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\begin{document}
\centerline{Math 201 lecture for Monday, Week 8}
\bigskip

\centerline{\textbf{Determinant of the transpose}}
\medskip

\textbf{Goal:} Our goal for the day is to show that the determinant of a matrix and
the determinant of the transpose of that matrix are equal.  It then immediately
follows that the determinant is not only a multilinear, alternating, normalized
function of the \emph{rows} of a matrix (by definition), but it is also a
multilinear, alternating, normalized function of its \emph{columns}.  So one may
use both row and column operations to compute the determinant.

\textbf{Elementary matrices.} An~$n\times n$ matrix is called an \emph{elementary
matrix} if it is obtained from the identity matrix,~$I_n$, through a single
elementary row operation (scaling a row by a nonzero scalar, swapping rows, or
adding one row to another).  

Here is why elementary matrices are interesting: Let~$E$ be an $n\times n$
elementary matrix corresponding to some row operation and let~$A$ be
any~$n\times k$ matrix.  Then~$EA$ is the matrix obtained from~$A$ by performing
that row operation.  Thus, you can perform row operations through multiplication
by elementary matrices.
\medskip

\textbf{Example.} Let
\[
A = 
\left(\begin{array}{cccc}
    1&2&3&4\\
    3&0&-1&2\\
    1&5&6&7
\end{array} \right).
\]
To find the elementary matrix that will subtract~$3$ times the first row of~$A$
from the second row, we do that same operation to the identity matrix:
\[
  \left(\begin{array}{ccc}
      1&0&0\\
      0&1&0\\
      0&0&1
  \end{array} \right)
  \quad
  \xrightarrow{r_2\to r_2-3r_1}
  \quad
  \left(\begin{array}{rrr}
      1&0&0\\
      -3&1&0\\
      0&0&1
  \end{array} \right)
  =: E_1.
\]
Multiplying by~$E_1$ on the left performs the same elementary row operation
on~$A$:
\[
  E_1A =
  \left(\begin{array}{rrr}
      1&0&0\\
      -3&1&0\\
      0&0&1
  \end{array} \right)
\left(\begin{array}{cccc}
    1&2&3&4\\
    3&0&-1&2\\
    1&5&6&7
\end{array} \right)
=
\left(\begin{array}{rrrr}
    1&2&3&4\\
    0&-6&-10&-10\\
    1&5&6&7
\end{array} \right).
\]
Next, let~$E_2$ be the elementary matrix corresponding to subtracting the first
row from the third:
\[
  \left(\begin{array}{ccc}
      1&0&0\\
      0&1&0\\
      0&0&1
  \end{array} \right)
  \quad
  \xrightarrow{r_3\to r_3-r_1}
  \quad
  =
  \left(\begin{array}{rrr}
      1&0&0\\
      0&1&0\\
      -1&0&1
  \end{array} \right)
  =:E_2.
\]
Multiplying~$E_1A$ on the left by~$E_2$ performs the corresponding row
operation:
\[
  E_2E_1A
  =
  E_2(E_1A)
  =
  \left(\begin{array}{rrr}
      1&0&0\\
      0&1&0\\
      -1&0&1
  \end{array} \right)
\left(\begin{array}{rrrr}
    1&2&3&4\\
    0&-6&-10&-10\\
    1&5&6&7
\end{array} \right)
=
\left(\begin{array}{rrrr}
    1&2&3&4\\
    0&-6&-10&-10\\
    0&3&3&3
\end{array} \right).
\]
Next, swap the second and third rows:
\[
  \left(\begin{array}{ccc}
      1&0&0\\
      0&1&0\\
      0&0&1
  \end{array} \right)
  \quad
  \xrightarrow{r_2\leftrightarrow r_3}
  \quad
  \left(\begin{array}{rrr}
      1&0&0\\
      0&0&1\\
      0&1&0
  \end{array} \right)
  =: E_3,
\]
and, thus,
\[
  E_3E_2E_1A
  =
  E_3(E_2E_1A)
  =
  \left(\begin{array}{rrr}
      1&0&0\\
      0&0&1\\
      0&1&0
  \end{array} \right)
\left(\begin{array}{rrrr}
    1&2&3&4\\
    0&-6&-10&-10\\
    0&3&3&3
\end{array} \right)
=
\left(\begin{array}{rrrr}
    1&2&3&4\\
    0&3&3&3\\
    0&-6&-10&-10
\end{array} \right).
\]
To continue row reduction, we would now scale the second row by~$\frac{1}{3}$:
\[
  \left(\begin{array}{ccc}
      1&0&0\\
      0&1&0\\
      0&0&1
  \end{array} \right)
  \quad
  \xrightarrow{r_2\leftrightarrow r_2/3}
  \quad
  \left(\begin{array}{rrr}
      1&0&0\\
      0&\frac{1}{3}&0\\
      0&0&1
  \end{array} \right)
  =: E_4,
\]
and, thus,
\[
  E_4E_3E_2E_1A
  =
  E_4(E_3E_2E_1A)
  =
  \left(\begin{array}{rrr}
      1&0&0\\
      0&\frac{1}{3}&0\\
      0&0&1
  \end{array} \right)
\left(\begin{array}{rrrr}
    1&2&3&4\\
    0&3&3&3\\
    0&-6&-10&-10
\end{array} \right)
=
\left(\begin{array}{rrrr}
    1&2&3&4\\
    0&1&1&1\\
    0&-6&-10&-10
\end{array} \right).
\]

