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Let $A$ be an $m \times n $ matrix, $b \in \mathbb{R}^n$, and consider the linear program $$\max\{ 0^Tx: Ax = b, x \ge 0\},$$ and its dual $$\operatorname{min}\{y^Tb : y^TA \ge 0 \}.$$ Here $x \in \mathbb{R}^n$ and $y \in \mathbb{R}^m$. I am trying to show that the first one is feasible if and only if the second one is bounded. I think I have the $(\implies)$ direction, since if there is some feasible $x$, then $y^TA x= y^Tb \ge 0$ for all $y$ such that $y^T A \ge 0$, so the bottom set is bounded below by 0. However I am having trouble with the converse. Any help would be appreciated.

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up vote 1 down vote accepted

I'd say there is no equally easy way to prove the reverse direction. If you want to go down the easy path, you can say that the statement (both directions) is an application of the strong duality theorem of linear programming and there is also an equivalent variant of Farkas's lemma. You can track down he whole proof in

Schrijver: Theory of Linear and Integer Programming

Corollary 7.1d is the corresponding form of the Farkas lemma and the proof of the difficult direction is in Theorem 7.1 (Fundamental theorem of linear inequalities).

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