For Euclidean norm.
If so, why?
If not, might $(I-N)^{-1}$ exist some other way?
This spins-off from here.
2026-03-29 16:00:52.1774800052
If the entries of an invertible matrix N are between -1 and 1, is its operator norm less than 1?
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To your first question, the answer is generally no.
The matrix $$ N= \frac 23 \pmatrix{1&1\\1&-1} $$ will have (Euclidean) operator norm $4/3 > 1$.
$(I - N)^{-1}$ will exists if $\rho(N)<1$ ($\rho(N)$ denotes the spectral radius of $N$). This in turn is true if and only if $\|N\| < 1$ for some multiplicative matrix norm $\|\cdot \|$.