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**Unformatted text preview: **Math 2940 Solutions, Fall 2011 Section 6 . 1 5 ) det 3- 1 1- = (3- )(1- ) has roots = 1 , 3. det 1- 1 3- = (1- )(3- ) has roots = 1 , 3. det 4- 1 1 4- = (4- ) 2- 1 = 2- 8 + 15 = ( - 5)( - 3) has roots = 5 , 3. The eigenvalues of a sum of matrices are not the sum of the eigenvalues of the matrices. Actually it doesnt even make sense - there are four possible sums of the eigenvalues of A and B but only two eigenvalues for A + B . 10 ) det . 6- . 2 . 4 . 8- = 2- 1 . 4 + . 4 = ( - 1)( - . 4) has roots = 1 , . 4. An eigenvector for = . 4 is (1 ,- 1) and an eigenvector for = 1 is (1 , 2). det (1 / 3)- 1 / 3 2 / 3 (2 / 3)- = 2- = ( - 1) has roots = 0 , 1. An eigenvector for = 0 is (1 ,- 1) and an eigenvector for = 1 is (1 , 2). So A 100 has eigenvalues 1 and ( . 4) 100 , which is close to 0. So A 100 and A have very close eigenvalues (and it is not a repeated eigenvalue) so they should have very close eigenvectors.eigenvalues (and it is not a repeated eigenvalue) so they should have very close eigenvectors....

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