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\begin{document}

\title{Math 413/513 Chapter 4 (from Friedberg, Insel, \& Spence)}
\author{David Glickenstein}
\maketitle

\section{Determinants}

\subsection{Definitions and main results}

\begin{definition}
Let $A=\left(  A_{ij}\right)  \in F^{n\times n}.$ The \emph{determinant}
$\det\left(  A\right)  $ is a scalar defined recursively as $\det\left(
A\right)  =A_{11}$ if $n=1$ and if $n\geq2,$
\[
\det A=\sum_{j=1}^{n}\left(  -1\right)  ^{1+j}A_{1j}\det\left(  \tilde{A}%
_{1j}\right)
\]
where $\tilde{A}_{ij}\in F^{\left(  n-1\right)  \times\left(  n-1\right)  }$
is the matrix obtained from $A$ by deleting the $i$th row and $j$th column.
Sometimes $\det A$ is denoted as $\left\vert A\right\vert .$ The scalar
$\left(  -1\right)  ^{i+j}A_{ij}\det\left(  \tilde{A}_{ij}\right)  $ is called
the $i,j$ \emph{cofactor}.
\end{definition}

We note that in the $2\times2$ case, we have
\[
\det\left(
\begin{array}
[c]{cc}%
a & b\\
c & d
\end{array}
\right)  =ad-bc.
\]


\begin{theorem}
\label{thm:det linear}The determinant is a linear function of each row if the
other rows are held fixed. That is, for any $r$ between $1$ and $n,$%
\[
\det\left(
\begin{array}
[c]{c}%
a_{1}\\
\vdots\\
a_{r-1}\\
u+kv\\
a_{r+1}\\
\vdots\\
a_{n}%
\end{array}
\right)  =\det\left(
\begin{array}
[c]{c}%
a_{1}\\
\vdots\\
a_{r-1}\\
u\\
a_{r+1}\\
\vdots\\
a_{n}%
\end{array}
\right)  +k\det\left(
\begin{array}
[c]{c}%
a_{1}\\
\vdots\\
a_{r-1}\\
v\\
a_{r+1}\\
\vdots\\
a_{n}%
\end{array}
\right)
\]
for $k$ a scalar and $a_{j},u,v$ row vectors.
\end{theorem}

Note that this tells us how the determinant is affected by one type of
elementary row operation.

\begin{corollary}
\label{cor:row of zeroes}If $A$ has a row of all zeroes, then $\det A=0$
\end{corollary}

\begin{theorem}
\label{thm:cofactor expansion}The determinant can be defined by a cofactor
expansion in any row, i.e., for any $i,$%
\[
\det A=\sum_{j=1}^{n}\left(  -1\right)  ^{i+j}A_{ij}\det\left(  \tilde{A}%
_{ij}\right)  .
\]

\end{theorem}

\begin{corollary}
\label{cor:identical rows}If $A$ has two identical rows, then $\det A=0.$
\end{corollary}

\begin{theorem}
\label{thm:exchange rows}If $B$ is obtained from $A$ by exchanging any two
rows, $\det B=-\det A.$
\end{theorem}

\begin{theorem}
\label{thm:det row op type 3}Let $B$ be obtained from $A$ by adding a multiple
of one row to another. Then $\det B=\det A.$
\end{theorem}

\begin{corollary}
\label{cor: lower rank}If $A\in F^{n\times n}$ has $\operatorname{rank}A<n,$
then $\det A=0.$
\end{corollary}

\begin{theorem}
\label{thm: transpose}For $A\in F^{n\times n},$ $\det\left(  A^{T}\right)
=\det A.$
\end{theorem}

\begin{corollary}
The determinant can be defined by a cofactor expansion in any column.
\end{corollary}

\begin{theorem}
\label{thm:product}For $A,B\in F^{n\times n},$ $\det\left(  AB\right)
=\left(  \det A\right)  \left(  \det B\right)  $
\end{theorem}

\begin{corollary}
\label{cor:invertible}A matrix $A\in F^{n\times n}$ is invertible if and only
if $\det A\neq0.$ If $A$ is invertible, $\det\left(  A^{-1}\right)  =\frac
{1}{\det A}.$
\end{corollary}

