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\begin{document}

\noindent
{\large \bf Additional notes \#4.}

\bigskip

\noindent
{\bf Things that ``everyone knows'' about the determinant.}

\bigskip

Let $V$ be an ${\BF}$-space with basis ${\mathcal B}=(e_1,\ldots,e_n)$.

\bigskip

\begin{enumerate}
\item[(1)] Definition: ${\rm det}= e_1^*\wedge \cdots \wedge e_n^* =
\sum_{\sigma\in S_n} {\rm sgn}(\sigma)\,e_1^*\otimes \cdots \otimes e_n^*$.
(This definition is the one used over any commutative ring.)
\bigskip
\item[(2)] The abstract characterization: the determinant
is the only alternating $n$-form on $V$ whose value on the
identity matix is 1.
\bigskip
\item[(3)] The permutation expansion:
${\rm det}([a_{ij}]) = 
\sum_{\sigma\in S_n} {\rm sgn}(\sigma)\,
a_{\sigma(1),1}\cdots a_{\sigma(n),n}$.
(This is valid over any commutative ring.)
\bigskip
\item[(4)] The Cofactor Expansion: 
$$
{\rm det}(A) = 
(-1)^{i+1}a_{i1} A_{i1} + \cdots + 
(-1)^{i+j}a_{ij} A_{ij} + \cdots + 
(-1)^{i+n}a_{in} A_{in}
$$
where $A_{ij}$ is the determinant of the matrix obtained 
from $A = [a_{ij}]$ by deleting the $i$-th row and the $j$-th column.
(This is valid over any commutative ring.)
\bigskip
\item[(5)] ${\rm det}(A) = {\rm det}(A^t)$.
(This is valid over any commutative ring.)
\bigskip
\item[(6)] ${\rm det}(AB) = {\rm det}(A){\rm det}(B)$.
(This is valid over any commutative ring.)
\bigskip
\item[(7)] The ``classical adjoint'': Given $A = [a_{ij}]$,
the classical adjoint of $A$ is defined to be:
$$
{\rm adj}(A) = [(-1)^{i+j} A_{ij}]^t
$$
where $A_{ij}$ is the determinant of the matrix obtained 
from $A$ by deleting the $i$-th row and the $j$-th column.
The main property of the classical adjoint that makes it
interesting is:
$$
({\rm adj}(A))\cdot A =A\cdot ({\rm adj}(A)) = {\rm det}(A)\cdot I.
$$
This formula is true over any commutative ring, not just 
over fields. (It follows immediately from the cofactor
expansion of the determinant.)
\bigskip
\item[(8)] Let $A$ be a matrix. Then:
\begin{enumerate}
\item[(i)] The columns of $A$ are independent
if and only if 
${\rm det}(A)$ is not a zero divisor.
\item[(ii)] $A$ is invertible
if and only if 
${\rm det}(A)$ is a unit.
\end{enumerate}
(Notes: Item (ii) follows from (7) above. In fact, (7) shows
that if the determinant is a unit, then 
$A^{-1} = ({\rm det}(A))^{-1}\cdot {\rm adj}(A)$.
The properties in 
items (i) and (ii) are equivalent over fields, but not
equivalent over commutative rings that are not fields.)
\bigskip
\item[(9)] Assume that $\varphi(x_1,\ldots,x_k)\in \Lambda^k(V)$.
Then we have shown that $\varphi$ is a linear combination
of $k$-forms of the form $e_{i_1}^*\wedge\cdots\wedge e_{i_k}^*$.
To compute the values of $\varphi$, one must know
how to compute the values of each 
$e_{i_1}^*\wedge\cdots\wedge e_{i_k}^*$. Note that the value of 
$
e_{i_1}^*\wedge\cdots\wedge e_{i_k}^*({\bf v}_1,\ldots,{\bf v}_k)
$
only depends on the $i_1, \ldots, i_k$-entries of
the ${\bf v}_j$'s. Striking out all other
entries leaves a $k\times k$ matrix. If one checks
the definition of 
$e_{i_1}^*\wedge\cdots\wedge e_{i_k}^*$, and compares it
to the permutation expansion of a $k\times k$
matrix, one
finds that the value of 
$
e_{i_1}^*\wedge\cdots\wedge e_{i_k}^*({\bf v}_1,\ldots,{\bf v}_k)
$
is the determinant of this matrix.

\medskip

\noindent
{\bf Example.} Let $\varphi = 2(e_1^*\wedge e_2^*) - 3(e_1^*\wedge e_3^*)$
be an element of $\Lambda^2({\BR}^3)$. Then 
$$
\begin{array}{rl}
\varphi\left(
\left[\begin{array}{c} a\\b\\c\end{array}\right],
\left[\begin{array}{c} d\\e\\f\end{array}\right]\right) &=
2\cdot {\rm det}\left(
\left[\begin{array}{cc}
a & d\\ b & e\end{array}\right]\right) - 
3\cdot {\rm det}\left(
\left[\begin{array}{cc}
a & d\\ c & f\end{array}\right]\right)\\
&= 2(ae-bd) + 3(af-bd).
\end{array}
$$
\end{enumerate}

\bigskip

\noindent
{\bf The interpretation of the determinant as 
a ``signed'' volume.}
\bigskip

Human beings are capable of detecting
with great accuracy whether a piece of cake they are served
is smaller than a piece served to their neighbor.
This means that we ``know'' that volume exists.
How is it to be defined?

It is reasonable to decide that $n$-dimensional
volume is a measure of size 
that is translation invariant
and additive on disjoint regions.
This is enough to force
the formula for $n$-dimensional
volume of $n$-dimensional rectangular solids
to be additive in each dimension and linear with respect
to positive scalars. This makes $n$-dimensional volume the
restriction to positive values of a uniquely
determined $n$-form. (In dimension $3$,
this is the $n$-form that 
computes the volume of a box with the formula
length~$\times$~width~$\times$~height.)
For algebraic convenience, we usually
extend the definition of volume
to this $n$-form, and call 
the result `signed volume'.

If we consider the $n$-dimensional rectangular solid
spanned by $n$ vectors, two of which are equal,
then a reasonable person would say that
these vectors span {\it zero} $n$-dimensional
volume. In fact, such a region can be surrounded
by a rectangular solid of arbitrarily small positive
$n$-dimensional volume. This forces signed 
$n$-dimensional volume to be an {\it alternating}
$n$-form. If we normalize signed volume
by defining the signed volume of the unit $n$-cube to be $1$,
then signed $n$-dimensional volume is given
by the determinant since the determinant is
the unique alternating $n$-form whose
value is $1$ on the $n$-cube.

\vfill

\end{document}

