Showing that if $\inf_{\lambda\in\mathbb{R}}\| I - \lambda A\| < 1 $ then A is invertible

Question

Original question: (I couldn't comment and it seems to be against the rules to answer as a reply to previous answers)

Proof about matrix invertibility using matrix norms and infima

In the answer of how to show A is invertible if $\inf_{\lambda\in\mathbb{R}}\| I - \lambda A\| < 1 $, daw writes that due to $I-(I-\lambda\ A)$ = $\lambda\ A$ being invertible as $\sum_{k=1}^\infty (I-\lambda A)^k$ converges, then A must be invertible too.

I understand that for values of $\lambda\in\mathbb{R}$, satisfying $| I - \lambda A\| < 1 $, the matrix $\lambda\ A$ must be invertible, which includes infimum in this case. But how does this convert to A also being invertible?

This would imply that $\sum_{k=1}^\infty (I-\ A)^k$ converges, due to $\sum_{k=1}^\infty (I-\lambda A)^k$ converging for certain values of $\lambda$ (correct?), but how is this connection established/seen?

Answer 1

The convergence of $\sum_{k=0}^\infty (I-\lambda A)^k$ implies that $I-(I-\lambda A)= \lambda A$ is invertible. Since $\lambda\ne0$ this implies $A$ is invertible with inverse $\lambda(\lambda A)^{-1}$.

This does not imply the convergence of $\sum_{k=0}^\infty (I- A)^k$: Take $A=-I$.

Answer 2

A simpler proof would be to suppose there is some $\lambda$ such that $\|I-\lambda A\| <1$.

Then if $Ax = 0$, with $x \neq 0$, we have $\|x-\lambda Ax\| = \|x\| \le \|I-\lambda A\| \|x\| < \|x\|$, which is a contradiction.

hence $\ker A = \{0\}$ and so $A$ is invertible.