Let $C = \{A \lambda \, | \, \lambda \geq 0\}$ be a finitely generated cone for some $A \in \mathbb{R}^{m \times n}$. Suppose the columns of $A$ do not span $\mathbb{R}^m$. Let $S$ denote the column space of $A$ and write the polar cone of $S$ as $S^{\star} := \{y \in \mathbb{R}^m \, | \, \langle y,x\rangle \leq 0, \, \forall \, x \in S\}$.
Show that $S^{\star} + C$ is a finitely generated cone, and the generators span $\mathbb{R}^m$.
One way that I'm trying to tackle this problem is to find some nonempty set $T = \{t_1, \dotsc, t_k\}$ such that $\text{cone}(T) = S^{\star} + C$ and $\text{span}(t_1, \dotsc, t_k) = \mathbb{R}^m$. The obvious set of vectors that span $\mathbb{R}^m$ would be the standard basis, but that got me nowhere.
A typical element in $S^{\star} + C$ has the form $s + \sum_{i=1}^n \lambda_i A_i$ with $\lambda_i \geq 0$ and $s$ having the property that $\sum_{i=1}^n c_i \langle y, A_i \rangle \leq 0$ for $c_i \in \mathbb{R}$. Given the form of this element, I am trying to come up with a set of generators, but I'm getting nowhere.
If someone could offer a hint on what I should consider, that would be amazing.
Consider the subset $S_0$ of $S^\prime,$ given by
$S_0 = \{y \in \mathbb{R}^m \left| (y, x) = 0, \forall x \in S\right.\}.$ Notice that it is nothing more (or less) than the orthogonal complement of $S.$