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SVD vs Eigenvalue Decompositon (Diagonalizability)

  1. Jun 15, 2011 #1
    Okay, I know that if I can't get n linearly independent eigenvectors out of a matrix A (∈ℝnxn), it is not diagonalizable
    (and that some necessary conditions for diagonalizability in this regard may be being symmetric and/or having distinct eigenvalues.)
    This is how things are for the usual eigenvalue decomposition (A=SΛS-1), right?

    However, if I am not mistaken, there is not such a restriction present on a matrix for being able to perform the SVD decomposition (A=UΣVT) on it. I mean, every matrix, irrespective of the state of its eigenvectors, can be decomposed into a three parts, one diagonal, two orthogonal, right? So doesn't this mean that every matrix is diagonizable, regardless of its eigenvectors?


    I know that the answer to the question is no but I don't understand why we don't consider Σ to be a diagonalized form of A.
    Please correct me whereever I am mistaken..
    Thanks again.. :)
    Last edited: Jun 15, 2011
  2. jcsd
  3. Jun 16, 2011 #2


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    U and V is different generally, so cannot be consider diagonalizable.
    Also, The diagonal entries of SVD is the square root of eigenvalue of A*A'(A'*A), so not the eigenvalue of A itself. They are the same iff A is normal.
  4. Jun 16, 2011 #3


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    Those are sufficient conditions, not necessary conditions.

  5. Jun 19, 2011 #4


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    sorry... should be positive semidefinite.
    Last edited: Jun 19, 2011
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