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Unformatted text preview: MATH 110 - SOLUTIONS TO PRACTICE FINAL LECTURE 1, SUMMER 2009 GSI: SANTIAGO CA ˜ NEZ 1. Axler, 6.18 (Hint: By 6.17 it is enough to show that every vector in null P is orthogonal to every vector in range P — use 6.2 to show this) Proof. First, we know that P is the projection operator corresponding to the decomposition V = range P ⊕ null P. Let u ∈ range P and w ∈ null P . We must show that h u,w i = 0. Now, for any a ∈ F . we have P ( u + aw ) = u since aw ∈ null P . Thus k u k ≤ k u + aw k for any scalar a by the assumption that k Pv k ≤ k v k for any v . By problem 6.2, this implies that h u,w i = 0 as required. We conclude that P is an orthogonal projection by problem 6.17. 2. Axler, 6.20 Proof. Suppose that U and U ⊥ are T-invariant. To show that P U T = TP U , we must show that both sides give the same result when applied to an arbitrary v ∈ V . To this end, let v ∈ V and write it as v = u + w according to the decomposition V = U ⊕ U ⊥ . Then Tu ∈ U and Tw ∈ U ⊥ since U and U ⊥ are T-invariant. Thus P U T ( v ) = P U ( Tu + Tw ) = Tu and TP U ( v ) = T ( P U ( u + w )) = Tu, so P U T = TP U as required. Conversely, suppose that P U T = TP U . Let u ∈ U and w ∈ U ⊥ . We must show that Tu ∈ U and Tw ∈ U ⊥ . We have P U T ( u + w ) = P U ( Tu + Tw ) and TP U ( u + w ) = T ( P U ( u + w )) = Tu. For these to be equal means that Tu itself is the piece of Tu + Tw which is in U and that Tw must then be in U ⊥ . Hence U and U ⊥ are T-invariant as claimed. 3. Suppose that V is an inner-product space and that U and W are subspaces of V . Show that U ⊆ W if and only if W ⊥ ⊆ U ⊥ . Date : August 6, 2009. 2 GSI: SANTIAGO CA ˜ NEZ Proof. Suppose that U ⊆ W and let v ∈ W ⊥ . Then h v,w i = 0 for any w ∈ W . In particular then, since U ⊆ W , h v,u i = 0 for any u ∈ U . Thus w ∈ U ⊥ so W ⊥ ⊆ U ⊥ . Conversely, suppose that W ⊥ ⊆ U ⊥ . By what we just shown, we have ( U ⊥ ) ⊥ ⊆ ( W ⊥ ) ⊥ , and hence U ⊆ W since taking two orthogonal complements gives back the original space. 4. Suppose that S and T are self-adjoint operators on an inner-product space V . Show that ST is self-adjoint if and only if S and T commute....
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This note was uploaded on 08/31/2009 for the course MATH 110 taught by Professor Gurevitch during the Spring '08 term at Berkeley.
- Spring '08