Irreducible C*-algebra $A$ implies that projection $p$ is rank one if $pAp=\Bbb C p$

Question

Let $A$ be an irreducible C*-subalgebra of $B(H)$ and $p$ be a nonzero projection in $B(H)$. Suppose $pAp=\Bbb C p$, show that $p$ is rank one.

I do not have any idea about it. Please give me a hint. Thanks.

Answer 1

Let $p\ne 0$. If $A$ is irreducible, then each nonzero vector $\xi \in H$ is cyclic (otherwise $\overline{A\xi}$ would be a proper nontrivial invariant subspace). Assume that $\xi \in H$ is a nonzero vector such that $p\xi=\xi$. Then $(Ap)\xi=A\xi=\{ x\xi; x\in A\}$ is a dense subset of $H$. Because of $pAp={\mathbb C}p$ we see that $p\bigl( (Ap)\xi\bigr)={\mathbb C}p\xi={\mathbb C}\xi$, i.e., $p$ is of rank one (because we actually have $p(H)={\mathbb C}\xi$).