[proofplan] We prove the two implications directly. If $\xi$ is cyclic and an operator in the commutant annihilates $\xi$, then it annihilates every vector of the form $S\xi$, hence a dense subspace of $H$, so it is zero. Conversely, if $\xi$ is not cyclic, the closed subspace generated by $M\xi$ is a proper reducing subspace for $M$; its [orthogonal projection](/theorems/437) therefore belongs to $M'$, and the complementary projection is a nonzero element of $M'$ annihilating $\xi$. [/proofplan] [step:Use cyclicity to force every commutant operator annihilating $\xi$ to vanish] Assume first that $\xi$ is cyclic for $M$. Let $T\in M'$ satisfy $T\xi=0$. For every $S\in M$, the defining commutation relation for $T\in M'$ gives $TS=ST$. Applying both sides to $\xi$ yields \begin{align*} T(S\xi) = S(T\xi) \end{align*} and since $T\xi=0$, we obtain \begin{align*} T(S\xi)=0. \end{align*} Thus $T$ vanishes on the subspace $\{S\xi:S\in M\}$. Since $\xi$ is cyclic for $M$, the closure of this subspace is all of $H$. The operator $T$ is bounded because $T\in\mathcal{L}(H)$, hence continuous. A continuous linear operator that vanishes on a dense subspace vanishes on its closure, so $T=0$ on $H$. Therefore $\xi$ is separating for $M'$. [guided] Assume that $\xi$ is cyclic for $M$. To prove that $\xi$ is separating for $M'$, we must start with an arbitrary operator $T\in M'$ such that $T\xi=0$ and prove $T=0$. The point of the commutant condition is that $T$ can be moved past every operator in $M$. Let $S\in M$. Since $T\in M'$, we have $TS=ST$. Applying this equality to $\xi$ gives $T(S\xi)=S(T\xi)=S0=0$. Thus $T$ annihilates every vector generated from $\xi$ by the action of $M$. By cyclicity, those generated vectors are dense: $\overline{\{S\xi:S\in M\}}=H$. Because $T\in\mathcal{L}(H)$, the operator $T$ is bounded and therefore continuous. If $h\in H$, choose a net or sequence $(h_i)$ in $\{S\xi:S\in M\}$ converging to $h$. Since $T h_i=0$ for every index $i$, continuity gives $T h=\lim_i T h_i=0$. Hence $T$ vanishes on every vector $h\in H$, so $T=0$. This is exactly the separating condition for $\xi$ with respect to $M'$. [/guided] [/step] [step:Show the cyclic subspace is reducing for $M$ when $M$ is self-adjoint] Assume conversely that $\xi$ is not cyclic for $M$. Define the closed subspace $K:=\overline{\{S\xi:S\in M\}}\subsetneq H$. For every $A\in M$ and every $S\in M$, the product $AS$ belongs to $M$, so $A(S\xi)=(AS)\xi\in \{R\xi:R\in M\}$. Since $A$ is bounded, it follows by continuity that $A K\subseteq K$. Because $M$ is self-adjoint, $A^*\in M$ for every $A\in M$. Applying the preceding invariance statement to $A^*$ gives $A^*K\subseteq K$. Now let $y\in K^\perp$ and $k\in K$. Using the [Hilbert space](/page/Hilbert%20Space) adjoint relation, $(Ay,k)_H=(y,A^*k)_H$. Since $A^*k\in K$ and $y\in K^\perp$, the right-hand side is $0$. Hence $Ay\in K^\perp$. Therefore both $K$ and $K^\perp$ are invariant under every $A\in M$, so $K$ reduces $M$. [/step] [step:Put the projection onto the proper cyclic subspace in the commutant] Let $I_H: H\to H$ denote the identity operator on $H$, so $I_H\in\mathcal{L}(H)$. Let $P\in\mathcal{L}(H)$ be the orthogonal projection onto $K$. Since $H=K\oplus K^\perp$ and both $K$ and $K^\perp$ are invariant under every $A\in M$, for each $h\in H$ there are unique vectors $k\in K$ and $y\in K^\perp$ such that $h=k+y$, and $P A h=P(Ak+Ay)=Ak$. Also, $A P h=A k$. Thus $PAh=APh$ for every $h\in H$, so $PA=AP$ for every $A\in M$. Hence $P\in M'$. Since $M$ is unital, $I_H\in M$, and therefore $\xi=I_H\xi\in \{S\xi:S\in M\}\subseteq K$. Thus $P\xi=\xi$, so $(I_H-P)\xi=0$. Because $K\subsetneq H$, the orthogonal complement $K^\perp$ is nonzero, and the projection $I_H-P$ onto $K^\perp$ is a nonzero operator. Since $P\in M'$ and $I_H\in M'$, we also have $I_H-P\in M'$. We have found a nonzero operator $I_H-P\in M'$ such that $(I_H-P)\xi=0$. Therefore $\xi$ is not separating for $M'$. [/step] [step:Conclude the equivalence] The first step proves that cyclicity of $\xi$ for $M$ implies that $\xi$ is separating for $M'$. The last two steps prove the contrapositive of the converse: if $\xi$ is not cyclic for $M$, then $\xi$ is not separating for $M'$. Therefore $\xi$ is cyclic for $M$ if and only if $\xi$ is separating for $M'$. [/step]