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# HW2 - MATH 108B HW 2 SOLUTIONS RAHUL SHAH Problem 1[6.27...

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MATH 108B HW 2 SOLUTIONS RAHUL SHAH Problem 1. [ § 6.27] Solution. Given z = ( z 1 , . . . z n ) , z 0 = ( z 0 1 , . . . z 0 n ) F n , notice that z , z 0 = * n X i =1 z i e i , n X i =1 z 0 i e i + = n X i =1 * z i e i , n X j =1 z 0 j e j + = n X i,j =1 z i z 0 j h e i , e j i = n X i =1 z i z 0 i Now define T * ( z 1 , . . . z n ) = ( z 2 , . . . z n , 0) and notice that T ( z ) , z 0 = n X i =2 z i - 1 z 0 i and z , T * ( z 0 ) = n - 1 X i =1 z i z 0 i +1 . Both these quantities are equal and thus T * is the adjoint of T . Problem 2. [ § 6.28] Solution. Let λ F s.t. Tv = λv for some v 6 = 0. Thus h Tv, v i = h λv, v i = λ h v, v i = v, λv Thus h v, T * v i = h Tv, v i = v, λv . Thus v, T * v - λv = 0. Since v 6 = 0, we find that T * v - λv = 0 and thus λ is an eigenvalue of T * . Now notice that since ( T * ) * = T , and λ = λ , we find that λ an eigenvalue of T * λ 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 . Hence U is invariant under T * . Since ( T * ) * = T and ( U ) = U for U a subspace, we find that the reverse implication also holds.

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HW2 - MATH 108B HW 2 SOLUTIONS RAHUL SHAH Problem 1[6.27...

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