Strongly convex cones A polyhedral cone is strongly convex if $\sigma \cap -\sigma =\{0\}$is a face. Then here is the following proposition.
Let $\sigma$ be a strongly convex polyhedral cone. Then the following are equivalent.
$\sigma$ contains no positive dimensional subspace of $N_\mathbb{R}$
$\sigma \cap (-\sigma)= \{0\}$ 
$\dim(\sigma^\vee)=n$
This proposition can be found in almost every paper I can find online which talks about strongly convex polyhedral cones, but I cant find a single one which includes the proof. Anyone who knows a rigorous proof who can help me, or who knows where to find one?
 A: You can likely find a proof in either of Cox, Little and Schenk's Toric Varieties or Fulton's Introduction to Toric Varieties.  Both are good references for these matters, particularly if you are interested in learning about toric varieties.  Ziegler's Lectures on Polytopes is another good source for this information.
Calling your statements $(a)$, $(b)$ and $(c)$, here's a quick sketch:
$(b) \Leftrightarrow (a)$: If $\sigma$ contained a subspace, then $-\sigma$ would contain the same subspace since a subspace is closed under negation.  Thus $\sigma \cap -\sigma = \{ 0 \}$ implies that $\sigma$ contains no subspace besides the $0$ subspace.  Conversely, if there was a non-zero vector $v \in \sigma \cap -\sigma$, then it follows easily that $\mathbb{R}v$ (the subspace spanned by $v$) belongs to $\sigma$.
$(c) \Leftrightarrow (b)$:  Suppose $v$ is a non-zero vector in $\sigma \cap -\sigma$.  Then we claim that $\sigma^{\vee}$ is contained in $M_{\mathbb{R}} \cap v^{\perp} := \{ w \in M_{\mathbb{R}} : \langle w, v \rangle = 0 \}$.  Since this is an $n-1$ dimensional subspace, it follows that $\sigma^{\vee}$ is not $n$-dimensional, proving $(c) \Rightarrow (b)$.  For the claim, if $w \in \sigma^{\vee}$, then $\langle w, v \rangle \geq 0$ by definition, while $\langle w, -v \rangle \geq 0$ since $-v \in \sigma$ by hypothesis.  Thus $\langle w, v \rangle \leq 0$, so $\langle w, v \rangle = 0$.
For the other direction, suppose $\sigma^{\vee}$ is not $n$-dimensional.  Then $\sigma^{\vee}$ is contained in a hyperplane $H \subseteq M_{\mathbb{R}}$, i.e. there exists $v \in N_{\mathbb{R}}$ such that $\sigma^{\vee} \subseteq M_{\mathbb{R}} \cap v^{\perp}$.  Then $\langle w, v \rangle = 0$ for all $w \in \sigma^{\vee}$ shows that both $v$ and $-v$ belong to $(\sigma^{\vee})^{\vee}$.  Then it is a non-trivial fact that $(\sigma^{\vee})^{\vee} = \sigma$, which gives $v \in \sigma \cap -\sigma$.
