How can we show that $(I-A)$ is invertible?

Question

$A$ is an $n\times n$ matrix with $\|A\|≤a<1 $ . I need to prove that the matrix $(I-A)$ is invertible with $\|(I-A)^{-1}\|\le\frac1{(1-a)}$.

It doesn't say anything more. The norm makes me confuse. How can we start to solve this. Could you please help?

Answer 1

You can also argue without using the geometric series. The matrix $I-A$ is invertible if and only if $\lambda = 1$ is not an eigenvalue of $A$.

For a contradiction, assume $\lambda = 1$ is an eigenvalue. Then $Ax = x$ for some $x$ with $\|x\| = 1$, so $\|A\| \ge 1$.

Answer 2

Hint: Show that a certain series converges in the norm $\|\cdot \|$ and that this is an inverse for $I-A$.

Answer 3

$\bf Hint$: Consider the series $\sum_{n=0}^\infty A^n$

Answer 4

Another way to show this is the following:

$I-A$ is invertible iff $\ker(I-A) = \{0\}$.

Now, let $x \in \mathbb{R}^n\backslash \{ 0 \}$. Then $$ \lVert x \rVert \leq \lVert x-Ax\rVert + \lVert Ax \rVert \leq \lVert (I - A) x \rVert +\lVert A \rVert \lVert x \rVert \underset{\text{Assumption}}{<} \lVert (I - A) x \rVert + \lVert x \rVert $$ So $0 < \lVert (I-A) x \rVert$ and thus $x \notin \ker(I-A)$.