I try to find conditions under which the linear system of inequalities $$ (S):\ Ax\ge0,\ x\ge0 $$
has a solution $x\in\mathbb{R}^n$ that is not zero ($x\neq0$). Here $A$ is a $n\times n$ square matrix with real entries. Of course I could combine the two inequalities in (S) into one and write it in the form of $\tilde{A}x\ge0$ where $\tilde{A}$ is $2n\times n$.
I assume this problem has long been addressed. Any suggestion on how to attack this problem is appreciated.
(S) is equivalent to $$ [A\ -I_n] \left[\begin{array}{c} x\\ b \end{array}\right] =0,\ \left[\begin{array}{c} x\\ b \end{array}\right]\ge0\ \mathrm{and}\ \left[\begin{array}{c} x\\ b \end{array}\right]\neq0\ (\Leftrightarrow\ x\neq0) $$ According to Gordan’s theorem if (S) does not have solutions then there is $y\in\mathbb{R}^n$ such that $\left[\begin{array}{c} A^{T}\\ -I_{n} \end{array}\right]y<0$ which comes to $y>0$ and $A^Ty<0$.