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Let $A$ be a square invertible matrix, and $\epsilon$ a small positive quantity. To first-order in $\epsilon$, what is the inverse of $A + \epsilon I$, where $I$ is the identity matrix?

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    $\begingroup$ You can probably do some series expansion. Like Taylor of $1/(1+x^{-1})$ or something of the sort. $\endgroup$ – mathreadler Sep 21 '18 at 16:13
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    $\begingroup$ Hint: geometric series. $\endgroup$ – Nate Eldredge Sep 21 '18 at 16:16
  • $\begingroup$ @NateEldredge The formula I typically see is for the inverse of $I + \epsilon A$, where you can apply the geometric series directly. Right after posting I found a way to apply that result in this case. $\endgroup$ – becko Sep 21 '18 at 16:18
  • $\begingroup$ Yep, factor out an $A$. $\endgroup$ – Nate Eldredge Sep 21 '18 at 16:21
  • $\begingroup$ Similar question asked long ago: math.stackexchange.com/questions/189750/… $\endgroup$ – StubbornAtom Sep 21 '18 at 16:30
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I found a way to do it right after posting.

$$(A + \varepsilon I)^{- 1} = A^{- 1} A (A + \varepsilon I)^{- 1} = A^{- 1} (I + \varepsilon A^{- 1})^{- 1} \approx A^{- 1} (I - \varepsilon A^{- 1})$$

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  • $\begingroup$ Or just check $(A^{-1}-\epsilon A^{-2})(A+\epsilon I) = I - \epsilon^2 A^{-2} = I + O(\epsilon^2)$. $\endgroup$ – Michael Sep 21 '18 at 16:21
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You can write it as

$\left((I+\varepsilon A^{-1})A\right)^{-1}=A^{-1}\sum \limits_{n=0}^\infty A^{-n}\varepsilon^n= \sum \limits_{n=0}^\infty A^{-n-1}\varepsilon^n,$

where $A^0=I$. ((I+B) is invertible if $B$ has norm, as a linear operator, less than 1, so for $\varepsilon$ small this works).

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