Let $M$ be a von Neumann algebra in $B(H)$. Assume that there exists a vector $\zeta\in H$ such that for any vector $h\in H$, one may find $\eta\in H$ and operators $T,S$ in $M$ with
$\bullet$ $T\eta=\zeta$ and $\eta\in \ker T^{\perp}$.
$\bullet$ $S\eta=h$.
Q. Can we conclude that $M$ is a cyclic Von Neumann algebra?
Remark. This property is valid in any cyclic Von Neumann algebra (see What is the idea behind the proof of the Theorem 7.2.1 in this book : Fundamental of the theory of operator algebras Vol II (Kadison and Ringros)).