5.1 (3,4,8,11)

# 5.1 (3,4,8,11) - Linear Algebra-115 Solutions to Tenth...

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Linear Algebra -115 Solutions to Tenth Homework Problem 3( d ) (Section 5 . 1) We have det ( A - λI ) = det 2 - λ 0 - 1 4 1 - λ - 4 2 0 - 1 - λ = (1 - λ )[(2 - λ )( - 1 - λ ) + 2] = (1 - λ )( λ 2 - λ ) , which has solutions λ = 0 , 1 , 1. We compute E 0 ,E 1 . 2 0 - 1 4 1 - 4 2 0 - 1 2 0 - 1 0 1 - 2 0 0 0 . So, x y z E 0 iﬀ 2 0 - 1 0 1 - 2 0 0 0 x y z = 0 iﬀ 2 x - z = 0 , y - 2 z = 0 iﬀ 2 x = z, y = 2 z iﬀ x y z = x 1 4 2 . I.e. E 0 = span { (1 , 4 , 2) } . Also, 1 0 - 1 4 0 - 4 2 0 - 2 1 0 - 1 0 0 0 0 0 0 , which means that x y z E 1 iﬀ 1 0 - 1 0 0 0 0 0 0 x y z = 0 iﬀ x = z. I.e. E 1 = { x y x = y 0 1 0 + x 1 0 1 } = span { (1 , 0 , 1) , (0 , 1 , 0) } . Combining these two we get that β = { (1 , 4 , 2) , (1 , 0 , 1) , (0 , 1 , 0) } is a basis of eigenvectors. 1

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Then for D = 0 0 0 0 1 0 0 0 1 ,Q = 1 1 0 4 0 1 2 1 0 , it holds that Q - 1 AQ = D . Note: D is just a diagonal that contains the eigenvalues. Q contains the vectors of the β basis (vertically). ± Problem 4( e ) (Section 5 . 1) We need a basis of eigenvectors. Let
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5.1 (3,4,8,11) - Linear Algebra-115 Solutions to Tenth...

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