$A \in \mathbb R^{m \times n}$. Suppose that the following system has no solution:$$Ax=0, x>0$$ Then the following system has a solution:$$A^Ty\le0,A^Ty\ne 0$$
My attempt:
First I supposed on the contrary that for all $y$ then $A^Ty\not \ge0$ and
if $A^Ty\le0$ then $A^Ty=0$. By this assumption is tried to find a vector $x>0$ such that $Ax=0$, but I got stuck.
Then I tried to manipulate the first system into a form the fits into the Farkas' Lemma and then to use the lemma, again with no success.
Aby ideas?
Here's a partial solution using Farkas' Lemma:
The system
\begin{equation} \begin{array}{l} Ax=0\\x>0 \end{array} \end{equation}
has a solution iff the system
\begin{equation} \begin{array}{l} Ax=0\\x\geqslant\mathbb{1} \end{array} \end{equation}
has a solution iff the system
\begin{equation} \begin{array}{l} Ax=0\\x-s=\mathbb{1}\\x,s\geqslant0 \end{array} \end{equation}
has a solution. The final system is equivalent to the system
\begin{equation} \begin{array}{l} \begin{bmatrix}A&0\\I&-I\end{bmatrix}\begin{bmatrix}x\\s\end{bmatrix}=\begin{bmatrix}0\\\mathbb{1}\end{bmatrix}\\ x,s\geqslant0 \end{array} \end{equation}
Apply Farkas' Lemma to this system and see if you can get it.