I need to show that if the matrix $A$ is invertible, and $\|B-A\| < \|A^{-1}\|^{-1}$, then $B$ is invertible.
I was able to prove that B is invertible $$(A^{-1}B)^{-1} = B^{-1}A = (I-A^{-1}B)^k$$, thus $$B^{-1} = \sum_{k=0}^{\infty} {(I-A^{-1}B)^k A^{-1}}$$
However, I am unsure how can I show that there also exists a $$B^{-1} = A^{-1} \sum_{k=0}^{\infty} {(I - BA^{-1})^k}$$ Any idea on how to do so?
Any feedback on this question would be greatly appreciated!
Hint: $$ \|I - B A^{-1})\| = \|(A - B) A^{-1}\| \le \ldots < 1$$ so your series converges in norm.
Observe that $ \| A^{-1}B- I \| = \|A^{-1}(B-A)\| \leq \|A^{-1}\|\|B-A\| < 1 $. Thus we have $$ \|\sum_{k=0}^\infty (I-A^{-1}B)^k\| \leq \sum_{k=0}^\infty \|(I-A^{-1}B)^k\|\leq \sum_{k=0}^\infty \|I-A^{-1}B\|^k < \infty $$( being a GP series with common ratio less than 1). So $ C = \sum_{k=0}^\infty (I-A^{-1}B)^k $ is well defined and we can verify that $$ A^{-1}BC = (I-(I-A^{-1}B)C = I $$ Thus $ A^{-1}B $ is invertible and hence $B$ is invertible.
