Dual of a polyhedral cone A general polyhedral cone $\mathcal{P} \subseteq \mathbb{R}^n$ can be represented as either $\mathcal{P} = \\{x \in \mathbb{R}^n : Ax \geq 0 \\}$ or $\mathcal{P} = \\{V x : x \in \mathbb{R}_+^k , V \in \mathbb{R}^{n \times k} \\}$. I am trying to do show the dual of $\mathcal{P}$, $\mathcal{P}^*$, is a polyhedral set. I start by writing $\mathcal{P}^* = \\{ y \in \mathbb{R}^n : y^T V x \geq 0$ $\forall x \in \mathbb{R}_+^k \\}$. (1) A polyhedral set is a set of the form $\\{x \in \mathbb{R}^n : Ax \leq b \\}$ (2). Any ideas how to go from (1) to (2). I also know that an image of a polyhedral set under a linear map is also polyhedral. PS: Similar questions like these have very advanced solution methods; I am looking for a much simpler way.

It's quite non-trivial to prove that your two definitions of polyhedral cone are equivalent. However, granted that, $$\\{y\in\mathbb{R}^n:y^TVx\ge 0\ \forall x \in \mathbb{R}_+^k\\} =\\{y\in\Bbb R^n:x^RV^Ty\ge0\ \forall x\in \Bbb R_+^k\\}=\\{y\in\Bbb R^n:V^Ty\ge0\\}$$ so that $\mathcal{P}^*$ meets your first definition of polyhedral cone.