Let $\mathcal{M}\subseteq B(\mathcal{H})$ be a von Neumann algebra with a given cyclic vector $\xi\in\mathcal{H}$, i.e. $[\mathcal{M}\xi]=\mathcal{H}$ where $[\mathcal{M}\xi]$ is the closed vector subspace generated by $\mathcal{M}\xi$. Let us consider the commutant $\mathcal{M}'$ of $\mathcal{M}$ defined by $\mathcal{M}'=\{y\in B(\mathcal{H}):xy=yx \text{ for all } x\in B(\mathcal{H})\}$.
Question: Is it true that the vector $\xi$ becomes cyclic with respect to $\mathcal{M}'$ as well, i.e. $[\mathcal{M}'\xi]=\mathcal{H}$?
No, this is not true. For example, every non-zero vector is cyclic for $B(H)$, yet $B(H)^\prime=\mathbb{C} I$ does not have any cyclic vectors (unless $\operatorname{dim} H\leq 1$).
But cyclic vectors can be characterized in terms of the commutant: A vector $\xi$ is cyclic for $\mathcal{M}$ if and only if it is separating for $\mathcal{M}^\prime$, that is, $x\xi=0$ implies $x=0$ for $x\in \mathcal{M}^\prime$.
I suggest you try and prove this yourself. If necessary, you can find the proof in the first volume of Takesaki's book on operator algebras.