This preview shows pages 1–5. 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 10: due April 4 Chapter 5, Section 8 and Chapter 6, Sections 12. 2 Week 10: 5.8 Diagonalization (Mon) 6.1, 6.2 Second Order Linear DE (Fri) [reprint, week09: Eigenvalues and Eigenvectors] + diagonaliization 1. 5.8 Eigenspaces, Diagonalization 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 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 , 3 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 vectorv 1 = ( 1 , 1) . For 2 = 2 , we start over with coef A + 2 I = parenleftbigg 3 + 2 1 5 1 + 2 parenrightbigg parenleftbigg 5 1 parenrightbigg . To find a spanning set without fractions, we anticipate that finding x 1 will use division by 5, and take x 2 = 5 b. Then 5 x 1 x 2 = 0 gives 5 x 1 = 5 b, so x 1 = b, and ( x 1 , x 2 ) = b (1 , 5) , spanned by vectorv 2 = (1 , 5) [or, if you must, ( 1 5 , 1)]. Example 1 (diagonalization): We write vectorv 1 ,vectorv 2 as columns, then S 1 AS = D with S = parenleftbigg 1 1 1 5 parenrightbigg and D = diag (4 , 2) = parenleftbigg 4 2 parenrightbigg . Problem 2. Find the eigenvalues and eigenvectors of A = 10 12 8 2 8 12 6 . Solution. Since A is 3by3 , the characteristic polynomial P ( ) is cubic, . . . . . . expansion on the 2nd row gives: we get P ( ) = (2 ) vextendsingle vextendsingle vextendsingle vextendsingle 10 8 8 6 vextendsingle vextendsingle vextendsingle vextendsingle = (2 )[(10 )( 6 ) + 64] = (2 )[ 2 4 60 + 64] = ( 2) 3 . 4 Anyway, this matrix only has one distinct eigenvalue, 1 = 2...
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

Click to edit the document details