# Thread: speeding up inverse calculation

1. ## 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:

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

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:

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

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

2. Ok.. after some digging i cam across a paper (Henderson, 1981) on deriving the inverse of a matrix sum:

$\displaystyle (C+SVD)^{-1}=C^{-1}-C^{-1}S(I+VDC^{-1}S)^{-1}VDC^{-1}$

Which in my case simplifies to:

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

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