M 340L CS Homework Set 12 Solutions. Note: Scale all eigenvectors so the largest component is +1.

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1 M 34L CS Homework Set 2 Solutions Note: Scale all eigenvectors so the largest component is +.. For each of these matrices, find the characteristic polynomial p( ) det( A I). factor it to get the eigenvalues:, 2,..., n. i i for i=,, n: find x the eigenvector corresponding.(that is, find a vector x in the nullspace of A I ). i i a. A The characteristic polynomial is ( )(2 ) 2 3 ( 5)( 2) so the / 4 eigenvalues are 5 and -2. The null space of A5I 4 3 is the vector (and its 3 3 multiples). The null space of A ( 2) I 4 4 is the vector (and its multiples). 3 / 4 Thus the eigenvectors are and. b. 8 2 A The characteristic polynomial is (8 )(9 )(3 ) ( 2)( 3)( 5) (8 )( 3)3 (3 )( 3)( 2) ( 8)( 2)( ) 2 so the eigenvalues are 8, 2, and. The null space of A8I 3 3 is the vector (and its multiples) since / 5 4 / 5. The null

2 4 2 space of A2I is the vector / 3 (and its multiples) since / The null space of A I is the vector 2/3 (and its multiples) since / / 2 3. Thus the eigenvectors are / 2 3 / 3, and 2/3. 2/3, 2. Two eigenvectors of an upper triangular matrix: a b Let U c be n by n and upper triangular. Assume a c. a. Show that the eigenvector corresponding to the eigenvalue a is e (i.e the first column of the n by n identity matrix). (Use this below.) a b a Ue c e ae. b. Show that the eigenvector corresponding to the eigenvalue c is ( c a) / b scaled version. below.) b / ( c a). (Use this or a

3 b / ( c a) b / ( c a) b / ( c a) / ( ) a b ab c a b U. c c c. 3. How do perturbations affect eigenvalues and eigenvectors? 2 a. Let A. 2 What are its eigenvalues and eigenvectors? (See the note above regarding the scaling of eigenvectors and make sure you do it throughout the homework.) The eigenvalues are 2 and 2. The null space of A2I is the vector (and its multiples). is only one linearly independent eigenvector. 2 b. Let B. 2 What are its eigenvalues and eigenvectors? (Your answers should be in terms of the perturbation parameter.) The eigenvalues are 2 and 2. The null space of B2I is the vector (and its multiples). The null space of B (2 ) I is the vector (and its multiples). Thus the eigenvectors are and. c. Describe the effect of the perturbation on eigenvalues and eigenvectors of A. Comment on the linear independence of the eigenvectors of B. The perturbation has introduced a second eigenvector but it is nearly linearly dependent upon the first. 2 d. Let C. 2 What are its eigenvalues and eigenvectors? (Choose linearly independent eigenvectors.)

4 The eigenvalues are 2 and 2. The null space of C2I is all of independent eigenvectors are and. 2. Two linearly e. Let 2 D. 2 What are its eigenvalues and eigenvectors? The eigenvalues are 2 and 2. The null space of D2I is the vector multiples). The only eigenvector is. (and its f. Describe the effect of the perturbation on eigenvalues and eigenvectors of C. The perturbation has left the eigenvalues unperturbed but has removed the second eigenvector. 4. Using the diagonal form to compute high powers: Let 2 A. 2 Feel free to express answers in parts c, d, and e using expressions involving powers. a. What are its eigenvalues and eigenvectors? (SeeProblem, if necessary.) 2 The characteristic polynomial is ( )( ) ( 3)( ) so the 2 2 eigenvalues are 3 and -. The null space of A3I 2 2 is the vector (and its 2 2 multiples). The null space of A ( ) I 2 2 is the vector (and its multiples). Thus the eigenvectors are and. b. Using part a., form D, a diagonal matrix of eigenvalues, form V whose columns are the associated eigenvectors, then compute V, and finally VDV. Compare VDV to A. / 2 / 2 Since, / 2 / / 2 / 2 A 2 / 2 / 2.

5 c. Using part b., what is clever way.) A y, for y 2? (Do not compute A - yet. Use associativity in a A y VD V y / 2 / 2 3 ( ) / 2 / 2 2 / 2 3 ( ) 3 / 2 3 / 2 3 / 2 (3 3 ) / 2 (3 3) / 2 d. Express your answer in part c as A y, where is such that the largest component of is +. Compare to the eigenvector corresponding to. 3 3 (3 3 ) / 2 (3 3) A y 3 3. The vector (3 3) / ) to the negative of the eigenvector. e. Using part b., what is A? is very close (within 3 / 2 / 2 A VD V ( ) / 2 / 2 3 / 2 3 / 2 / 2 / 2 ( 3 ) / 2 (3 ) / 2 (3 ) / 2 ( 3 ) / 2.

6 5. A Markov process: Repeat all five parts of Problem 4 with A 24 / 25 4 / 25 / 25 2/ 25 except in part c. use / 2 y / 2. 6% 96% A 4% a. What are its eigenvalues and eigenvectors? B 84% The characte3ristic polynomial is 2 2 (24 / 25 )(2/ 25 ) 4 / 25 9 / 5 4 / 5 ( )( 4 / 5) so the eigenvalues / 25 4 / 25 are and 4/5. The null space of AI. / 25 4 / 25 is the vector / 4 (and its 4 / 25 4 / 25 multiples). The null space of A4 / 5I / 25 / 25 is the vector (and its multiples). Thus the eigenvectors are / 4 and. b. Using part a. express A VDV (where the columns of V are the eigenvectors and D is a diagonal matrix containing the associated eigenvalues.) 4 / 5 4 / 5 Since, / 4 / 5 4 / 5 24 / 25 4 / 25 4 / 5 4 / 5 A / 25 2/ 25 / 4 4 / 5 / 5 4 / 5. / 2 c. Using part b., what is A y, for y / 2? (Do not compute A - yet.)

7 A y VD V y 4 / 5 4 / 5 / 2 / 4 (4 / 5) / 5 4 / 5 / 2 4 / 5 / 4 (4 / 5) 3 / 4 / 5 / 4 3 (4 / 5) / 4 / 5 3 (4 / 5) / / 5 3 (4 / 5) / d. Express your answer in part c as A y, where is such that the largest component of is +. Compare to the eigenvector corresponding to. 4 / 5 3 (4 / 5) / A y (4 / 5 3 (4 / 5) /) 5 (4 / 5) / / 5 3 (4 / 5) /. The 4 5 (4 / 5) / vector 5 (4 / 5) / is very close (within - ) to the eigenvector / (4 / 5) /. e. Using part b., what is A? 4 / 5 4 / 5 A VD V / 4 (4 / 5) / 5 4 / 5 4 / 5 4 / 5 / 4 (4 / 5) / 5 (4 / 5) 4 / 5 (4 / 5) / 5 4 / 5 (4 / 5) / 5 (4 / 5) / 5 / 5 (4 / 5). 6. All zero eigenvalues: Find a simple non-zero matrix having all zero eigenvalues. The matrix A is a non-zero matrix having all zero eigenvalues

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