\subsection{Proofs of theorems and corollaries}

\begin{proof}
[Proof of Theorem \ref{thm:det linear}]The proof is by induction on $n.$
Clearly this is true for $n=1.$ Now suppose it is true for $n$ and let
$A,B,C\in F^{\left(  n+1\right)  \times\left(  n+1\right)  }$ be of the form%
\[
A=\left(
\begin{array}
[c]{c}%
a_{1}\\
\vdots\\
a_{r-1}\\
u+kv\\
a_{r+1}\\
\vdots\\
a_{n}%
\end{array}
\right)  ,~~~B=\left(
\begin{array}
[c]{c}%
a_{1}\\
\vdots\\
a_{r-1}\\
u\\
a_{r+1}\\
\vdots\\
a_{n}%
\end{array}
\right)  ,~~C=\left(
\begin{array}
[c]{c}%
a_{1}\\
\vdots\\
a_{r-1}\\
v\\
a_{r+1}\\
\vdots\\
a_{n}%
\end{array}
\right)
\]
Notice that by the inductive hypothesis, $\det\tilde{A}_{1j}=\det\tilde
{B}_{1j}+k\det\tilde{C}_{1j}$ for each $j$ and also note that if $i\neq r,$
$A_{ij}=B_{ij}=C_{ij}$ Then if $r>1,$ then
\begin{align*}
\det A  &  =\sum_{j=1}^{n}\left(  -1\right)  ^{1+j}A_{1j}\det\left(  \tilde
{A}_{1j}\right) \\
&  =\sum_{j=1}^{n}\left(  -1\right)  ^{1+j}A_{1j}\left(  \det\tilde{B}%
_{1j}+k\det\tilde{C}_{1j}\right) \\
&  =\sum_{j=1}^{n}\left(  -1\right)  ^{1+j}B_{1j}\det\tilde{B}_{1j}%
+k\sum_{j=1}^{n}\left(  -1\right)  ^{1+j}C_{1j}\det\tilde{C}_{1j}\\
&  =\det B+k\det C.
\end{align*}
We leave the case of $r=1$ as an exercise.
\end{proof}

\begin{proof}
[Proof of Corollary \ref{cor:row of zeroes}]Exercise.
\end{proof}

\begin{proof}
[Proof of Theorem \ref{thm:cofactor expansion}]We do induction on $n.$ The
base case is easy. Notice that since we can expand in any row in  $\tilde
{A}_{ij},$ we can compute the following if we let $\tilde{A}_{ij,k\ell}$ be
the matrix obtained by removing the $i$th and $k$th rows and $j$th and $\ell
$th columns from $A.$%
\begin{align*}
\det A  & =\sum_{j=1}^{n}\left(  -1\right)  ^{1+j}A_{1j}\det\left(  \tilde
{A}_{1j}\right)  \\
& =\sum_{j=1}^{n}\left(  -1\right)  ^{1+j}A_{1j}\left(  \sum_{k=1}%
^{j-1}\left(  -1\right)  ^{k+i-1}A_{ik}\det\left(  \tilde{A}_{1j,ik}\right)
+\sum_{k=j+1}^{n}\left(  -1\right)  ^{k+i}A_{ik}\det\left(  \tilde{A}%
_{1j,ik}\right)  \right)  \\
& =\sum_{j=1}^{n}\sum_{k=1}^{j-1}\left(  -1\right)  ^{j+k+i}A_{1j}A_{ik}%
\det\left(  \tilde{A}_{1j,ik}\right)  +\sum_{j=1}^{n}\sum_{k=j+1}^{n}\left(
-1\right)  ^{j+k+i+1}A_{1j}A_{ik}\det\left(  \tilde{A}_{1j,ik}\right)  \\
& =\sum_{k<j}\left(  -1\right)  ^{j+k+i}A_{1j}A_{ik}\det\left(  \tilde
{A}_{1j,ik}\right)  +\sum_{k>j}\left(  -1\right)  ^{j+k+i+1}A_{1j}A_{ik}%
\det\left(  \tilde{A}_{1j,ik}\right)
\end{align*}
Similarly,%
\begin{align*}
& \sum_{k=1}^{n}\left(  -1\right)  ^{i+k}A_{ik}\det\left(  \tilde{A}%
_{ik}\right)  \\
\;\;\;  & =\sum_{k=1}^{n}\left(  -1\right)  ^{i+k}A_{ik}\left(  \sum
_{j=1}^{k-1}\left(  -1\right)  ^{1+j}A_{1j}\det\left(  \tilde{A}%
_{ik,1j}\right)  +\sum_{j=k+1}^{n}\left(  -1\right)  ^{1+j-1}A_{1j}\det\left(
\tilde{A}_{ik,1j}\right)  \right)  \\
& =\sum_{k=1}^{n}\sum_{j=1}^{k-1}\left(  -1\right)  ^{1+j+k+i}A_{ik}A_{1j}%
\det\left(  \tilde{A}_{ik,1j}\right)  +\sum_{k=1}^{n}\sum_{j=k+1}^{n}\left(
-1\right)  ^{j+k+i}A_{ik}A_{1j}\det\left(  \tilde{A}_{ik,1j}\right)  \\
& =\sum_{j<k}\left(  -1\right)  ^{1+j+k+i}A_{ik}A_{1j}\det\left(  \tilde
{A}_{ik,1j}\right)  +\sum_{j>k}\left(  -1\right)  ^{j+k+i}A_{ik}A_{1j}%
\det\left(  \tilde{A}_{ik,1j}\right)  .
\end{align*}

