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Consider a matrix $A$ which we subject to a small perturbation $\partial A$. If $\partial A$ is small, then we have $(A + \partial A)^{-1} \approx A^{-1} - A^{-1} \partial A A^{-1}$

I came across this approximation in some notes and I am trying to understand where it comes from. This answer seems related, but I am having trouble translating the results from the cited paper into the provided equation.

Gaussian0617
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  • If you want to use the cited answer you just need to neglect $B$ in $(A+B)$ in the RHS. Anyway I'd prefer a formal binomial expansion. – N74 Oct 19 '18 at 06:59

2 Answers2

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The usual argument is that, if you perturb $A$ by a small $X$ and get $(A+X)^{-1}=A^{-1}+Y+O(\|X\|^2)$, where $Y$ is the first-order (i.e. linear) change in $A^{-1}$, then by comparing the first-order terms on both sides of $\left(A^{-1}+Y+O\left(\|X\|^2\right)\right)(A+X)=I$, you get $YA+A^{-1}X=0$. Hence $Y=-A^{-1}XA^{-1}$ and $$(A+X)^{-1}=A^{-1}+Y+O\left(\|X\|^2\right)\approx A^{-1}+Y=A^{-1}-A^{-1}XA^{-1}. $$

Edit. The above is a rigorous argument provided that $A\mapsto A^{-1}$ is differentiable in the first place, but this is indeed the case because $A^{-1}=\frac1{\det(A)}\operatorname{adj}(A)$ is a rational function in the entries of $A$.

user1551
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  • How do we show that the effect of the perturbation is in fact a sum of first-order and second-order changes in $A^{-1}$? – Gaussian0617 Oct 19 '18 at 22:16
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    @Gaussian0617 See my edit. I thought you only wanted to know how $-A^{-1}XA^{-1}$ arises, and so some details were omitted. – user1551 Oct 20 '18 at 05:00
  • Do you have a reference for this result? I'd like to use it, but I'm not sure how well known it is in my area, and including a proof seems unnecessary. – Eddy May 22 '19 at 09:19
  • @Eddy I don't. It's a well-known result and it isn't a difficult one. I think most authors would feel no need to include any proofs in their books. It may have been proven somewhere, but in the relevant books that I've read (e.g. Stewart and Sun's Matrix Perturbation Theory), the result is only mentioned without proof. – user1551 May 22 '19 at 18:46
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    Ah, found it! Stewart and Sun section 2.4, thanks :) – Eddy May 23 '19 at 09:44
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In fact, you can understand the problem by the solution, given by mjqxxxx, of the related problem you give. The primary problem is to find a way to estimate the term in the right hand. Why not consider the determinant of the term. The determinant of the product equals the product of the corresponding determinant. Then if we consider epsilon as being infinitesimal, then the higher term is approaching zero. $\square$

weijun yin
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  • In fact, if you have studied the multi-scale analysis, you will find that the higher term is always omitted in the problem. We often do it in multi-scale analysis. – weijun yin Oct 19 '18 at 10:48