Let $H$ be a Hilbert space and denote as $B(H)$ the bounded linear operators on $H$. Let $M$ be a subset of $B(H)$, s.t. for $A \in M$, also $A^* \in M$.
How can one show that if the commutant has the form $M'=\{\lambda \mathbb{1} : \lambda \in \mathbb{C}\}$, then $M$ is irreducible, i.e. under the action of $M$, the only closed invariant subspaces $G\subset H$ are $G=\{0\}$ and $G=H$?
Suppose $M$ is reducible and $G\subset H$ is a nontrivial closed subspace invariant under all $A\in M.$ Since $M=M^*,$ $G^\perp$ is also invariant under $M$. Denote by $P:H\to G$ the orthogonal projection. Then $P$ is nonscalar and $P\in M'.$ Indeed, let $A\in M.$ Then
if $\varphi\in G$ then $A\varphi\in G$ and $AP\varphi=A\varphi=PA\varphi,$
if $\varphi\in G^\perp$ then $A\varphi\in G^\perp$ and $AP\varphi=0=PA\varphi.$
Hence $AP\varphi=PA\varphi$ for all $\varphi\in H.$