[proofplan] We order all mutually orthogonal families of nonzero closed $C^*(T)$-cyclic reducing subspaces by inclusion and apply [Zorn's lemma](/theorems/1226) to obtain a maximal such family. The union of a chain is still an admissible family because orthogonality and the reducing-cyclic property are properties of the individual members. We then take the closed orthogonal direct sum of a maximal family and show that its orthogonal complement must be zero: otherwise a nonzero vector in the complement generates a new closed cyclic reducing subspace, contradicting maximality. [/proofplan] [step:Handle the zero Hilbert space by the empty decomposition] If $H=\{0\}$, take $I=\varnothing$. Then the Hilbert-space orthogonal direct sum over the empty family is $\{0\}$, and there are no cyclicity conditions to verify. Hence assume for the rest of the proof that $H\ne \{0\}$. [/step] [step:Partially order mutually orthogonal cyclic reducing families] Call a family $\mathcal{F}$ admissible if every member $K\in\mathcal{F}$ is a nonzero closed subspace of $H$ reducing $T$, distinct members of $\mathcal{F}$ are orthogonal, and for each $K\in\mathcal{F}$, with \begin{align*} T_K:=T|_K\in\mathcal{L}(K), \end{align*} there exists $x_K\in K$ such that \begin{align*} K=\overline{\operatorname{span}}\{p(T_K,T_K^*)x_K: p \text{ is a complex } *\text{-polynomial in two variables}\}. \end{align*} Let $\mathscr{P}$ be the collection of all admissible families, ordered by inclusion of families. The collection $\mathscr{P}$ is nonempty. Indeed, choose $0\ne x\in H$ and define \begin{align*} K_x:=\overline{\operatorname{span}}\{p(T,T^*)x: p \text{ is a complex } *\text{-polynomial in two variables}\}. \end{align*} Then $K_x\ne \{0\}$ because $x=1(T,T^*)x\in K_x$. The subspace $K_x$ is closed by definition. We verify that $K_x$ reduces $T$. If $y$ belongs to the algebraic span of the vectors $p(T,T^*)x$, then both $Ty$ and $T^*y$ again belong to that algebraic span, because multiplying a $*\text{-polynomial}$ on the left by the variables corresponding to $T$ or $T^*$ gives another $*\text{-polynomial}$ in $T$ and $T^*$. Since $T$ and $T^*$ are bounded operators, they preserve the closure of this span. Hence $K_x$ is invariant under both $T$ and $T^*$. Because $K_x$ is closed, this implies that $K_x$ reduces $T$: if $z\in K_x^\perp$ and $k\in K_x$, then \begin{align*} (Tz,k)_H=(z,T^*k)_H=0 \end{align*} and \begin{align*} (T^*z,k)_H=(z,Tk)_H=0, \end{align*} since $Tk,T^*k\in K_x$. Thus $K_x^\perp$ is also invariant under both $T$ and $T^*$. Therefore $\{K_x\}$ is an admissible family, so $\mathscr{P}\ne\varnothing$. [guided] We first set up the object to which Zorn's lemma will be applied. An admissible family is a collection of mutually orthogonal pieces, where each piece is already one of the cyclic reducing subspaces we want in the final decomposition. The only point requiring verification is that at least one such piece exists when $H\ne\{0\}$. Choose a vector $x\in H$ with $x\ne 0$ and form the closed subspace generated from $x$ by all $*\text{-polynomials}$ in $T$ and $T^*$: \begin{align*} K_x:=\overline{\operatorname{span}}\{p(T,T^*)x: p \text{ is a complex } *\text{-polynomial in two variables}\}. \end{align*} This subspace is nonzero because the constant polynomial $1$ gives $x=1(T,T^*)x\in K_x$. Why is $K_x$ reducing? The construction includes both $T$ and $T^*$, not only $T$. If $p$ is a $*\text{-polynomial}$, then $T p(T,T^*)x$ and $T^*p(T,T^*)x$ are again obtained by applying $*\text{-polynomials}$ in $T$ and $T^*$ to $x$. By linearity, the algebraic span is invariant under both $T$ and $T^*$. Since $T$ and $T^*$ are bounded, this invariance passes to the closure $K_x$. For a closed subspace, invariance under both $T$ and $T^*$ implies reducing. Indeed, let $z\in K_x^\perp$ and $k\in K_x$. Since $T^*k\in K_x$, we have \begin{align*} (Tz,k)_H=(z,T^*k)_H=0. \end{align*} Since $Tk\in K_x$, we also have \begin{align*} (T^*z,k)_H=(z,Tk)_H=0. \end{align*} Thus $Tz,T^*z\in K_x^\perp$, so $K_x^\perp$ is invariant under both $T$ and $T^*$. Hence $K_x$ reduces $T$. Therefore the singleton family $\{K_x\}$ is admissible. [/guided] [/step] [step:Apply Zorn's lemma to obtain a maximal admissible family] Let $\mathcal{C}$ be a chain in $\mathscr{P}$, and define \begin{align*} \mathcal{U}:=\bigcup_{\mathcal{F}\in\mathcal{C}}\mathcal{F}. \end{align*} Every member of $\mathcal{U}$ is a nonzero closed cyclic reducing subspace, because it belongs to some admissible family in the chain. If $K,L\in\mathcal{U}$ with $K\ne L$, then there exist $\mathcal{F}_K,\mathcal{F}_L\in\mathcal{C}$ such that $K\in\mathcal{F}_K$ and $L\in\mathcal{F}_L$. Since $\mathcal{C}$ is totally ordered by inclusion, one of these two families contains the other. Hence both $K$ and $L$ lie in a single admissible family, so $K\perp L$. Thus $\mathcal{U}$ is an admissible family and is an upper bound for $\mathcal{C}$ in $\mathscr{P}$. By Zorn's lemma, there exists a maximal admissible family. Denote it by \begin{align*} \mathcal{F}_{\max}=\{H_i:i\in I\}, \end{align*} where $I$ is an index set. [/step] [step:Show the closed sum of the maximal family reduces $T$] Define \begin{align*} M:=\overline{\operatorname{span}}\bigcup_{i\in I}H_i. \end{align*} Since the family $(H_i)_{i\in I}$ is mutually orthogonal, $M$ is the Hilbert-space orthogonal direct sum of the family: \begin{align*} M=\bigoplus_{i\in I}H_i. \end{align*} We verify that $M$ reduces $T$. For every $i\in I$, the subspace $H_i$ reduces $T$, so $T(H_i)\subset H_i$ and $T^*(H_i)\subset H_i$. Therefore the algebraic span of $\bigcup_{i\in I}H_i$ is invariant under both $T$ and $T^*$. Since $T$ and $T^*$ are bounded, its closure $M$ is invariant under both $T$ and $T^*$. As before, because $M$ is closed and invariant under both $T$ and $T^*$, the orthogonal complement $M^\perp$ is also invariant under both $T$ and $T^*$. Hence $M$ reduces $T$, and $M^\perp$ reduces $T$. [/step] [step:Generate a new cyclic reducing summand from any nonzero orthogonal complement] Assume for contradiction that $M^\perp\ne\{0\}$. Choose $0\ne x\in M^\perp$. Let \begin{align*} S:=T|_{M^\perp}\in\mathcal{L}(M^\perp). \end{align*} This operator is well-defined because $M^\perp$ is invariant under $T$. Its Hilbert-space adjoint is \begin{align*} S^*=T^*|_{M^\perp}, \end{align*} because $M^\perp$ also reduces $T^*$. Define the closed cyclic subspace \begin{align*} K:=\overline{\operatorname{span}}\{p(S,S^*)x: p \text{ is a complex } *\text{-polynomial in two variables}\}\subset M^\perp. \end{align*} Then $K\ne\{0\}$ because $x\in K$. The same polynomial-multiplication argument used above shows that $K$ is invariant under both $S$ and $S^*$. Since $K$ is closed in $M^\perp$, it reduces $S$. Because $M^\perp$ reduces $T$ and $S=T|_{M^\perp}$, the