# Upper and lower bounds on the entries of a matrix power

Say I have a non-negative square $$n\times n$$ irreducible stochastic matrix $$A$$ (i.e., each column sums to 1), for which the following holds: $$A_{ij} > 0 \iff A_{ji} > 0.$$ I know that no more than 15 entries per row are positive, and if $$A_{ij} > 0$$, then $$\frac{1}{15} \leq A_{ij} \leq \frac{6}{11}$$ (this last bound is less important, the point is that it's $$< 1$$). For instance, this implies that $$A^2_{ii} \geq 1/15$$.

It's easy to prove that, for each $$k > 0$$, the smallest positive entry of $$A^k$$ is $$\geq \frac{1}{15^k}$$. It is also easy to prove that the greatest entry of $$A^k$$ will be $$\leq \frac{6}{11}$$, as $$A_{ij}^k \leq A^{k-1}_i(A^{k-1})^j \leq \max A_i^{k-1}.$$

However I would like to either a) give a better bound on the smallest positive entry, b) give a bound on the greatest entry that also varies in the exponent. The reason to do this is to give a good bound on the principal ratio $$\gamma$$ (defined below) of the Perron vector $$x$$ of $$A$$, i.e., the positive vector satisfying $$Ax=x$$ and $$||x||_1 = 1$$.

We know that there's some exponent $$d < n$$ such that $$A^d$$ is positive. By Minc's Nonnegative matrices, Theorem 3.1, one has $$\gamma := \max_{i,j}\frac{x_i}{x_j} \leq \max_{s,t,j}\frac{A^d_{sj}}{A^d_{tj}},$$ but the best we have with the previous discussion is $$\rho \leq 6\cdot 15^d / 11.$$

So, apart from a better entrywise bound, is it possible that I can get a (perhaps constant?) bound on $$\gamma$$?

Some of the references I'm working with are:

There are a number of related questions here on MO:

• Welcome to MO. I wonder if $t$ is the best variable name to use for an exponent of a matrix -- at first I got confused because it looked like the transpose ... – gmvh Apr 17 '20 at 15:14
• @gmvh I fixed it – Enric Florit Apr 17 '20 at 15:22