Is there another representation for $\left(\operatorname{Id}-A\right)^{-1}$, when the Neumann series does not converge? Consider a matrix $A \in \mathbb{R}^{n \times n}$. It is well known, that if the operatornorm of $A$ fulfills $\vert \vert A \vert \vert <1$, then the inverse of $\operatorname{Id}-A$ exists and we can express it as a Neumann series
$$\left(\operatorname{Id}-A\right)^{-1} = \sum_{k=0}^{\infty} A^k.$$
Now if $A$ does not fulfill the above condition, then still $\operatorname{Id}-A$ might be invertible. My question is, can one similarly find another expression of the inverse, maybe as a series?
Of course, there is one special case that essentially is the same as the standard case: if $A^{-1}$ exists and has $\vert \vert A^{-1} \vert \vert <1$ then
$$\left(\operatorname{Id}-A\right)^{-1}= A^{-1}\sum_{k=0}^{\infty} A^{-k}$$
but this is not really the interesting case here. So one should assume that $A$ has some eigenvalues $\vert \lambda \vert >1 > \vert \lambda' \vert $.
Edit:
Thinking about it, one can of course do the following:
If $A = O D O^{-1}$, where $D$ is a diagonal matrix of eigenvalues, then by defining $\tilde{A} = O \tilde{D} O^{-1}$, where $$\tilde{D}_{ii} = \begin{cases} & D_{ii}, &&\vert D_{ii}\vert <1  \\ &\frac{1}{D_{ii}}, &&else \end{cases}$$
then we should have
$$ \left(\operatorname{Id}-A\right)^{-1}= \sum_{k=0}^{\infty} \tilde{A}^k.$$
I guess this is the correct generalization?
 A: If there is a line through $1$ such that all the eigenvalues of $A$ are on one side of the line (and off the line,) we can find a $w$ such that $|w-\lambda|<|w-1|$ for all eigenvalues $\lambda.$
Then, $$\frac1{1-z}=\frac1{1-w}\frac{1}{1-\frac{z-w}{1-w}}=\sum_{n=0}^{\infty}\frac{(z-w)^n}{(1-w)^{n+1}}$$
So you can compute:
$$(I-A)^{-1}=\sum_{n=0}^{\infty}\frac{1}{(1-w)^{n+1}}(A-wI)^n$$
And you need $\|A-wI\|<|1-w|.$

In general, a power series of matrices converges if an only if the power series converges for all the matrix's eigenvalues. (This is sort of obvious if $A$ is diagonalizable, but a little less obvious if $A$ is not diagonalizable.)

Another possibility is to use a Laurent series. If we pick $w$ such that $|w-1|>|w-\lambda|$ for all eigenvalues, then you can write:
$$\frac1{1-z}=-\dfrac{(z-w)^{-1}}{1-(w-1)(z-w)^{-1}}=-\sum_{n=1}^{\infty}(1-w)^{n-1}(z-w)^{-n}$$
So you get:
$$(I-A)^{-1}=-\sum_{n=1}^\infty (1-w)^{n-1}(A-wI)^{-n}$$
The obvious problem with that is that we've solved inverting $I-A$ by inverting $I-wA,$ and then applying a power series to that, which is hardly much help.
Indeed, if we set $w=1,$ then, by the convention that $0^0=1,$ you just get the identity.
