TITLE: Determining existence of a solution for a system of linear inequalities
QUESTION [5 upvotes]: 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.

REPLY [2 votes]: Sounds really similar to Farkas's lemma. In order to prove that your problem has a solution, you must show that the problem
$$
(P)\quad A\underline{x}=\underline{b}, \quad \underline{x}\geq \underline{0}
$$
has no solution (in your case $\underline{b}=\underline{0}$ and $A=V$). To this end, you can use the simplex method to verify that there is no admissible solution to problem $(P)$.
Edit: this is how you use Linear Programming (LP) to determine if there is a feasible solution. Consider the LP problem
$$
\min f(\underline{y})=\sum y_i\\
\\
s.t. A\underline{x} + \underline{y} = \underline{b}\\
x_i\geq 0,y_i\geq 0,b_i\geq 0
$$
Then, if (one of) the optimal solution(s) verifies $f(\underline{y})=0$, then that means that $y_i=0$ for all $i$, that is, your original system is feasible. This is the first phase of the so called two-phases simplex method, where you try to determine an initial solution for a LP problem. Notice that this problem (the first phase) always has the trivial initial solution $\underline{y}=\underline{b}$.