I would like to invert a square matrix $L$.
One can write it as a sum of two matrices, one containing the diagonal terms ($D$) and the other the off-diagonal ones ($A$).
$$L = D+A$$
I would like to know, under which conditions can one make the following expansion of the inverse of $L$:
$$L^{-1} = (D+A)^{-1} = (D(I+D^{-1}A))^{-1} \approx (I-D^{-1}A)D^{-1} = D^{-1}-D^{-1}AD^{-1}$$
Rephrasing the question, I realise that this somehow implies that the terms in the off-diagonal part $A$ must be "small", but I do not know how to write that condition formally. When is one allowed to make such an expansion and what would be the error?
Let me write the matrix in question in the form $L:=D-A$ (instead of $D+A$ to avoid alternating signs). Let $D$ be nonsingular and $\|D^{-1}A\|=:\epsilon<1$ for some operator matrix norm. We can write $$ L=D(I-D^{-1}A). $$ Using the Neumann series and with $B:=D^{-1}-D^{-1}AD^{-1}$, we have $$ L^{-1}=(I-D^{-1}A)^{-1}D^{-1}=[I+D^{-1}A+(D^{-1}A)^2+\cdots]D^{-1}=B+(D^{-1}A)^2(I-D^{-1}A)^{-1}D^{-1}. $$ Hence the error can be bounded by using the same norm as $$ \begin{split} \|L^{-1}-B\|\leq\|D^{-1}A\|^2\|(I-D^{-1}A)^{-1}\|\|D^{-1}\| \leq \frac{\|D^{-1}A\|^2\|D^{-1}\|}{1-\|D^{-1}A\|}=\|D^{-1}\|\frac{\epsilon^2}{1-\epsilon}. \end{split} $$