Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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

share|cite|improve this answer
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 $(A + O(N^{-1}))^{-1} = A^{-1} + O(N^{-1})$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.