This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Math 205, Spring 2011 Homework 9: due March 28 Chapter 5, Sections 6 and 7 (5.5 is NOT on the syllabus) 2 Week 9: 5.6 Eigenvalues and Eigenvectors 5.7 Eigenspaces, nondefective matrices A vector vectorv negationslash = vector 0 in R n (or in C n ) is an eigenvector with eigenvalue of an nby n matrix A if Avectorv = vectorv. We rewrite the vector equation as ( A I n ) vectorv = vector , which is a homogeneous system with coef matrix ( A I ) , and we want so that the system has a nontrivial solution. We see that the eigenvalues are the roots of the characteristic polynomial P ( ) = 0 , where P ( ) = det( A I ) . To find the eigenvectors we find the distinct roots = i , and for each i solve ( A i I ) vectorv = vector . Problem 1. Find the eigenvalues and eigenvectors of A = parenleftbigg 3 1 5 1 parenrightbigg 3 Solution. P ( ) = det ( A I 2 ) = vextendsingle vextendsingle vextendsingle vextendsingle 3 1 5 1 vextendsingle vextendsingle vextendsingle vextendsingle = ( 1 )(3 ) 5 = 2 2 3 5 = 2 2 8 = ( 4)( + 2) , so 1 = 4 , 2 = 2 . For 1 = 4 , we reduce the coef matrix of the system ( A 4 I ) vectorx = vector , A 4 I = parenleftbigg 3 4 1 5 1 4 parenrightbigg = parenleftbigg 1 1 5 5 parenrightbigg parenleftbigg 1 1 parenrightbigg , so for x 2 = a we get x 1 = a, and the eigenvectors for 1 = 4 are the vectors ( x 1 , x 2 ) = a ( 1 , 1) with a negationslash = 0 , from the space with basis ( 1 , 1) ....
View
Full
Document
This note was uploaded on 11/24/2011 for the course MATH 205 taught by Professor Zhang during the Spring '08 term at Lehigh University .
 Spring '08
 zhang
 Eigenvectors, Vectors, Matrices

Click to edit the document details