Polyhedral cone as conic hull of a finite set I am reading notes on optimization and it was claimed that all polyhedral cones in $K\subseteq \mathbb{R}^n$ can be written Cone(R) where $R\subseteq \mathbb{R}^n$ is a finite set.  That is, if K is a polyhedral cone, then $K=\{\sum_{i=1}^k a_i h_i: a_i\geq 0 \}$ for some finite set $R:=\{h_1,...,h_k\}\subseteq \mathbb{R}^n$. 
I would like to convince myself this is true.  It seems intuitively obvious, but I don't know how to go about showing it.  I already know that all bounded, polyhedral sets can be expressed as the convex hull of a finite # of pts.  One direction I am considering is how to use that result here.
 A: The polyhedral cone $K$ is defined as an intersection of a finite number of half-spaces, i.e. $K=\{x\in\mathbb{R}^n\colon Ax\ge 0\}$, where $A\in\mathbb{R}^{m\times n}$. Since $\text{Im}\,A$ is a subspace, it can be represented as a kernel of some matrix $M$, that is $\ker M=\text{Im} A$. Hence, we have
$$
y=Ax,\ x\in K\qquad\Leftrightarrow\qquad y\in Y=\{y\in\mathbb{R}^m\colon\ My=0,\ y\ge 0\}.\tag1
$$
Introduce the set
$$
P=\{z\in Y\colon\ (\matrix{1 & 1 & \ldots & 1})z=1\}.
$$
It is a bounded polyhedral set, thus, finitely generated (according to what you know)
$$
\exists z_1,z_2,\ldots,z_N\in P\colon\ P=\text{conv}\{z_1,z_2,\ldots,z_N\}.
$$
Therefore, even $Y$ is a finitely generated (positive) cone
$$
Y=\text{cone}\{z_1,z_2,\ldots,z_N\}
$$
since any $y\in Y$ is a non-negative scaling of some $z\in P$. 
Now by $(1)$ we can pick $x_k\in K$ such that $z_k=Ax_k$, and we are almost done finding a finite generating set for $K$. The minor trouble left is $\ker A$. Actually, it is quite easy to see that
$$
K=\ker A+\text{cone}\{x_1,x_2,\ldots,x_N\}.
$$
I leave it as an exercise (together with the fact that $\ker A$ is a finitely generated cone).
