\documentclass[12pt]{amsart}

\usepackage{241}

\title{21-241 Matrices and Linear Transformations \\
       Summer 1 2012 \\
       Course notes \\
       June 7}
\author{Paul McKenney}


\begin{document}

\maketitle

\begin{defn*}
  $M_n(\CC)$ is the set of $n\times n$ matrices with entries from $\CC$.  $M_n(\CC)$ is the set of $n\times n$ matrices with entries from $\CC$.  For today we will be identifying a matrix $A\in M_n(\CC)$ with its sequence of rows, $\rho_1,\ldots,\rho_n\in\CC^n$.

  A \emph{multilinear map} is a function $T : M_n(\CC)\to \CC$ such that
  \begin{enumerate}
    \item  For all $\rho_1,\ldots,\rho_n\in\CC^n$ and $\sigma\in\CC^n$,
    \[
      T(\rho_1,\ldots,\rho_i + \sigma,\ldots,\rho_n) = T(\rho_1,\ldots,\rho_i,\ldots,\rho_n) + T(\rho_1,\ldots,\sigma,\ldots,\rho_n)
    \]
    \item  For all $\rho_1,\ldots,\rho_n\in\CC^n$ and $t\in\CC$,
    \[
      T(\rho_1,\ldots,t\rho_i,\ldots,\rho_n) = tT(\rho_1,\ldots,\rho_i,\ldots,\rho_n)
    \]
  \end{enumerate}
  In other words, $T$ is a linear map on its $i$th argument when all others are fixed.

  A multilinear map $T$ is \emph{alternating} if in addition we have
  \[
    T(\rho_1,\ldots,\rho_j,\ldots,\rho_i,\ldots,\rho_n) = -T(\rho_1,\ldots,\rho_i,\ldots,\rho_j,\ldots,\rho_n)
  \]
  whenever $\rho_1,\ldots,\rho_n\in\CC^n$ and $i < j$.
\end{defn*}

\begin{lemma}
  Suppose $T : M_n(\CC)\to\CC$ is an alternating multilinear map, and $A\in M_n(\CC)$ is some matrix with two rows which are the same, or a row of all zeroes.  Then $T(A) = 0$.
\end{lemma}
\begin{proof}
  Suppose $A$ has two rows which are the same, ie $\rho_i = \rho_j$ for some $i\neq j$.  Then
  \[
    T(\rho_1,\ldots,\rho_i,\ldots,\rho_j,\ldots,\rho_n) = -T(\rho_1,\ldots,\rho_j,\ldots,\rho_i,\ldots,\rho_n) = -T(\rho_1,\ldots,\rho_i,\ldots,\rho_j,\ldots,\rho_n)
  \]
  The only real (or complex) number $t$ satisfying $t = -t$ is $t = 0$.

  Now suppose $A$ has a zero row in the $i$th place.  Then by linearity,
  \[
    T(\rho_1,\ldots,0_{1\times n},\ldots,\rho_n) = T(\rho_1,\ldots,0\cdot 0_{1\times n},\ldots,\rho_n) = 0\cdot T(\rho_1,\ldots,0_{1\times n},\ldots,\rho_n) = 0
  \]
\end{proof}

\begin{thm}
  \label{rowops}
  Suppose $T : M_n(\CC)\to\CC$ is an alternating multilinear map, and $A$ and $B$ are $n\times n$ matrices such that $B$ is the result of applying a single row operation to $A$.  Then $T(A)$ and $T(B)$ are related in the following way depending on the row operation in question;
  \[
    \begin{matrix}
      i\neq j & \rho_i\swap\rho_j &  T(B) = -T(A) \\
      i\neq j & \rho_i\to \rho_i + \lambda\rho_j &  T(B) = T(A) \\
              & \rho_i\to \lambda\rho_i &  T(B) = \lambda T(A) \\
    \end{matrix}
  \]
\end{thm}
\begin{proof}
  The swap and scaling cases are simply part of the definition of an alternating multilinear map.  The row-combination case follows from the lemma above;
  \begin{align*}
    T(\rho_1,\ldots,\rho_i + \lambda\rho_j,\ldots,\rho_j,\ldots,\rho_n) & = T(\rho_1,\ldots,\rho_i,\ldots,\rho_j,\ldots,\rho_n) + \lambda T(\rho_1,\ldots,\rho_j,\ldots,\rho_j,\ldots,\rho_n) \\
      & = T(\rho_1,\ldots,\rho_i,\ldots,\rho_j,\ldots,\rho_n)
  \end{align*}
\end{proof}

\begin{thm}
  If $T$ is an alternating multilinear map, and $A$ is not invertible, then $T(A) = 0$.
\end{thm}
\begin{proof}
  Let $R$ be the reduced row-echelon form of $A$.  By the above theorem, $T(A) = 0$ if and only if $T(R) = 0$.  Since $A$ is not invertible, $R$ cannot be $I$; but then $R$ has some zero row.  So $T(R) = 0$, and hence $T(A) = 0$.
\end{proof}

\begin{thm}
  If $S,T : M_n(\CC)\to \CC$ are alternating multilinear maps, and $S(I) = T(I)$, then $S = T$.
\end{thm}
\begin{proof}
  We've already seen that if $A$ is not invertible then $S$ and $T$ both send $A$ to zero.  If $A$ is invertible, then it is row-reducible to $I$.  Theorem~\ref{rowops} (along with a routine induction) shows that
  \[
    T(I) = (-1)^k t_1\cdots t_\ell T(A)\qquad S(I) = (-1)^k t_1\cdots t_\ell S(A)
  \]
  where $k$ is the number of swaps used, $\ell$ is the number of scaling operations used, and $t_1,\ldots,t_\ell$ are the scaling factors.  Hence if $S(I) = T(I)$, then
  \[
    S(A) = (-1)^k\frac{1}{t_1}\cdots \frac{1}{t_\ell} S(I) = (-1)^k\frac{1}{t_1}\cdots\frac{1}{t_\ell} T(I) = T(A)
  \]
\end{proof}

\begin{defn*}
  The \emph{determinant} $\det : M_n(\CC)\to\CC$ is the unique alternating multilinear map such that $\det(I) = 1$.
\end{defn*}

\begin{thm}
  Let $A$ be a square matrix.  Then the following are equivalent.
  \begin{enumerate}
    \item  $A$ is invertible.
    \item  $\det(A) \neq 0$.
  \end{enumerate}
\end{thm}

\begin{example*}
  Let's calculate the determinants of the following matrices.
  \[
    \begin{pmatrix}
          1       & \frac{1}{2} & \frac{1}{3} \\
      \frac{1}{2} & \frac{1}{3} & \frac{1}{4} \\
      \frac{1}{3} & \frac{1}{4} & \frac{1}{5} \\
    \end{pmatrix}
    \qquad\qquad
    \begin{pmatrix}
      a & b \\
      c & d \\
    \end{pmatrix}
    \qquad\qquad
    \begin{pmatrix}
      1 & 2 & 3 \\
      4 & 6 & 10 \\
      6 & 11 & 17 \\
    \end{pmatrix}
    \qquad\qquad
    \begin{pmatrix}
      -2 & 2 &  3 \\
      -1 & 1 &  3 \\
       2 & 0 & -1 \\
    \end{pmatrix}
  \]
\end{example*}

\end{document}
