Determinant nonvanishing implies local invertibility lemma

Let $A:\mathbb{R}^n\to\mathbb{R}^n$ be a linear map . Saying $\det A\neq 0$ is equivalent to saying $A$ is invertible.

Stability of invertibility (Neumann series lemma): If $A$ is invertible and $B$ is another linear map such that $\|A^{-1}(B-A)\|<1,$ then $B$ is invertible and $B^{-1}=\sum_{k=0}^\infty \bigl(-A^{-1}(B-A)\bigr)^k\,A^{-1}.$ Moreover, $\|B^{-1}\|\le \frac{\|A^{-1}\|}{1-\|A^{-1}(B-A)\|}.$ In particular, if $\|B-A\|\le \frac{1}{2\|A^{-1}\|}$ then $B$ is invertible and $\|B^{-1}\|\le 2\|A^{-1}\|$.

This lemma is a key linear-algebraic ingredient in the inverse function theorem : once $Df(a)$ is invertible, $Df(x)$ remains invertible for all $x$ sufficiently close to $a$ (because $Df$ is continuous ).

Proof sketch: Write $B=A\bigl(I+E\bigr)\quad\text{where}\quad E=A^{-1}(B-A).$ If $\|E\|<1$, then $(I+E)^{-1}=\sum_{k=0}^\infty (-E)^k$ converges in operator norm , so $B^{-1}=(I+E)^{-1}A^{-1}$. The norm bound follows from the geometric series estimate.