State space of finite dimensional, abelian C*-algebra is a simplex. I am looking for a proof that the state space of a finite dimensional C*-algebra is a simplex and, vice versa, if the state space is a simplex, the C*-algebra is abelian. I've found one proof, but it involves representations and the GNS construction, which is way beyond the scope of what I need.
I was hoping that because I only need the result in finite dimensions, there would be a simple proof. Does anyone have an idea?
 A: For the finite-dimensional case: 
You have $\mathcal A=\bigoplus_{k=1}^m M_{n_k}(\mathbb C)\subset M_n(\mathbb C)$, where $n=n_1+\cdots+n_m$. 


*

*Any linear functional on $\mathcal A$ is of the form $f(X)=\sum_{k,j}\beta_{k,j}X_{k,j}$. This is precisely $$f(X)=\operatorname{Tr}(BX),$$ where $B=[\beta_{k,j}]$. We may assume that $B\in \mathcal A$. 

*If $f$ is a state, it follows that $B\geq0$, $\operatorname{Tr}(B)=1$. 

*Within this picture, one can show that the pure states (the extreme points of the state space) are given by the rank-one projections. If $n_k\geq 2$ for at least one index, then there are uncountably many of them. 
When $\mathcal A$ is abelian, $n_1=\cdots=n_m=1$, and there are finitely many rank-one projections. Namely, $\mathcal A=\mathbb C^n$, and the pure states are given by the canonical basis (so, for instance, $\mathbb C^2$ has the pure state $(1,0)$ and $(0,1)$). 
If the state space is a simplex (finitely many pure states), then $\mathcal A$ is (finite-dimensional and) abelian. First let us show that $\mathcal A$ is finite-dimensional. Let $\varphi_1,\ldots,\varphi_n$ be the pure states. Note that the pure states separate points (if $\varphi_1(a)=\cdots=\varphi_n(a)$ for some $a$, then all linear combinations of those numbers will still agree, and so will limits; this means that $\phi(a)=\psi(a)$ for all states, contradicting the fact that states separate points). Suppose that $\dim\mathcal A\geq n+1$. Then there exist $a_1,\ldots,a_{n+1}$, linearly independent. Consider the $n\times(n+1)$ matrix $B=[\varphi_k(a_j)]_{kj}$. Since it has more columns than rows, it has a nontrivial kernel. That is, there exists nonzero $\alpha\in\mathbb C^{n+1}$ such that $B\alpha=0$. Explicitly, $$\sum_j\alpha_j\varphi_k(a_j)=0$$ for all $k$. Since the $\varphi_k$ are linear, this gives $$\varphi_k\left(\sum_j\alpha_ja_j\right)=0$$ for all $k$. But the pure states separate points, so $\sum_j\alpha_ja_j=0$, a contradiction. Thus $\dim\mathcal A\leq n$. 
Once we know that $\mathcal A$ is finite-dimensional, by the bullets above the only way to have finitely many pure states is to have $n_1=\cdots=n_m=1$, so $\mathcal A$ is abelian. 
