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Show that this orthogonal diagonalization is a singular value decomposition.

  1. Nov 22, 2009 #1
    1. The problem statement, all variables and given/known data

    Prove that if A is an nxn positive definite symmetric matrix, then an orthogonal diagonalization A = PDP' is a singular value decomposition. (where P' = transpose(P))


    2. The attempt at a solution.

    I really don't know how to start this problem off. I know that the singular value decomposition is of the form A = UEV' where E will be an nxn matrix containing the singular values of A, and in this case the eigenvalues of A as well. But that's about it. Any help would be greatly appreciated!
     
  2. jcsd
  3. Nov 22, 2009 #2

    lanedance

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    is A a real matrix? if A is symmetric, how are the eigenvectors of A related to that of A^T
     
  4. Nov 22, 2009 #3
    Well, A has an orthonormal set of n eigenvectors, which would therefore be the same as A^T, but I don't know how to use this in the proof.
     
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