A7_soln

# A7_soln - Math 235 Assignment 7 Solutions 1 For each of the...

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Math 235 Assignment 7 Solutions 1. For each of the following symmetric matrices, ﬁnd an orthogonal matrix P and diagonal matrix D such that P T AP = D . (a) A = ± 1 - 2 - 2 1 ² Solution: We have C ( λ ) = det( A - λI ) = ³ ³ ³ ³ 1 - λ - 2 - 2 1 - λ ³ ³ ³ ³ = λ 2 - 2 λ - 3 = ( λ - 3)( λ + 1) Hence, the eigenvalues are λ 1 = 3 and λ 2 = - 1. For λ 1 = 3 we get A - λ 1 I = ± - 2 - 2 - 2 - 2 ² ± 1 1 0 0 ² . Thus, a basis for E λ 1 is ´± - 1 1 ²µ . For λ 2 = - 1 we get A - λ 2 I = ± 2 - 2 - 2 2 ² ± 1 - 1 0 0 ² . Thus, a basis for E λ 2 is ´± 1 1 ²µ . After normalizing, the basis vectors for the eigenspaces form an orthonormal basis for R 2 . Hence, P = ± - 1 / 2 1 / 2 1 / 2 1 / 2 ² is orthogonal and P T AP = ± 3 0 0 - 1 ² . (b) A = 0 1 1 1 0 1 1 1 0 Solution: We have C ( λ ) = ³ ³ ³ ³ ³ ³ - λ 1 1 1 - λ 1 1 1 - λ ³ ³ ³ ³ ³ ³ = ³ ³ ³ ³ ³ ³ - λ 1 2 1 - λ 1 - λ 0 1 + λ 0 ³ ³ ³ ³ ³ ³ = (1 + λ )( λ 2 - λ - 2) = ( λ + 1) 2 ( λ - 2) Hence, the eigenvalues are λ 1 = - 1 and λ 2 = 2. For λ 1 = - 1 we get A - λ 1 I = 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 Thus, a basis for E λ 1 is - 1 1 0 , - 1 0 1 . Since the basis is not orthogonal, we need to apply the Gram-Schmidt procedure. We pick

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## This note was uploaded on 09/30/2011 for the course MATH 235 taught by Professor Celmin during the Spring '08 term at Waterloo.

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A7_soln - Math 235 Assignment 7 Solutions 1 For each of the...

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