Why does $(I-A)$ has inverse when $\|A \| < 1$ I've seen a proof using the convergence of numerical series, but it is too large, could you please tell me if there is a shorter proof. Thank you!

We need the norm to be submultiplicative, i.e. $\|XY\|\le\|X\|\|Y\|$ for every pair of matrices $X$ and $Y$, otherwise the statement isn't true. For instance, let $\epsilon>0$ and define a matrix norm $\|X\|=\epsilon\sum_{i,j}|x_{ij}|$ for real matrices. Then $\|I\|<1$ when $\epsilon$ is sufficiently small, but $I-I=0$ is not invertible.
Suppose $\|A\|<1$ for some submultiplicative matrix norm $\|\cdot\|$. If $I-A$ is singular, then $(I-A)x=0$ for some nonzero vector $x$. Therefore $\|xx^T\|=\|Axx^T\|\le\|A\|\|xx^T\|<\|xx^T\|$, which is a contradiction.
