 Solution

If $q\in \mathcal{Q}$ and $\varphi \in \mathcal{A}$, thenbut strict positivity of $q$ and the nonzero vector $P\varphi \ge 0$ make the left side positive. Thus the two sets cannot both be nonempty.

Conversely, suppose $\mathcal{A}=\varnothing$. The subspace $L=\mathrm{Im}P$ is disjoint from the compact simplexStrict separation gives a vector $q$ that vanishes on $L$ and is positive on all of $\Delta$. The first property says $P^{T}q=0$; testing the vertices of $\Delta$ shows every $q_i>0$. After normalization ${\sum }_i{q}_i=1$, this gives $q\in \mathcal{Q}$. This is Stiemke theorem in normalized form. Therefore