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HW7_solutions - Math 167 homework 7 solutions 5.1.12 Find...

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Unformatted text preview: Math 167 homework 7 solutions November 17, 2010 5.1.12: Find the eigenvalues and eigenvectors of A = bracketleftbigg 3 4 4 − 3 bracketrightbigg and B = bracketleftbigg a b b a bracketrightbigg . The characteristic polynomial for A is p ( λ ) = (3 − λ )( − 3 − λ ) − 16 = λ 2 − 25 . Thus the eigenvalues of A are λ 1 = 5 and λ 2 = − 5; corresponding eigenvec- tors are x 1 = bracketleftbigg 2 1 bracketrightbigg and x 2 bracketleftbigg − 1 2 bracketrightbigg , respectively. The characteristic polynomial for B is p ( λ ) = ( a − λ ) 2 − b 2 = λ 2 − 2 aλ + ( a − b )( a + b ) = 0 . Thus the eigenvalues of B are λ 1 = a + b and λ 2 = a − b ; corresponding eigenvectors are x 1 bracketleftbigg 1 1 bracketrightbigg and bracketleftbigg − 1 1 bracketrightbigg , respectively. 5.1.14 Find the rank and all four eigenvalues for both the matrix of ones and the checkerboard matrix: A = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 and C = 0 1 0 1 1 0 1 0 0 1 0 1 1 0 1 0 . 1 Which eigenvectors correspond to nonzero eigenvalues? Since all columns of A are identical and A is not the zero matrix, the rank of A is 1. Thus A has only one nonzero eigenvalue, and the other three eigenvalues are all 0. Since Ae = 4 e , where e is the vector of all 1’s, the nonzero eigenvalue of A is 4. Note that the columns of the matrix C have the pattern C = bracketleftbig c 1 c 2 c 1 c 2 bracketrightbig , where c 1 = 1 1 and c 2 = 1 1 . Since c 1 and c 2 are linearly independent, the rank of C is 2. Thus C has two nonzero eigenvalues, and the other two eigenvalues are both 0. Since Cc 1 = 2 c 2 and Cc 2 = 2 c 1 , it follows that C ( c 1 + c 2 ) = 2( c 1 + c 2 ) , c 1 + c 2 = 1 1 1 1 negationslash = 0...
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