In $\mathbb{R}^n$ ($n\ge 1$) endowed with the usual dot product, for any linear subspace $F$, does there exist a non-null vector with non-negative coordinates in $F\cup F^\perp$?



Existence of a non-null and non-negative vector in $F\cup F^\perp$
Yes. For any two closed convex cones $C,D$ we have $(C+ D)^\circ=C^\circ\cap D^\circ$, where $C^\circ$ is the dual cone: $C^\circ=\{y\in\mathbf{R}^n:\forall c\in C:\langle y,c\rangle\ge 0\}$. A standard duality theorem is $C^{\circ\circ}=C$ for every closed convex cone $C$. Hence, the dual equality holds: $(C\cap D)^\circ=C^\circ+ D^\circ$.


Let $P$ be the cone of non-negative vectors. Apply this to $P$ and $F$. Assume that $P\cap F=\{0\}$. Then $(P\cap F)^\circ =\mathbf{R}^n$. Since $P^\circ=P$ and $F^\circ=F^\bot$, this yields $P+F^\bot=\mathbf{R}^n$. Fix any nonzero $\xi\in P$ (this requires $n\ge 1$ which I discretely added to the assumptions). Then $-\xi=p+\eta$ for some $p\in P$ and $\eta\in F^\bot$. Hence $-\eta=p+\xi$ is a nonzero element in $P\cap F^\bot$.


