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Math Help - [SOLVED] Det A and singular values

  1. #1
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    [SOLVED] Det A and singular values

    There is a rule that says:
     |det(A)| = \prod_{i=1}^m\sigma_i for an m x m matrix.
    where \sigma_i
    are singular values. It doesn't specify whether they are just singular values or non-zero singular values.

    Does this mean just the non-zero singular values?

    Because that means a determinant I'm working on is zero, not 12.
    can someone please tell me?

    Thanks
    Last edited by Bucephalus; April 13th 2010 at 03:14 AM. Reason: Forgot to add the singular values into the equation
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  2. #2
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    Quote Originally Posted by Bucephalus View Post
    There is a rule that says:
     |det(A)| = \prod_{i=1}^m for an mxm matrix.


    What did you write here?? The product...of what?

    Tonio



    Does this mean just the non-zero singular values?

    Because that means a determinant I'm working on is zero, not 12.
    can someone please tell me?

    Thanks
    .
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  3. #3
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    I edited it up above...

    Thanks.
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  4. #4
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    I just thought about it some more though.
    If for example, a 2 x 2 matrix had only one non-zero singular value it means that it's rank one, not full rank, which mean's it's not invertible, which means that it has a 0 determinant.
    So I'm thinking that you have to go with how the equation reads and include all singular values non-zero or not.

    Do you agree?
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  5. #5
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    Quote Originally Posted by Bucephalus View Post
    There is a rule that says:
     |det(A)| = \prod_{i=1}^m\sigma_i for an m x m matrix.
    where \sigma_i
    are singular values. It doesn't specify whether they are just singular values or non-zero singular values.

    Does this mean just the non-zero singular values?

    Because that means a determinant I'm working on is zero, not 12.
    can someone please tell me?

    Thanks

    Googling this stuff I found this SVD and LSI Tutorial 2: Computing Singular Values

    Look at 1 just before figure 6 (almost at the bottom): it seems to be the determinant is the product of all the singular values.

    Tonio
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  6. #6
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    Thanks for that

    Yes, I agree with you.
    Thanks for your help.
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  7. #7
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    "Singular values" are just the eigenvalues of the matrix. Any matrix is similar to a matrix in either diagonal form or Jordan normal form where the determinant is just the product of the values on the diagonal. And the values on the diagonal are the eigenvalues.

    And if a matrix has any 0 eigenvalue, then its determinant is 0.

    (If the determinant were the product of the "non-zero" singular values, then NO matrix would have 0 determinant!)
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  8. #8
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    thanks for your help.

    Thanks.
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  9. #9
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    Quote Originally Posted by HallsofIvy View Post
    "Singular values" are just the eigenvalues of the matrix.


    This isn't accurate: the singular values of a matrix A (in general of an operator in a Hilbert space that blah-blah) are the eigenvalues of the normal matrix (operator) A^{*}A which can be very different from the eigenvalues of A itself.

    That the determinant of any matrix is the product of its eigenvalues (in some field extension of the definition field) is basic stuff in linear algebra. Singular values, imo, aren't and, besides, belong a little more to functional analysis than to just linear algebra.

    Tonio


    Any matrix is similar to a matrix in either diagonal form or Jordan normal form where the determinant is just the product of the values on the diagonal. And the values on the diagonal are the eigenvalues.

    And if a matrix has any 0 eigenvalue, then its determinant is 0.

    (If the determinant were the product of the "non-zero" singular values, then NO matrix would have 0 determinant!)
    .
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  10. #10
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    Hey, yeah you're right.
    The singular values are the square roots of the eigenvalues of A^HA

    Good pick up Tonio.
    Thanks.
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