I have $a^\intercal M b > 0$, where $\forall a_i > 0$, $\forall b_j > 0$, and M is known. I'd like to find a tight linear constraint on $b$ which is independent of $a$ (other than the positivity constraint on $a$). That is, I'd like to find a vector $v$ such that $v^\intercal b > 0$. (It's possible, of course, that such a vector may not exist for some $M$.)

Example: For the matrix $\left[\begin{matrix}9 & -2 \\ 8 & -2\end{matrix}\right]$, $v^\intercal = [4,-1]$ is a solution, because $[4,-1] b > 0$ implies $a^\intercal M b > 0$ for all positive $a$.

This feels like it should be reducible to a constrained linear optimization problem, but for the life of me I cannot figure out how. I'm not even sure how to typify $v$ as "tight"; for the example above, $v^\intercal = [5,-1]$ is also a solution but one which is weaker than $v^\intercal = [4,-1]$. For that matter, I wonder if there are multiple tight constraints when the size of $b$ grows beyond 2.

Incidentally, I'm not sweating $> 0$ versus $\geq 0$. Whichever makes the problem easier.


Hmmm.... I'm not 100% certain of this, but it looks like $v^\intercal$ may just be composed of the minimum element from each column of $M$. As for "tight", that seems to mean "can't subtract a nonnegative vector from $v$ without violating the constraint". And there I was gearing up with the LCP...


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