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

\centerline{\textbf{Existence and uniqueness of the determinant}}  
\medskip

\textbf{Laplace expansion of the determinant.} Let~$A$ be an~$n\times n$ matrix.
For each~$i,j\in\left\{ 1,2,\dots,n \right\}$, define~$A^{ij}$ to be the matrix
formed by removing the~$i$-th row and~$j$-th column from~$A$. Fix~$k\in\left\{
1,2,\dots,n \right\}$.  Then
\[
  \det(A) = \sum_{j=1}^{n}(-1)^{k+j}A_{kj}\det(A^{kj}).
\]
This expresses~$\det(A)$ in terms of an alternating sum of determinants of
$(n-1)\times(n-1)$ matrices.  We call this {\em expanding $\det(A)$
along the~$k$-th row}.  Applying the formula recursively leads to
a complete evaluation of~$\det(A)$.  Since,~$\det(A)=\det(A^t)$, you can also
calculate the determinant by recursively expanding along {\em columns}.
\medskip

\textbf{Example.} Let
\[
A = 
\left(\begin{array}{ccc}
    1&2&3\\
    2&0&1\\
    1&1&1
\end{array} \right).
\]
Let's calculate the determinant by expanding along the first row:
\begin{align*}
  \det(A)
  &=
  1\cdot\det
    \left(\begin{array}{cc}
	0&1\\1&1
    \end{array} \right)
  -2\cdot\det
    \left(\begin{array}{cc}
	2&1\\1&1
    \end{array} \right)
  +3\cdot\det
    \left(\begin{array}{cc}
	2&0\\1&1
    \end{array} \right)
    \\[12pt]
    &= (-1)-2(1)+3(2) = 3.
\end{align*}
To check, let's expand along the second row, instead, noting the signs:
\begin{align*}
  \det(A)
  &=
  -2\cdot\det
    \left(\begin{array}{cc}
	2&3\\1&1
    \end{array} \right)
  0\cdot\det
    \left(\begin{array}{cc}
	1&3\\1&1
    \end{array} \right)
  -1\cdot\det
    \left(\begin{array}{cc}
	1&2\\1&1
    \end{array} \right)
    \\[12pt]
    &= -2(-1)+0(-2)-1(-1) = 3.
\end{align*}
Finally, let's expand along the third {\em column}:
\begin{align*}
  \det(A)
  &=
  3\cdot\det
    \left(\begin{array}{cc}
	2&0\\1&1
    \end{array} \right)
  -1\cdot\det
    \left(\begin{array}{cc}
	1&2\\1&1
    \end{array} \right)
  +1\cdot\det
    \left(\begin{array}{cc}
	1&2\\2&0
    \end{array} \right)
    \\[12pt]
    &= 3(2)-1(-1)+1(-4) = 3.
\end{align*}
Note: if your matrix has a particular row or column with a lot of~$0$s in it,
you might want to expand along that row or column since a lot of the terms will
be~$0$.  For example, to compute
\[
\det
\left(\begin{array}{ccc}
    1&3&0\\
    3&2&3\\
    1&4&0
\end{array} \right),
\]
expand along the third column:
\[
  0\,(\text{blah})
  -3\det
  \left(\begin{array}{cc}
      1&3\\1&4
  \end{array} \right)
  +0\,(\text{blah})
  =-3(1)=-3.
\]
The ``blah''s are there instead of explicit determinants since they are being
multiplied by~$0$.  Their exact values don't matter, so we don't need to waste
time calculating them.
We will not prove the formula for the Laplace expansion.  It is very similar to
that for the permutation expansion.
\medskip

\textbf{Existence and uniqueness of the determinant.} Recall the definition that
started our discussion of the determinant:
\medskip

\textbf{Definition.} The \emph{determinant} is a multilinear, alternating
function $\det\colon M_{n\times n}(F)\to F$ of the rows of square matrix,
normalized so that its value on the identity matrix is~$1$.  
\medskip

The definition says ``the determinant'', but for all we knew, there could be
several different functions~$M_{n\times n}(F)\to F$ all satisfying the criteria
of being multilinear, alternating, and normalized.
Or, it is possible there are no functions that satisfy the criteria?  So the
definition requires us to prove that, in fact, there exists exactly one
determinant function (for each~$n$).

