chapter4solutions

chapter4solutions - Math 152 Chapter 4 solutions 1 Problem...

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Unformatted text preview: Math 152 Chapter 4 solutions 1 Problem 4.1: Show that 1 1 and 1- 1 are eigenvectors for the matrix 1 1 1 1 . What are the corresponding eigenvalues? Answer: We compute 1 1 1 1 1 1 = 2 2 = 2 1 1 , so 1 1 is an eigenvector with eigenvalue 2. Also 1 1 1 1 1- 1 = = 0 1- 1 , so 1- 1 is an eigenvector with eigenvalue 0. Problem 4.2: Suppose P is a projection matrix. What are the eigenvalues and eigenvectors of P ? Answer: If P projects onto some line, then a vector x lying on that line doesn’t get changed by P so P x = x and x is an eigenvector with eigenvalue 1. On the other hand, if x is perpendicular to the line, then P x = = 0 x so x is an eigenvector with eigenvalue 0. Problem 4.3: Find the eigenvalues and eigenvectors for a ) 3 3 b )- 2- 8 4 10 c ) 29- 10 105- 36 d )- 9- 14 7 12 Answer: a) det( A- λI ) = λ 2- 9 = ( λ- 3)( λ + 3) so the eigenvalues are λ 1 = 3 and λ 2 =- 3. To find the eigenvector for λ 1 = 3 we must solve the homegeneous equation with matrix- 3 3 3- 3 . The matrix reduces to- 3 3 and the eigenvector is x 1 = 1 1 . To find the eigenvector for λ 2 =- 3 we must solve the homegeneous equation with matrix 3 3 3 3 . The matrix reduces to 3 3 and the eigenvector is x 2 = 1- 1 . b) det( A- λI ) = λ 2- 8 λ + 12 = ( λ- 2)( λ- 6) so the eigenvalues are λ 1 = 3 and λ 2 =- 3. To find the eigenvector for λ 1 = 2 we must solve the homegeneous equation with matrix- 4- 8 4 8 . The matrix reduces to- 4- 8 and the eigenvector is x 1 = 2- 1 . To find the eigenvector for λ 2 = 6 we must solve the homegeneous equation with matrix- 8- 8 4 4 . The matrix reduces to- 8- 8 and the eigenvector is x 2 = 1- 1 . c) det( A- λI ) = λ 2 + 7 λ + 6 = ( λ + 6)( λ + 1) so the eigenvalues are λ 1 =- 6 and λ 2 =- 1. To find the eigenvector for λ 1 =- 6 we must solve the homegeneous equation with matrix 35- 10 105- 30 . The matrix reduces to 35- 10 and the eigenvector is x 1 = 2 7 . To find the eigenvector for λ 2 =- 1 we must solve the homegeneous equation with matrix 30- 10 105- 35 . The matrix reduces to 30- 10 and the eigenvector is x 2 = 1 3 . 2 Math 152 Chapter 4 solutions d) det( A- λI ) = λ 2- 3 λ- 10 = ( λ + 2)( λ- 5) so the eigenvalues are λ 1 =- 2 and λ 2 = 5. To find the eigenvector for λ 1 =- 2 we must solve the homegeneous equation with matrix- 7- 14 7 14 . The matrix reduces to- 7- 14 and the eigenvector is x 1 = 2- 1 . To find the eigenvector for λ 2 = 5 we must solve the homegeneous equation with matrix- 14- 14 7 7 . The matrix reduces to- 14- 14 and the eigenvector is x 2 = 1- 1 . Problem 4.4: Find the eigenvalues and eigenvectors for a ) - 1 1 1 2 2 2 b ) 1 1 1 1- 2 1- 1 1 c ) 7- 9- 15 4 3- 9- 11 d ) 31- 100 70 18- 59 42 12- 40 29 Answer: a) det( A- λI ) =- λ 3 + 2 λ 2 + λ- 2 =- ( λ- 1)( λ- 2)( λ + 1) so the eigenvalues are λ 1 = 1 and λ 2 = 2 and λ 3 =- 1....
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This note was uploaded on 04/19/2008 for the course MATH 152 taught by Professor Caddmen during the Spring '08 term at UBC.

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chapter4solutions - Math 152 Chapter 4 solutions 1 Problem...

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