hw10solutions

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

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Unformatted text preview: Math 215 HW #10 Solutions 1. Problem 5.2.14. Suppose the eigenvector matrix S has S T = S- 1 . Show that A = S 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 | . . . | = h ~v 1 ,~v 1 i . . . h ~v 1 ,~v n i . . . . . . h ~v n ,~v 1 i . . . h ~v n ,~v n i , so we see that h ~v i ,~v j i = 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 S- 1 ) T = ( S S T ) T = ( S T ) T T S T = S S T = A since T = and ( S T ) T = S , so we see that A is symmetric. 2. Problem 5.2.30. Find and S to diagonalize A in Problem 29 ( A = . 6 . 4 . 4 . 6 ). What is the limit of k as k ? What is the limit of S 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- I ) = . 6- . 4 . 4 . 6- = ( . 6- ) 2- . 4 2 = 2- 1 . 2 + . 2 = ( - 1)( - . 2) , so the eigenvalues of A are 1 = 1 and 2 = . 2. The eigenvector associated to 1 = 1 is the generator of the nullspace of A- I =- . 4 . 4 . 4- . 4 , which row-reduces to- . 4 . 4 , so the nullspace consists of multiples of ~v 1 = 1 1 . On the other hand, the eigenvector associated to 2 = . 2 is the generator of the nullspace of A- . 2 I = . 4 . 4 . 4 . 4 , which row-reduces to . 4 . 4 , so the nullspace consists of multiples of ~v 2 =- 1 1 . 1 Therefore, we see that A = S S- 1 where S = 1- 1 1 1 and = 1 . 2 . Since k = 1 . 2 k = 1 k ( . 2) k = 1 1 5 k , we see that k 1 0 0 0 as k . Therefore, A k = S k S- 1 = 1- 1 1 1 1 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 1 = 1 of A . 3. Problem 5.2.32. Diagonalize A and compute S 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- I ) = 2- 1 1 2- = (2- ) 2- 1 = 2- 4 + 3 = ( - 3)( - 1) , so the eigenvalues of 1 = 3 and 2 = 1. Hence, the eigenvector corresponding to 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 2 = 1 is the generator of the nullspace of A- I = 1 1 1 1 , namely ~v 2 =- 1 1 ....
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hw10solutions - Math 215 HW #10 Solutions 1. Problem...

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