Math 462; Assignment 8 - Solutions 1. Suppose that T 2 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 2 ker T ¤ if and only if w 2 (range T )?. Hence, ker T ¤ = f0g 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 2 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 2 U for all u 2 U, and hT u; vi = 0 for all u 2 U and v 2 U ?. It follows that hu; T ¤vi = 0 for all u 2 U and all v 2 U ?. Therefore T ¤v 2 U ? for all v 2 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 De¯ne T 2 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 0 1 0 1 0 0 0 @ A 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 2 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 2 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 2 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 2 L(V ) be a normal operator. Show that range T = range T ¤ Solution: Since kT vk = kT ¤vk for all v 2 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. ..