A cone with apex $0$ is said to be pointed if it does not contain any non-trivial subspace. Let C be a closed convex cone with apex $0$. Show that $C$ is full-dimensional if and only if its dual cone $C^*$ is pointed.
I'm doing the forward direction of the proof and need some help. Here's my attempt.
Let $C \subset \mathbb{R}^d$. Assume $S^*$ is a non-trivial subspace of $C^*$. Then for all $z \in S^*$, we have $-z \in S^*$. Since both $z$ and $-z$ are in $C^*$, for all $x \in C$, we have $\langle x,z\rangle \geq 0$ and $\langle x,-z\rangle \geq 0$. So $\langle x,z\rangle = 0$ (1).
It's here that I got stuck. I think I need to either show that, if $x$ is written as $x = \lambda_1i_1 + \dotsb + \lambda_di_d$, where $i_1,...,i_d$ are unit vectors of $\mathbb{R}^d$, then by (1), $\lambda_1z_1 + ... + \lambda_dz_d = 0$, and somehow deduce that at least one of the $\lambda_i = 0$. Or I need to prove my suspection that $S^*$ is not full-dimensional. Both cases would lead to the contradiction that $C$ is not full-dimensional.
Sketch: Now you know that $C$ is contained in the hyperplane $\langle y,z\rangle=0$ (here, $z$ is fixed and $y$ is a variable). Since a hyperplane is of lower dimension than the ambient space, it follows that $C$ is not full dimensional.