Let $a_1,\dots, a_m$ be vectors in $\mathbb R^n$, let $A\in\mathbb R^{n\times m}$ whose columns are $a_i$s. The cone generated by $A$ (denoted $\operatorname{cone}A$) is $C=\{ Aw: 0\leq w \}$, its dual cone is \begin{align*} C^o=\{ y: \forall i\in[1:m], 0\leq \langle a_i, y\rangle\} \end{align*}
I am trying to study the basis of this cone, I am interested in practical ways of computing it, and the maximal size of this basis.
As an example, if all pairs of columns of $A$ have positive inner product, then $C\subseteq C^o$ and therefore the basis is exactly $\{ a_1,\dots, a_m\}$.
Another extreme case is when $C^o$ is strictly contained in $C$ which happens for instance when all inner products between columns of $A$ are negative, suppose further that $A$ has rank $m$. In this case, the basis can be built by taking any $i\in [1:m]$, then some vector in the set $\{ y : \forall j\neq i, \langle a_j, y \rangle = 0 \}$ would work, and in particular $p_{C^o}(a_i)$ works, i.e. the projection of $A_i$ onto $C^o$. Note that in the case $C\subseteq C^o$ above, $p_{C^o}(a_i)=a_i$ and so in both cases $p_{C^o}(a_i)$ is extreme.
I'm wondering if this always hold and if $\{ p_{C^o}(a_i) : i\in[1:m] \}$ generates the cone $C^o\cap C$.