Prove that $A+I_n$ is invertible, where $\left\lVert A\right\rVert<1$

Note that for any matrix $M$: if $\lambda$ is an eigenvalue, then $|\lambda| < \|M\|$. Thus, all eigenvalues of $A$ satisfy $|\lambda| < 1$.

Now, if $\mu$ is an eigenvalue of $A + I$, then $(\mu - 1)$ is an eigenvalue of $A + I$, which tells us that $|\mu - 1| < 1$. We can conclude that $A + I$ does not have zero as an eigenvalue. It follows that $A+I$ is invertible.

Given a matrix such that $\lVert \mathbf{A} \rVert < 1$, the matrix $\mathbf{I} \color{red}{-} \mathbf{A}$ is nonsingular with $$ \left( \mathbf{I} - \mathbf{A} \right)^{-1} = \sum_{k=0}^{\infty}\mathbf{A}^{k}, $$ and $$ \lVert \left( \mathbf{I} \color{red}{-} \mathbf{A} \right)^{-1} \rVert \le \frac{1}{1-\lVert \mathbf{A}\rVert}. $$

Proof by contradiction

Let $\mathbf{I} - \mathbf{A}$ be singular. $\exists$ a nonzero $x$ such that $\left( \mathbf{I} - \mathbf{A} \right)x = 0.$ Then we have $$ \lVert x \rVert = \lVert \mathbf{A} x \rVert $$ which implies $\lVert \mathbf{A} \rVert \ge 1.$ $\color{red}{\Rightarrow \Leftarrow}$

Derivation

Start with the telescopic identity $$ \left( \sum_{k=0}^{N}\mathbf{A}^{k} \right) % \left( \mathbf{I} - \mathbf{A} \right) % = % \mathbf{I} - \mathbf{A}^{N+1} % $$ Knowing the property of submultiplicative norms $\lVert \mathbf{A}^{k} \rVert \le \lVert \mathbf{A} \rVert^{k}$ and given $\lVert \mathbf{A} \rVert < 1$ we see $\lim_{k\to\infty}\mathbf{A}^{k} = 0$. This implies $$ \left( \lim_{N\to \infty} \sum_{k=0}^{N}\mathbf{A}^{k} \right) % \left( \mathbf{I} - \mathbf{A} \right) % = % \mathbf{I}, % $$ and $$ \left( \mathbf{I} - \mathbf{A} \right)^{-1} = \left( \lim_{N\to \infty} \sum_{k=0}^{N}\mathbf{A}^{k} \right). $$ At last, $$ \lVert \left( \mathbf{I} - \mathbf{A} \right)^{-1} \rVert \le \sum_{k=0}^{\infty}\lVert \mathbf{A} \rVert^{k} = \frac{1}{1-\lVert \mathbf{A}\rVert} $$