If A is invertible and $||B-A|| < ||A^{-1}||^{-1}$ prove $B$ is invertible. Just having really hard time trying to proof :
If $A$ is invertible and $||B-A|| < ||A^{-1}||^{-1}$ prove $B$ is invertible.
It is related to Neumann Series but i don't understand how to proof with math.
Thanks for your help and time.
Brian Ignacio

This is an old question, but I think it deserves a simple solution which does not need a convergence argument. So here we go.
By assumption, we have $\| A-B \| \| A^{-1}\| < 1$. Moreover because the induced $2$-norm is consistent, we have $\| (A-B)A^{-1}\|\leq \| A-B \| \| A^{-1}\|$. Therefore, $\| (A-B)A^{-1}\|=\|I-BA^{-1} \|<1$. Now we show that $BA^{-1}$ is invertible. For the sake of contradiction, suppose that $BA^{-1}$ is singular. Then there exists $v\neq 0$ such that $BA^{-1}v = 0$. Without loss of generality, we assume $\|v\|_2=1$ (if not, we can take $v/\|v\|_2$). Now by the triangle property, we have
$$1=\|v\|_2\leq \|v-BA^{-1}v\|_2+\|BA^{-1}v\|_2 = \|(I-BA^{-1})v \|_2\leq \| I- BA^{-1}\|,
$$
where the second equality follows from $BA^{-1}v = 0$, and the last inequality follows from the definition of the induced matrix norm and $\|v\|_2=1$. This is a contradiction, because previously, we showed that $\|I-BA^{-1} \|<1$. Therefore, the invertibility of $BA^{-1}$ follows from this contradiction. Now one can easily see that $A^{-1}(BA^{-1})^{-1}$ is the inverse of $B$.
