2130_Chapter7_Notes

Two independent eigenvectors v1 and v2 x1 v1 et and x2

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Unformatted text preview: nd Sol (slide 3) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exa (solution) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Summary of Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 2 3 4 5 6 7 8 9 Introduction Remember: x′ = 011 101 110 x. det(A − λI) = (2 − λ)(λ + 1)2 = 0 =⇒ λ = 2, −1 (repeated) λ = −1 =⇒ A − λI = A + I = 111 111 111 −→ 111 000 000 =⇒ v1 + v2 + v3 = 0 =⇒ v3 = −v1 − v2 =⇒ v = c1 c2 −c 1 −c 2 = c1 1 0 −1 +c2 0 1 −1 , (c1 or c2 = 0) v(2) v(1) The repeated eigenvalue λ = −1 has two linearly independent eigenvectors: v(1) and v(2) . If this happens, you can construct two linearly independent solutions: x(1)...
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This note was uploaded on 03/19/2014 for the course APMA 2130 taught by Professor Bernardfulgham during the Fall '09 term at UVA.

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