Math 462; Assignment 8 - Solutions

1. Suppose that T ∈ L(V, W ). Prove that

(a) T is surjective if and only if T is injective. (b) T is injective if and only if T is surjective.

Solution: Observe that hv, T wi = 0 if and only if hT v, wi = 0, i.e. w ∈ ker T if and only if w ∈ (range T )⊥. Hence, ker T = {0} if and only if range T = W , which is the desired result. The second statement is proven analogously by switching the roles of T and T .

2. Suppose that T ∈ L(V ) and that U is a subspace of V . Prove that U is invariant under T if and only if U ⊥ is invariant under T . Solution: Suppose U is invariant under T . Then T u ∈ U for all u ∈ U, and hT u, vi = 0 for all u ∈ U and v ∈ U ⊥. It follows that hu, T vi = 0 for all u ∈ U and all v ∈ U ⊥. Therefore T v ∈ U ⊥ for all v ∈ U ⊥ and U ⊥ is invariant under T . The opposite direction is proven by the analogous argument (i.e. read the above argument backwards).

3. Consider P2(R) with the inner product

1 [p, q] = p(x)q(x) dx Z0

2 Dene T ∈ L(P2(R) by T (a0 + a1x + a2x ) = a1x.

(a) Show that T is not self-adjoint. (b) Compute the matrix of T in the basis (1, x, x2) and show that it equals its conjugate transpose. (c) Explain that this is not a contradiction.

Solution: To show that T is not selfadjoint observe that:

1 1 2 2 hT p, qi = T (p(x))q(x) dx = T (p0 + p1x + p2x )(q0 + q1x + q2x ) dx Z0 Z0 1 2 3 1 1 1 = (p1q0x + p1q1x + p1q2x ) dx = p1q0 + p1q1 + p1q2 Z0 2 3 4 1 1 2 2 hp, T qi = p(x)T (q(x)) dx = (p0 + p1x + p2x )T (q0 + q1x + q2x ) dx Z0 Z0 1 2 3 1 1 1 = (q1p0x + q1p1x + q1p2x ) dx = q1p0 + q1p1 + q1p2 Z0 2 3 4 and that the two are not equal. The matrix for T is given by

0 0 0 M(T ) =  0 1 0  0 0 0   The matrix is diagonal despite the fact that T is not self-adjoint. This does not contradict the theory, since the basis used is not orthonormal.

4. Let P ∈ L(V ) such that P 2 = P . Prove that P is an orthogonal projection if and only if P is selfadjoint. Solution: P is an orthogonal projection if and only if hu P u, P ui = 0 for all u ∈ V . This holds if and only if hu, P ui = hP u, P ui and hence

hu, P ui = hu, P P ui

for all u ∈ V . This holds if and only if P P = P = P 2, i.e if and only if P is self-adjoint.

5. Let T ∈ L(V ) be a normal operator. Show that

range T = range T

Solution: Since kT vk = kT vk for all v ∈ V it follows that kT vk = 0 if and only if kT vk = 0. Therefore ker T = ker T . Since hw, T vi = 0 if and only if hT w, vi = 0 it follows that range T = (ker T )⊥ and analogously range T = (ker T )⊥. The result follows immediately.

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