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 = 4 2 1 1 · 0 0 0 - 2 · 4 2 1 1 - 1 = 4 2 1 1 · 0 0 0 - 2 · 1 / 2 - 1 - 1 / 2 2 Section 6.3, Problem 3(c).
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