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Explain how any square matrix A can be written as

  1. Nov 29, 2012 #1
    1. The problem statement, all variables and given/known data
    a) Explain how any square matrix A can be written as

    [itex] A = QS [/itex]
    where Q is orthogonal and S is symmetric positive semidefinite.

    b) Is it possible to write

    [itex] A = S_1 Q_1 [/itex]
    Where Q1 is orthogonal and S1 is symmetric positive definite?




    2. Relevant equations

    [itex] A = U \Sigma V^T [/itex]

    3. The attempt at a solution

    For a) I've gotten to the point where I've written:

    [itex] A = U V^T V \Sigma V^T [/itex]

    Which is just a rearrangement of the single value decomposition. From this I believe there is some logic as to why U V^T is orthogonal, and VΣV^T is symmetric positive definite, but I can't seem to figure out the reasoning. Any pointers?

    For b) I've surmised this is possible given that it is simply the "left polar decomposition" (http://en.wikipedia.org/wiki/Polar_decomposition) But again, I can't think about how to show this mathematically.
     
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  3. Nov 29, 2012 #2

    Dick

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    Try and show the product of two orthogonal transformations is orthogonal. What's the definition of orthogonal? Ditto for the second part. What's the definition of positive semidefinite? Definitions are a great place to start.
     
    Last edited: Nov 29, 2012
  4. Nov 29, 2012 #3
    Okay well can I start by saying U and V^T are both orthogonal? I think they are, right? If so, does this work?:
    [itex](UV^T)^T=VU^T [/itex]
    [itex](UV^T)((UV^T)^T) = UV^T VU^T = UIU^T=UU^T=I[/itex]

    I'm not sure what to do about a good definition for positive semidefinite :/. The only one I know is that the eigenvalues are all 0 or positive. How can I use that on the second series of matrices?
     
  5. Nov 29, 2012 #4

    Dick

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    Yes, that shows your first factor is orthogonal. For the second one the definition I'm thinking of is that a matrix Q is positive semidefinite if x^TQx>=0 for all vectors x.
     
  6. Nov 29, 2012 #5
    So I have to show that [itex] x^T(V\Sigma V^T x) \geq 0[/itex]?
    hmm... I'm not sure how to show this.
    I know [itex]\Sigma[/itex] has all diagonal entries... and I know that [itex]V, V^T[/itex] are orthogonal. I can't seem to think of how that helps me though :/
     
  7. Nov 29, 2012 #6

    Dick

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    The singular value decomposition tells you can pick [itex]\Sigma[/itex] to be positive semidefinite, since it has all nonnegative entries along the diagonal. Try to use the definition I gave you.
     
  8. Nov 30, 2012 #7
    Oh, right. So it would become a matter of showing that the [itex]V\Sigma V^T [/itex] yields a matrix that's also positive semidefinite?
     
  9. Nov 30, 2012 #8

    Dick

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    That would be correct.
     
  10. Nov 30, 2012 #9
    Well I don't know how to show this, but I have a feeling that the product of an orthogonal matrix and a positive semidefinite matrix is positive semi definite... But alas I can't think of a way to reason this out. I'm no good at these matrix manipulation proofs it seems.
     
  11. Nov 30, 2012 #10

    Dick

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    You want to show x^T(VΣV^T)x≥0. Regroup that. I think you can do it. (V^T)x is just 'any vector'.
     
  12. Nov 30, 2012 #11
    OH wait I think I may have something...
    I know that [itex]V\Sigma V^T = \Sigma[/itex] (I can't figure how to show this though...)
    So that makes my inequality [itex]x^T(\Sigma x) \geq 0 [/itex]
    Was that at least one of the right steps? Though I can't justify the first part...
     
  13. Nov 30, 2012 #12

    Dick

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    No, they aren't equal. But if [itex]x^T(\Sigma x) \geq 0 [/itex] is true for ANY x, then it must also be true if you replace x with [itex]V^T x[/itex], right?
     
  14. Nov 30, 2012 #13

    Ray Vickson

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    The _definition_ of positive semidefinite is that x^T A x >= 0 for any x in R^n. One *consequence* of that is: a symmetric matrix is positive semidefinite if and only if its eigenvalues are >= 0. This is a theorem, not a definition.
     
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