# ass6 - MATH 223, Linear Algebra Fall, 2007 Assignment 6...

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Unformatted text preview: MATH 223, Linear Algebra Fall, 2007 Assignment 6 Solutions 1. Diagonalize the following matrices over R : (a) 4 2 3- 1 (b) 1 1 2- 1 Solution: For each matrix A above, we compute the characteristic polynomial to find the eigenval- ues, then for each eigenvalue, we find a basis of the corresponding eigenspace. Finally we determine the matrix P such that P- 1 AP is diagonal. (a) We have A ( t ) = det 4- t 2 3- 1- t = t 2- 3 t- 10 = ( t- 5)( t + 2) , so the eigenvalues are 5 ,- 2. To determine bases of the corresponding eigenspaces, we find E 5 = nullsp 4 2 3- 1 - 5 I 2 = nullsp - 1 2 3- 6 = span 2 1 and E- 2 = nullsp 4 2 3- 1 + 2 I 2 = nullsp 6 2 3 1 = span 1- 3 . It follows that B := 2 1 , 1- 3 is a basis for R 2 consisting of eigenvectors for A . Therefore, if we set P = 2 1 1- 3 we have P- 1 AP = 5- 2 . (b) Here we find A ( t ) = t 2- 3 = ( t- 3)( t + 3) , so the eigenvalues are 3 ,- 3. We compute E 3 = nullsp 1 1 2- 1 - 3 I 2 = nullsp 1- 3 1 2- 1- 3 = span 1 + 3 2 . and E- 3 = nullsp 1 1 2- 1 + 3 I 2 = nullsp 1 + 3 1 2- 1 + 3 = span 1- 3 2 . 1 It follows that B := 1 + 3 2 , 1- 3 2 is a basis for R 2 consisting of eigenvectors for A . Therefore, if we set P = 1 + 3 1- 3 2 2 we have P- 1 AP = 3- 3 ....
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## This note was uploaded on 04/29/2008 for the course MATH 223 taught by Professor Loveys during the Fall '07 term at McGill.

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ass6 - MATH 223, Linear Algebra Fall, 2007 Assignment 6...

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