A finite matrix within unit distance of the identity is invertible

By induction using [F1], $\mathrm{\parallel }{X}^k{\mathrm{\parallel }}_2\le q^k$ for every $k\ge 0$. For $N\ge 0$, set $S_N={\sum }_{k=0}^N(-X{)}^k$. If $M>N$, the geometric sum bound gives $\mathrm{\parallel }{S}_M-S_N{\mathrm{\parallel }}_2\le {\sum }_{k=N+1}^M{q}^k\le q^{N+1}\mathrm{/}(1-q)$, so $(S_N)$ is Cauchy.

For each entry, $\mathrm{\mid }(S_M-S_N{)}_{ij}\mathrm{\mid }\le \mathrm{\parallel }(S_M-S_N)e_j{\mathrm{\parallel }}_2\le \mathrm{\parallel }{S}_M-S_N{\mathrm{\parallel }}_2$ by [F2]; hence every entry converges to a finite limit. Let $S$ be the matrix of those limits. For any matrix $Y$ and unit vector $x$, put $c={\max }_{ij}\mathrm{\mid }{Y}_{ij}\mathrm{\mid }$ and $a_j=\mathrm{\mid }{x}_j\mathrm{\mid }$. Then $\mathrm{\mid }(Yx{)}_i\mathrm{\mid }\le c{\sum }_j{a}_j$. The identity $n{\sum }_j{a}_j^2-({\sum }_j{a}_j{)}^2={\sum }_{j<k}(a_j-a_k{)}^2\ge 0$ gives ${\sum }_j{a}_j\le \sqrt{n}$, so each coordinate of $Yx$ has magnitude at most $c\sqrt{n}$. Summing the $n$ coordinate squares gives $\mathrm{\parallel }Yx{\mathrm{\parallel }}_2\le nc$, and [F2] gives $\mathrm{\parallel }Y{\mathrm{\parallel }}_2\le n{\max }_{ij}\mathrm{\mid }{Y}_{ij}\mathrm{\mid }$. Entrywise convergence of $S_N$ to $S$ now implies $\mathrm{\parallel }{S}_N-S{\mathrm{\parallel }}_2\to 0$.

Finite multiplication gives $(I+X)S_N=S_N(I+X)=I-(-X{)}^{N+1}$. By step 1.1, $\mathrm{\parallel }{X}^{N+1}{\mathrm{\parallel }}_2\le q^{N+1}\to 0$. Submultiplicativity [F1] makes multiplication by $I+X$ continuous, so taking limits using step 2.1 yields $(I+X)S=S(I+X)=I$. Hence $S$ is a two-sided inverse.