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 $\lambda\_{\textrm{PF}}$.



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> 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?
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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.



Does small Perron-Frobenius eigenvalue imply small entries for integral matrices?
This is true. Indeed, you can estimate the sum of *all $n^2$* elements of $A$ rather than individual elements. (Thanks to [thomashennecke](https://mathoverflow.net/users/37855/thomashennecke) for observing this, my original answer dealt with the row sums of $A$ instead of the sum of all $n^2$ elements of $A$.) Moreover, you do not need the elements to be integer; it suffices to have them uniformly bounded away from $0$. 


Write $A=(a\_{ij})$ and let $x=(x\_1,\ldots,x\_n)$ be an eigenvector corresponding to the eigenvalue $\lambda\_{\rm PF}$. Suppose, for instance, that $a\_{ij}\ge 1$ for all $i,j\in[n]$. For each $i\in[n]$ you then have
 $$ \lambda\_{\rm PF} x\_i=\sum\_{j=1}^n a\_{ij}x\_j \ge \|x\|\_1, $$
whence $x\_i\ge\lambda\_{\rm PF}^{-1}\|x\|\_1$. As a result, letting $\xi:=\min\{x\_j\colon j\in[n]\}$,
 $$ \lambda\_{\rm PF}\|x\|\_1 
 = \sum\_{i=1}^n\lambda\_{\rm PF}x\_i 
 = \sum\_{i=1}^n\sum\_{j=1}^n a\_{ij}x\_j 
 \ge \xi \sum\_{i,j=1}^n a\_{ij} 
 \ge \lambda\_{\rm PF}^{-1} \|x\|\_1 \sum\_{i,j=1}^n a\_{ij}. $$
Consequently,
 $$ \sum\_{i,j=1}^n a\_{ij} \le \lambda\_{\rm PF}^2. $$


Equality is attained, for instance, for the all-$1$ matrix.


