2130_Chapter7_Notes

K1 dene a v1 v2 v3 k k2 then k3 k1 v1 k2

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Unformatted text preview: − 3λ) = (1 − λ)λ(λ − 3) det(A − λI) = 0 =⇒ λ = 0, 1, 3. 9 / 13 5 Exa 3 (slide 2) Now find the eigenvectors by solving the equation (A − λI)v = 0 for λ = 0, 1, 3. In general, solve (A − λI)v = 0 by reducing the augmented matrix A − λI | 0 . Since row operations don’t affect column vector 0, we usually drop it from the calculations. Note: λ = 0 =⇒ A − λI = A =⇒ (A − λI)v = 0 ⇐⇒ Av = 0. This equation was solved in Example 2: λ1 = 0 =⇒ v(1) −1 = c 1 , 1 c = 0. 10 / 13 Exa 3 (slide 3) λ2 = 1 =⇒ A − λI = A − I = = 0 2 A − I = 2 0 1 −1 1 R −R2 −3− → 0 −− R1 + R2 0 1−1 2 −1 2 1−1 1 1 −1 2−1 0 2 −1 201 1 −1 1 −1 1 −1 1 R −2R3 1 −2 − → 0 −− − 2 −1 R1 ↔ R3 1 0 2 −1 1 0 2 −1 0 0 v1 + v2 = 0 v = − v2 =⇒ 1 2v2 − v3 = 0 v3 = 2v2 −c −1 (2) = c = c 1 , c = 0 =⇒ v 2c 2 =⇒ 11 / 13 6 Exa 3 (slide 4) λ3 = 3 =⇒ A − λI = A − 3I = = 1−3 2 −1 2 1−3 1 1 −1 2−3 −2 2 ...
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