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anonymous

  • 4 years ago

Can someone help me with eigenvalues and eigenvectors?

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  1. anonymous
    • 4 years ago
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  2. anonymous
    • 4 years ago
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    I need help with problem number 6

  3. anonymous
    • 4 years ago
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    Use: \[det(A-\lambda I) = 0\] to find the eigenvalues. For two by two matrices I think you will get two values of lambra (as a quadratic in lambda will come out).

  4. anonymous
    • 4 years ago
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    I basically need help with part C

  5. anonymous
    • 4 years ago
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    Lambda=0 and 4

  6. anonymous
    • 4 years ago
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    Is that correct?

  7. anonymous
    • 4 years ago
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    You have to find a vector X such that \[ AX=\lambda X \]

  8. TuringTest
    • 4 years ago
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    \[\det(A-I\lambda)=\det\left[\begin{matrix}2-\lambda & 2 \\ 2 & 2-\lambda\end{matrix}\right]=4-2\lambda+\lambda^2-4\]sorry gotta eat!

  9. anonymous
    • 4 years ago
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    LOL Go Eat:D

  10. anonymous
    • 4 years ago
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    For the corresponding eigenvectors, I think it's a case of finding X in \[A-\lambda I = X\] when you know the two values of lambda...

  11. anonymous
    • 4 years ago
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    Turing its the opposite

  12. TuringTest
    • 4 years ago
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    you have I*Iambda-A I'm guessing, it's the same it is the same

  13. anonymous
    • 4 years ago
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    oh sorry \[(A-\lambda I)X = 0\] (the ero matrix)

  14. anonymous
    • 4 years ago
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    zero*

  15. anonymous
    • 4 years ago
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    ohhh ya that is the one i have in my textbook

  16. anonymous
    • 4 years ago
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    Check out this example : http://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors#Worked_example

  17. anonymous
    • 4 years ago
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    Thanks I figured it out :D

  18. mathmate
    • 4 years ago
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    For \( \lambda=0 \) reduced matrix is \[\left[\begin{matrix}1 & 1 \\ 0 & 0\end{matrix}\right]\]which gives an eigenvector of (1,-1). For \( \lambda=4 \), the reduced matrix is\[\left[\begin{matrix}-1 & 1 \\ 0 & 0\end{matrix}\right]\]whose eigenvector is (1,1).

  19. anonymous
    • 4 years ago
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    Thanks Guys :D I really appreciate ur help

  20. mathmate
    • 4 years ago
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    You're welcome! :)

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