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Product of singular values = determinant proof

  1. Jul 11, 2011 #1

    Vai

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    1. The problem statement, all variables and given/known data

    So I'm working on this proof. Given an n x n (square) matrix, prove that it's determinant is equal to the product of it's singular values.

    2. Relevant equations

    We are given A = U*E*V as a singular value decomposition of A.

    3. The attempt at a solution

    I was thinking that det(A) = det(U) * det(E) * det(V)

    and since E is the diagonal matrix with singular values on it's diagonal, it's determinant is the product of those singular values.

    But then what to do about det(U) and det(V)? I guess it's logical that the product of their determinants is 1, but how do I show that?
     
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  3. Jul 11, 2011 #2

    micromass

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    Hi Vai! :smile:

    Do your singular values always need to be positive?? Does det(E) always need to be positive and det(A)??

    U and V are unitary, what do you know about the determinant of unitary matrices?
     
  4. Jul 11, 2011 #3

    Vai

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    Thanks for the quick response.

    I'm not sure about the sign on the singular values. Since they are the square roots of the eigenvalues of A' * A, then I assume that they are all positive. So then that means det(E) is also positive.

    I wasn't aware that U and V are unitary matrices. But your comment made me think, and according to the definition of the singular value decomposition, they are orthogonal matrices. So then I can say their determinants are +/- 1.

    I think that is enough then to show the det(A) is the product of the singular values since:

    det(A) = det(U) * det(E) * det(V)
    = (+/- 1) * (product of singular values) * (+/- 1)
     
    Last edited: Jul 11, 2011
  5. Jul 11, 2011 #4

    micromass

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    How can det(A) equal det(E) if det(E) is always positive, but if det(A) is not always positive??
     
  6. Jul 11, 2011 #5

    Vai

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    Whoops, I'm really sorry. I had to prove that the absolute value of the determinant of A is equal to the product of the singular values.

    then:

    |det(A)| = |det(U) * det(E) * det(V)|
    = | (+/- 1) * (product of singular values) * (+/- 1) |
    = product of singular values
     
  7. Jul 11, 2011 #6

    micromass

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    That is correct! :smile:
     
  8. Jul 11, 2011 #7

    Vai

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    Ok, thank you very much for your help; saved me a lot of time there.
     
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