Union of a subspace and its orthogonal complement The following statement seems to be true, but I am not sure:
For any subspace $A$ of $\mathbb{R}^n$, there is a nonzero vector $\vec{x}$ such that $\vec{x}\in A\cup A^\bot$ and each entry of $x$ is nonnegative. 
I am asking for a formal proof or a counterexample. Moreover, does it hold for $A\subseteq \mathbb{C}^n$?
 A: Hint. Show that if $A$ has no nonzero $x$ with all nonnegative entries, then it must have a vector $x$ with at least one coordinate of each sign. Then show there is a vector orthogonal to such an $x$ with all coordinates nonnegative.
A: The cases $A=(\text{the whole space})$ and $A=\{0\}$ are trivial, so I shall assume that $\dim A>0$ and $A$ is proper.
Complex case: No, because there's no guarantee that $(A\cup A^\perp)\cap \Bbb R^n\ne\{0\}$. For instance, consider $\Bbb C^2$, $A=\operatorname{Span}\begin{pmatrix}1\\i\end{pmatrix}$, $A^\perp=\operatorname{Span}\begin{pmatrix}1\\-i\end{pmatrix}$
Real case: Yes. Pick a matrix $Q\in \Bbb R^{n\times (n-\dim A)}$ the columns of which span $A^\perp$ and an analogous matrix $P\in\Bbb R^{n\times\dim A}$ for $A$. Then \begin{cases}x\in A&\iff Q^Tx=0\\ x\in A^\perp&\iff P^Tx=0\end{cases}
Now, there's this corollary of Motzkin's theorem:

Theorem (Gordan): Let $M$ be a matrix. The following are equivalent

*

*The system $\begin{cases} Mx=0\\ x\ge0\\ x\ne0\end{cases}$ has a solution


*The inequation $M^Ty>0$ does not have a solution.

where $x\ge0$ and $M^Ty>0$ are meant as component-wise inequalities (but $x\ne0$ has the usual meaning).
In light of this, negating your claim is equivalent to claiming that the system $$\begin{cases} Qy>0\\ Pz>0\end{cases}$$ has solutions $y,z$.
But, by hypothesis, $Qy\in A^\perp\setminus\{0\}$ and $Pz\in A\setminus\{0\}$, which is not consistent with the assumption that $(A\cup A^\perp)\cap \Bbb R_+^n=\{0\}$
