $(x, y) \mapsto xy$, is called a group, if the following conditions hold.
\begin{enumerate}
\item For all $x, y, z \in G$ we have $(xy)z = x(yz)$.

\item There is an element $e \in G$ so that $ex = xe = x$ for all $x \in
    G$.

\item For any $x \in G$, there is an element $y \in G$ so that $xy = yx =
    e$.
\end{enumerate}

If in addition, $xy = yx$ for all $x, y \in G$, then $G$ is called an abelian
group.
\end{definition}


Let $G$ and $H$ be groups. A homomorphism $f: G \to H$ is a map so that
    \[
    f(xy) = f(x)f(y)
    \]
for all $x, y \in G$. In particular this implies that $f(e_G) = e_H$. The
homomorphism is called an isomorphism if it is bijective. In this case $G$
and $H$ are said to be isomorphic and denoted by $G \simeq H$. Isomorphic
groups are usually treated to be the same group in group theory.

\begin{example}
Here are basic examples of groups.
\begin{itemize}
\item Cyclic group, i.e. $\Z$ or $\Z/n\Z$.

\item Symmetric groups $S_n$ and alternating groups $A_n$. By definition
    $S_n$ is the set of all permutations on a set with $n$ elements and
    $A_n$ is a subgroup of $S_n$ consisting of even permutations.

\item Matrix groups, e.g. $\GL_n(\R)$ the group of all invertible $n \times
    n$ matrices, or $\SL_n(\R)$ the subgroup consist of all invertible
    matrices with determinant one.
\end{itemize}
\end{example}




\subsection{Subgroups}

\begin{definition}
Let $G$ be a group and $H \subset G$ a subset.

\begin{enumerate}
\item $H$ is called a subgroup if it is a group under the group operator in
    $G$.

\item Assume that $H$ is a subgroup. For any $g \in G$, the subset of $G$
    of the form
    \[
    gH = \{ g h \mid h \in H \}
    \]
    is called a left coset. Similarly we have the right coset $Hg$.

\item $H$ is called normal subgroup if $Hg = gH$ for all $g \in G$.

\item A group is called simple if the only normal subgroups of $G$ are
    $\{1\}$ and $G$ itself.
\end{enumerate}
\end{definition}

\begin{prop}
Let $G$ be a group and $H$ be a subgroup, then two left cosets either
coincide or are disjoint. In particular $G$ is a disjoint union of left
cosets of $H$, and hence $\abs{H} \mid \abs{G}$.
\end{prop}

The ratio $\abs{G}/\abs{H}$ is an integer, which is called the index of $H$
in $G$ and is usually denoted by $[G:H]$.

Let $G$ be a group and $H$ be a subgroup. We denote by $G/H$ the set of all
left cosets. The main point of $H$ being a normal subgroup is that $G/H$
would again have a natural group structure which we now explain. Assume now
that $H$ is a normal subgroup. We define
    \[
    g_1 H \cdot g_2 H = g_1 g_2 H.
    \]
The claim is that this is a well-defined group operation on $G/H$. First why
is this a statement that needs justification? The point is that we can have
$g_2, g_2' \in G$ so that $g_2 H = g_2' H$. Then in the above definition, if
we would have
    \[
    g_1 H \cdot g_2 H = g_1 g_2' H.
    \]
What needs to be justified is that $g_1 g_2 H = g_1 g_2' H$, for otherwise
the operation on $G/H$ would make no sense. Note that for any $x, y \in G$,
$xH = yH$ if and only if $x^{-1} y \in H$. Therefore
    \[
    (g_1 g_2)^{-1} g_1 g_2' = g_2^{-1} g_1^{-1} g_1 g_2' = g_2^{-1} g_2' \in H.
    \]
It follows that $g_1 H \cdot g_2 H = g_1 g_2' H$. Similarly one can show that
if $g_1H = g_1'H$, then $g_1 g_2 H = g_1' g_2 H$. One then checks easily that
the operation
    \[
    g_1 H \cdot g_2 H = g_1 g_2 H
    \]
defines a natural group structure on $G/H$. This group is called the quotient
group (of $G$ by $H$).




\subsection{Simple groups}
A group $G$ is called a simple if it does not have any normal subgroups apart
from $\{e\}$ and $G$ itself. Simple groups are the ``building blocks'' of all
groups. The classification theorem of all finite simple groups has been
called one of the great intellectual achievements of humanity.

The most obvious simple groups are of the shape $\Z/p\Z$, $p$ a prime. They
are all abelian. One classical result is the following.

\begin{theorem}
If $n \geq 5$, then $A_n$ is simple. Moreover $A_5$ is the smallest
nonabelian simple group.
\end{theorem}

According to the finite classification theorem, every finite simple group is
either cyclic, or alternating, or it belongs to a broad infinite class called
the groups of Lie type, or else it is one of twenty-seven exceptions, called
sporadic groups. We would not have enough room defining all of these terms,
but just point out that many of these subjects are still active research
areas today.


\subsection{Finitely generated abelian groups}

If $G_1$ and $G_2$ are groups, there is a natural group structure on the
Cartesian product $G_1 \times G_2$ given by
    \[
    (x_1, y_1)(x_2, y_2) = (x_1x_2, y_1y_2).
    \]
This is called the product of $G_1$ and $G_2$.

We now discuss abelian groups. The convention is that for an abelian group
$G$, the group operation is usually denoted by $x+y$, the identity $e$ is
usually denoted by $0$, and the inverse of $x$ is denoted by $(-x)$. So we
have the intuitively correct identity $x +0 = 0+x = x$ and $x+(-x) = 0$.
Moreover we write $nx$ for $x+x+ \cdots +x$ ($n$ times). If $G_1$ and $G_2$
are abelian group, then we denote their product by $G_1 \oplus G_2$ and call
it the direct sum of $G_1$ and $G_2$. Similarly we can define the direct sum
of finitely many abelian groups.

Let $G$ be an abelian group. We say that $G$ is finitely generated if we can
find $x_1, \cdots, x_n \in G$ so that for all $g \in G$, we can find integers
$a_1, \cdots, a_n$ so that
    \[
    g = a_1 x_1 + \cdots + a_n x_n.
    \]
(Note that this looks like $x_1, \cdots, x_n$ ``span'' the group $G$.)

The following theorem completely describes the structure of a finitely
abelian group.

\begin{theorem}
Finitely abelian groups are (finite) direct sum of cyclic groups. More
precisely let $G$ be a finitely generated abelian group. Then there is an
integer $r$ and finite cylic groups of prime power order $\Z/p_i^{e_i}\Z$, $i
= 1, \cdots, n$, so that
    \[
    G \simeq \Z^{\oplus r} \oplus \left( \Z/p_1^{e_1}\Z \oplus \cdots  \Z/p_n^{e_n}\Z \right).
    \]
Moreover the prime powers $p_1^{e_1}, \cdots, p_n^{e_n}$ are uniquely
determined by $G$.
\end{theorem}

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