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feldy90

  • 3 years ago

How do I find the eigenvectors of the following matrix?

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  1. feldy90
    • 3 years ago
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    \[A=\left[\begin{matrix}1 & 2 \\ 2 & -2\end{matrix}\right]\]

  2. Coolsector
    • 3 years ago
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    \[Av = \lambda v\]

  3. hesam
    • 3 years ago
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    a=-6

  4. Coolsector
    • 3 years ago
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    first find eigenvalues :det(A - λI) = 0 λI - A = \[\left[\begin{matrix}1-λ & 2 \\ 2 & -2-λ\end{matrix}\right]\] so Det : (1−λ)(−2−λ) - 4 = 0 λ^2 - λ + 2λ -2 -4 = 0 λ^2 +λ -6 = 0 λ^2 +3λ - 2λ - 6 = 0 λ(λ+3) - 2(λ+3) (λ-2)(λ+3) = 0 λ=-3,2 now Av=λv Av = 2v v1+2v2 = 2v1 2v1 -2v2 = 2v2 any vector of the form : 2v2 = v1 Av = -3v v1 + 2v2 = -3v1 2v1 - 2v2 = -3v2 2v2 = -4v1 any vector of the form : v2 = -2v1

  5. Coolsector
    • 3 years ago
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    by the way .. the eigenvector cant be zero.

  6. Coolsector
    • 3 years ago
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    example for the eigenvectors are : 2v2 = v1 -> (2,1) v2 = -2v1 ->(-1,2)

  7. Coolsector
    • 3 years ago
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    please tell me if it is fine..

  8. feldy90
    • 3 years ago
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    Sorry - I have actually left my books for a moment... I will definitely check this when I get back to them - thanks so much!!

  9. feldy90
    • 3 years ago
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    Totally forgot about this! Thank you so much - that was perfect!

  10. Coolsector
    • 3 years ago
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    oh im glad that you answered eventually :)

  11. feldy90
    • 3 years ago
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    Haha, later is better than never, right? :P Quick question, for the eigenvalue -3, is the final eigenvector\[v=\left(\begin{matrix}-2 \\ 1\end{matrix}\right)\] Since \[v_2=-2v_1\]??

  12. Coolsector
    • 3 years ago
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    if you take v1 = 1 then v2 = -2 (1,-2) its the opposite from yours

  13. feldy90
    • 3 years ago
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    Oh I see, thanks!

  14. feldy90
    • 3 years ago
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    So would the eigenspace of eigenvalue 2, be: \[span \left\{ \left(\begin{matrix}2 \\ 0\end{matrix}\right) ,\left(\begin{matrix}0 \\ 1\end{matrix}\right)\right\}\] or just \[span \left\{ \left(\begin{matrix}2 \\ 1\end{matrix}\right) \right\}\] Thanks for the ongoing help :)

  15. feldy90
    • 3 years ago
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    Never mind, I got it :)

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