\end{proof}

\begin{proof}
[Proof of Corollary \ref{cor:identical rows}]We leave this as an exercise if
$n\leq2.$ For $n\geq3,$ if we assume rows $1$ and $j$ are the same, then we
can expand in a row other than those two rows. We see that each of the
determinants in the expansion also has two identical rows, and by induction
these determinants are all zero. Hence the determinant is zero.
\end{proof}

\begin{proof}
[Proof of Theorem \ref{thm:exchange rows}]Let the rows of $A$ be labeled
$a_{1},\ldots,a_{n}.$ We see that by linearity in the rows we get%
\[
0=\det\left(
\begin{array}
[c]{c}%
a_{1}\\
\vdots\\
a_{i}+a_{j}\\
\vdots\\
a_{i}+a_{j}\\
\vdots\\
a_{n}%
\end{array}
\right)  =\det\left(
\begin{array}
[c]{c}%
a_{1}\\
\vdots\\
a_{i}\\
\vdots\\
a_{i}\\
\vdots\\
a_{n}%
\end{array}
\right)  +\det\left(
\begin{array}
[c]{c}%
a_{1}\\
\vdots\\
a_{i}\\
\vdots\\
a_{j}\\
\vdots\\
a_{n}%
\end{array}
\right)  +\det\left(
\begin{array}
[c]{c}%
a_{1}\\
\vdots\\
a_{j}\\
\vdots\\
a_{i}\\
\vdots\\
a_{n}%
\end{array}
\right)  +\det\left(
\begin{array}
[c]{c}%
a_{1}\\
\vdots\\
a_{j}\\
\vdots\\
a_{j}\\
\vdots\\
a_{n}%
\end{array}
\right)  .
\]
The first and last determinants are zero, and so if $A^{\prime}$ is gotten by
exchanging rows $i$ and $j,$ then $\det A+\det A^{\prime}=0.$
\end{proof}

\begin{proof}
[Proof of Theorem \ref{thm:det row op type 3}]By linear in the rows, we get
that
\[
\det\left(
\begin{array}
[c]{c}%
a_{1}\\
\vdots\\
a_{i}+ka_{j}\\
\vdots\\
a_{j}\\
\vdots\\
a_{n}%
\end{array}
\right)  =\det\left(
\begin{array}
[c]{c}%
a_{1}\\
\vdots\\
a_{i}\\
\vdots\\
a_{j}\\
\vdots\\
a_{n}%
\end{array}
\right)  +k\left(
\begin{array}
[c]{c}%
a_{1}\\
\vdots\\
a_{j}\\
\vdots\\
a_{j}\\
\vdots\\
a_{n}%
\end{array}
\right)  =\det A+0
\]
since the last matrix has two of the same rows.
\end{proof}

\begin{proof}
[Proof of Corollary \ref{cor: lower rank}]We now know how row operations
affect the calculation of the determinant. If the rank is less than $n,$ we
can perform row operations to get a row of all zeros. This matrix will have
determinant zero and the row operations will all show that the determinant of
$A$ is still zero.
\end{proof}

\begin{proof}
[Proof of Theorem \ref{thm:product}]We know that if $A$ or $B$ is not rank
$n,$ then its determinant is zero. Also, if $A$ or $B$ is not rank $n,$ $AB$
is not rank $n$ (why? Show $L_{A}$ is not onto implies $L_{AB}$ is not onto,
and $L_{B}$ is not one-to-one implies $L_{AB}$ is not one-to-one). Hence the
theorem is true if $A$ or $B$ is not rank $n.$ If both $A$ and $B$ are rank
$n,$ so is $AB.$ Using what we know about how row operations affect
determinants, we can easily see that for any matrix $C$ and elementary matrix
$E,$ $\det\left(  EC\right)  =\left(  \det E\right)  \left(  \det C\right)  .$
Since any invertible matrix is a product of elementary matrices, we can show
that $\det\left(  AB\right)  =\det\left(  E_{k}\cdots E_{1}B\right)
=\det\left(  E_{k}\cdots E_{1}\right)  \left(  \det B\right)  =\left(  \det
A\right)  \left(  \det B\right)  .$
\end{proof}

