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Math Help - Eigenvector and eigenvalue. Need help please.

  1. #1
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    Eigenvector and eigenvalue. Need help please.

    Let A be an n x n symmetric matrix, and let λ be an eigenvalue of A with corresponding eigenvector x.

    Show that λ is also an eigenvalue of the matrix P^T AP where P is an orthogonal matrix. State the corresponding eigenvector of P^T AP.


    How is the result modified if P^T AP is diagonal matrix D?
    Last edited by mr fantastic; December 8th 2009 at 01:31 AM. Reason: Edited post title
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  2. #2
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    Could you show it using the determinate,i.e.,
    \det(P^{T}AP) = \det(P^TP)\det(A) = \det(I)\det(A) = \det(A).
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  3. #3
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    I am unclear of this, could you explain further please?
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  4. #4
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    Well, the eigenvalues of a matrix are given by \det(\lambda I - A) = 0, but
    P^T(\lambda I - A)P = (\lambda P^TP - P^{T}AP) = (\lambda I - P^{T}AP). However, we can prove that \det(\lambda I - P^{T}AP) = \det(P^T(\lambda I - A)P) = \det(\lambda I - A) by the same reasoning above. Since these have the same solution, i.e., \lambda, then the eigenvalues of P^{T}AP\text{ and } A are the same. Is that a bit clearer? It is not a formal proof, but that should be left up to you to write out.
    Last edited by lvleph; December 6th 2009 at 05:00 PM.
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  5. #5
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    Thanks, but do you have any idea of how the the result is modified if P^T AP is a diagonal matrix D?
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  6. #6
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    The only thing I can think of is that the eigenvalues will be the diagonal.
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  7. #7
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    I am sorry but your above explanation is unclear to me still. Can you tell me how i can write this out as a proper answer. I am very weak with matrices so find it hard to follow.

    Thank you very much for your help.
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