homework #10 answers

Linear Algebra with Applications

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Math 20F - Homeworks 9 & 10 - Selected answers Samuel R. Buss - Winter 2003 - UC San Diego Revision 1.0. – March 13, 2003 Section 6.1, Problem 19. We did almost exactly this problem as a theorem proved in class. Section 6.1, Problem 22. Since u T j u i = δ i,j , we have A u i = n X j =1 c j u j u T j u i = n X j =1 δ i,j c j u j = c i u i . Hence u i is an eigenvector for the eigenvalue c i . Section 6.3, Problem 1(a). We did this as an example in class on Wednesday. Section 6.3, Problem 1(c). det( A - λI )=(2 - λ )( - 4 - λ )+8=2 λ + λ 2 =(2+ λ )( λ ). The roots of the characteristic polynomial are λ 1 = 1 and λ 2 = - 2: these are the eigenvalues of A . Solving A x = 0 for a nontrivial x , we find that x 1 =(4 , 1) T is a eigenvector corresponding to the eigenvalue λ 1 =0. Solving ( A +2 I ) x = 0, we find that x 2 =(2 , 1) T is an eigenvector corresponding to the eigen value λ 2 = - 2. The eigenvalues are distinct, hence x 1 and x 2 are linearly independent. Therefore, A is diagonalizable in the following form: A = ± 42 11 · ± 00 0 - 2 · ± - 1 = ± · ± 0 - 2 · ± 1 / 2 - 1 - 1 / 22 Section 6.3, Problem 3(c).
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This homework help was uploaded on 01/23/2008 for the course MATH 20F taught by Professor Buss during the Winter '03 term at UCSD.

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homework #10 answers - Math 20F Homeworks 9 10 Selected...

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