Existence of a non-null and non-negative vector in $F\cup F^\perp$ 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 $F\cup F^\perp$?
 A: 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 $P\cap F^\bot$.
A: Yes. This follows from Farkas' lemma. Let $\vec{v}_1$, $\vec{v}_2$, ..., $\vec{v}_k$ be a basis of $F$ and let $\vec{e}_1$, $\vec{e}_2$, ..., $\vec{e}_n$ be the standard basis of $\mathbb{R}^n$. Suppose that there is no nonzero nonnegative vector in $F^{\perp}$. In other words, suppose there is no nonzero solution to the linear inequalities:
$$\vec{v}_i \cdot \vec{x} = 0, \ 1 \leq i \leq k \qquad \qquad  \vec{e}_j \cdot \vec{x} \geq 0, \ 1 \leq j \leq n.$$
Then Farkas' lemma tells us that there must be some linear relation
$$\sum a_i \vec{v}_i + \sum b_j \vec{e}_j = 0$$
with nonnegative coefficients (and not all coefficients zero). Then $- \sum a_i \vec{v}_i$ is an element of $F$ with nonnegative entries. Moreover, we can't have $- \sum a_i \vec{v}_i = 0$, as the $\vec{v}_i$ are linearly independent and the $\vec{e}_j$ are as well.
A: Would you like an elementary proof ? By this, I mean a calculus proof, one which does not invoque Hahn-Banach (Farkas Lemma involves HB).
Let $P$ be the orthogonal projector over $F$, and $K$ be the cone of non-negative vectors. Denote $S$ the unit sphere. The continuous function $x\mapsto\|Px\|$ achieves its maximum over the non-void compact subset $S\cap K$, at some vector $a$. It amounts to saying that $a$ maximizes
$$f(x):=\frac{\|Px\|^2}{\|x\|^2}$$
over $K\setminus\{0\}$.
Comparing $f(a)$ with $f(a+t\vec e_i)$, we find that $\partial_{x_i}f(a)$ vanishes if $a_i>0$, and is $\le0$ if $a_i=0$. This gives either $(Pa)_i=\lambda^2a_i$, where $\lambda=\|Pa\|/\|a\|\le1$, or $(Pa)_i\le0$. We infer that the vector $b:=a-Pa$ belongs to $K\cap F^\bot$.
If $b\ne0$, we are done. If instead $b=0$, then $a=Pa$, meaning that $a\in F$, and we are done too.
The proof is similar when $K$ is replaced by a self-dual convex cone.
A: That is proved in Cor. 3', p. 309 of [1]. In fact, Cor. 3' is even stronger:
Let $P$ be the nonnegative cone.
If $F\cap P=\{0\}$, then $F^\perp\cap P^o\ne\emptyset$; i.e., $v_n>0\ \forall n$ for some $v\in F^\perp$.
[1] https://core.ac.uk/download/pdf/82596353.pdf
Notes on Linear Inequalities, I:
The intersection of the Nonnegative Orthant
with Complementary Orthogonal Subspaces*
ADI BEN-ISRAEL, JOURNAL OF MATHEMATICAL ANALYSIS AND APPLICATIONS 9, 303-314 (1964)