Just after defining the determinant, we showed that if~$d\colon M_{n\times
n}(F)\to F$ is any multilinear, alternating, normalized function, then a choice
of a row reduction for~$A\in M_{n\times n}(F)$ determines the value of $d(A)$.
The subtlety here is that, there are many different sequence of row operations
that would produced the row echelon form for~$A$.  Do each of these produce the
same value for $d(A)$ (in other words, is $d$ \emph{well-defined})?  We never proved that they would.

So let's begin again and consider the particular function $d\colon
M_{n\times n}(F)\to F$ defined recursively as the Laplace expansion of a matrix
along its first row:
\begin{equation}\label{eqn:08fri}
  d(A):=\sum_{j=1}^{n}(-1)^{1+j}A_{1j}\,d(A^{1j})
\end{equation}
if~$n\geq 1$, and by~$d(A)=a$ if~$A=[a]$ is a~$1\times 1$ matrix.  This
function~$d$ is well-defined---there are no choices to be made is in
calculation.  
\medskip

\textbf{Exercise.} Prove that~$d$ is multilinear, alternating, and normalized (i.e., its value at~$I_n$ is~$1$).
\medskip

Thus, we see there exists at least one determinant function. 

Having defined~$d$ by~\eqref{eqn:08fri}, now note that in addition to calculating~$d$
using the given recursive formula, since~$d$ is multilinear,
alternating, and normalize, its value can be determined via row reductions, just
as before.  What's new now is that we see that no matter which choices are made
in the row reduction, we must get the value determined by
\eqref{eqn:08fri}.  

In sum, we have shown that a multilinear, alternating, normalized function
exists and is unique.  Its value is completely determined by choosing any
sequence of row operations reducing a matrix to its row echelon form, and the
choice of the sequence of row operations does not matter.
So far, we have three different methods for calculating the determinant: using row operations, summing
over permutations, and via Laplace expansion along any row or column.  
\bigskip

\centerline{\textsc{bonus content}}
\medskip

\textbf{Generalized Laplace expansion.}  Let~$A\in M_{n\times n}(F)$, and
fix a subset of~$k$ rows~$r_{i_1},\ldots,r_{i_{k}}$ of~$A$ where $1\leq  k\leq n$.
Let $I=\left\{i_{1},\ldots,i_{k}\right\}$ be the indices of these rows.   For
any subset~$J\subseteq\left\{j_1,\ldots,j_k\right\}$, define
$|J|:=j_{1}+\cdots+j_{k}$, and define
\begin{align*}
  A^{IJ} = \parbox[t]{10cm}{\raggedright the~$k\times k$ submatrix of~$A$ formed by the intersection of
  rows indexed by~$I$ and the columns indexed by~$J$}\\[4pt]
  \bar{A}^{IJ} = \parbox[t]{10cm}{\raggedright the~$(n-k)\times(n-k)$ submatrix of~$A$ formed by the intersection of
  rows indexed by $\left\{1,\ldots,n\right\}\setminus I$ and the columns indexed by~$\left\{1,\ldots,n\right\}\setminus J$.}
\end{align*}
Then
\[
  \det(A)=\sum_{J}(-1)^{|I|+|J|}\det(A^{IJ})\det(\bar{A}^{IJ})
\]
where the sum is over all $k$-element subsets~$J$ of
$\left\{1,\ldots,n\right\}$.\footnote{Since~$\det(A)=\det(A^{t})$, there is a similar formula for expansion
along a fixed set of~$k$ columns.}
\medskip

\textbf{Example.} The case where~$k=1$ is the ordinary Laplace expansion
formula.
\medskip

