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hw10solutions

# hw10solutions - Math 215 HW#10 Solutions 1 Problem 5.2.14...

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Math 215 HW #10 Solutions 1. Problem 5.2.14. Suppose the eigenvector matrix S has S T = S - 1 . Show that A = S Λ S - 1 is symmetric and has orthogonal eigenvectors. Proof. Suppose S = [ v 1 . . . v n ], where v i are the eigenvectors of A . Then, since S T = S - 1 , we know that I = S T S = v T 1 . . . v T n | . . . | v 1 . . . v n | . . . | = v 1 , v 1 . . . v 1 , v n . . . . . . v n , v 1 . . . v n , v n , so we see that v i , v j = 0 unless i = j ; hence, the eigenvectors of A are orthogonal. Also, again using the fact that S T = S - 1 , we know that A T = ( S Λ S - 1 ) T = ( S Λ S T ) T = ( S T ) T Λ T S T = S Λ S T = A since Λ T = Λ and ( S T ) T = S , so we see that A is symmetric. 2. Problem 5.2.30. Find Λ and S to diagonalize A in Problem 29 ( A = . 6 . 4 . 4 . 6 ). What is the limit of Λ k as k → ∞ ? What is the limit of S Λ k S - 1 ? In the columns of this limit matrix you see the . Answer: To diagonalize A , we need to find the eigenvalues and eigenvectors of A . To that end, we want to solve 0 = det( A - λI ) = . 6 - λ . 4 . 4 . 6 - λ = ( . 6 - λ ) 2 - . 4 2 = λ 2 - 1 . 2 λ + . 2 = ( λ - 1)( λ - . 2) , so the eigenvalues of A are λ 1 = 1 and λ 2 = . 2. The eigenvector associated to λ 1 = 1 is the generator of the nullspace of A - I = - . 4 . 4 . 4 - . 4 , which row-reduces to - . 4 . 4 0 0 , so the nullspace consists of multiples of v 1 = 1 1 . On the other hand, the eigenvector associated to λ 2 = . 2 is the generator of the nullspace of A - . 2 I = . 4 . 4 . 4 . 4 , which row-reduces to . 4 . 4 0 0 , so the nullspace consists of multiples of v 2 = - 1 1 . 1

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Therefore, we see that A = S Λ S - 1 where S = 1 - 1 1 1 and Λ = 1 0 0 . 2 . Since Λ k = 1 0 0 . 2 k = 1 k 0 0 ( . 2) k = 1 0 0 1 5 k , we see that Λ k 1 0 0 0 as k → ∞ . Therefore, A k = S Λ k S - 1 = 1 - 1 1 1 1 0 0 1 5 k 1 / 2 1 / 2 - 1 / 2 1 / 2 = 1 - 1 1 1 1 2 1 2 - 1 2 · 5 k 1 2 · 5 k = 1 2 + 1 2 · 5 k 1 2 - 1 2 · 5 k 1 2 - 1 2 · 5 k 1 2 + 1 2 · 5 k . Hence, A k 1 / 2 1 / 2 1 / 2 1 / 2 as k → ∞ . Each column of this matrix is an eigenvector of A corresponding to the eigenvalue λ 1 = 1 of A . 3. Problem 5.2.32. Diagonalize A and compute S Λ k S - 1 to prove this formula for A k : A = 2 1 1 2 has A k = 1 2 3 k + 1 3 k - 1 3 k - 1 3 k + 1 . Answer: To diagonalize A , first find the eigenvalues: 0 = det( A - λI ) = 2 - λ 1 1 2 - λ = (2 - λ ) 2 - 1 = λ 2 - 4 λ + 3 = ( λ - 3)( λ - 1) , so the eigenvalues of λ 1 = 3 and λ 2 = 1. Hence, the eigenvector corresponding to λ 1 = 3 is the generator of the nullspace of A - 3 I = - 1 1 1 - 1 , namely v 1 = 1 1 . 2
Likewise, the eigenvector corresponding to λ 2 = 1 is the generator of the nullspace of A - I = 1 1 1 1 , namely v 2 = - 1 1 . Therefore, S and S - 1 are just as in Problem 2 above, so we have that A k = S Λ k S - 1 = 1 - 1 1 1 3 0 0 1 k 1 / 2 1 / 2 - 1 / 2 1 / 2 = 1 2 1 - 1 1 1 3 k 0 0 1 1 1 - 1 1 = 1 2 1 - 1 1 1 3 k 3 k - 1 1 = 1 2 3 k + 1 3 k - 1 3 k - 1 3 k + 1 , as desired. 4. Problem 5.2.34. Suppose that A = S Λ S - 1 . Take determinants to prove that det A = λ 1 · λ 2 · · · λ n = product of the λ s . This quick proof only works when A is . Proof. Since the determinant of a product is the product of the determinants, we know that det A = det( S Λ S - 1 ) = (det S )(det Λ)(det S - 1 ) .

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hw10solutions - Math 215 HW#10 Solutions 1 Problem 5.2.14...

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