Tutorial 11 sol.pdf - THE UNIVERSITY OF HONG KONG DEPARTMENT OF MATHEMATICS MATH2102 Linear Algebra II Tutorial 11 Solutions 1 We find that \u2223A \u2212

Tutorial 11 sol.pdf - THE UNIVERSITY OF HONG KONG...

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THE UNIVERSITY OF HONG KONG DEPARTMENT OF MATHEMATICS MATH2102 Linear Algebra II Tutorial 11 Solutions 1. We find that A - λI = ( λ - 3 ) 3 ( λ - 7 ) . For λ = 3, we solve ( A - 3 I ) v = 0 to find three linearly independent eigenvectors, say, ( 1 , 1 , 0 , 0 ) T , ( 1 , 0 , 1 , 0 ) T , ( 1 , 0 , 0 , - 1 ) T . Using the Gram-Schmidt process and normalization, we get an orthonormal basis { v 1 , v 2 , v 3 } for E 3 where v 1 = 1 2 ( 1 , 1 , 0 , 0 ) T , v 2 = 1 6 ( 1 , - 1 , 2 , 0 ) T , v 3 = 1 2 3 ( 1 , - 1 , - 1 , - 3 ) T . For λ = 7, we solve ( A - 7 I ) v = 0 to find a unit eigenvector v 4 = 1 2 ( 1 , - 1 , - 1 , 1 ) T . Let Q be the matrix with columns v 1 , v 2 , v 3 , v 4 . Then Q is orthogonal and Q T AQ is diagonal. 2. (a) Suppose A = UBU * . Then tr ( A * A ) = tr (( UBU * ) * ( UBU * )) = tr ( UB * U * UBU * ) = tr ( UB * BU * ) . Recall that tr ( XY ) = tr ( Y X ) for any square matrices X,Y (Q8(a) of tutorial 5). So tr ( U ( B * BU * )) = tr (( B * BU * ) U ) = tr ( B * B ) . Thus, tr ( A * A ) = tr ( B * B ) . Remark. One can show that tr ( A * A ) is the sum of the moduli of all the entries of A . (b) Let A = 1 2 2 i and B = i 4 1 1 . We find that A * A = 1 2 2 - i 1 2 2 i = 5 2 + 2 i 2 - 2 i 5 . Its trace is 5 + 5 = 10. Also, we have B * B = - i 1 4 1 i 4 1 1 = 2 1 - 4 i 1 + 4 i 17 . Its trace is 2 + 17 = 19. As tr ( A * A ) tr ( B * B ) , A and B are not unitarily equivalent from (a). 3. Suppose A v = λ v for some v 0 . This implies A v = λ v . As A is a real matrix, we have A = A . This shows A v = λ v . Thus, λ is an eigenvalue of A . Since A is orthogonal, it is also unitary when regarded as a complex matrix. So the eigenvalues of A have moduli 1. It follows that 1 λ = λ is an eigenvalue of A .

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