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Unformatted text preview: SSM: Linear Algebra Section 8.1 Chapter 8 8.1 1. ~e 1 , ~ e 2 is an orthonormal eigenbasis. 3. 1 √ 5 2 1 , 1 √ 5 1 2 is an orthonormal eigenbasis. 5. Eigenvalues 1, 1, 2 Choose ~v 1 = 1 √ 2  1 1 in E 1 and ~v 2 = 1 √ 3 1 1 1 in E 2 and let ~v 3 = ~v 1 × ~v 2 = 1 √ 6 1 1 2 . 7. 1 √ 2 1 1 , 1 √ 2 1 1 is an orthonormal eigenbasis, so S = 1 √ 2 1 1 1 1 and D = 5 1 . 9. 1 √ 2 1 1 , 1 √ 2  1 1 , 1 is an orthonormal eigenbasis, with λ 1 = 3, λ 2 = 3, and λ 3 = 2, so S = 1 √ 2 1 1 √ 2 1 1 and D = 3 3 2 . 11. 1 √ 2 1 1 , 1 √ 2  1 1 , 1 is an orthonormal eigenbasis, with λ 1 = 2, λ 2 = 0, and λ 3 = 1, so S = 1 √ 2 1 1 √ 2 1 1 and D = 2 1 . 13. Yes; if ~v is an eigenvector of A with eigenvalue λ , then ~v = I 3 ~v = A 2 ~v = λ 2 ~v , so that λ 2 = 1 and λ = 1 or λ = 1. Since A is symmetric, E 1 and E 1 will be orthogonal complements, so that A represents the reflection about E 1 . 15. Yes, if A~v = λ~v , then A 1 ~v = 1 λ ~v , so that an orthonormal eigenbasis for A is also an orthonormal eigenbasis for A 1 (with reciprocal eigenvalues). 199 Chapter 8 SSM: Linear Algebra 17. If A is the n × n matrix with all 1’s, then the eigenvalues of A are 0 (with multiplicity n 1) and n . Now B = qA + ( p q ) I n , so that the eigenvalues of B are p q (with multiplicity n 1) and qn + p q . Thus det( B ) = ( p q ) n 1 ( qn + p q ). 19. Let L ( ~x ) = A~x . Then A T A is symmetric, since ( A T A ) T = A T ( A T ) T = A T A , so that there is an orthonormal eigenbasis ~v 1 , . . . , ~v m for A T A . Then the vectors A~v 1 , . . . , A~v m are orthogonal, since A~v i · A~v j = ( A~v i ) T A~v j = ~v T i A T A~v j = ~v i · ( A T A~v j ) = ~v i · ( λ j ~v j ) = λ j ( ~v i · ~v j ) = 0 if i 6 = j . 21. For each eigenvalue there are two unit eigenvectors: ± ~v 1 , ± ~v 2 , and ± ~v 3 . We have 6 choices for the first column of S , 4 choices remaining for the second column, and 2 for the third. Answer: 6 · 4 · 2 = 48. 23. The eigenvalues are real (by Fact 8.1.3), so that the only possible eigenvalues are ± 1 (by Fact 7.1.2). Since A is symmetric, E 1 and E 1 are orthogonal complements. Thus A represents a reflection about E 1 . 25. Note that A is symmetric an orthogonal, so that the eigenvalues of A are 1 and 1. E 1 = span 1 1 , 1 1 , 1 , E 1 = span 1 1 , 1 1 The columns of S must form an eigenbasis for A : S = 1 √ 2 1 1 1 1 √ 2 1 1 1 1 is one possible choice....
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 Spring '08
 BRETSCH
 Linear Algebra, Algebra, Matrices, Eigenvalue, eigenvector and eigenspace, Singular value decomposition, Orthogonal matrix, orthonormal eigenbasis

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