> 
> If $A$ is an $n\times n$ matrix such that $\|A\|<1$,
>  show that $(I-A)$ is invertible and 
>  $$
> \frac{1}{1+\|A\|} < \|(I-A)^{-1}\| <\frac{1}{1-\|A\|}
> $$
> 
> 
> 

My trial:

First to show $I-A$ is invertible:
if not, then there exists nonzero vector $x$ such that 
$(I-A)x =0 $
Hence 
$Ax=x$
So $\|x\|=\|Ax\|\le\|A\| \|x\|$ and since $x$ is a non zero vector we get $\|A\|\ge\|x\|$ which contradicts the assumption.

To prove the second part, by the triangle inequality we have
$$
\|I\|-\|A\| \le \|I-A\|\le\|I\|+\|A\|
$$
But the norm of the identity is equal to 1.

I stopped here since I know $\|(I-A)^{-1}\|$ doesn't equal $\|I-A\|^{-1}$

Any help?

*The norm here is the spectral norm or the operator norm.*