Given an invertible $m \times m$ matrix $A$, is there a way to find the inverse of $A'$ where $A'$ is an infinitesimal perturbation of $A$?

Question

Given a $m\times m$ matrix $A$ with a known $A^{-1}$ is there a simple way to find $A'^{-1}$ where $A'$ is a $m\times m$ matrix infinitesimally different from $A$?

Answer 1

It kinda depends on your objective, but you can use the power series expansion $$(A+\varepsilon B)^{-1}=A^{-1}\sum_{k=0}^\infty (-\varepsilon BA^{-1})^k=\left(\sum_{k=0}^\infty (-\varepsilon A^{-1}B)^k\right)A^{-1}$$

Which is valid as long as all the eigenvalues of $\varepsilon BA^{-1}$ (or of $\varepsilon A^{-1}B$) have absolute value $<1$.

Answer 2

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

If the spectral radius of $A^{-1} B$ satisfies $\rho (A^{-1} B) \ll |\varepsilon|^{-1}$, then

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

Answer 3

Using matrix derivatives:

For matrix $A$ with nilsquare infinitesimal increment $\mathrm{d}A$, we have

$$ \mathrm{d}(A^{-1}) = - A^{-1} \,\mathrm{d}A \, A^{-1} $$

so for $A' = A + \mathrm{d}A$,

$$ \begin{align} A'^{-1} &= A^{-1} + \mathrm{d}(A^{-1}) \\ &= A^{-1} - A^{-1} \,\mathrm{d}A \, A^{-1} \\ &= A^{-1} (I - \mathrm{d}A \, A^{-1}) \\ &= (I - A^{-1} \,\mathrm{d}A) A^{-1}. \end{align} $$