As illustrated in the example, above, performing a sequence of row operations to a matrix is equivalent to
multiplying on the left by a sequence of elementary matrices.  In particular,
if~$\widetilde{A}$ is the reduced row echelon form of~$A$, then there are elementary
matrices~$E_1,\dots,E_{\ell}$ such that
\[
  \widetilde{A}=E_{\ell}\cdots E_2 E_1A.
\]
\medskip

\textbf{Determinant of the transpose.} If~$A$ is an~$m\times n$ matrix, recall that
its \emph{transpose} is the matrix~$A^t$ defined by
\[
  (A^{t})_{ij}:=A_{ji}.
\]
Thus, the rows of~$A^t$ are the columns of~$A$.
\medskip

\textbf{Example} If
\[
  A =
  \left(\begin{array}{ccc}
  1 & 2 & 3 \\
4 & 5 & 6
  \end{array}\right)
  \quad\text{and}\quad
  B = 
  \left(\begin{array}{cc}
  5 & 2 \\
1 & 3
  \end{array}\right),
\]
then
\[
  A^{t}=
  \left(\begin{array}{cc}
  1 & 4 \\
2 & 5 \\
3 & 6
  \end{array}\right)
  \quad\text{and}\quad
  B^{t}=
  \left(\begin{array}{cc}
  5 & 1 \\
2 & 3
  \end{array}\right).
\]


\textbf{Our goal} now is to prove the non-obvious fact that for an~$n\times n$
matrix~$A$, 
\[
  \det(A)=\det(A^t).
\]
\medskip

\textbf{Example.} We have seen that 
\[
  \det\left(\begin{array}{cc}
	a&b\\c&d
    \end{array} \right)=ad-bc.
\]
Note that we also have
\[
  \det\left(\left(\begin{array}{cc}
	a&b\\c&d
  \end{array} \right)^t\right)
    =
  \det \left(\begin{array}{cc}
      a&c\\b&d
  \end{array} \right)=ad-bc.
\]
Recall that we can compute the determinant of~$A$ by performing row operations
and keeping track of swaps and scalings of rows.  Once we have shown that
$\det(A)=\det(A^t)$, it follows that, in order compute the determinant
of~$A$, we may also use column operations (again keeping track of swaps and
scalings).  That's because row operations applied to~$A^t$ are the same as
column operations applied to~$A$.  

To prove this fact about determinants of transposes, we need the following
theorem, proposition, and lemma:
\medskip

\textbf{Theorem.} Let~$A$ and~$B$ be~$n\times n$ matrices.  Then
\[
  \det(AB)=\det(A)\det(B).
\]
\begin{proof}
  Upcoming homework.
\end{proof}

\textbf{Proposition.} Let~$A$ and~$B$ be~$n\times n$ matrices.  Then
\begin{enumerate}
  \item $(AB)^t=B^tA^t$.
  \item If~$A$ is invertible, then $(A^t)^{-1}=(A^{-1})^t$.
\end{enumerate}
\begin{proof}
  Upcoming homework.
\end{proof}

\textbf{Lemma.} Let~$E$ be an elementary matrix.  Then~$\det(E)=\det(E^t)\neq0$.  

\begin{proof}
  There are three cases to consider\footnote{The reader is strongly encouraged
  to create examples of each of these three cases.}:
  \begin{enumerate}
    \item Suppose~$E$ is formed from~$I_n$ by swapping rows~$i$ and~$j$.  In this
      case,~$E^t$ is also formed from~$I_n$ by swapping rows~$i$ and~$j$.
      Thus,~$E=E^t$, and~$\det(E^t)=\det(E)=-1$.
    \item Suppose~$E$ is formed from~$I_n$ by scaling row~$i$ by~$\lambda\neq0$.  In this
      case,~$E^t$ is also formed from~$I_n$ by scaling row~$i$ by~$\lambda$.
      So in this case,~$\det(E^t)=\det(E)=\lambda$.
    \item Suppose~$E$ is formed from~$I_n$ by adding~$\lambda r_i$ to~$r_j$ for
      rows~$r_i\neq r_j$.  Then~$E^t$ is formed from~$I_n$ by adding~$\lambda
      r_j$ to~$r_i$. So in this case,~$\det(E^t)=\det(E)=\det(I_n)=1$.
  \end{enumerate}
\end{proof}

