This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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....
View
Full
Document
This note was uploaded on 04/19/2008 for the course MATH 152 taught by Professor Caddmen during the Spring '08 term at UBC.
 Spring '08
 Caddmen
 Eigenvectors, Vectors

Click to edit the document details