The height of the Perron-Frobenius eigenvector Does the height of a real symmetric matrix with non-negative entries control the height of its Perron-Frobenius eigenvector, under some reasonable definition of heights?
Just as an example of what kind of heights I would be happy with. For the matrix, consider the model case where all of its entries are integer, and take the largest of the $L^1$-norms of its rows. For the eigenvector, represent the set of its coordinates as $U\cup V$ with $\max U\le \min V$ and $\sum_{u\in U}u^2=\sum_{v\in V}v^2$, and define the height to be the ratio of the largest and the smallest elements of $V$.
 A: Grrr, I am not used to this yet:
"Birkhoff norm" places the problem in a less elementary setting:
What is the Birkhoff norm of a Perron vector?
Second edit: I just found
concentration for eigenvectors
which refers to the original paper of Minc he cited in the reference I have given below. The reactions are not encouraging, but some references indicated there may be of help. This question is essentially duplicate ... 
See also
Calculating the Perron-Frobenius eigenvector of a positive matrix from limited information
for a more recent reference.
First edit: I have edited due to the comment concerning non-negativity.
This answer refers to the following book:
"Nonnegative matrices" by Henryk Minc, Wiley Interscience Series in discrete mathematics and optimization, 1988.
A certain control of some "height" of the Perron-Frobenius vector for general symmetric non-negative integer matrices can not be expected, as the Identity matrix impressively shows.
A (presumably "The" but I am not an expert) relevant concept is Irreducibility (section 1.2, p. 5). Any non-negative eigenvector of a non-negative irreducible matrix is strictly positive (section 1.2, Theorem 2.2, p. 7).
A non-negative matrix is irreducible if and only if for all i,j there is some power k of the matrix, such that the entry $(a_{ij})^k$ is positive (section 1.2, Theorem 2.3, p. 7), but it is not necessarily possible to choose this k to be equal for all entries.
One then derives that the Perron-Frobenius eigenvalue of an irreducible non-negative matrix is simple (section 1.4, Theorem 4.3, p.14) and it has a strictly positive eigenvector (section 1.4, Theorem 4.1, p.11).
Under this condition, there seems a chance, that controls for some "height" given below might be carried over from the positive to the non-negative case.
A more strict concept for non-negative matrices is primitivity, section 3.1, p. 47. A non-negative matrix is primitive if and only if some power of it is positive (section 3.2,Theorem 2.1, p.49). One gets a feeling for the difference between irreducibility and primitivity by looking at section 3.3, Theorem 3.1, p. 51.
Finally, we have a control on the height of the Perron-Frobenius (maximal) eigenvector for positive matrices in section 2.3, Theorem 3.1, p. 41 - 43, including the proof; the references cited there can be found on page 46.

section 2.3, Theorem 3.1:
Let $A \ = \ (a_{ij})$ be a positive matrix, with maximal eigenvector $x = (x_1, \ldots, x_n)$ and let $\gamma \ = \ \max_{i,j} (x_i/x_j)$ . Then
$ \sqrt{R/r} \ \leq \ \gamma \ \leq \ \max_{j,s,t} (a_{sj}/a_{tj}) \, ,$
where $R$ and $r$ are the greatest and least row sums of $A$, respectively. The left inequality is an equality if and only if $R = r$. Equality holds on the right-hand side if and only if the pth row of $A$ is a multiple of the qth row, for some pair of indices $p$ and $q$ satisfying $a_{ph}/a_{qh} \ = \ \max_{j,s,t} (a_{sj}/a_{tj}) \, .$ 

I just copied this Theorem with minor notational changes and did no proofreading. 
