Suppose $A\in\mathbb{R}^{m\times n}$, $b\in\mathbb{R}^m$, and $b\in \mathcal{R}(A)$. Show that there exists an $x$ satisfying $x \succcurlyeq 0$, $Ax = b$ if and only if there exists no $\lambda$ with $A^\top \lambda\succcurlyeq0$, $A^\top \lambda\ne0$ and $b^\top \lambda\le0$.
I solved one part of the proof by defining $$p^*=\min_{Ax=b, x\succcurlyeq 0} 0^\top x$$ and thus $$d^*=\max_{A^\top \lambda \succcurlyeq 0, A^\top \lambda \ne 0} -\lambda^\top b$$
If $\mathcal P\iff \mathcal Q$ is the required proof in the problem, $!Q\implies !P$ can be proved by weak duality in the above optimisation. How do we prove the other side $\mathcal Q\implies \mathcal P$? How does one define $!Q$, since there are three conditions involved?
This is weak duality and Farkas Lemma. See http://en.wikipedia.org/wiki/Farkas%27_lemma