s11week10 - Math 205, Spring 2011 Homework 10: due April 4...

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Unformatted text preview: Math 205, Spring 2011 Homework 10: due April 4 Chapter 5, Section 8 and Chapter 6, Sections 1-2. 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 n-by- n matrix A if Avectorv = vectorv. We re-write 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 non-trivial 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 3-by-3 , 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...
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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 .

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s11week10 - Math 205, Spring 2011 Homework 10: due April 4...

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