I have been trying to solve the problem, but wasn't able to move after applying Neumann expansion. The square matrix F satisfies $||F||<1$.
I am trying to show the following.
$$ ||(I-F)^{-1}|| \le \frac{1}{1-||F||} $$
I think I could convert the term in the left to the following given $||F||$ is a scalar. Using the geometric series: $$ ||\sum_{i=0}^{\infty}F^{i}|| \le 1+||F||+||F^{2}||+...+||F^{k}||= \frac{1}{1 - ||F||} $$ $$ ||\sum_{i=0}^{\infty}F^{i}|| \le \frac{1}{1 - ||F||} $$
Can I make the reasoning above?