A6_soln

# A6_soln - Math 235 Assignment 6 Solutions 1 For each of the following symmetric matrices find the orthogonal matrix R that diago nalizes the given

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Unformatted text preview: Math 235 Assignment 6 Solutions 1. For each of the following symmetric matrices, find the orthogonal matrix R that diago- nalizes the given matrix and the corresponding diagonal matrix. a) A = 7 3 3- 1 Solution: The characteristic polynomails is C ( λ ) = det( A- λI ) = det 7- λ 3 3- 1- λ = ( λ + 2)( λ- 8) . Hence, the eigenvalues of A are λ 1 =- 2 and λ 2 = 8. For λ 1 =- 2 we have A- λ 1 I = 9 3 3 1 ∼ 1 1 / 3 . Thus, an eigenvector corresonding to λ 1 is 1- 3 . For λ 2 = 8 we have A- λ 2 I =- 1 3 3- 9 ∼ 1- 3 . Thus, an eigenvector corresonding to λ 2 is 3 1 . Normalizing the eigenvectors, we find that A is diagonalied by the orthogonal matrix P = 1 √ 10 1 3- 3 1 and P T AP =- 2 0 8 = D. 2 b) B = 1- 1 1 1- 1 1 Solution: The characteristic polynomails is C ( λ ) = det( B- λI ) = det - λ 1- 1 1- λ 1- 1 1- λ = det 1- λ 1- 1 1- λ- λ 1 1- λ = det 1- λ 1- 1- 1- λ 2 1- λ =- ( λ- 1)( λ 2 + λ- 2) =- ( λ- 1) 2 ( λ + 2) . Hence, the eigenvalues of B are λ 1 = 1 and λ 2 =- 2. For λ 1 = 1 we have B- λ 1 I = - 1 1- 1 1- 1 1- 1 1- 1 ∼ - 1 1- 1 . Thus, linearly independent eigenvectors corresonding to λ 1 are ~v 1 = 1 1 , and ~v 2 = - 1 1 . For λ 2 =- 2 we have B- λ 2 I = 2 1- 1 1 2 1- 1 1 2 ∼ 1 0- 1 0 1 1 0 0 . Thus, an eigenvector corresonding to λ 2 is 1- 1 1 . We first observe that eigenvectors for λ 1 are not orthogonal. So, applying the Gram-Schmidt procedure to { ~v 1 ,~v 2 } : Let ~w 1 = (1 , 1 , 0); then ~w 2 = (- 1 , , 1)- (- 1 , ,...
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## This note was uploaded on 10/12/2010 for the course MATH 235 taught by Professor Celmin during the Fall '08 term at Waterloo.

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A6_soln - Math 235 Assignment 6 Solutions 1 For each of the following symmetric matrices find the orthogonal matrix R that diago nalizes the given

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