Lower bound for the Perron eigen-value of a non-negative, irreducible, integer matrix

Rob10

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[FONT=MathJax_Math]A is a non-negative, integer, irreducible, [FONT=MathJax_Math]m[/FONT] by [FONT=MathJax_Math]m[/FONT] matrix. It is well known (Perron-Frobenius) that [FONT=MathJax_Math]A[/FONT] has a positive eigen value (denote it by [FONT=MathJax_Math]λ[/FONT]) with a positive eigen vector ([FONT=MathJax_Math]x[/FONT]). It is needed to prove that:[/FONT]
[FONT=MathJax_Math]x[FONT=MathJax_Math]m[/FONT][FONT=MathJax_Math]a[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]/[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Math]m[/FONT][FONT=MathJax_Math]i[/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Main]≤[/FONT][FONT=MathJax_Math]λ[/FONT][FONT=MathJax_Math]m[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]1[/FONT][/FONT]
[FONT=MathJax_Math]x[FONT=MathJax_Math]m[/FONT][FONT=MathJax_Math]a[/FONT][FONT=MathJax_Math]x[/FONT] denotes the largest element of [FONT=MathJax_Math]x[/FONT], [FONT=MathJax_Math]x[/FONT][FONT=MathJax_Math]m[/FONT][FONT=MathJax_Math]i[/FONT][FONT=MathJax_Math]n[/FONT] denotes the smallest element of [FONT=MathJax_Math]x[/FONT].[/FONT]
I proved that when considering A as an adjacency matrix, where there is an edge from state [FONT=MathJax_Math]u to state [FONT=MathJax_Math]v[/FONT], the following holds:[/FONT]
[FONT=MathJax_Math]x[FONT=MathJax_Math]v[/FONT][FONT=MathJax_Main]≤[/FONT][FONT=MathJax_Math]λ[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Math]u[/FONT][/FONT]
(it was the hint in this question). I don't see how to proceed from here.
 
Found elsewhere online:

Well, I found the answer. It is easy to prove that xvλxu\displaystyle x_v\,\leq \,\lambda x_u simply by writing down Ax=λx.\displaystyle Ax\,=\, \lambda x. Now, since A\displaystyle A is irreducible, there must exist a power km\displaystyle k\, \leq\, m such that state u\displaystyle u is connected to state v\displaystyle v by a path of length k.\displaystyle k. Ak\displaystyle A^k represents a graph with paths of length k\displaystyle k between the states of the A\displaystyle A-graph. This matrix has eigenvalue λk\displaystyle \lambda^k with eigenvector x.\displaystyle x. Thus, we must be able to find indices u\displaystyle u and v\displaystyle v such that u\displaystyle u corresponds to xmin\displaystyle x_{min} and v\displaystyle v corresponds to xmax\displaystyle x_{max} for some k\displaystyle k, so:

xmaxxminλkλm1\displaystyle \frac{x_{max}}{x_{min}}\, \leq\, \lambda^k\, \leq\, \lambda^{m-1}
 
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