Exercise 1Let be the vector space of continuous real-valued functions on the interval . Then, for any ,

*Proof:* Let be the measure on . Then is a probability space, is Lebesgue-integrable on and is a convex function . By Jensen’s inequality,

Multiplying throughout by we get the claimed inequality.

Exercise 2Let be a linear operator on a finite-dimensional vector space . (a) Prove that if every one-dimensional subspace of is -invariant, then is a scalar multiple of the identity operator. (b) Prove that if every codimension-one subspace of is -invariant, then is a scalar multiple of the identity operator.

*Proof:* (a) The hypothesis means that every nonzero vector of is an eigenvector of . Suppose are eigenvectors of with eigenvalues , . Since, by assumption is also an eigenvector, and and are independent, we can read off the eigenvalue of off of either coefficient in the equation , and therefore . Therefore is a multiple of the identity operator.

(b) Let be the dual operator on . We claim that satisfies the condition of . First, we have the following:

Lemma 1Two functionals (where is the ground field) have the same kernel if and only if they are multiples of each other.

*Proof:* Indeed, it is trivial if either of or is (in which case both are zero), so suppose neither is . Recall that if and we define , then we have a canonical isomorphism , which in particular implies . If we apply this to , we have, under assumption,

which has codimension since . Therefore has dimension , and and are scalar multiples of each other.

Now, back to . S suppose that . Then has codimension in , and therefore, under the hypothesis of (b), . This implies ; indeed, if , then since . Since is codimension , we either have equality, or . If there is equality, then and have the same kernel and therefore they are proportional, i.e. is an eigenvector of . If then is trivially an eigenvector of . In every case, we see that is an eigenvector of . By (a), , and therefore , is a multiple of the identity operator.

Exercise 3Let be a linear operator on a finite-dimensional inner product space .

- (a) Define what is meant by the adjoint of .
- (b) Prove that .
- (c) If is normal, prove that . Give an example when the equality fails (and, of course, is not normal).

*Proof:*

- (a) It is the unique linear operator on such that for every .
- (b) Indeed,
- (c) A normal operator is one which commutes with its adjoint, i.e. . Thus,
An example where the equality fails is supplied by the operator acting on in the standard way. The vector is in the kernel of but not of .