Let $A$ be a non-negative, non-zero, square matrix and define $S_n=\sum$(entries of $A^n$). Let $\lambda$ be the eigenvalue of $A$ with the largest modulus. How do I prove that $$\lim_{n\to\infty}\frac{1}{n}\log(S_n)=\log(\lambda)?$$
I tried to do something with the following fact, but I got stuck: If $\lambda_1,\ldots,\lambda_k$ are the eigenvalues of $A$, then $\lambda_1^n,\ldots,\lambda_k^n$ are the eigenvalues of $A^{n}$.
(Also: what if $\lambda\in\mathbb{C}\setminus\mathbb{R}$?)
Let us assume that $A \in \mathbb{R}^{p \times p}$ is diagonalizable, so that we can write the eigenvalue-eigenvector decomposition:
$$A=PDP^{-1} \ \ implying \ \ A^n=PD^nP^{-1}\tag{1}$$
where $D=\operatorname{diag}(\lambda_1,\ldots, \lambda_p)$ with
$$|\lambda_1|>|\lambda_2|\geq\cdots\geq |\lambda_p|.$$
The equality you have to establish is equivalent to this one :
$$\text{for } n \to \infty, \ \ \ S_n \approx K \lambda_1^n \tag{$*$}$$
(where $K$ is a certain constant).
Let $C_i \ \ (i=1, \ldots,p)$ be the columns of $P$ and $L_j \ \ (j=1,\ldots,p)$ be the lines of $P^{-1}$.
Relationship $(1)$ can be written as the following combination of rank-one matrices:
$$A^n=\sum_{i=1}^p \lambda_i^nC_iL_i\tag{3}$$
As the first matrix "dwarves" the other ones when $n \to \infty$,
$$A^n \approx \lambda_1^n\,\underbrace{C_1L_1}_{M_1}\tag{4}$$
If $K$ is the sum of all elements of $M_1$, we obtain in this way relationship $(*).$