How to determine whether the span of a set of vectors contains a non-negative vector Let $U\subset\mathbb R^n$ be a proper subspace. What is the best way to determine whether there exists a positive vector $v>0$ with positive non-zero entries such that $v\in U^\perp$?
Edit: maybe it would be good to mention my attempt. It is an iterative algorithm. Let $\{u_1,\dots,u_k\}$ be a basis for $U$. Assuming $u_1$ has a negative and a positive entry, let $L_1\in\mathbb R^{(n-1)\times n}$ be a full-rank matrix with non-negative entries such that $L_1u_1\in\mathbb R^{n-1}$ is the zero vector (e.g. one can construct $L_1$ using pairs of entries of $u_1$ whose signs are different). Then the existence of a positive vector $v\in U^\perp$ is equivalent to the existence of a positive vector in the orthogonal complement of the linear span of $\{L_1u_2,\dots,L_1u_k\}\subset \mathbb R^{n-1}$. Now iterate.
So that's my attempt, but what is the best way?
 A: An algorithm for determining whether $U^\perp$ contains a positive vector is given here:
https://mathoverflow.net/questions/363181/intersection-of-a-vector-subspace-with-a-cone/363188
I present below a necessary condition ("mixed") and a sufficient condition ("mixed & dominating"), but first I discuss your proof.
You assumed that $u_1$ is mixed (i.e., has both positive and negative entries). Even if every $u_k$ was mixed (which is clearly a necessary condition), $v$ need not exist:
Example: Let $(1,-1,1)$ and $(1,-1,0)$ span $U$. Then $U^\perp = \{tv:\, t\in{\mathcal R}\}$, where $v=(1,1,0)$, so no $w\in U^\perp$ is positive. (Positive means: $w_k>0\ \forall k$.)
However, if every $u_k$ is mixed and the matrix formed by them is dominating, then the answer is positive, by Proposition 4.1 of https://www.sciencedirect.com/science/article/pii/S0024379597002395
I don't know if "dominating" is necessary (to prove that, I'd try to use Theorem 2.9 or just to construct a counter-example).
Page 198 contains an algorithm for recognizing mixed dominating matrices.
In your proof, even if $u_k$ is mixed, $L_1 u_k$ need not be mixed. Otherwise your proof would have worked.
