Different versions of theorem of the alternative? I am looking for help to find necessary and sufficient conditions for a solution $x\in \mathbb{R}^n, x>0$ to exist to the following linear system:
$Ax = b$ with where $A$ is $m\times n$ and $Cx>0$ where $C$ is $k\times n$ and 0 means a vector of zeros of dimension $k$.  
I am guessing there is some version of Farka's Lemma or Theorem of the alternative that may help here.  Thanks for any guidance you can provide!
 A: *

*Assume first that all inequalities are non-strict to see how to treat this problem in the canonical context of Farkas lemma.


So we have $Ax=b$, $Cx\ge 0$, $x\ge 0$.
Introduce the slack variable $w\ge 0$: $Cx\ge 0$ $\Leftrightarrow$
$Cx-w=0$, $w\ge 0$. Thus the system can be written as
$$
\left[\matrix{A & 0\\C & -I}\right]\left[\matrix{x\\w}\right]=\left[\matrix{b\\0}\right], \quad 
\left[\matrix{x\\w}\right]\ge 0.
$$
It looks exactly like the first Farkas alternative. Now we make the second alternative
$$
\left[\matrix{A^T & C^T\\0 & -I}\right]\left[\matrix{y\\v}\right]\le 0,\quad \left[\matrix{b^T & 0}\right]
\left[\matrix{y\\v}\right]>0.
$$
Then we can get rid of block matrices to see what we get.


*If the inequalities are strict, it is a bit harder. It means that the inequalities for $x$, $w$ are strict, i.e. $x,w>0$.


Let as look at the first Farkas alternative with the strict inequality for the variable $Ax=b$, $x>0$. We can write it as $\exists\epsilon>0\colon \hat x=x-\epsilon\mathbf{1}\ge 0$ (here $\mathbf{1}$ is the vector of all ones). So the original system is equivalent to 
$A\hat x=b-\epsilon A\mathbf{1}=\hat b$, $\hat x\ge 0$. Then we can construct the second alternative as usual.
The result will be messier because of the epsilon. It will take some effort to make it look "nicer". For example, the KKT necessary condition has this extra Constraint Qualification condition exactly because of a strict inequality in Farkas alternative. I will leave it for you to fight it further :-)
A: Lemma 1 in Matzkin and Richter (1991) has a version of the Theorem of the Alternative for this case. They have a small explanation about how they get the result, but do not show it explicitly.
