# 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?

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!)