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Unformatted text preview: Math 416  Abstract Linear Algebra Fall 2011, section E1 Homework 11 solutions Section 6.2 2.13. (1 pt check) Let A be a normal operator on C n (without loss of generality) whose eigenvalues all have modulus 1. Since A is normal, it can be unitarily diagonalized: A = UDU * where U is unitary and D is diagonal, with the eigenvalues 1 ,..., n of A on the diagonal. The adjoint of A is A * = ( U * ) * D * U * = UD * U * = U DU * and thus we have A * A = ( U DU * )( UDU * ) = U DDU * = U 1 ... 2 ... . . . . . . . . . . . . ... n 1 ... 2 ... . . . . . . . . . . . . ... n U * = U  1  2 ...  2  2 ... . . . . . . . . . . . . ...  n  2 U * = UIU * = I. Likewise, we have AA * = UD DU * = UIU * = I which means A * = A 1 , in other words A is unitary. Section 6.5 A6.5.1 (2 pts) Every vector x R n can be written (uniquely) as x = p + p + w with p P,p P ,w ( P + P ) . Let us check the R and R commute when restricted to those three subspaces. For p P , we have RR p = Rp since p P P R Rp = Rp since Rp P P . 1 For p P , we have RR p = R p since R p P P R Rp = R p since p P P....
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 Fall '11
 FRANKLAND
 Math, Linear Algebra, Algebra

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