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Unformatted text preview: MATH 108B HW 2 SOLUTIONS RAHUL SHAH Problem 1. [ 6.27] Solution. Given z = ( z 1 ,...z n ) , z = ( z 1 ,...z n ) F n , notice that z , z = * n X i =1 z i e i , n X i =1 z i e i + = n X i =1 * z i e i , n X j =1 z j e j + = n X i,j =1 z i z j h e i , e j i = n X i =1 z i z i Now define T * ( z 1 ,...z n ) = ( z 2 ,...z n , 0) and notice that T ( z ) , z = n X i =2 z i 1 z i and z ,T * ( z ) = n 1 X i =1 z i z i +1 . Both these quantities are equal and thus T * is the adjoint of T . Problem 2. [ 6.28] Solution. Let F s.t. Tv = v for some v 6 = 0. Thus h Tv,v i = h v,v i = h v,v i = v, v Thus h v,T * v i = h Tv,v i = v, v . Thus v,T * v v = 0. Since v 6 = 0, we find that T * v v = 0 and thus is an eigenvalue of T * . Now notice that since ( T * ) * = T , and = , we find that an eigenvalue of T * an eigenvalue of T . Problem 3. [ 6.29] Solution. Assume that T ( U ) U , for U a subspace of V . Let w U . Let u U . h u,T * w i = h Tu,w i = 0, since T ( U ) U and w U . Thus T * w U and thus T ( U ) U...
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This document was uploaded on 08/27/2011.
 Winter '09
 Math

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