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Math Help - speeding up inverse calculation

  1. #1
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    Feb 2010
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    inverse of matrix sums

    Hi

    First of all apologies if this is in the wrong forum. I 'm not sure if this problem counts as basic or advanced.

    I am running several iterations in which I calculate:

    Q(j)=(C+Ia(j))^{-1}

    Where C is a complex, positive-definite symmetrical matrix and is constant.
    I is the identity matrix and a(j) is a scalar which varies with each iteration i.

    Computing the inverse for each iteration is time consuming and i want to, if possible, find a way of computing the inverse of C one time only, and then use some operator F, such that:

    Q(j)=F(C^{-1},Ia(j)).

    I assume with a being scalar such as function will be a lot quicker.

    Does anyone know if there is a way of doing this?

    Many thanks

    Mark
    Last edited by mrkdsmith; February 10th 2010 at 03:19 PM.
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  2. #2
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    Joined
    Feb 2010
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    Ok.. after some digging i cam across a paper (Henderson, 1981) on deriving the inverse of a matrix sum:

    <br />
(C+SVD)^{-1}=C^{-1}-C^{-1}S(I+VDC^{-1}S)^{-1}VDC^{-1}<br />

    Which in my case simplifies to:

    (C+Ia(j))^{-1}=a(j)C^{-1}(I+a(j)C^{-1})^{-1}C^{-1}

    now i can calculate my inverse C just once and seems to be giving me the same results... although i'm sill having to calculate a different inverse in each iteration, so not a complete solution...

    I found another similar method (Miller, 1981) which seems to over come this problem but it requires one of the matrices to be rank 1, which neither C nor aI(j) is.

    so still a bit stuck.

    Thanks

    Mark
    Last edited by mrkdsmith; February 10th 2010 at 02:25 PM.
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