\textbf{Example.} Let
\[
A = 
\left(\begin{array}{rrrr}
1 & 7 & 0 & 5 \\
2 & 2 & 2 & 2 \\
5 & 1 & 4 & 6 \\
0 & 6 & 7 & 3
\end{array}\right).
\]
We will compute~$\det(A)$ using the generalize Laplace expansion along the first
two rows of~$A$.  So, using the notation from above,
$I=\left\{1,2\right\}\subset\left\{1,2,3,4\right\}$.  There are six choices for
a pair of columns: 
\[
  J\in \left( \left\{1,2\right\}, \left\{1,3\right\},
\left\{1,4\right\},\left\{2,3\right\},\left\{2,4\right\},\left\{3,4\right\}\right).
\]
The term in the expansion corresponding to~$J=\left\{1,3\right\}$ would be
\begin{align*}
(-1)^{|I|+|J|}\det(A^{IJ})\det(\bar{A}^{IJ})
&=(-1)^{(1+2)+(1+3)}
\det
\left(\begin{array}{cc}
1 & 0 \\
2 & 2
\end{array}\right)
\det
\left(\begin{array}{cc}
1 & 6 \\
6 & 3
\end{array}\right)
\\[8pt]
&=(-1)(1\cdot2-0\cdot2)(1\cdot3-6\cdot6)= 66.
\end{align*}
The entire expansion is
\begin{align*}
  \det(A) &= \sum_{J}(-1)^{|I|+|J|}\det(A^{IJ})\det(\bar{A}^{IJ})\\[6pt]
  &=\sum_{J}(-1)^{3+|J|}\det(A^{\left\{1,2\right\}J})\det(\bar{A}^{\left\{1,2\right\}J})\\[6pt]
  &=
  (-1)^{3+(1+2)}\det(A^{\left\{1,2\right\}\left\{1,2\right\}})\det(\bar{A}^{\left\{3,4\right\}\left\{3,4\right\}})
  +(-1)^{3+(1+3)}\det(A^{\left\{1,2\right\}\left\{1,3\right\}})\det(\bar{A}^{\left\{3,4\right\}\left\{2,4\right\}})\\[6pt]
  &\quad+
  (-1)^{3+(1+4)}\det(A^{\left\{1,2\right\}\left\{1,4\right\}})\det(\bar{A}^{\left\{3,4\right\}\left\{2,3\right\}})
  +(-1)^{3+(2+3)}\det(A^{\left\{1,2\right\}\left\{2,3\right\}})\det(\bar{A}^{\left\{3,4\right\}\left\{1,4\right\}})\\[6pt]
  &\quad+
  (-1)^{3+(2+4)}\det(A^{\left\{1,2\right\}\left\{2,4\right\}})\det(\bar{A}^{\left\{3,4\right\}\left\{1,3\right\}})
  +(-1)^{3+(3+4)}\det(A^{\left\{1,2\right\}\left\{3,4\right\}})\det(\bar{A}^{\left\{3,4\right\}\left\{1,2\right\}})\\[6pt]
  &=
  \det(A^{\left\{1,2\right\}\left\{1,2\right\}})\det(\bar{A}^{\left\{3,4\right\}\left\{3,4\right\}})
  -\det(A^{\left\{1,2\right\}\left\{1,3\right\}})\det(\bar{A}^{\left\{3,4\right\}\left\{2,4\right\}})\\[6pt]
  &\quad+
  \det(A^{\left\{1,2\right\}\left\{1,4\right\}})\det(\bar{A}^{\left\{3,4\right\}\left\{2,3\right\}})
  +\det(A^{\left\{1,2\right\}\left\{2,3\right\}})\det(\bar{A}^{\left\{3,4\right\}\left\{1,4\right\}})\\[6pt]
  &\quad
  -\det(A^{\left\{1,2\right\}\left\{2,4\right\}})\det(\bar{A}^{\left\{3,4\right\}\left\{1,3\right\}})
  +\det(A^{\left\{1,2\right\}\left\{3,4\right\}})\det(\bar{A}^{\left\{3,4\right\}\left\{1,2\right\}})\\[6pt]
\end{align*}
\newpage

