# Convex hull and projective matrices

If $$\mathcal{A}$$ is a unital $$C^*$$-algebra then I want to show that $$\operatorname{conv}(\operatorname{Proj}(\mathcal{A}))=\mathcal{P}_1(\mathcal{A}):= \lbrace x \in \mathcal{A^+} : \Vert x \Vert \leq 1 \rbrace$$ when $$\mathcal{A}= M_n(\mathbb{C})$$ for some $$n \geq 2$$.

Previously I have shown that the set of extreme points of $$\mathcal{P}_1(\mathcal{A})$$ is equal to $$\operatorname{Proj}(\mathcal{A})$$. Thus I can write $$\operatorname{Ext}(\mathcal{P}_1(\mathcal{A})) = \operatorname{Proj}(\mathcal{A})$$ which then implies that

$$\operatorname{conv}(\operatorname{Proj}(\mathcal{A}))= \operatorname{conv} (\operatorname{Ext}(\mathcal{P}_1(\mathcal{A}))$$

Where $$\operatorname{conv}$$ is the convex hull. By the Krein-Milman Theorem, I then have that

$$\overline{\operatorname{conv}(\operatorname{Ext}(\mathcal{P}_1(\mathcal{A}))}= \mathcal{P}_1(\mathcal{A})$$

So now I want to proceed from here. If $$\operatorname{conv}(\operatorname{Ext}(\mathcal{P}_1(\mathcal{A}))$$ is a closed set then it coincides with the closure (right?) but I am not sure how to prove this.

Because you are in finite dimension and $$\operatorname{Proj}\mathcal A$$ is closed and bounded, it is compact. Caratheodory's Theorem then guarantees that you can use at most $$n^2+1$$ terms in your convex combinations; a consequence of that is that $$\operatorname{conv}(\operatorname{Proj}\mathcal A)$$ is closed