# Matrix inverse of $A + \epsilon I$, where $A$ is invertible

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?

• You 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
• Hint: geometric series. – Nate Eldredge Sep 21 '18 at 16:16
• @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. – becko Sep 21 '18 at 16:18
• Yep, factor out an $A$. – Nate Eldredge Sep 21 '18 at 16:21
• Similar question asked long ago: math.stackexchange.com/questions/189750/… – StubbornAtom Sep 21 '18 at 16:30

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