We can now prove our main result:
\medskip

\textbf{Theorem.} Let~$A$ be an~$n\times n$ matrix.  Then~$\det(A)=\det(A^t)$.
\smallskip

\textbf{Proof.} Let~$\widetilde{A}$ be the reduced echelon form for~$A$.
Then~$\widetilde{A}=I_n$ if and only if~$\rk{A}=n$.  So if~$\widetilde{A}\ne
I_n$, we have~$\rk(A)<0$, which means that~$\det(A)=0$.  Since row rank and
column rank are equal, we would then have~$\rk(A^t)=\rk(A)<n$, which
means~$\det(A^t)=0$, too.  So the theorem holds in that case.

Now consider that case where~$\widetilde{A}=I_n$.  Thus, by applying row
operations to~$A$, we arrive at the identity matrix.
Choose
elementary matrices~$E_i$ such that
\begin{equation}\label{eqn:08mon1}
  E_{\ell}\cdots E_2 E_1 A=I_n.
\end{equation} 
Taking determinants and using the fact that determinants preserve products
yields:
\begin{equation}\label{eqn:08mon2}
  \det(E_{\ell})\cdots\det(E_2)\det(E_1)\det(A)=1.
\end{equation} 

Taking transposes in equation~(\ref{eqn:08mon1}) gives
\[
  A^t E_1^t\cdots E_{\ell}^t=I_n^t=I_n.
\]
Take determinants and use the fact from the Lemma that~$\det(E)=\det(E^{t})$
if~$E$ is an elementary matrix to get
\begin{equation}\label{eqn:08mon3}
  \det(A^t)\det(E_1)\cdots \det(E_{\ell})=1.
\end{equation} 
The result follows by equating~\eqref{eqn:08mon2} and~\eqref{eqn:08mon3} and using the
fact that the determinant of an elementary matrix is nonzero (so that we may
cancel).
\hfill$\Box$
\medskip

\textbf{Corollary.} The determinant is a multilinear, alternating, normalized
function of the \emph{columns} of a square matrix.
\medskip

For instance, the above corollary says the in order to compute the determinant,
one may use column operations to simplify the calculation, just as we used row
operations.  Further, one may mix row and column operations, when convenient, as in the following
example.
\medskip

\textbf{Example.}  Let
\[
  M =
\left(\begin{array}{rrrr}
1 & -1 & -1 & -1 \\
-2 & 3 & 0 & 1 \\
-2 & 1 & 4 & -1 \\
-5 & 1 & 1 & 5
\end{array}\right).
\]
To compute the determinant of~$M$, add the first row to each other row to get
\[
  M' =
  \left(\begin{array}{rrrr}
1 & -1 & -1 & -1 \\
-1 & 2 & -1 & 0 \\
-1 & 0 & 3 & -2 \\
-4 & 0 & 0 & 4
\end{array}\right),
\]
then, in $M'$, add columns~$2$, $3$, and~$4$ to column~$1$ to get
\[
 M''
 =
 \left(\begin{array}{rrrr}
-2 & -1 & -1 & -1 \\
0 & 2 & -1 & 0 \\
0 & 0 & 3 & -2 \\
0 & 0 & 0 & 4
\end{array}\right).
\]
The types of row and column operations we used to get from~$M$ to $M''$ have no
effect on the determinant, and thus,
\[
  \det(M)=\det(M')=\det(M'') = -2\cdot 2\cdot 3\cdot 4 = -48.
\]
\medskip

\textbf{Remark.} Start with $I_n$ and perform a sequence of row
operations to get a matrix~$B\in M_{n\times n}(F)$. Then multiplying~$A\in
M_{n\times k}(F)$ on the \emph{left} by~$B$ gives the matrix~$BA\in M_{n\times k}(F)$ formed from~$A$ by performing the same
sequence of row operations.  Similarly, one can start with~$I_n$, perform column
operations to produce a matrix~$C\in M_{n\times n}(F)$, then multiplying~$D\in
M_{k\times n}(F)$ on the \emph{right} by~$C$ give the matrix~$DC\in M_{k\times n}(F)$ obtained by
performing the column operations on~$D$.
\end{document}
