TITLE: Does small Perron-Frobenius eigenvalue imply small entries for integral matrices?
QUESTION [8 upvotes]: Suppose that $M$ is an $n \times n$ matrix where each entry is a positive integer. Then $M$ is Perron-Frobenius and so has unique largest real eigenvalue $\lambda_{\textrm{PF}}$.

Does an upper bound on $\lambda_{\textrm{PF}}$ give an upper bound on each of the entries of $M$? That is, does $\lambda_{\textrm{PF}}$ being small imply that each entry of $M$ is too?

Note that when $n = 2$:
$$ \lambda_{\textrm{PF}} = \frac{1}{2}\left(a + d + \sqrt{a^2 - 2ad + d^2 + 4bc}\right) $$
and so $a, b, c, d \leq \lambda_{\textrm{PF}}^2$.
Additionally, if we remove the requirement that the entries of $M$ be integers then the answer is no as for every $k \geq 1$ the matrix
$$ \left( \begin{matrix} 1 & k \\ k^{-1} & 1 \end{matrix} \right) $$
has Perron-Frobenius eigenvalue 2.

REPLY [4 votes]: This is a little late in the game, but here is another estimate. Let $N$ be the maximal entry. Then the minimal entry of $A^3$ is at least $N + n^2-1$: the $ij$th entry of $A^3$ is the sum of terms of the form $A_{ik}A_{kl} A_{lj}$; if the max occurs at $kl$,  we obtain the result (since the entries are all at least one).
Thus every entry of $A^3$ is at least $N + n^2-1$, so its Perron eigenvalue is at least $n(N+n^2-1)$ (where $n$ is the size of the matrix). Hence $\lambda_{PF} \geq (n(N+n^2-1))^{1/3} \geq \max \{n, (nN)^{1/3}\}$.