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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2You can probably do some series expansion. Like Taylor of $1/(1+x^{-1})$ or something of the sort. – mathreadler Sep 21 '18 at 16:13
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1Hint: geometric series. – Nate Eldredge Sep 21 '18 at 16:16
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@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. – a06e Sep 21 '18 at 16:18
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Yep, factor out an $A$. – Nate Eldredge Sep 21 '18 at 16:21
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Similar question asked long ago: https://math.stackexchange.com/questions/189750/does-mathbf-a-epsilon-mathbf-i-1-always-exist-why – StubbornAtom Sep 21 '18 at 16:30
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@StubbornAtom That raises the question, is there a similar approximation if $A$ is not invertible? – a06e Sep 21 '18 at 17:38
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I found a way to do it right after posting.
$$(A + \varepsilon I)^{- 1} = A^{- 1} A (A + \varepsilon I)^{- 1} = A^{- 1} (I + \varepsilon A^{- 1})^{- 1} \approx A^{- 1} (I - \varepsilon A^{- 1})$$
a06e
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Or just check $(A^{-1}-\epsilon A^{-2})(A+\epsilon I) = I - \epsilon^2 A^{-2} = I + O(\epsilon^2)$. – 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).
Daniel R
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