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Math Help - Is there a way to find matrix from characteristic polynomial?

  1. #1
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    Is there a way to find matrix from characteristic polynomial?

    i.e., p(lambda) = (1 - lamba)^3. Find A^25.
    It was kind of my instinct to choose [1, 0, 0; 0, 1, 0; 0, 0, 1] as my answer, but I'm not sure the proper way of going about it.
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  2. #2
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    Re: Is there a way to find matrix from characteristic polynomial?

    Hey biggerleaffan.

    You need to know the eigen-vectors in addition to the eigen-values to know the original matrix.

    The eigen-values give the information about how the matrix scales a vector with respect to some eigen-value but it doesn't give the direction.
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  3. #3
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    Re: Is there a way to find matrix from characteristic polynomial?

    Besides the size of the matrix (3x3), no other information was given.
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  4. #4
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    Re: Is there a way to find matrix from characteristic polynomial?

    If the matrix A were diagonalizable then we could write A= B^{-1}DB where D is the diagonal matrix having the eigenvalues of A along its diagonal and B is the matrix havig the eigenvectors of A as columns. That is why we cannot determine A itself from its eigenvalues alone- we need to know the corresponding eigenvectors also.

    However, whatever B is, in that case we can write A^{25}= (B^{-1}DB)^{25}= (B^{-1}AB)(B^{-1}AB)...(B^{-1}AB)= B^{-1}D^{25}B which, if D is the identity matrix, which is certainly possible since the three eigenvalues are "1", then A^{25}= B^{-1}IB= B^{-1}B= I.

    However this is, as I said, "If the matrix A were diagonalizable". Since 1 is a triple eigenvalue it is possible that there do NOT exist three independent eigenvectors. The "eigenspace" may be a subspace of dimension 1 or a subspace of dimension 2 in which we cannot diagonalize A. The best we could do would be the "Jordan Normal Form"

    \begin{bmatrix}1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 &  0 & 1\end{bmatrix}
    if the eigenspace is of dimension 1 or

    \begin{bmatrix}1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}
    if the eigenspace is of dimension 2.

    And finding powers of those is much harder.

    Added: Well, not that much harder! A little calculation shows that
    \begin{bmatrix}1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix}^n= \begin{bmatrix}1 & n & 1 \\ 0 & 1 & n \\ 0 & 0 & 1\end{bmatrix}
    and
    \begin{bmatrix}1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}^n= \begin{bmatrix}1 & n & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}

    Of course you would still have the problem that, since these are NOT the identity matrix, B and B^{-1} will not cancel so the A^n will still depend on the eigenvectors.
    Last edited by HallsofIvy; June 6th 2013 at 05:37 AM.
    Thanks from topsquark
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  5. #5
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    Re: Is there a way to find matrix from characteristic polynomial?

    Makes sense, thank you!
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