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Is there a formula to \begin{equation} (I+\varepsilon A)^{-1} \end{equation} in terms of $A^{-1}$ or $A$, where $I$ is the identity matrix $A$ is an invertible matrix and $\varepsilon$ is a constant?

Ayman Hourieh
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user29999
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3 Answers3

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Assuming that $ \epsilon$ is a small quantity you have \begin{equation} (I+\epsilon A)^{-1} =I-\epsilon A +O(\epsilon ^2) \end{equation}. This can be easily verified directly: \begin{equation} (I+\epsilon A)(I-\epsilon A)= I +\epsilon A-\epsilon A+ \epsilon ^2 A^2=I+\epsilon ^2 A^2 \end{equation} so the answer is indeed accurate up to order $\epsilon$.

GFR
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    To get the answer to higher order use the geometric series as Squid suggested: \begin{equation} (I+\epsilon A)^{-1}=\sum _{k=0}^n (-1)^k\epsilon ^k A^k +O \left(\epsilon^{n+1}\right)=I-\epsilon A+ \epsilon ^2 A^2 -\epsilon^3 A^3 + \ldots +O \left(\epsilon^{n+1}\right) \end{equation} – GFR Jan 16 '14 at 16:05
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It's not precisely what you're asking about but if $A$ is any matrix with standard norm $\| A \| < 1$ then geometric series essentially applies

$$(I - A)^{-1} = I + A + A^2 + A^3 + ...$$

Squid
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there is example of article

http://dx.doi.org/10.1016/j.aml.2005.04.004

which talks about pertubation matrix inverse,it may help you,please try to download it,it is free,it is given that matrix is perturbed by some another matrix,but it may help you

Asaf Karagila
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dato datuashvili
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