[proofplan] We compare three ways of saying that $T$ respects the [orthogonal decomposition](/theorems/436) $H=M\oplus M^\perp$. First, the adjoint identity transfers invariance of $M^\perp$ under $T$ into invariance of $M$ under $T^*$, and conversely transfers invariance of $M$ under $T^*$ into invariance of $M^\perp$ under $T$. Then we show that commuting with the [orthogonal projection](/theorems/437) $P_M$ is exactly the operator equation expressing that both summands $M$ and $M^\perp$ are preserved by $T$. [/proofplan]

[step:Transfer invariance between $T$ and $T^*$ using orthogonality]Assume first that $M$ reduces $T$. By definition, $T(M)\subset M$ and $T(M^\perp)\subset M^\perp$. Let $x\in M$ and $y\in M^\perp$. By the defining identity of the [Hilbert space](/page/Hilbert%20Space) adjoint, with the [inner product](/page/Inner%20Product) linear in the first variable, \begin{align*} (T^*x,y)_H=(x,Ty)_H. \end{align*} Since $Ty\in M^\perp$ and $x\in M$, the right-hand side is $0$. Therefore $T^*x$ is orthogonal to every element of $M^\perp$. Because $M$ is closed, $(M^\perp)^\perp=M$, so $T^*x\in M$. Hence $T^*(M)\subset M$, and condition (ii) holds. Conversely, assume condition (ii): $T(M)\subset M$ and $T^*(M)\subset M$. To prove that $M$ reduces $T$, it remains only to show that $T(M^\perp)\subset M^\perp$. Let $y\in M^\perp$ and $m\in M$. The adjoint identity gives \begin{align*} (Ty,m)_H=(y,T^*m)_H. \end{align*} Since $T^*m\in M$ and $y\in M^\perp$, the right-hand side is $0$. Thus $Ty$ is orthogonal to every $m\in M$, so $Ty\in M^\perp$. Therefore $T(M^\perp)\subset M^\perp$, and $M$ reduces $T$.[/step]

[guided]The point of this step is that orthogonal complements are detected by inner products. We first assume that $M$ reduces $T$, so by definition both $M$ and $M^\perp$ are invariant under $T$: \begin{align*} T(M)\subset M \quad \text{and} \quad T(M^\perp)\subset M^\perp. \end{align*} We must prove that $M$ is invariant under $T^*$. Let $x\in M$ be arbitrary. To prove $T^*x\in M$, it is enough to prove that $T^*x$ is orthogonal to $M^\perp$, because $M$ is closed and hence $(M^\perp)^\perp=M$. Let $y\in M^\perp$. By the defining property of the adjoint, using the convention that $(\cdot,\cdot)_H$ is linear in the first variable, \begin{align*} (T^*x,y)_H=(x,Ty)_H. \end{align*} Since $M^\perp$ is invariant under $T$, we have $Ty\in M^\perp$. Since $x\in M$ and $M$ is orthogonal to $M^\perp$, it follows that \begin{align*} (x,Ty)_H=0. \end{align*} Therefore $(T^*x,y)_H=0$ for every $y\in M^\perp$, so $T^*x\in(M^\perp)^\perp=M$. Since $x\in M$ was arbitrary, $T^*(M)\subset M$. Together with $T(M)\subset M$, this proves condition (ii). Now assume condition (ii), namely \begin{align*} T(M)\subset M \quad \text{and} \quad T^*(M)\subset M. \end{align*} To prove that $M$ reduces $T$, the missing part is invariance of $M^\perp$ under $T$. Let $y\in M^\perp$. We prove $Ty\in M^\perp$ by testing against an arbitrary element $m\in M$. The adjoint identity gives \begin{align*} (Ty,m)_H=(y,T^*m)_H. \end{align*} Because $T^*(M)\subset M$, the vector $T^*m$ belongs to $M$. Since $y\in M^\perp$, we get \begin{align*} (y,T^*m)_H=0. \end{align*} Thus $(Ty,m)_H=0$ for every $m\in M$, which is precisely the statement that $Ty\in M^\perp$. Hence $T(M^\perp)\subset M^\perp$. Combining this with the assumed invariance $T(M)\subset M$, we conclude that $M$ reduces $T$.[/guided]

[step:Show that reduction implies commutation with the orthogonal projection] Assume that $M$ reduces $T$. Let $x\in H$. Since $M$ is closed, the [orthogonal projection theorem](/theorems/4916) gives a unique decomposition \begin{align*} x=m+n \end{align*} with $m\in M$ and $n\in M^\perp$, and $P_Mx=m$. Since $T(M)\subset M$ and $T(M^\perp)\subset M^\perp$, we have $Tm\in M$ and $Tn\in M^\perp$. Therefore \begin{align*} P_MTx=P_M(Tm+Tn)=Tm. \end{align*} Also, \begin{align*} TP_Mx=T m. \end{align*} Thus $P_MTx=TP_Mx$ for every $x\in H$, so $P_MT=TP_M$. [/step]

[step:Show that commutation with the orthogonal projection preserves both summands] Assume that $P_MT=TP_M$. Let $m\in M$. Since $P_Mm=m$, we obtain \begin{align*} P_MTm=TP_Mm=Tm. \end{align*} A vector $u\in H$ satisfies $P_Mu=u$ exactly when $u\in M$, so $Tm\in M$. Hence $T(M)\subset M$. Let $n\in M^\perp$. Since $P_Mn=0$, we obtain \begin{align*} P_MTn=TP_Mn=0. \end{align*} A vector $u\in H$ satisfies $P_Mu=0$ exactly when $u\in M^\perp$, so $Tn\in M^\perp$. Hence $T(M^\perp)\subset M^\perp$. Therefore $M$ reduces $T$. [/step]

[step:Conclude the equivalence of the three characterisations] The first step proves that condition (i) is equivalent to condition (ii). The second and third steps prove that condition (i) is equivalent to condition (iii). Therefore conditions (i), (ii), and (iii) are all equivalent. [/step]