11hw8 - Homework Assignment#8 23.5 Suppose that A is an...

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Homework Assignment #8 23.5 Suppose that A is an invertible matrix. Show that ( A - rI ) v = 0 implies that ( A - 1 - 1 r I ) v = 0 . Conclude that for an invertible matrix A , r is an eigenvalue of A if and only if 1 /r is an eigenvalue of A - 1 . Answer: Since A is invertible, r 6 = 0. Now apply A - 1 to ( A - rI ) v = 0 to obtain ( I - rA - 1 ) v = 0 . Then divide by - 1 r to find ( A - 1 - 1 r I ) v = 0 . The rest just restates this. 23.7 For each of the following matrices A , find nonsingular matrix P and diagonal matrix D so that P - 1 AP = D . a ) ± 3 0 1 2 ² , b ) ± 0 1 - 1 5 ² , c ) ± 1 - 1 2 4 ² , d ) 3 - 1 0 - 1 2 - 1 0 - 1 3 , e ) 4 - 2 - 2 0 1 0 1 0 1 . Answer: a ) The eigenvalues obey (3 - λ )(2 - λ ) = 0, so σ ( A ) = { 2 , 3 } . Corresponding eigenvectors are v 2 = (0 , 1) T and v 3 = (1 , 1) T . Then P = [ v 3 , v 2 ] = ± 1 0 1 1 ² and P - 1 = ± 1 0 - 1 1 ² yielding D = ± 3 0 0 2 ² . b ) The eigenvalues obey λ 2 - 5 λ +1 = 0, so σ ( A ) = { 1 2 (5 ± 21) } . Corresponding eigenvectors are v + = (1 , 1 2 (5 + 21)) T and v - = (1 , 1 2 (5 - 21)) T . Then P = [ v + , v - ] = ± 1 1 1 2 (5 + 21) 1 2 (5 - 21) ² and P - 1 = 1 21 ± 1 2 ( - 5 + 21) 1 1 2 (5 + 21) - 1 ² yielding D = ± 1 2 (5 + 21) 0 0 1 2 (5 - 21) ² . c ) The eigenvalues obey λ 2 - 5 λ + 6 = 0, so σ ( A ) = { 2 , 3 } . Corresponding eigenvectors are v 2 = (1 , - 1) T and v 3 = (1 , - 2) T . Then P = [ v 3 , v 2 ] = ± 1 1 - 2 - 1 ² and P - 1 = ± - 1 - 1 2 1 ² yielding D = ± 3 0 0 2 ² .
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11hw8 - Homework Assignment#8 23.5 Suppose that A is an...

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