Perron Frobenius with one negative pair of entries Suppose you have a real symmetric matrix $A$ which is positive except for $a_{ij},a_{ji}$, who are negative. 
While it is not generally true that the eigenvector of the dominant eigenvalue of $A$ is positive (I have a  7x7 counterexample which I can post if somebody cares for it), I wonder if under some additional assumptions (maybe on the value of the negative entries), this eigenvector can be proved to be positive?
 A: The following paper (and the large number of references cited therein) provides some general sufficient conditions to ensure the "Perron-Frobenius property," thereby offering a set of useful answers to your question.
Reference


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*A. Elhashash, D. B. Szyld. On general matrices having the Perron-Frobenius property. Electronic J. Linear Algebra, vol. 17, pp. 389--413. (2008).

A: I think I found something! After long and hard googling, I stumbled upon this:
MR2311036 (2008e:15011) 15A18 (15A29 15A48)
Chen, Jianbiao; Xu, Zhaoliang The inverse eigenvalue problem for real eventually positive matrices. Filomat 21 (2007), no. 1, 1–16. 
http://www.doiserbia.nb.rs/img/doi/0354-5180/2007/0354-51800701001C.pdf
The result that interests me is a structure theorem for eventually positive matrices (Theorem 3.2):
A real $n \times n$ matrix $A$ is eventually positive iff there exist positive vectors $\alpha,\beta$ (of length $n$) and a $n \times n$ matrix $Y$ so that:
$A=\frac{1}{(\beta^{T}\alpha)^{2})}\alpha\beta^{T}+\frac{1}{\beta^{T}\alpha}(I-\frac{1}{\beta^{T}\alpha}\alpha\beta^{T})Y(I-\frac{1}{\beta^{T}\alpha}\alpha\beta^{T})$ and the spectral radius of $(I-\frac{1}{\beta^{T}\alpha}\alpha\beta^{T})Y(I-\frac{1}{\beta^{T}\alpha}\alpha\beta^{T})$ is $<1$.
