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Homework Help: Find Eigenvalue/Eigenvector

  1. Apr 22, 2013 #1
    1. The problem statement, all variables and given/known data

    If invertible matrices A and B have eigenvalues α and β resp. wrt to a common eigenvector x, what is the eigenvalue of

    4A3B-4-17AB2 + ∏I

    (We'll call this equation (1))

    2. Relevant equations

    Ax = αx
    Bx = βx

    3. The attempt at a solution

    I think I'm ok for the actual eigenvalue. Basically we "exchange" α for A and β for B in the equation above, since Ax = αx implies A2x = α2x and so on and so forth. ∏I is just an identity matrix multiplied by ∏, so it's eigenvalue is ∏ (with multiplicity two).

    So I get 4α24 - 17αβ2 + ∏

    But I can't seem to figure out how to approach getting the corresponding Eigenvector. I know the characteristic equation

    det(A - λI) = 0

    may tell me something, but if I plug in (1) for A and the eigenvalue above for λ I just get a complete mess.

    -Dave K
  2. jcsd
  3. Apr 22, 2013 #2


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    You can't really find the eigenvector unless you have actual values for the A and B matrices. Do you?
  4. Apr 22, 2013 #3

    I like Serena

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    Hi Dave!

    Let's call C = 4A3B-4-17AB2 + ∏I
    Then, as you said, Cx = (4α24 - 17αβ2 + ∏)x

    Let's call λ=4α24 - 17αβ2 + ∏.
    Then Cxx.

    You had already found the eigenvalue λ.
    What do you think the corresponding eigenvector is? :wink:
  5. Apr 23, 2013 #4
    I did try that but I'm not sure how to "solve" for x. As the other poster pointed out, I don't even know what the values of C are... I don't even know what dimension so I can't use some identity matrix and use λ (the mess) as a scalar... Not sure what I'm supposed to do.
  6. Apr 23, 2013 #5


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    You can't solve for x any more than you can solve for α and β. x is the eigenvector. I think you are done.
  7. Apr 23, 2013 #6

    I like Serena

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    The vector x is given as the common eigenvector of A and B.
    None of them is specified other than that they are given to exist.
    C has the same common eigenvector x as A and B.
  8. Apr 24, 2013 #7
    Yeah, we went over this in class, and the eigenvector is just x. Ok then. Thanks everyone.
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