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In $\Bbb R^n$, suppose $V$ is a linear subspace such that $\forall v\in V\setminus \{0\}$, $v\nsucceq 0$ (inequality w.r.t. the nonnegative orthant, or componentwise inequality equivalently), then $\exists u\succ 0$ s.t. $u\in V^\perp$.

Incomplete functional analysis proof:
Let $X_+=\{x\mid x\succ0\}$ be the positive orthant, a convex and open subset. Then $V\cap X_+=\varnothing$. By Separation theorem II we can find a $u\in\Bbb R^n$ and $c\in\Bbb R$ s.t. $u^Tv\le c$ for all $v\in V$ and $u^Tx>c$ for all $x\in X_+$. It follows easily that $c=0$ and $u\succeq 0$ (we do not yet have $u\succ 0$ at this stage though).

I do not know how to attain strict componentwise inequality.

Complete elementary proof:
Please fill out this hungry blank with your brilliant ideas. Thanks guys!

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    $V$ is not a subspace. Take $v=(-1,\dots,-1)$. Then $-v$ is not in $V$. – 2017-02-28
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    @CapimMatinho You have a misunderstanding here perhaps. A subspace doesn't have to contain nontrivial non-negative elements. Consider $\{(-t,t)\mid t\in\Bbb R\}\subset \Bbb R^2$. – 2017-02-28
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    You specified that the inequality is componentwise, so $v^{i} < 0$ for all $v \in V\setminus 0$, $i=1,\dots,n$. Perhaps you could explain a little more about this inequality. – 2017-02-28
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    @CapimMatinho no. $\nsucceq$ doesn't mean $\prec$... componentwise inequality isn't like the usual inequality. – 2017-02-28

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The linear subspace $V$ is an intersection of hyperplanes, i.e. can be described by $V=\{x\in{\Bbb R}^n\colon Ax=0\}$ with some matrix $A$. The condition $V\cap{\Bbb R}^n_+=\{0\}$ is equivalent to $$ Ax=0,\ x\ge 0\quad\Rightarrow\quad x=0, $$ which, in turn, is the same as to say that the system $$ Ax=0,\ x\ge 0,\ e^Tx=1 $$ has no solution. Here $e=(1,1,\ldots,1)^T$. (Clearly it has no solution with $e^Tx>0$, then one can scale $x$ to get no solution with $e^Tx=1$.) Combine the equalities together to conclude that the original information is equivalent to the system $$ \begin{bmatrix}A\\e^T\end{bmatrix}x=\begin{bmatrix}0\\1\end{bmatrix},\ x\ge 0 $$ having no solution.

Now Farkas' lemma says that there must be a solution to $$ [A^T\ e]y\ge 0,\ [0\ 1]^Ty<0. $$ Partitioning $y=\begin{bmatrix}z\\w\end{bmatrix}$ correspondingly gives $$ A^Tz\ge -e w,\ w<0 $$ or with the notaion $v=z/(-w)$ $$ A^Tv\ge e>0. $$ Existence of $v$ implies existence of $u=A^Tv$ that clearly has positive components and orthogonal to $V$ as $u^Tx=v^TAx=0$ for all $x\in V$.

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    Absolutely nice. Farkas lemma still has a flavour of functional analysis or convex analysis though. Perhaps this can't be helped. (I'm trying to give a proof to a freshman so the simpler the better :) – 2017-02-28
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    @Vim Alternatively I can think of proving by Hahn-Banach theorem, dual cones or separation as you suggested ($X_+$ is no good, you better use separation of $V$ and the *closed* $\Bbb R^n_+$, but you have to work it out), but all that seems to be more elaborated. – 2017-03-01