2040hw12sol - MATH2040 HW 12 Solution Tsang Chi Shing Yu...

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Mathematics: A Discrete Introduction
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Chapter 5 / Exercise 24.24
Mathematics: A Discrete Introduction
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MATH2040 HW 12 Solution Tsang Chi Shing, Yu Rongfeng Sec. 7.1 : 2, 4, 5, 8, 10; Sec. 7.2 : 2, 3, 4, 6, 7, 11, 12, 13, 14, 15, 16, 17. Sec. 7.1: 2) (a) The characteristic polynomial of A is: det( A - tI ) = det 1 - t 1 - 1 3 - t = ( t - 2) 2 Hence λ = 2 is the only eigenvalue of A with multiplicity 2 (so dim( K λ ) = 2). We have A - λI = - 1 1 - 1 1 So dim( E λ ) = 2 - rank ( A - λI ) = 1. We have ( A - λI ) 2 = 0 0 0 0 Thus we can choose v = 1 0 such that ( A - λI ) v = - 1 - 1 6 = 0 and ( A - λI ) 2 v = 0. We have β = - 1 - 1 , 1 0 is a Jordan canonical basis for A . Therefore a Jordan canonical form of A is J = [ L A ] β = 2 1 0 2 (b) The characteristic polynomial of A is: det( A - tI ) = det 1 - t 2 3 2 - t = ( t + 1)( t - 4) Hence λ 1 = - 1 and λ 2 = 4 are eigenvalues of A with the same multiplicity 1 (so dim( K λ 1 ) = dim( K λ 2 ) = 1). Thus we have E λ 1 = 1
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Mathematics: A Discrete Introduction
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Chapter 5 / Exercise 24.24
Mathematics: A Discrete Introduction
Scheinerman
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N ( A - λ 1 I ) = K λ 1 and E λ 2 = N ( A - λ 2 I ) = K λ 2 . Consider that 1 - 1 is an eigenvector of A corresponding to - 1, and 2 3 is an eigenvector of A corresponding to 4, therefore β 1 = 1 - 1 is a basis for K λ 1 , and β 2 = 2 3 is a basis for K λ 2 . We have β = β 1 β 2 = 1 - 1 , 2 3 is a Jordan canonical basis for A . Therefore a Jordan canonical form of A is J = [ L A ] β = - 1 0 0 4 (c) The characteristic polynomial of A is: det( A - tI ) = det 11 - t - 4 - 5 21 - 8 - t - 11 3 - 1 - t = - ( t + 1)( t - 2) 2 Hence λ 1 = - 1 and λ 2 = 2 are eigenvalues of A with multiplicities 1 and 2 respectively (so dim( K λ 1 ) = 1 and dim( K λ 2 ) = 2). For λ 1 = - 1, we have E λ 1 = N ( A - λ 1 I ) = K λ 1 . Consider that 1 3 0 is an eigenvector of A corresponding to - 1, therefore β 1 = 1 3 0 is a basis for K λ 1 . For λ 2 = 2, we have A - λ 2 I = 9 - 4 - 5 21 - 10 - 11 3 - 1 - 2 So dim( E λ 2 ) = 3 - rank ( A - λ 2 I ) = 1. We have ( A - λ 2 I ) 2 = - 18 9 9 - 54 27 27 0 0 0 Thus we can choose v = 1 2 0 such that ( A - λ 2 I ) v = 1 1 1 6 = 0 and ( A - λ 2 I ) 2 v = 0. We have β 2 = 1 1 1 , 1 2 0 2
is a basis for K λ 2 . We have β = β 1 β 2 = 1 3 0 , 1 1 1 , 1 2 0 is a Jordan canonical basis for A . Therefore a Jordan canonical form of A is J = [ L A ] β = - 1 0 0 0 2 1 0 0 2 (d) The characteristic polynomial of A is: det( A - tI ) = det 2 - t 1 0 0 0 2 - t 1 0 0 0 3 - t 0 0 1 - 1 3 - t = - ( t - 2) 2 ( t - 3) 2 Hence λ 1 = 2 and λ 2 = 3 are eigenvalues of A with the same multi- plicity 2 (so dim( K λ 1 ) = dim( K λ 2 ) = 2). For λ 1 = 2, we have A - λ 1 I = 0 1 0 0 0 0 1 0 0 0 1 0 0 1 - 1 1 So dim( E λ 1 ) = 4 - rank ( A - λ 1 I ) = 1. We have ( A - λ 1 I ) 2 = 0 0 1 0 0 0 1 0 0 0 1 0 0 1 - 1 1 Thus we can choose v = 0 1 0 - 1 such that ( A - λ 1 I ) v = 1 0 0 0 6 = 0 and ( A - λ 2 I ) 2 v = 0. We have β 1 =

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