Is there a version of Gordan's Theorem in which $Ax=0$ has been replaced by $Ax=b$? That is, I want a condition (possibly including conditions on $A$) for when $Ax=b$ has a solution $x \neq 0$, for $A \in \mathbb{R}^{m \times n}$ and $b \in \mathbb{R}\_+^m \setminus \{0\}$


Gordan's Theorem says that for all $A \in \mathbb{R}^{m \times n}$ we have
$
\text{either} \quad
\exists x \in \mathbb{R}\_+^m\setminus{0} \centerdot Ax = 0,
\quad\text{or}\quad
\exists y\in\mathbb{R}^n\centerdot A^\top y > 0,
$$


In particular, I cannot apply Farkas' Lemma to my problem because I want a condition with non-zero solutions.



Gordan's Theorem with $Ax=b$
All these results follow from the duality theory of linear programming. 


Let $e$ be the vector in $R^n$ with all entries 1.


$Ax = b$, $x \ge 0$, $x \ne 0$ is solvable iff the problem P: 


maximize $e^T x$ subject to $Ax = b$, $x \ge 0$


is unbounded. This implies that the dual problem D: 


minimize $b^T y$ subject to $A^T y \ge e$ 


is infeasible, which is equivalent to saying that $A^T y > 0$ is unsolvable. However, it is possible for both problems P and D to be infeasible, so you don't have "if and only if". What's special about $b=0$ is that in that case you know that the problem P is feasible, namely $x=0$ is a feasible solution.