Continuing the calculation:
\[
A = 
\left(\begin{array}{rrrr}
1 & 7 & 0 & 5 \\
2 & 2 & 2 & 2 \\
5 & 1 & 4 & 6 \\
0 & 6 & 7 & 3
\end{array}\right)
\]
\begin{align*}
\det(A) &=
  \det(A^{\left\{1,2\right\}\left\{1,2\right\}})\det(\bar{A}^{\left\{3,4\right\}\left\{3,4\right\}})
  -\det(A^{\left\{1,2\right\}\left\{1,3\right\}})\det(\bar{A}^{\left\{3,4\right\}\left\{2,4\right\}})\\[6pt]
  &\quad+
  \det(A^{\left\{1,2\right\}\left\{1,4\right\}})\det(\bar{A}^{\left\{3,4\right\}\left\{2,3\right\}})
  +\det(A^{\left\{1,2\right\}\left\{2,3\right\}})\det(\bar{A}^{\left\{3,4\right\}\left\{1,4\right\}})\\[6pt]
  &\quad
  -\det(A^{\left\{1,2\right\}\left\{2,4\right\}})\det(\bar{A}^{\left\{3,4\right\}\left\{1,3\right\}})
  +\det(A^{\left\{1,2\right\}\left\{3,4\right\}})\det(\bar{A}^{\left\{3,4\right\}\left\{1,2\right\}})\\[6pt]
  &=
  \det
  \left(\begin{array}{rr}
1 & 7 \\
2 & 2
\end{array}\right)
  \det
  \left(\begin{array}{rr}
4 & 6 \\
7 & 3
\end{array}\right)
  -
  \det
  \left(\begin{array}{rr}
1 & 0 \\
2 & 2
\end{array}\right)
\det
\left(\begin{array}{rr}
1 & 6 \\
6 & 3
\end{array}\right)
\\[6pt]
  &\quad+
  \det
  \left(\begin{array}{rr}
1 & 5 \\
2 & 2
\end{array}\right)
\det
\left(\begin{array}{rr}
1 & 4 \\
6 & 7
\end{array}\right)
  +\det
  \left(\begin{array}{rr}
7 & 0 \\
2 & 2
\end{array}\right)
\det
\left(\begin{array}{rr}
5 & 6 \\
0 & 3
\end{array}\right)
\\[6pt]
  &\quad
  -\det
  \left(\begin{array}{rr}
7 & 5 \\
2 & 2
\end{array}\right)
\det
\left(\begin{array}{rr}
5 & 4 \\
0 & 7
\end{array}\right)
  +\det
  \left(\begin{array}{rr}
0 & 5 \\
2 & 2
\end{array}\right)
\det
\left(\begin{array}{rr}
5 & 1 \\
0 & 6
\end{array}\right)
\\[6pt]
  &=(-12)(-30)-(2)(-33)+(-8)(-17)+(14)(15)-(4)(35)+(-10)(30)\\[6pt]
  &=332.
\end{align*}
\medskip

\textbf{Example.}  Let
\[
  A = 
  \left(\begin{array}{ccccc}
      3 & 2 &1&0&0\\
      1 & 2 &4&0&0\\
      0 & 1 &7&0&0\\
      1 & 2 & 1 & 4&7 \\
      3 & 4 & 2 & 9 &3\\
  \end{array}\right).
\]
The generalized Laplace expansion along the first three rows has only two nonzero
terms, yielding
\[
  \det(A)=
  \det
  \left(\begin{array}{ccc}
      3&2&1 \\
      1&2 & 4\\
      0&1&7
  \end{array}\right)
  \det
  \left(\begin{array}{cc}
      4&7\\
      9&3
  \end{array}\right).
\]
\end{document}
