$(I-A)^{-1}=\sum_{i=0}^\infty A^i$

Question

Let $V$ be a finite dimentional normed vector space and let $A$ a linear transformation from $V$ to $V$ such that $\left \| A \right \|<1$. Show that the linear tranformation $I-A$ is invertible and $$ (I-A)^{-1}=\sum_{i=0}^\infty A^i. $$

In the previous question we've proved that if $(v_n)$ is a sequence in $V$ such that $\sum_{n=1}^\infty \left \| v_n \right \|<\infty$ then $\sum_{n=1}^\infty v_n$ converges.

My solution is too short and doesn't relate to the previous question so I want to know where is my mistake: $$ \ \left \| A^n \right \| \leq \left \| A \right \| ^n \xrightarrow[n\to\infty]{}0 \\ \ A^n\xrightarrow[n\to\infty]{}0\\ \ (I-A)\cdot\sum_{i=0}^n A^i=I-A^{n+1}\xrightarrow[n\to\infty]{}I\\ \ (I-A)\cdot \sum_{i=0}^\infty A^i=I \\ $$

Answer 1

Abridged solution. Consider the function $B(f, g) = f \circ g,$ where $f$ and $g$ are linear functions $V \to V.$ Then $B$ is a linear and continuous (in fact, $\|B\| \leq 1$). Then, as the element $\sum\limits_{n=0}^\infty A^n$ exists (as you already proved), $(I -A)\sum\limits_{n = 0}^\infty A^n = B\left(I - A, \sum\limits_{n = 0}^\infty A^n \right) = \lim\limits_{N \to \infty} B\left(I - A, \sum\limits_{n = 0}^N A^n \right)=\lim\limits_{N \to \infty}(I-A^{N+1})=I$ and similarly for the product with exchanged factors. Q.E.D.

Answer 2

Let us set $B=1-A$. Then $||B||$ is finite if $||A||$ is less than 1. Thus, if $\sum_{n=0}^{\infty} ||A_n||$ exists then $\sum_{n=0}^{\infty}|B|| \times ||A^n|| \geq \sum_{n=0}^{\infty} ||BA^n||$ exists which implies $\lim_{N \rightarrow \infty} \sum_{n=0}^N BA^n$ $=\lim_{N \rightarrow \infty} B\left(\sum_{n=0}^N A^n\right)$ $=B \left(\lim_{N \rightarrow \infty} \sum_{n=0}^{N} A^n\right)$ $=B\left(\sum_{n=0}^{\infty} A^n\right)$ exists.

On the one hand,

$\lim_{N \rightarrow \infty} B\left(\sum_{n=0}^N A^n\right)$ $=\lim_{N \rightarrow \infty}( I-A^{N+1}) = I$.

So this implies $I=\lim_{N \rightarrow \infty} B\left(\sum_{n=0}^N A^n\right) = B \left(\lim_{N \rightarrow \infty} \sum_{n=0}^{N} A^n\right)$. As we have already shown that$\lim_{N \rightarrow \infty} \sum_{n=0}^N A_n$ $=\sum_{n=0}^{\infty} A^n$ exists, writing $C = \sum_{n=0}^{\infty} A^n$, it follows that $BC=I$, which implies that $C=B^{-1}$, giving the desired result.