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

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? $$(A+O(N^{-1}))^{-1} = A^{-1}+O(N^{-1})?$$

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What is $O$ (the big $O$-notation)? What is $N$? – Anon Jan 28 '13 at 20:13
whats A and $O(N^{-1})$ clarify your question ? – Maisam Hedyelloo Jan 28 '13 at 20:13
Yes, the big O notation. That is the inverse of an invertible matrix A plus some matrix which converge to 0 as N trend to infinity. – ANuo Jan 28 '13 at 20:20

you have to use the von Neumann series, see

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

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Is the following equality true? (A+O(N^-1))^-1=A^-1+O(N^-1) – ANuo Jan 28 '13 at 20:36
Neumann is not von Neumann here. – 1015 Jan 28 '13 at 21:22

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 $(A + O(N^{-1}))^{-1} = A^{-1} + O(N^{-1})$.

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many thanks! Could you give me a reference about the second equality of the second line? – ANuo Jan 29 '13 at 8:07
See Nils's answer. – Robert Israel Jan 29 '13 at 18:44
OK, I will check. Thank you for your help – ANuo Jan 30 '13 at 7:56