301_F10_hw7_soln

# 301_F10_hw7_soln - AME 301 Homework#7 Solutions | Fall 2010...

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AME 301 Homework #7 Solutions | Fall 2010 1. Similar matrices A and ~ A = P ± 1 AP have the same eigenvalues. Verify this property for A = 2 4 1 0 1 0 1 1 1 1 0 3 5 and P = 2 4 1 0 0 0 0 ± 1 0 ± 1 0 3 5 : Noting that P is an orthogonal matrix, we have P ± 1 = P T , P ± 1 = 2 4 1 0 0 0 0 ± 1 0 ± 1 0 3 5 : Then ~ A = P ± 1 AP = 2 4 1 0 0 0 0 ± 1 0 ± 1 0 3 5 2 4 1 0 1 0 1 1 1 1 0 3 5 2 4 1 0 0 0 0 ± 1 0 ± 1 0 3 5 = 2 4 1 ± 1 0 ± 1 0 1 0 1 1 3 5 : Next compare the eigenvalues. The eigenvalues of A are found by det( A ± ± I ) = ± ± ± ± ± ± 1 ± ± 0 1 0 1 ± ± 1 1 1 ± ± ± ± ± ± ± ± = ± ± 3 + 2 ± 2 + ± ± 2 = ± ( ± ± 2)( ± ± 1)( ± + 1) = 0 : ± 1 = 2 ; ± 2 = 1 ; ± 3 = ± 1 : The eigenvalues of ~ A are found by det( ~ A ± ± I ) = ± ± ± ± ± ± 1 ± ± ± 1 0 ± 1 ± ± 1 0 1 1 ± ± ± ± ± ± ± ± = ± ± 3 + 2 ± 2 + ± ± 2 = ± ( ± ± 2)( ± ± 1)( ± + 1) = 0 : ± 1 = 2 ; ± 2 = 1 ; ± 3 = ± 1 : Note that A and ~ A have the same characteristic equation and the same eigenvalues, as required for similar matrices. 2. Consider the real symmetric matrix A = 2 4 2 1 1 1 2 1 1 1 2 3 5 : 1

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(a) Find the eigenvalues and eigenvectors: det( A ± ± I ) = ± ± ± ± ± ± 2 ± ± 1 1 1 2 ± ± 1 1 1 2 ± ± ± ± ± ± ± ± = (2 ± ± ) ² (2 ± ± ) 2 ± 1 ³ ± [(2 ± ± ) ± 1] + [1 ± (2 ± ± )] = (2 ± ± )((2 ± ± ) ± 1)((2 ± ± ) + 1) ± 2((2 ± ± ) ± 1) = (1 ± ± ) [(2 ± ± )(3 ± ± ) ± 2] = (1 ± ± ) ² 4 ± 5 ± + ± 2 ³ = ± ( ± ± 1)( ± ± 1)( ± ± 4) = 0 ± 1 = 4 ; ± 2 = 1 ; ± 3 = 1 : For ± 1 = 4, ( A ± ± 1 I ) = 2 4 ± 2 1 1 1 ± 2 1 1 1 ± 2 3 5 ) U 1 = 2 4 ± 2 1 1 0 ± 3 = 2 3 = 2 0 0 0 3 5 ) s 1 = ² 2 4 1 1 1 3 5 ; where we set the free variable x 3 = ² , then solved for x 1 and x 2 . For ± 2 = 1, ( A ± ± 2 I ) = 2 4 1 1 1 1 1 1 1 1 1 3 5 ) U 2 = 2 4 1 1 1 0 0 0 0 0 0 3 5 ) s 2 = ³ 2 4 ± 1 0 1 3 5 + ´ 2 4 ± 1 1 0 3 5 ; where we set the free variables x 3 = ³ and x 2 = ´ , then solved for x 1 .
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301_F10_hw7_soln - AME 301 Homework#7 Solutions | Fall 2010...

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