Based on the paper the cyclic operator is defined as:
A bounded linear operator $T$ on a complex Hilbert space $H$ is cyclic if there is a vector $x$ in $H$ such that $H$ is the closed linear span of the vectors $x$, $Tx$, $T^2x$, ... (in this case, $x$ is a cyclic vector of $T$).
In Proposition 1.1. of this paper, it is stated that:
Every operator $T$ can be expressed in a triangular form $$ \begin{bmatrix} T_1 & & &* \\ & T_2 \\ & &\ddots \\ 0 & & &T_n \end{bmatrix}, $$ where $n$ is the multiplicity of $T$ and the $T_j$'s are all cyclic.
Moreover, it is denoted in Proposition 1.2. that:
An operator is cyclic if and only if it has the matrix representation $$ \begin{bmatrix} a_1 & & & * \\ b_1 & a_2 \\ & b_2 & \ddots \\ & & \ddots & \\ 0 & & & \end{bmatrix}, $$ with all $b_n$'s nonzero.
Moreover, we know that the cyclic shift operator is defined as $$ A = \begin{bmatrix} 0 & 1 \\ & 0 & 1 \\ & & \ddots & & 1\\ 1&&&&0 \end{bmatrix}. $$ Now, I jut confused about this definition and its relation with cyclic shift operator. My question is that:
- Does the definition of cyclic operator above for $T$ also the case for cyclic shift operator and this shift operator can be represented by the form in Proposition 1.2.?
- Can you provide tangible elaboration about the term "cyclic" in the definition of cyclic operator (please also elaborate on the term "cyclic" onthe cyclic shift operator based on the definition of cyclic operator above)?
Suppose that $(e_n)_{n \in \mathbb N}$ is the standard basis of $\mathbb R^n$. The cyclic shift operator $S_1$ as defined by you will act as $$ S_1(e_k) = \begin{cases} e_{k-1} & k \ge 2 \\ e_n & k = 1. \end{cases} $$ By renaming the basis elements (eg. reversing their order), we can achieve that the index is increased instead of decreased, which puts the operator into the form of theorem 1.2.
Regarding the term "cyclic", it commonly means that whatever object one is considering is generated by one element. Here, one may consider the (non-invertible) topological dynamical system given by $T$ and $H$, and speak of a subset of $H$ that includin its arbitrary iterations by $T$ is dense in $H$ as a generating set.