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

$\begingroup$ @gmvh this has nothing to do with ringsandalgebras, which you added. $\endgroup$– YCorOct 11, 2020 at 11:14

$\begingroup$ ... but the key is to use convex geometry (this is why I added the tag). $\endgroup$– YCorOct 11, 2020 at 11:23

$\begingroup$ @YCor: well, linear algebra is listed as a subtopic of ra.ringsandalgebras, which is why I added it; which toplevel tag would you suggest? $\endgroup$– gmvhOct 11, 2020 at 12:07

1$\begingroup$ @gmvh if any, mg.metricgeometry fits better. To avoid filling comments here, I replied more in general in the Editor's lounge (particular message here) $\endgroup$– YCorOct 11, 2020 at 12:14
4 Answers
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 nonnegative 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 $P\cap F^\bot$.

1$\begingroup$ Note that this works equally when $P$ is replaced with any selfdual cone. $\endgroup$– YCorOct 11, 2020 at 15:07

$\begingroup$ In the same space, if for $d\geq 1$, $(F_i)_{1\leq i\leq d}$ are linear subspaces such that for $i\neq j$, $F_i\perp F_j$ and $\oplus_{1\leq i\leq d}F_i=\mathbb{R}^n$, is it possible to adapt this proof to find a nonnull nonnegative vector in one of these spaces? $\endgroup$– G. PanelOct 17, 2020 at 21:15

1$\begingroup$ @G.Panel No, in dimension 3 just rotate the three main axes by an orthogonal rotation of angle $\pi/3$ with axis the diagonal. That is, take $F_1,F_2,F_3$ as the lines spanned by $(1,2,2)$, $(2,1,2)$, $(2,2,1)$. $\endgroup$– YCorOct 17, 2020 at 21:30
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.

$\begingroup$ This argument should be rewritten more carefully; it is right in concept but needs work on the details. $\endgroup$ Oct 11, 2020 at 15:23
Would you like an elementary proof ? By this, I mean a calculus proof, one which does not invoque HahnBanach (Farkas Lemma involves HB).
Let $P$ be the orthogonal projector over $F$, and $K$ be the cone of nonnegative vectors. Denote $S$ the unit sphere. The continuous function $x\mapsto\Px\$ achieves its maximum over the nonvoid 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 $b:=aPa$ 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 selfdual convex cone.

$\begingroup$ It's a finitedimensional HahnBanach that is used in the previous arguments, and this proved in roughly the same lines (still it's good to have the whole result starting from very basic level). $\endgroup$– YCorOct 15, 2020 at 16:25

$\begingroup$ @YCor. I fully agree. My post doesn't raise any doubt about AC. It is just that even finitedimensional HB is not taught in Calculus, at least in France. $\endgroup$ Oct 15, 2020 at 16:50

$\begingroup$ Unless I missed something, fortunately in France we don't have any math course named with the depressing name "calculus" :) (But I see, it's not part of any of analysis//geometry 1st/2nd year course at least.) $\endgroup$– YCorOct 15, 2020 at 16:55
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 BENISRAEL, JOURNAL OF MATHEMATICAL ANALYSIS AND APPLICATIONS 9, 303314 (1964)