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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.
 Fall '07
 Loveys
 Linear Algebra, Algebra, Matrices

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