Given an nxn matrix N and $I=I_n$, under what conditions does $(I-N)^{-1}$ exist?
On one hand $(I-N)(I + N + N^2 + ...) = (I + N + N^2 + ...) - (N + N^2 + ...) = I?$
On the other hand, $(I-N)(I + N + N^2 + ...) = \lim_{m \to \infty} (I-N) \sum_{i=0}^{m} N^i = \lim_{m \to \infty} \sum_{i=0}^{m} (N^i - N^{i+1}) = \lim_{m \to \infty} (I - N^{m+1}) = I - \lim_{m \to \infty} (N^{m+1}) = I?$
I think $I - \lim_{m \to \infty} (N^m) = I$ if (and only if?) the entries of N are between -1 and 1?
Iirc, one of my professors stated it without specifying why it is true in that case if it was true in that case but not true in other cases or why it is true in all cases. This arose in the context of Markov chains which iirc deal with matrices whose entries are between -1 and 1.
So, does $(I-N)^{-1}$ always exist? If so, why? If not, under what conditions?
Of course $I-N$ is not always invertible. Clearly, if $A$ is a non-invertible function, then $N=I-A$ causes $I-N=I - (I-A) = A$ to be non-invertible.
The thing is that the expression $$(I+N+N^2+\cdots)$$
makes no sense if $N$ is not some special matrix. For example, if $N=I$, what is $I+N+N^2+\dots$? It equals an infinite sum of identity matrices and it does not converge.
There is a lemma which tells you that if the operator norm of $N$ is smaller than one, then $I+N+N^2+\cdots$ converges, and it converges to an inverse of $I-N$, so in that case, $I-N$ is invertible.