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Math Help - Any rank one matrix is diagonalizable?

  1. #1
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    Any rank one matrix is diagonalizable?

    How to show that any rank one matrix is diagonalizable?

    I've no idea how to start with.
    Last edited by wesleybrown; November 21st 2012 at 03:54 AM.
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  2. #2
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    Re: Any rank one matrix is diagonalizable?

    An n by n matrix is "diagonalizable" if and only if there exist a basis for R^n consisting entirely of eigenvectors of the matrix. If the matrix has "rank 1" then it has one non-zero eigenvalue. An eigenvector corresponding to that eigenvalue will span a one dimensional subspace. The matrix, applied to any vector in the n-1 dimensional orthogonal subspace, will give 0 so any basis for that subspace will be a basis consisting of eigenvectors corresponding to eigenvalue 0. Adding the first eigenvector to that will give a basis for the entire space consisting of eigenvectors.
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  3. #3
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    Re: Any rank one matrix is diagonalizable?

    Thank you,
    "If the matrix has "rank 1" then it has one non-zero eigenvalue."
    This is exactly the point where I stuck, can you explain more on this sentence?
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