# Exercises: Chapter 4, Section 1

1. Let be the usual basis of and let be the dual basis.
1. Show that . What would the right hand side be if the factor did not appear in the definition of ?

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 .

2. Show that is the determinant of thee minor of obtained by selecting columns .

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.

2. If is a linear transformation and , then must be multiplication by some constant . Show that .

Let . Then by Theorem 4-6, one has for , . So .

3. If is the volume element determined by and , and , show that

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.

4. If is the volume element of determined by and , and is an isomorphism such that and such that , show that .

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.

5. If is continuous and each is a basis for , show that .

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.

1. If , what is ? is the cross product of a single vector, i.e. it is the vector such that for every . Substitution shows that works.
2. If are linearly independent, show that is the usual orientation of .

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.

6. Show that every non-zero is the volume element determined by some inner product and orientation for .

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 .

7. If is a volume element, define a ``cross product" in terms of .

The cross product is the such that for all .

8. Deduce the following properties of the cross product in :
1. All of these follow immediately from the definition, e.g. To show that , note that for all .

2. .

Expanding out the determinant shows that:

3. , where , and .

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.

4. 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:

5. .

See the proof of part (c).

9. If , show that

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.

10. If is an inner product on , a linear transformation is called self-adjoint (with respect to ) if for all . If is an orthogonal basis and is the matrix of with respect to this basis, show that .

One has for each . Using the orthonormality of the basis, one has: But , which shows the result.

11. If , define by . Use Problem 2-14 to derive a formula for when are differentiable.

Since the cross product is multilinear, one can apply Theorem 2-14 (b) and the chain rule to get: