Only one system of equation has solution

Question

Let $A$ be a $m\times n$ real matrix. Show that only one of the following systems has solution:

(I): $Ax > 0$

(II): $Ay = 0, y \geq 0, y \neq \theta$, where $\theta$ is zero vector.

I have no idea to proceed the proof. Can anyone give me some hints? Thanks

Answer 1

Are you sure this problem is correct as stated. Consider the following counterexample. Let

$$ A = \left[ \begin{array}{cc} 1 & 0\\ 1 & 0\\ \end{array} \right],\quad x = \left[ \begin{array}{c} 1\\ 0\\ \end{array}\right], \quad\text{and}\quad y = \left[ \begin{array}{c} 0\\ 1\\ \end{array}\right] $$

Then $Ax > 0$ and $Ay = 0$, $y \geq 0$, $y \neq 0$, i.e., both systems have a solution.