Kevin asked in class: in the “geometric” proof of duality we talked about, did we need the fact that existed a BFS, i.e., that there were tight constraints? (Recall we were considering the LP ; since we were not talking about equational form LPs, the optimum could be achieved at a non-BFS.) He was right: we don’t. Here’s a proof.
Pick an optimal feasible point for the LP, call this point . Let for be all the constraints tight at —there may be linearly independent constraints in there, or there may be fewer, we don’t care. Now the claim is that the objective function vector is contained in the cone generated by these vectors . If we prove this claim, we’re done–the rest of the argument follows that in lecture.
(Indeed, this was precisely where we were being hand-wavy earlier, even when we assumed was a BFS. So this also answers Jamie’s request.)
Ok, back to business at hand. Suppose does not lie in the cone generated by the vectors . Then there must be a separating hyperplane between and : i.e., there exists a vector such that for all , but . So now consider the point for some tiny . Note the following:
- We claim that for small enough , the point satisfies the constraints . Consider for : since this constraint was not tight, we won’t violate it if we choose small enough. And for with ? See, , since and .
- What about the objective function value? .
But this contradicts the fact that was optimal.
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