Proving a vector state is pure Let $\mathcal{M}$ be a until $*$-algebra of 3x3 complex matrices. We have the general form of a vector state $\omega_{\psi} : \mathcal{M} \to \mathbb{C}$ over $\mathbb{C}^3$ as given by
$$\omega_{\psi}(A) := \langle \psi, A\psi \rangle, \quad A\in \mathcal{M}, \psi \in \mathbb{C}^3, \|\psi \|=1.$$
I need to pick an $\psi$ such that the above is a pure state (i.e. it cannot be written as a convex combination $\lambda \omega_{\sigma} + (1-\lambda)\omega_{\phi}$ of pure states $\omega_{\sigma}, \omega_{\phi}$ and $\lambda \in (0,1)$).
The proof I have formulated is to pick a basis of $\mathbb{C}^3$ over $\mathbb{C}$ and take $\psi$ to be one of these basis vectors, for example 
$$\psi = \begin{pmatrix} 1\\0\\0 \end{pmatrix}.$$
In this case a convex combination would be of the form
$$\langle \sqrt{\lambda}\sigma + \sqrt{1-\lambda}\phi, A\left(\sqrt{\lambda}\sigma + \sqrt{1-\lambda}\phi \right)\rangle,$$
which implies that for $\omega_{\psi}$ to be mixed, $\psi$ must be a linear combination of the form $\sqrt{\lambda}\sigma + \sqrt{1-\lambda}\phi$, but since it is a basis vector, this isn't the case and we have a contradiction. 
I have a feeling this isn't totally sufficient as a proof - does anyone have any ideas/hints?
 A: I don't think your argument works. For instance you can write 
$$
\begin{bmatrix} 1\\ 0\\0\end{bmatrix} 
= \sqrt{\frac12}\,\begin{bmatrix} 1/\sqrt2\\ 1/\sqrt2\\0\end{bmatrix} 
+
\sqrt{1-\frac12}\,\begin{bmatrix} 1/\sqrt2\\ -1/\sqrt2\\0\end{bmatrix} 
$$
Now, any state of the form $\omega_\psi$ is pure. Suppose that $\omega_\psi=\lambda\omega_\sigma+(1-\lambda)\omega_\phi$. Take $A=P_\psi$, the projection onto $\mathbb C\psi$. Then
$$\tag1
1=\langle \psi,P_\psi\psi\rangle=\lambda\langle \sigma,P_\psi\sigma\rangle+(1-\lambda)\langle \phi,P_\psi\phi\rangle. 
$$
Since $0\leq \langle \sigma,P_\psi\sigma\rangle\leq1$ and $0\leq \langle \phi,P_\psi\phi\rangle\leq1$, it follows that 
$$
1=\langle \sigma,P_\psi\sigma\rangle=\langle \phi,P_\psi\phi\rangle.
$$
Then, since $P_\psi$ is a projection, $1=\langle \sigma,P_\psi\sigma\rangle=\langle P_\psi\sigma,P_\psi\sigma\rangle=\|P_\psi\sigma\|^2.$ From 
$$
1=\|\sigma\|^2=\|P_\psi\sigma+(1-P_\psi)\sigma\|^2=\|P_\psi\sigma\|^2+\|(1-P_\psi)\sigma\|^2=1+\|(1-P_\psi)\sigma\|^2,
$$
we conclude that $(1-P_\psi)\sigma=0$. In other words, $P_\psi\sigma=\sigma$. So 
$$
\sigma=P_\psi\sigma=\langle\psi,\sigma\rangle\,\psi.
$$
That is, $\sigma=\alpha\,\psi$, with $|\alpha|=1$. Similarly, $\phi=\beta\psi$ with $|\beta|=1$. Thus $\omega_\sigma=\omega_\phi=\omega_\psi$. 
The above reasoning can be made more general. You can consider states of the form $A\longmapsto \operatorname{Tr}(XA)$ for some $X$ with $X\geq0$ and $\operatorname{Tr}(X)=1$. Your $\omega_\psi$ is obtained when $X=P_\psi$. Even in this setting, the pure states are precisely the point states $\omega_\psi$ for some $\psi$, and the proof is basically the same as above. 
