0

I'm trying to figure out a way of seperating out my $n\times n$-dimensional matrix $A$ (to use in my Finite Difference scheme) from the following equation:

$(cA + I)^{-1}$,

where $c$ is a scalar and $I$ a $n\times n$ identity matrix.

Does anyone have a (reference to a) clue how to proceed and, if possible, under which assumptions a solution can be obtained?

Pim
  • 9
  • What exactly is your question? There is clear mismatch between what you say in the title and what you say later. – Kavi Rama Murthy Jul 22 '19 at 23:26
  • @Kavi Rama Murthy, I indeed forgot the inverse sign in the equation. I updated my question just now. – Pim Jul 23 '19 at 05:32

1 Answers1

-1

I'm a bit confused; if you have a matrix $M$ and you want to write it in the form $$M = cA + I$$ you simply directly solve for $A$: $$A = (M-I)/c$$ where if you're not given the value of $c$, you can pick it arbitrarily ($c=1$, say).

If you're trying to compute the matrix inverse of $cA+I$, that is a different problem (and one without a closed-form solution, though you can use some common approximations, e.g. $$(I+cA)^{-1} \approx I - cA,$$ if $cA$ is small).

user7530
  • 50,625
  • The latter is what I’m trying to achieve, I edited my question accordingly. Do you maybe know of a reference to a proof for your statement? How small does $cA$ have to be? For example, in my Finite Difference scheme, c would be the step size in time (a value $\in(0,1]$) and $A$ would be a sparse matrix of dimension up to $40.000\times40.000$. – Pim Jul 23 '19 at 05:26
  • I feel like the approximate formula just doesn't make a lot of sense. Can you please reply by giving a reference to a proof or article such that I can read into this. Thanks in advance. Obviously, if $c=0$, I get it. But I don't see any other case for which the approximation would be useful... – Pim Jul 24 '19 at 14:36
  • @Pim It's the truncation of the Taylor expansion of $(I + c A)^{-1}$: $$(I+cA)^{-1} = I - c A + c^2 A^2 - c^3 A^3 + o(c^3)$$ etc. You can use the same expansion to bound the error in the estimate (you will want $c$ well smaller than the largest entry, or largest singular value, of $A$). – user7530 Feb 10 '21 at 18:56