inclusions $S(K)\subset K$ and $S^*(K)\subset K$ give \begin{align*} T(K)\subset K \end{align*} and \begin{align*} T^*(K)\subset K. \end{align*} Thus $K$ is a closed subspace of $H$ invariant under both $T$ and $T^*$, hence $K$ reduces $T$. Also $K\subset M^\perp$, so $K$ is orthogonal to every $H_i$. By construction, $K$ is $C^*(S|_K)$-cyclic with cyclic vector $x$, equivalently $C^*(T|_K)$-cyclic because $T|_K=S|_K$ and $(T|_K)^*=S^*|_K$. Therefore \begin{align*} \mathcal{F}_{\max}\cup\{K\} \end{align*} is an admissible family strictly larger than $\mathcal{F}_{\max}$, contradicting maximality. [guided] This is the key maximality argument. We have built a maximal orthogonal family of cyclic reducing subspaces, and we now show that there cannot be any vector left outside their closed orthogonal sum. Assume that the remaining space $M^\perp$ is nonzero, and choose $0\ne x\in M^\perp$. Since $M^\perp$ reduces $T$, the restriction \begin{align*} S:=T|_{M^\perp}\in\mathcal{L}(M^\perp) \end{align*} is a bounded operator on the [Hilbert space](/page/Hilbert%20Space) $M^\perp$. The adjoint of this restricted operator is \begin{align*} S^*=T^*|_{M^\perp}, \end{align*} because both $T$ and $T^*$ leave $M^\perp$ invariant. Now generate from $x$ the smallest closed subspace of $M^\perp$ that is stable under applying $S$ and $S^*$: \begin{align*} K:=\overline{\operatorname{span}}\{p(S,S^*)x: p \text{ is a complex } *\text{-polynomial in two variables}\}. \end{align*} The constant polynomial shows that $x\in K$, so $K\ne\{0\}$. The inclusion of both $S$ and $S^*$ in the generating algebra is exactly what gives reducing, not merely invariance for $S$. If $y$ is a finite linear combination of vectors $p(S,S^*)x$, then $Sy$ and $S^*y$ are again finite linear combinations of such vectors, because left multiplication by $S$ or $S^*$ produces another $*\text{-polynomial}$ in $S$ and $S^*$. Boundedness of $S$ and $S^*$ lets this invariance pass to the closure, so $K$ is invariant under both $S$ and $S^*$. Since $K$ is closed in $M^\perp$, invariance under both $S$ and $S^*$ implies that $K$ reduces $S$. As $M^\perp$ itself reduces $T$, the relations $S=T|_{M^\perp}$ and $S^*=T^*|_{M^\perp}$ imply \begin{align*} T(K)\subset K \end{align*} and \begin{align*} T^*(K)\subset K. \end{align*} Hence $K$ also reduces $T$ as a subspace of $H$. Finally, $K\subset M^\perp$, while every $H_i$ is contained in $M$. Thus $K$ is orthogonal to every $H_i$. Adding $K$ to the maximal family gives a strictly larger admissible family: \begin{align*} \mathcal{F}_{\max}\cup\{K\}. \end{align*} This contradicts maximality. Therefore the assumption $M^\perp\ne\{0\}$ must be false. [/guided] [/step] [step:Conclude that the maximal family gives the required orthogonal direct sum] The contradiction proves that \begin{align*} M^\perp=\{0\}. \end{align*} Since $M$ is closed, it follows that $M=H$. Therefore \begin{align*} H=\bigoplus_{i\in I}H_i. \end{align*} Each $H_i$ is nonzero, closed, reducing for $T$, and $C^*(T|_{H_i})$-cyclic by the definition of admissibility. Writing \begin{align*} T_i:=T|_{H_i}\in\mathcal{L}(H_i), \end{align*} we obtain for each $i\in I$ a vector $x_i\in H_i$ satisfying \begin{align*} H_i=\overline{\operatorname{span}}\{p(T_i,T_i^*)x_i: p \text{ is a complex } *\text{-polynomial in two variables}\}. \end{align*} This is exactly the asserted cyclic reducing decomposition. [/step]