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\section*{Lecture 18:  Properties of determinants}

\subsection*{Determinants}

Now halfway through the course, we leave behind rectangular matrices
and focus on square ones.  Our next big topics are determinants and
eigenvalues.

The {\em determinant} is a number associated with any square matrix;
we'll write it as $\det \mb A$ or $|\mb A|$.  The determinant encodes
a lot of information about the matrix; the matrix is
invertible exactly when the determinant is non-zero.

\subsection*{Properties}

Rather than start with a big formula, we'll list the properties
of the determinant.  We already know that $\twobytwodet abcd = ad-bc$; these properties will give us a formula for the determinant of square matrices of all sizes.
\begin{enumerate}

\item $\det \mb I = 1$

\item If you exchange two rows of a matrix, you reverse the sign of
  its determinant from positive to negative or from negative to
  positive.

\item \begin{enumerate}
\item If we multiply one row of a matrix by $t$, the determinant is multiplied by $t$:  $ \twobytwodet {ta}{tb}cd = t \twobytwodet abcd$.

\item The determinant behaves like a linear function on the rows of the matrix: $$\twobytwodetc {a+a'} {b+b'} c d = \twobytwodet abcd + \twobytwodet {a'}{b'}cd.$$
\end{enumerate}
\end{enumerate}

Property $1$ tells us that $ \small \twobytwodet 1001 = 1$.  Property $2$ tells us that $\small \twobytwodet 0110 = -1$. The determinant of a permutation matrix $\mb P$ is $1$ or $-1$ depending on whether $\mb P$ exchanges an even or odd number of rows.

From these three properties we can deduce many others:
\begin{enumerate}
\setcounter{enumi}{3}
\item If two rows of a matrix are equal, its determinant is zero.

This is because of property $2$, the exchange rule.  On the one hand, exchanging the two identical rows does not change the determinant.  On the other hand, exchanging the two rows changes the sign of the determinant.  Therefore the determinant must be $0$.

\item If $i \neq j$, subtracting $t$ times row $i$ from row $j$ doesn't change the determinant.

In two dimensions, this argument looks like:
\begin{eqnarray*}
\twobytwodetc ab {c-ta}{d-tb} &=& \twobytwodet abcd - \twobytwodet ab {ta}{tb} 
\quad \mbox{property 3(b)}\\
 &=& \twobytwodet abcd - t \twobytwodet ab {a}{b} 
\quad \mbox{property 3(a)}\\
 &=& \twobytwodet abcd \quad \mbox{property 4.}
\end{eqnarray*}
The proof for higher dimensional matrices is similar.

\item If $\mb A$ has a row that is all zeros, then $\det \mb A = 0$.

We get this from property $3$ (a) by letting $t=0$.

\item The determinant of a triangular matrix is the product of the
  diagonal entries (pivots) $d_1, d_2, ..., d_n$.

Property 5 tells us that the determinant of the triangular matrix won't change if we use elimination to convert it to a diagonal matrix with the entries $d_i$ on its diagonal.  Then property $3$ (a) tells us that the determinant of this diagonal matrix is the product $d_1 d_2 \cdots d_n$ times the determinant of the identity matrix.  Property $1$ completes the argument.

Note that we cannot use elimination to get a diagonal matrix if one of the $d_i$ is zero.  In that case elimination will give us a row of zeros and property 6 gives us the conclusion we want.

\item $\det \mb A = 0$ exactly when $\mb A$ is singular.

If $\mb A$ is singular, then we can use elimination to get a row of zeros, and property 6 tells us that the determinant is zero.

If $\mb A$ is not singular, then elimination produces a full set of
pivots $d_1, d_2, ..., d_n$ and the determinant is $d_1 d_2
\cdots d_n \neq 0$ (with minus signs from row exchanges).

\end{enumerate}

We now have a very practical formula for the determinant of a
non-singular matrix.  In fact, the way computers find the determinants
of large matrices is to first perform elimination (keeping track of
whether the number of row exchanges is odd or even) and then multiply
the pivots:

$$\twobytwo abcd \longrightarrow \twobytwoc a b 0 {d-\frac{c}{a} b}, \mbox{ if } a \neq 0, \mbox{ so}$$

$$\twobytwodet abcd = a(d-\frac{c}{a}b) = ad-bc.$$

\begin{enumerate}
\setcounter{enumi}{8}

\item $\det \mb A \mb B = (\det \mb A)(\det \mb B)$

This is very useful.  Although the determinant of a sum does not equal
the sum of the determinants, it is true that the determinant of a
product equals the product of the determinants.

For example: $$\det \mb A^{-1} = \frac{1}{\det \mb A},$$ because $\mb
A^{-1} \mb A = 1$.  (Note that if $\mb A$ is singular then $\mb
A^{-1}$ does not exist and $\det \mb A^{-1}$ is undefined.)  Also,
$\det \mb A^2 = (\det \mb A)^2$ and $\det 2\mb A = 2^n\det A$ (applying property 3 to each row of the matrix).  This reminds us of volume -- if we double the
length, width and height of a three dimensional box, we increase its
volume by a multiple of $2^3=8$.

\item $\det \mb A^T = \det \mb A$

$$\twobytwodet abcd = \twobytwodet acbd = ad-bc.$$

This lets us translate properties (2, 3, 4, 5, 6) involving rows into
statements about columns.  For instance, if a column of a matrix is
all zeros then the determinant of that matrix is zero.

To see why $|\mb A^T| = |\mb A|$, use elimination to write $\mb A =
\mb L \mb U$.  The statement becomes $|\mb U^T \mb L^T | = |\mb L \mb U|$.  Rule 9
then tells us $|\mb U^T||\mb L^T| = |\mb L||\mb U|$.

Matrix $\mb L$ is a lower triangular matrix with $1$'s on the
diagonal, so rule 5 tells us that $|L| = |L^T| = 1$.  Because $\mb U$
is upper triangular, rule 5 tells us that $|U| = |U^T|$.  Therefore
$|\mb U^T||\mb L^T| = |\mb L||\mb U|$ and $|\mb A^T| = |\mb A|$.

\end{enumerate}

\bigskip

We have one loose end to worry about.  Rule 2 told us that a row exchange changes the sign of the
determinant.  If it's possible to do
seven row exchanges and get the same matrix you would by doing ten row exchanges, then we could
prove that the determinant equals its negative.  To complete the proof that the
determinant is well defined by properties 1, 2 and 3 we'd need
to show that the result of an odd number of row exchanges (odd
permutation) can never be the same as the result of an even number of
row exchanges (even permutation).

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