Quiz 6 - Solutions

Quiz 6 - Solutions - P ( x ) = ax 2 + bx + c . If A is an n...

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Math 54 Quiz 6 Mike Hartglass November 6, 2010 1.) Find P and D with P invertible and D diagonal such that A = PDP - 1 or explain why this is impossible. A = ± 1 3 3 5 ² . The characteristic polynomial for A is t 2 - 6 t - 4 = ( t - 3) 2 - 13 so we see that it has roots t = 3 ± 13. If t = 3 + 13 then A - tI = ± - 2 - 13 3 3 2 - 13 ² which is row equivalent to ± - 2 - 13 3 0 0 ² so we see that null space consists of vectors of the form ((2 + 13) / 3 x 2 ,x 2 ) so an eigenvector of A with this eigenvalue is ((2+ 13) / 3 , 1). Similarly, an eigenvector with eigenvalue 3 - 13 is ((2 - 13) / 3 , 1) so we see that D = ± 3 + 13 0 0 3 - 13 ² and P = ± (2 + 13) / 3 (2 - 13) / 3 1 1 ² . 1
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2.) Let P be a quadratic polynomial i.e.
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Unformatted text preview: P ( x ) = ax 2 + bx + c . If A is an n × n matrix, define P ( A ) = aA 2 + bA + cI . Prove that if λ is an eigenvalue of A then P ( λ ) is an eigenvalue of P ( A ). If λ is an eigenvalue of A then there is a nonzero vector v satisfying Av = λv . Thus, P ( A ) v = aA 2 v + bAv + cIv = λaAv + bλv + cv = aλ 2 v + bλv + cv = ( aλ 2 + bλ + c )( v ) = P ( λ ) v . This shows that v is also an eigenvector of P ( A ) with eigenvalue P ( λ ) proving that P ( λ ) is an eigenvalue of P ( A ). 2...
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This note was uploaded on 11/19/2010 for the course LECTURE 1 taught by Professor Yildiz during the Fall '10 term at Berkeley.

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Quiz 6 - Solutions - P ( x ) = ax 2 + bx + c . If A is an n...

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