A proof of the Continuity of the inverse matrix function

Question

I would like to see a proof to this fact.

If $A$ is an invertible matrix and $B \in \mathcal{L}(\mathbb{R}^n,\mathbb{R}^n)$, that is an bounded linear opertor in $\mathbb{R}^n$. Then, if there holds $$ \|B-A\| \|A^{-1}\| <1, $$ we have that B is invertible.

Moreover, if possible, how to use it to prove that the aplication $A \rightarrow A^{-1}$ is continuous?

Answer 1

Since $$\|I - BA^{-1}\| = \|(A - B)A^{-1}\| \le \|B-A\|\,\|A^{-1}\| < 1,$$ the Neumann series \begin{align*} \sum_{k=0}^{\infty} (I-BA^{-1})^k \end{align*} converges and therefore (compare https://en.wikipedia.org/wiki/Neumann_series) \begin{align*} (I - (I-BA^{-1}))^{-1} = (BA^{-1})^{-1} \end{align*} exists. Therefore $B$ is invertible.

Answer 2

To prove the continuity, consider, if $A,B$ are two invertible $n \times n$ matrices then $$\|A^{-1} - B^{-1}\| \le\| A^{-1}B - I\| \|B^{-1}\| \le \|A^{-1}\| \| B-A \| \|B^{-1}\|.$$ This shows that if $\| B- A\|$ is small, then so is $\|A^{-1} - B^{-1}\|.$