The inverse of $A+O(N^{-1})$

Question

Assume $A$ is invertible and I want to calculate $(A+O(N^{-1}))^{-1}$

I want to know if there exist any formula for it?

$O(N^{-1})$ is the big $O$ notation. That is the inverse of an invertible matrix $A$ plus some matrix which converge to $0$ as $N$ tends to infinity.

Is the following equality true? $$(A+O(N^{-1}))^{-1} = A^{-1}+O(N^{-1})?$$

Answer 1

you have to use the von Neumann series, see

http://en.wikipedia.org/wiki/Neumann_series

Answer 2

If $A$ is an invertible matrix, then so is $A + B = A (I + A^{-1} B)$ when $\|B\| < \|A^{-1}\|^{-1}$, and $(A + B)^{-1} = (I + A^{-1} B)^{-1} A^{-1} = A^{-1} - A^{-1} B A^{-1} + \ldots$. In particular, $\|(A+B)^{-1} - A^{-1}\| \le \dfrac{\|A^{-1}\|^2 \|B\|}{1 - \|A^{-1}\| \|B\|}$, which you can write as $(A + O(N^{-1}))^{-1} = A^{-1} + O(N^{-1})$.