Cyclic vector and cyclic normal operator

Assume AC. Let $T$ be a bounded normal operator on a nonzero complex Hilbert space $H$, with continuous functional calculus $f\mapsto f(T)$, a unital star-isomorphism $C(\sigma (T))\to C^{\ast }(I,T)$ (Continuous functional calculus for bounded normal operators).

A vector $x\in H$ is cyclic for $T$ when the closed linear span of the set $\{f(T)x:f\in C(\sigma (T))\}$ is all of $H$; equivalently, when $\stackrel{}{\mathrm{span}}\{f(T)x:f\in C(\sigma (T))\}=H\mathrm{.}$ For a normal operator this is the same as cyclicity for the unital $\ast$-algebra generated by $T$: because the calculus is a unital $\ast$-isomorphism onto $C^{\ast }(I,T)$ and $C^{\ast }(I,T)$ is the norm closure of the unital $\ast$-algebra generated by $T$ and $T^{\ast }$ (C star algebra generated by a normal operator), the set $\{f(T)x:f\in C(\sigma (T))\}$ is precisely the set $\{ax:a\in C^{\ast }(I,T)\}$, and cyclicity uses the normal operator $T$ together with its adjoint, never only the nonnegative powers of $T$. A normal operator is called cyclic when it has a cyclic vector.

For an arbitrary vector $x\in H$ define its cyclic subspace $H_x:=\stackrel{}{\mathrm{span}}\{f(T)x:f\in C(\sigma (T))\}\mathrm{.}$

Well-definedness and elementary properties. $H_x$ is a closed linear subspace by definition, it contains $x=\mathbf{1}(T)x$, and it is invariant under $T$ and $T^{\ast }$: for $f$ continuous one has $Tf(T)x=(zf)(T)x$ and $T^{\ast }f(T)x=(\bar{z}f)(T)x$, both again of the form $g(T)x$ with $g$ continuous (Continuous functional calculus properties). Hence $H_x$ is a reducing subspace for $T$. If $x\mathrm{\ne }0$, then $H_x\mathrm{\ne }\{0\}$ and the restriction $R:=T{\mathrm{\mid }}_{H_x}$ is a bounded normal operator. It is cyclic with cyclic vector $x$: every $f(T)\in C^{\ast }(I,T)$ is a norm limit of star-polynomials $q_n(T,T^{\ast })$, and restriction to $H_x$ gives $q_n(R,R^{\ast })\to f(T){\mathrm{\mid }}_{H_x}$, so $f(T){\mathrm{\mid }}_{H_x}\in C^{\ast }(I,R)$. The continuous calculus for $R$ is onto $C^{\ast }(I,R)$, so its orbit of $x$ contains the original orbit $\{f(T)x:f\in C(\sigma (T))\}$, whose closed span is $H_x$ by definition; therefore $x$ is cyclic for $R$. If $x=0$, then $H_x=\{0\}$; the restriction to this zero space is not fed to the library's nonzero-space functional calculus, and $0$ is not cyclic for the original nonzero $H$.

Finally, $x$ is cyclic for $T$ if and only if no nonzero $y\in H$ is orthogonal to every $f(T)x$; this is the standard description of a closed span as the orthogonal complement of its annihilator (Orthogonality and the orthogonal complement, Hilbert space, Self-adjoint, positive, unitary and normal operators).