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Math Help - [SOLVED] Generalizing the n-th power of a matrix

  1. #1
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    [SOLVED] Generalizing the n-th power of a matrix

    Question here, I'm interested in generalizing the n-th power of a matrix P, P^n. The vector in question is P=\left(\begin{array}{ccc}1&0&0\\0.2&0.5&0.3\\0&0.  7&0.3\end{array}\right)

    I've found two approaches to doing this, z-transform and Diagonalization but I'm getting conflicting answers with both. What answer are you getting? also, is there a program that can calculate this for any matrix (perhaps a project I could pick up later if not)? Thanks
    Last edited by bidii; May 7th 2010 at 03:40 AM.
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  2. #2
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    Quote Originally Posted by bidii View Post
    Question here, I'm interested in generalizing the n-th power of a vector P, P^n. The vector in question is P=\left(\begin{array}{ccc}1&0&0\\0.2&0.5&0.3\\0&0.  7&0.3\end{array}\right)

    I've found two approaches to doing this, z-transform and Diagonalization but I'm getting conflicting answers with both. What answer are you getting? also, is there a program that can calculate this for any matrix (perhaps a project I could pick up later if not)? Thanks
    Your language is a bit confusing. You titled this "Generalizing the n-th power of a matrix" but then talk about a vector- except that the example you show is a matrix! And why talk about "generalizing the n-th power"? What you are really asking about is calculating the n-th power of a matrix, there is no "generalization" involved.

    I don't know what you mean by using the "z-transform" to find powers but diagonalizing certainly will work. Since the matrix you give it has three distinct eigenvalues and can be diagonalized.
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    Quote Originally Posted by HallsofIvy View Post
    Your language is a bit confusing. You titled this "Generalizing the n-th power of a matrix" but then talk about a vector- except that the example you show is a matrix! And why talk about "generalizing the n-th power"? What you are really asking about is calculating the n-th power of a matrix, there is no "generalization" involved.

    I don't know what you mean by using the "z-transform" to find powers but diagonalizing certainly will work. Since the matrix you give it has three distinct eigenvalues and can be diagonalized.
    Oh, I was a bit sleepy, meant "matrix" there. I also meant the use of z-transforms, not "using the z-transform". By "generalizing" i meant having a solution that can be used to calculate the nth power of the matrix, given any n.
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    I am not sure what you mean by z transform but if you do as HallsofIvy suggested, you can diagonalize the matrix.

    A^n=XD^nX^{-1}

    X is the matrix of the eigenspace of the eigenvectors, D is the diagonal matrix with corresponding eigenvalues, and X^{-1} is self-explanatory.
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    Quote Originally Posted by dwsmith View Post
    I am not sure what you mean by z transform but if you do as HallsofIvy suggested, you can diagonalize the matrix.

    A^n=XD^nX^{-1}

    X is the matrix of the eigenspace of the eigenvectors, D is the diagonal matrix with corresponding eigenvalues, and X^{-1} is self-explanatory.
    I was referring to a method detailed in Kleinrock's Queueing Theory book, volume one, which used z-transforms. I chose the Diagonalization method, though.
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    I don't have that books so I am not sure what that method is.
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    No I was just mentioning my source, I wasn't expecting you to be familiar with the approach
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