Existence of $V \subset \mathbb{R}^n : V, V^\perp \cap \mathbb{R}_{\geq0}^n = 0$ Is there a subspace $V \subset \mathbb{R}^n$ with $V \cap \mathbb{R}_{\geq0}^n = V^\perp \cap \mathbb{R}_{\geq0}^n = 0$?
My geometric intuition tells me that there is no such $V$ but i'am completely stuck on proof.

Suppose $V\cap \mathbb{R}^n_{\geq 0} = \{0\}$, and let $P$ denote orthogonal projection onto $V$.  A vector $w$ is in $V^\perp$ iff $P(w)=0$.
Let $H$ be the convex hull of $\{e_1,e_2,\dots,e_n\}$, where $e_i$ is a standard basis vector.  Note that $H\subset \mathbb{R}^n_{\geq 0}$.  The projection $P(H)$ is a convex subset of $V$.  For it to not contain $0$, there must be a hyperplane separating $0$ and $P(H)$, which means there exists some non-zero $v\in V$ such that $\forall(w\in P(H))\: w\cdot v > 0$.  
But the condition that $V\cap \mathbb{R}^n_{\geq 0} = 0 $ means that for any non-zero $v\in V$ the components of $v$ cannot all be non-negative (nor all non-positive, else $-v\in\mathbb{R}^n_{\geq 0}$).  Hence for any $v\in V$ there is some $e_i$ such that $v\cdot e_i < 0$.  Thus $v\cdot P(e_i) = v\cdot e_i < 0$, and since $P(e_i)\in P(H)$, so there cannot be a hyperplane separating $0$ and $P(H)$.
Hence $0\in P(H)$, and thus $H\subset \mathbb{R}^n_{\geq}$ intersects $V^\perp$.  
