Let $A\in \mathbb {R}^{n\times n}$, and $\|A\|=\lambda$, where $\lambda$ is the largest eigenvalue of $A$, and $0<\lambda<1$. Let $\vec b$ be the eigenvector of $A$ which corresponds to $\lambda$.
It's true that $\vec b$ is also an eigenvector which corresponds to $\sum_{i=1}^{\infty}A^{i}$.
I know that $\left\|\sum_{i=1}^{\infty}A^{i}\vec b\right\|=\left\|\sum_{i=1}^{\infty}\lambda^{i}\vec b\right\|=\left\|\sum_{i=1}^{\infty}\lambda^{i}\right\|\|\vec b\|$. I want to know whether I can have $\left\|\sum_{i=1}^{\infty}\lambda^{i}\right\|\|\vec b\|=\left\|\sum_{i=1}^{\infty}A^{i}\right\|\|\vec b\|$, which is to say $\left\|\sum_{i=1}^{\infty}\lambda^{i}\right\|$ is the largest eigenvalue of $\left\|\sum_{i=1}^{\infty}A^{i}\right\|$. Intuitively, I think it is true and might be some theorem.
$\sum\limits_{k=1}^{\infty}A^{i}$ is nothing but $A(I-A)^{-1}$ and its eigen values are of the form $\frac c {1+c}$ where $c$ is an eigen value of $A$. From this it follows that that the largest eigen value of $\sum\limits_{k=1}^{\infty}A^{i}$ is $\frac {\lambda} {1+\lambda}$ which is $\sum\limits_{k=1}^{\infty}\lambda^{i}$.