\begin{proof}
[Proof of Corollary \ref{cor:invertible}]If $A$ is not invertible, then it has
rank less than $n$ and so $\det A=0.$ If $A$ is invertible, then $1=\det
I=\det\left(  A^{-1}A\right)  =\left(  \det A^{-1}\right)  \left(  \det
A\right)  $ so $\det A\neq0$ and $\det A^{-1}=\frac{1}{\det A}.$
\end{proof}

\begin{proof}
[Proof of Theorem \ref{thm: transpose}]If $A$ is not invertible, then neither
is $A^{T},$ and so $\det A^{T}=\det A=0.$ If $A$ is invertible, then
$A=E_{1}\cdots E_{k}$ for some elementary matrices. We then have that
$A^{T}=E_{k}^{T}\cdots E_{1}^{T}.$ We now see that $\det A^{T}=\left(  \det
E_{k}^{T}\right)  \cdots\det\left(  E_{1}^{T}\right)  $ and the result follows
from checking that $\det E_{k}^{T}=\det E_{k}$ for each type of elementary matrix.
\end{proof}

\subsection{Remarks on determinants as volumes}

If $v_{1},\ldots,v_{n}$ are vectors in $\mathbb{R}^{n},$ then it turns out
that $\det A,$ where the rows of $A$ are the vectors $v_{1},\ldots,v_{n}$ is
equal to $\pm$ the volume of the parelelopiped determined by the vectors.
Notice that this is zero if the vectors form a degenerate paralellopiped
(lower dimensional), which is geometrically the same as saying the vectors are
linearly dependent. The sign has to do with the ordering of the vectors, and
obeys the right hand rule for $n=2,3,$ and gives a way of defining an analogue
of the right hand rule in higher dimensions. This is called a choice of
orientation and is important in algebraic topology and differential geometry.

\section{Problems}

\begin{itemize}
\item FIS Section 4.1 exercises 2, 3, 6, 7, 10

\item FIS Section 4.2 exercises 3, 5-25,27, 29 

\item FIS Section 4.3 exercises 9-13, 17, 21, 28
\end{itemize}

\section{Characterization of the determinant (Comprehensive/Graduate option)}

\begin{definition}
A function $\delta:F^{n\times n}\rightarrow F$ is called an $n$\emph{-linear
function} if it is a linear function of each row when the remaining rows are
held fixed.
\end{definition}

\begin{definition}
An $n$-linear function $\delta:F^{n\times n}\rightarrow F$ is called
\emph{alternating} if $\delta\left(  A\right)  =0$ whenever two rows are identical.
\end{definition}

Note that the determinant satisfies both of these properties. In fact, it is
essentially the only such function.

\begin{theorem}
\label{thm:determinant characterization}Let $\delta:F^{n\times n}\rightarrow
F$ be an alternating $n$-linear function such that $\delta\left(  I\right)
=1.$ Then $\delta\left(  A\right)  =\det A$ for all $A\in F^{n\times n}.$
\end{theorem}

\begin{proof}
[Proof (sketch)]If you carefully look at our proofs from the previous section,
all we used is $n$-linearity and alternating to get the characterization of
what row operations do. It then follows that $\delta\left(  A\right)  =0$ if
$\operatorname{rank}A<n,$ and that if $\operatorname{rank}A=n,$ then we can
perform row operations to row reduce $A$ to $I,$ and so $\delta\left(
I\right)  $ determines $\delta\left(  A\right)  .$
\end{proof}

\begin{corollary}
We can write the determinant in the following way%
\[
\det A=\sum_{\sigma\in S_{n}}\operatorname{sgn}\left(  \sigma\right)
a_{1~\sigma\left(  1\right)  }\cdots a_{n~\sigma\left(  n\right)  }%
\]
where $S_{n}$ is the set of permutations of $\left\{  1,\ldots,n\right\}  $
and $\operatorname{sgn}\left(  \sigma\right)  $ is the sign of the
permutations ($-1$ to the number of transpositions to form the permutation
$\sigma$).
\end{corollary}

\begin{proof}
It is easy to see that the function on the right satisfy the properties of
Theorem \ref{thm:determinant characterization}.
\end{proof}

\section{Problems}

\begin{itemize}
\item FIS Section 4.5 exercises 16, 18, 19
\end{itemize}


\end{document}