The result is false if the
are not distinct; in that case, the value
is zero. Assume therefore that the
are distinct.
One has using Theorem 4-1(3):
because all
the summands except that corresponding to the identity permutation are zero.
If the factor were not in the definition of
, then the right hand
side would have been
.
Assume as in part (a) that the
are all distinct.
A computation similar to that of part (a) shows that
if some
for all
. By multilinearity, it follows that we need only verify
the result when the
are in the subspace generated by the
for
.
Consider the linear map
defined by
. Then
. One has for all
:
This shows the result.
Let
. Then by Theorem 4-6, one has for
,
. So
.
where
.
Let
be an orthonormal basis for V with respect to
,
and let
where
. Then we have
by blinearity:
; the right hand sides
are just the entries of
and so
.
By Theorem 4-6,
.
Taking absolute values and substituting gives the result.
One has
by
the definition of
and the fact that
is the volume element
with respect to
and
. Further,
for some
because
is of dimension 1. Combining, we
have
,
and so
as desired.
The function
is a continuous function,
whose image does not contain 0 since
is a basis for every t. By the
intermediate value theorem, it follows that the image of
consists of
numbers all of the same sign. So all the
have the same orientation.
By the definition, we have
. Since the
are linearly
independent, the definition of cross product with
completing the basis
shows that the cross product is not zero. So in fact, the determinant is
positive, which shows the result.
Let
be the volume element determined by some inner product
and orientation
, and let
be an orthornormal basis
(with respect to
) such that
. There is
a scalar
such that
. Let
,
,
, and
for
. Then
are an orthonormal
basis of
with respect to
, and
. This shows that
is the volume element of
determined by
and
.
The cross product is the
such that
for all
.
All of these follow immediately from the definition, e.g. To show that
,
note that
for all
.
Expanding out the determinant shows that:
The result is true if either
or
is zero. Suppose that
and
are both non-zero. By Problem 1-8,
and since
, the first
identity is just
. This is easily
verified by substitution using part (b).
The second assertion follows from the definition since the determinant of a square matrix with two identical rows is zero.
For the first assertion, one has
and
.
For the second assertion, one has:
So, one needs to
show that
for all
.
But this can be easily verified by expanding everything out using the formula
in part (b).
The third assertion follows from the second:
See the proof of part (c).
where
.
Using the definition of cross product and Problem 4-3, one has:
since the matrix
from Problem 4-3 has the form
.
This proves the result in the case where
is
not zero. When it is zero, the
are linearly dependent, and the
bilinearity of inner product imply that
too.
One has
for each
. Using the orthonormality
of the basis, one has:
But
,
which shows the result.
Since the cross product is multilinear, one can apply Theorem 2-14 (b) and the chain rule to get: