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

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Math 205, Spring 2011 Homework 10: due April 4 Chapter 5, Section 8 and Chapter 6, Sections 1-2.
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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 λ of an n -by- n matrix A if Avectorv = λvectorv. We re-write the vector equation as ( A λI n ) vectorv = vector 0 , which is a homogeneous system with coef matrix ( A λI ) , and we want λ so that the system has a non-trivial solution. We see that the eigenvalues are the roots of the characteristic polynomial P ( λ ) = 0 , where P ( λ ) = det( A λI ) . To find the eigenvectors we find the distinct roots λ = λ i , and for each i solve ( A λ i I ) vectorv = vector 0 . ————— Problem 1. Find the eigenvalues and eigenvectors of A = parenleftbigg 3 1 5 1 parenrightbigg Solution. P ( λ ) = det ( A λI 2 ) = vextendsingle vextendsingle vextendsingle vextendsingle 3 λ 1 5 1 λ vextendsingle vextendsingle vextendsingle vextendsingle = ( 1 λ )(3 λ ) 5 = λ 2 2 λ 3 5 = λ 2 2 λ 8 = ( λ 4)( λ + 2) , so λ 1 = 4 , λ 2 = 2 . For λ 1 = 4 , we reduce the coef matrix of the system ( A 4 I ) vectorx = vector 0 ,
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3 A 4 I = parenleftbigg 3 4 1 5 1 4 parenrightbigg = parenleftbigg 1 1 5 5 parenrightbigg parenleftbigg 1 1 0 0 parenrightbigg , so for x 2 = a we get x 1 = a, and the eigenvectors for λ 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 λ 2 = 2 , we start over with coef A + 2 I = parenleftbigg 3 + 2 1 5 1 + 2 parenrightbigg parenleftbigg 5 1 0 0 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 0 0 2 parenrightbigg . Problem 2. Find the eigenvalues and eigenvectors of A = 10 12 8 0 2 0 8 12 6 . Solution. Since A is 3-by-3 , the characteristic polynomial P ( λ ) is cubic, . . . . . . expansion on the 2nd row gives: we get P ( λ ) = (2 λ ) vextendsingle vextendsingle vextendsingle vextendsingle 10 λ 8 8 6 λ vextendsingle vextendsingle vextendsingle vextendsingle = (2 λ )[(10 λ )( 6 λ ) + 64] = (2 λ )[ λ 2 4 λ 60 + 64] = ( λ 2) 3 .
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4 Anyway, this matrix only has one distinct eigenvalue, λ 1 = 2 . To find the eigenvectors, we reduce the coef matrix A 2 I = 10 2 12 8 0 2 2 0 8 12 6 2 2 3 2 0 0 0 0 0 0 .
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