How does one show that irreducibility is equivalent to a vector state being pure?
In what follows I will fill in the details of the question: Let $\mathcal{H}$ denote a Hilbert space and let $\mathfrak{A}$ be the set of all linear transformations defined on all of $\mathcal{H}$. The latter is just the set of bounded operators on $\mathcal{H}$.
Definition. A vector state $\omega: \mathfrak{A} \to \Bbb F$ is given by the expectation of $a \in \mathfrak{A}$ in a vector $\Omega \in \mathcal{H},$ $$\omega (a) = \langle \Omega, a \Omega \rangle_{\mathcal{H}}$$ where $\Omega$ is a unit vector.
We say that a vector state is pure if it has no non-trivial decomposition into a convex sum of other states. When is a vector state pure? It turns out that purity is linked to the following algebraic property, namely is the algebra $\pi(\mathfrak{A})$ irreducible? Irreducibility means that the only element in the algebra commuting with the whole algebra is a multiple of the identity.
I'm stuck with the proof that shows equivalence of the aforementioned algebraic property and purity of a vector state. Help along these lines would be greatly appreciated.
Furthermore, can I phrase irreducibility in terms of the algebra's commutant, i.e. the algebra is irreducible iff the commutant contains the identity element and scalar multiples of it?