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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. –  Xijia Jan 28 '13 at 20:20
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2 Answers

you have to use the von Neumann series, see


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Is the following equality true? (A+O(N^-1))^-1=A^-1+O(N^-1) –  Xijia Jan 28 '13 at 20:36
Neumann is not von Neumann here. –  1015 Jan 28 '13 at 21:22
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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? –  Xijia 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 –  Xijia Jan 30 '13 at 7:56
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