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 .
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).
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: