Engineering Calculus Notes 393

Engineering Calculus Notes 393 - 2 − 3 v 3 = 0 2 v 1 −...

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3.9. THE PRINCIPAL AXIS THEOREM 381 so v 1 + v 3 = 0 or v 3 = v 1 and then v 2 = 2 v 1 + v 3 = 2 v 1 v 1 = v 1 . Setting v 1 = 1 leads to −→ v = (1 , 1 , 1) and the unit eigenvector −→ u 1 = 1 3 (1 , 1 , 1) . The eigenvectors for λ = 4 must satisfy 4 v 1 2 v 2 +2 v 3 = 4 v 1 2 v 1 v 2 3 v 3 = 4 v 2 2 v 1 3 v 2 v 3 = 4 v 3 or 8 v 1 2 v 2 +2 v 3 = 0 2 v 1 +3 v
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Unformatted text preview: 2 − 3 v 3 = 0 2 v 1 − 3 v 2 +3 v 3 = 0 . The second and third equations are (essentially) the same; the Frst divided by 2 is 4 v 1 − v 2 + v 3 = 0 so v 2 = 4 v 1 + v 3 ....
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