I am struggling with this question from my Differential Equations course:
If $T$ is a linear transformation on $\mathbb{R}^n$ with $||T-I||<1$, prove that $T$ is invertible and that the series S= $\sum_{k=0}^{\infty}(I-T)^k$ converges absolutely to $T^{-1}$.
It's safe to assume $T$ is square ($n \times n$) matrix.
From what I understand:
- A series $\sum_0^{\infty}a_n$ is called absolutely convergent if $\sum_0^{\infty}|a_n|$ is convergent.
- A need to first prove that $T$ is invertible. Once I do, I need to show $||S-T^{-1}||<\epsilon$ for some $k$ greater than $N$ large.
For simplicity, let's define $A = T-I$. By hypotheses, we can assume $||T-I||<1$ is true. Now
\begin{array} a T - I = \begin{bmatrix} t_{11}-1&t_{12} & \dots & t_{1n}\\ t_{21} & t_{22} - 1 & \dots & t_{2n}\\ \vdots\\ t_{n1} & t_{n2} & \dots & t_{nn}-1 \end{bmatrix} \end{array}
Now in this case, the operator norms of this linear transformation: $$||A||=\text{max}_{|\vec{x}|\leq 1}||A\vec{x}|| \leq \sqrt{n}\mathcal{l} < 1$$ Where $\mathcal{l}$ is the maximum length of the rows of $A$.
This is where I get confused. Where would I continue from here to prove $T$ is invertible?
Now, to show that $S$ converged absolutely, I wanted to use a geometric series since $||A||<1$. Thus $$\sum_{k=0}^{\infty}||A||^k=\frac{1}{1-||A||}$$ This implies that $S$ converges. The problem is, I am not sure how to prove that $S$ does indeed converge to $T^{-1}$.
Any suggestions or comments? If you need me to clarify something, please let me know.
Thank you once again for reading this post. I humbly thank you in advance for any assistance or guidance you may provide.
Hint.
Note $$S_n= \sum_{k=0}^{n}(I-T)^k.$$ As $\Vert T-I \Vert <1$, $S_n$ converges absolutely to $S$. Now you have $$S_n . (I-(I-T))= S_n . T= I -(I-T)^{n+1}$$ And $(I-T)^n \to 0$ as $n \to \infty$.
You finally get $$S.T=I$$ as the map $A \mapsto A.T$ is continuous.