$A$ is a non-negative, integer, irreducible, $m$ by $m$ matrix. It is well known (Perron-Frobenius) that $A$ has a positive eigen value (denote it by $\lambda$) with a positive eigen vector ($x$). It is needed to prove that:
$x_{max}/x_{min} \le \lambda^{(m-1)}$
$x_{max}$ denotes the largest element of $x$, $x_{min}$ denotes the smallest element of $x$.
I proved that when considering A as an adjacency matrix, where there is an edge from state $u$ to state $v$, the following holds:
$x_v \le \lambda x_u$
(it was the hint in this question). I don't see how to proceed from here.
Well, I found the answer. It is easy to prove that $x_v \le \lambda x_u$ simply by writing down $Ax=\lambda x$. Now, since $A$ is irreducible, there must exist a power $k \le m$ such that state $u$ is connected to state $v$ by a path of length $k$. $A^k$ represents a graph with paths of length $k$ between the states of the $A$-graph. This matrix has eigenvalue $\lambda^k$ with eigenvector $x$. Thus, we must be able to find indices $u$ and $v$ such that $u$ corresponds to $x_{min}$ and $v$ corresponds to $x_{max}$ for some $k$, so:
$x_{max}/x_{min} \le \lambda^k \le \lambda ^{m-1}$