Determining existence of a solution for a system of linear inequalities I have a set of vectors $\mathbf{v}_1,\mathbf{v}_2,...,\mathbf{v}_m\in\mathbb{R}^n$ and I want to know if there exists a nonzero vector $\mathbf{x}$ such that $\mathbf{x}\cdot\mathbf{v}_i\le0$ for any $i$. This is the same as saying the equation $\mathbf{x}^TV\leqq\mathbf{0}$ has a nonzero solution, where $V$ is an $n\times m$ matrix whose columns are $\mathbf{v}_i$.
Geometrically, a solution exists if a plane can be drawn through the origin such that all of the vectors are on one side (on contained within the plane).
One can show using Stiemke's Theorem that a solution exists if and only if $V\mathbf{y}=\mathbf{0}\ ,\ \mathbf{y}>\mathbf{0}$ does not have a solution.
I have found a few sources on feasibility problems in linear programming, but none that have allowed me to find a solution to this problem. Any help would be appreciated.

An alternative approach based on Stiemke's theorem would involve the solution of a single somewhat larger LP.
Solve the LP
$\max z$
Subject to 
$Vy=0$
$y_{i} \geq z, i=1, 2, \ldots, n$
If the optimal solution has $z=0$, then $Vy=0$, $y > 0$ has no solution.
