Elementary and short proof for this simple linear algebra fact, perhaps involving some convex analysis too 
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!
 A: 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 $x\in V$.
