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practice-final-solns

practice-final-solns - MATH 110 SOLUTIONS TO PRACTICE FINAL...

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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.
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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. Proof. We have ( ST ) * = T * S * = TS . Thus ( ST ) * = ST if and only if ST = TS , which is the desired claim. (Easiest proof ever!) 5. Axler, 7.6 Proof. Recall that if T is normal, then null T = null T * . This follows from the fact that k Tv k = k T * v k for all v . Thus range T = (null T * ) = (null T ) = range T * as required.
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