Reducing Spaces: Characterization Given a Hilbert space $\mathcal{H}$.

Consider an operator:
$$T:\mathcal{D}(T)\to\mathcal{H}:\quad\mathcal{D}:=\mathcal{D}(T)$$

Regard a subspace:
$$\mathcal{S}\leq\mathcal{H}:\quad\mathcal{H}=\mathcal{S}\oplus\mathcal{S}^\perp$$

Then one has for reducibility:
$$PT\subseteq TP\iff T\mathcal{S}^{(\perp)}\subseteq\mathcal{S}^{(\perp)}\quad(P\mathcal{D}\subseteq\mathcal{D})$$

In general inclusion will be strict:
$$\mathcal{D}(PT)=\mathcal{D}\subsetneq P^{-1}\mathcal{D}=\mathcal{D}(TP)$$

How to prove this from scratch?
This answer is community wiki.

Preparation

Before it always holds:
$$\mathcal{D}\supseteq\mathcal{D}\cap\mathcal{S}+\mathcal{D}\cap\mathcal{S}^\perp$$

Especially one has:
$$P\mathcal{D}\subseteq\mathcal{D}\iff\mathcal{D}=\mathcal{D}\cap\mathcal{S}+\mathcal{D}\cap\mathcal{S}^\perp$$
(That will be crucial below!)

Characterization

On the one hand one has:
$$PT\subseteq TP\implies P\mathcal{D}\subseteq\mathcal{D}$$

Concluding that they're invariant as:
$$\varphi\in\mathcal{D}\cap\mathcal{S}:\quad T\varphi=TP\varphi=PT\varphi\in\mathcal{S}$$
$$\psi\in\mathcal{D}\cap\mathcal{S}^\perp:\quad T\psi=T(1-P)\psi=(1-P)T\psi\in\mathcal{S}^\perp$$

On the other hand one has:
$$P\mathcal{D}\subseteq\mathcal{D}\implies\mathcal{D}(PT)\subseteq\mathcal{D}(TP)$$

Concluding that they commute as:
$$\varphi\in\mathcal{D}:\quad PT\varphi=PT\varphi_\parallel+ PT\varphi_\perp=T\varphi_\parallel=TP\varphi$$
(Here equality above was crucial!)

Strictness

Consider an operator:
$$T:\mathcal{D}(T)\to\mathcal{H}:\quad\mathcal{D}(T)\subsetneq\mathcal{H}$$

Regard any closed subspace:
$$P^2=P=P^*:\quad\mathcal{R}(P)\subseteq\mathcal{D}(T)$$

Then extension will be strict:
$$\mathcal{D}(PT)=\mathcal{D}(T)\subsetneq\mathcal{H}=\mathcal{D}(TP)$$
(Note that happens in most cases!)
This answer is community wiki.

Preparation

Before it always holds:
$$\mathcal{D}\supseteq\mathcal{D}\cap\mathcal{S}+\mathcal{D}\cap\mathcal{S}^\perp$$

Especially one has:
$$P\mathcal{D}\subseteq\mathcal{D}\iff\mathcal{D}=\mathcal{D}\cap\mathcal{S}+\mathcal{D}\cap\mathcal{S}^\perp$$
(That will be crucial below!)

Characterization

On the one hand one has:
$$PT\subseteq TP\implies P\mathcal{D}\subseteq\mathcal{D}$$

Concluding that they're invariant as:
$$\varphi\in\mathcal{D}\cap\mathcal{S}:\quad T\varphi=TP\varphi=PT\varphi\in\mathcal{S}$$
$$\psi\in\mathcal{D}\cap\mathcal{S}^\perp:\quad T\psi=T(1-P)\psi=(1-P)T\psi\in\mathcal{S}^\perp$$

On the other hand one has:
$$P\mathcal{D}\subseteq\mathcal{D}\implies\mathcal{D}(PT)\subseteq\mathcal{D}(TP)$$

Concluding that they commute as:
$$\varphi\in\mathcal{D}:\quad PT\varphi=PT\varphi_\parallel+ PT\varphi_\perp=T\varphi_\parallel=TP\varphi$$
(Here equality above was crucial!)

Strictness

Consider an operator:
$$T:\mathcal{D}(T)\to\mathcal{H}:\quad\mathcal{D}(T)\subsetneq\mathcal{H}$$

Regard any closed subspace:
$$P^2=P=P^*:\quad\mathcal{R}(P)\subseteq\mathcal{D}(T)$$

Then extension will be strict:
$$\mathcal{D}(PT)=\mathcal{D}(T)\subsetneq\mathcal{H}=\mathcal{D}(TP)$$
(Note that happens in most cases!)