Re: matrix inverse of the sum of two inverse matrices

From: Dave Rusin (rusin_at_vesuvius.math.niu.edu)
Date: 12/21/04 

Date: 21 Dec 2004 17:02:04 GMT

In article <1103477827.788120.263890@z14g2000cwz.googlegroups.com>,
 <arenart@umh.es> wrote:

>-The most general problem I need to solve is to find the inverse of the
>sum of the inverses of these 2 matrices.

How do you "find" the inverse of 1/a+1/b where a and b are real numbers?

>-In a slightly more restricted problem, one of the matrices is diagonal
>in 'real' space, and the other
>is diagonal in frequency space (i.e., depends only on |i-j|, where i,j
>are indices over real space.

That's an interesting restriction, but even in the 3x3 case it does not
seem to make the problem much less complicated than inverting a general
3x3 matrix. (In higher dimensions, of course, there has to be _some_
simplification -- your sums lie in a 2N-dimensional subspace of an
N^2-dimensional space.)

>-In an even more restricted case (this is the one where I really hope
>to get some help), all the elements
>of one of the two matrices are multiplied by a very small number, i.e.,
>Inv(C) = a_i*delta_ij + epsilon*b_ij
>I can assume b_ij=f(|i-j|), so this matrix is diagonal in frequency
>space, epsilon is very small and I would like to know C. There must be
>some kind of power expansion one can do...

Well, sure. You want to invert M = A + B with A diagonal; it
suffices to be able to invert M' = A^{-1} M = I + A^{-1} B, and that
one can do with a power-series expansion: (I+X)^{-1} = I - X + X^2 - ...
In your case X = A^{-1} B will be small because B is, so this series
will converge ("quickly", since epsilon is "very small"). Of course
this X does not have the special structures of either A or B.
(As noted above, it's not a _completely_ structureless matrix, but I
don't think you can get much mileage out of the structure that it does have.)

I'm guessing these remarks are not very helpful to you but I am not
optimistic that there _is_ a good answer here.

dave

Relevant Pages

  • Re: abundance of irrationals!)
    ... over all n> m is an infinite sum. ... but for each n you must calculate the partial sum Sfrom 1 to ... special epsilon. ... about n is true for all n, given that arbitrary epsilon. ...
    (sci.math)
  • Re: Can someone please tell me where I am going wrong??
    ... in the interval, the program then gives the natural logarithm of ... int main{ ... float sum, term; ... double a, b, c, epsilon; ...
    (comp.lang.c)
  • Re: Continuos functions from [0,1] to R
    ... the sum of a finite number of bounded functions is bounded. ... typo.. ... delta <= epsilon provided delta>0 ...
    (sci.math)
  • Re: abundance of irrationals!)
    ... >>> Because for every positive real epsilon, there is a positive integer m ... but for each n you must calculate the partial sum Sfrom 1 to ... > The finite sums all exist, but there is no need for any infinite sum. ...
    (sci.math)