Let $A \in \mathbb{R}^{n \times n}$, with elements $a\_{ij}$


What conditions on $A$ are required for the following to be true?


There exists some vector $x \in \mathbb{R}^n\_+$, $x \neq 0$ such that for all $i=1\dots n$, 


$$\sum\_{j=1}^n x\_j(a\_{ij} - a\_{ji}) \geq 0$$


Obviously if $A$ is symmetric this is true. I believe it may be true for all $A$ but I haven't been able to prove it.


(sorry really wasn't sure what to title this post)


I've tried the usual Farkas' Lemma-style trick but the problem is self-dual



Sum of Difference of anti-diagonal matrix elements
Lawrence and Spingarn, On fixed points of non-expansive piecewise isometric mappings, Proc. London Math. Soc. (3) 55 (1987), no. 3, 605–624, MR0907234 (89d:58063), "obtain a constructive and simple new proof of Tucker's Theorem which states that for any antisymmetric real $n\times n$ matrix $A$, there exists $x\ge0$ such that $Ax\ge0$ and $x+Ax\gt0$." They also write, "This theorem is equivalent to the Farkas Lemma and the linear programming duality theorem in the sense that each may be easily derived from the other." 


