This is a real life problem. I have a matrix $A$ which is $m\times n$. I want to check for the conditions on the existence a vector $x\in\mathbb{R}^n$ such that $A x \geq 0$.
The Farkas's Lemma, as I have it, reads like this:
If $A$ is $m\times n$ and $b\in \mathbb{R}^m$ then only one of this conditions holds:
(1) $\exists$ $x\in\mathbb{R}^n$ such that $A x \leq b$.
(2) $\exists$ $y\in\mathbb{R}^m$ such that $y\geq0$, $yA = 0$ and $yb<0$.
I have been trying to use the Farkas Lemma. However, I don't see how can I use the lemma as I have $b=0$. If $b=0$ then no such $y$ exists such that $yb<0$. Is there an ammendment to the FL that I can use for existence?
Update: I think I have come to a solution, but I would be thrilled to know someone else's opinion. So remember, matrix $A$ is $m\times n$ and I am looking to be able to guarantee that there is an $x\in\mathbb{R}^n$ such that $A x \geq 0$. I will define $B=-A$ and $b=(-1,\dots,-1)'\in\mathbb{R}^m$. Now, apply Farkas's Lemma on matrix $B$ and vector $b$. Now (2) is fairly possible. And what happens if (2) is never true? Then (1) must be true, so that $\exists$ $x\in\mathbb{R}^n$ such that $B x \leq b$, and then $-A x \leq b <0$ so we have that $Ax>0$.
So all I have to do to prove existence is to deny (2). Now that is a very difficult problem, but it is a complete different subject.