If we have 2 linear equality and inequatlity systems:
$(1)$ $A x < = b $ $ x >= 0$.
$(2)$ $b^T y < 0, A^T y > = 0, y >= 0$
where A $in$ $R^mxn$ and c $in$ $R^n$ are given and x $in$ $R^n$ and y $in$ $R^m$ are decision variables. Prove that exactly one of the two systems (1) and (2) has a feasible solution. Which means that 1 has a feasible solution iff (2) does not have a feasible solution.
I have troubles doing the forward part of the proof and understanding it geometrically.
I imagine the forward part of the proof to be something like if there exists, $x>=0$ that satisfies the constraint of the primal problem and is thus feasible primal problem.. I don't know how to proceed further. Any guidance would be much appreciated.
Assume $(1)$ is feasible. Suppose on the contrary that $(2)$ is feasible as well.
We have $Ax \le b$ and $y \ge 0$.
Hence $y^TAx\le y^Tb$.
Since $y^TA \ge 0$ and $x \ge 0$, hence $y^TAx \ge 0$ but this contradicts $y^Tb < 0$.
Edit:
For the other direction, if the primal is infeasible, then the dual is either unbounded or infeasible. However, since $y=0$ is a feasible solution to $(D)$, we conclude that it is unbounded. Hence we can find a $y$ that satisfy $(2)$.