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Math Help - approximating (I+Ab)^-1

  1. #1
    kpc
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    approximating (I+Ab)^-1

    I am sorry if this question is asked in the wrong section. It looks to me as if matrix theory is not educated pre-university in the US?

    Given A, a complex symmetric large matrix (N>>1000), with elements near the diagonal |r-k| < approx. 10. values decaying as we go further from the diagonal.

    Is it possible to symbolically write the following
    (I+A)^-1
    in a manner without the inverse?
    Could it be approximated using a series development?

    Edit: just forget about the b in the title, it is just a constant.
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  2. #2
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    Quote Originally Posted by kpc View Post
    I am sorry if this question is asked in the wrong section. It looks to me as if matrix theory is not educated pre-university in the US?

    Given A, a complex symmetric large matrix (N>>1000), with elements near the diagonal |r-k| < approx. 10. values decaying as we go further from the diagonal.

    Is it possible to symbolically write the following
    (I+A)^-1
    in a manner without the inverse?
    Could it be approximated using a series development?

    Edit: just forget about the b in the title, it is just a constant.
    You can rewrite (1+ A)^{-1} as \frac{1}{1- (-A)} which is the sum of the geometric series \sum (-A)^n= \sum (-1)^n A^n. Stopping that at a finite value of n should give an approximation.

    Some examples:
    n= 1, I- A
    n= 2, I- A+ A^2
    n= 3, I- A+ A^2- A^3
    are all approximations to (I+ A)^{-1}.
    Last edited by HallsofIvy; July 25th 2009 at 01:18 PM.
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  3. #3
    kpc
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    Sorry, I did not realize it was this simple. Almost too embarrassing.
    Thanks.
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  4. #4
    Member alunw's Avatar
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    Schultz's method?

    I don't think this can really be right except under very restricted circumstances. Surely, if the matrix has any eigenvalue with an absolute value > 1 then A^n will tend to have large entries, so the series will be divergent, just as it would be in the case of 1*1 matrices (i.e. ordinary numbers). So unless you are sure in advance that all the eigenvalues are bound to be small this technique won't work. However, there is a technique based on this idea called Schultz's method which will work if you can make a sufficiently good guess at the inverse. It is described in chapter 2 of the "Numerical recipes" books, where the authors comment "before you get too excited about it, note that it requires two full matrix multiplications at every iteration". For big complex matrices this is likely to be very slow unless you have access to a good linear algebra package, and if you do you can probably use that to invert the matrix anyway.
    Last edited by alunw; July 26th 2009 at 01:15 PM. Reason: typo corrected
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  5. #5
    kpc
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    @alunw,

    The restriction you mention applies here. For my final result to be correct within 1% using typical data, I only need the first three terms.
    When only using the first two terms, for typical data stays within 5%, so suficiently accurate for a qualitative analysis.
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