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# s11week09 - Math 205 Spring 2011 Homework 9 due March 28...

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Math 205, Spring 2011 Homework 9: due March 28 Chapter 5, Sections 6 and 7 (5.5 is NOT on the syllabus)

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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 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
3 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 , 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 ( 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 .

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s11week09 - Math 205 Spring 2011 Homework 9 due March 28...

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