Cyclic Vectors

An operator $A$ on a Hilbert space $\mathbf{H}$ has a cyclic vector $f$ if the vectors $f, Af, A^{2}f, \ldots$ span $\mathbf{H}$; equivalently, $f$ is a cyclic vector for $A$ in case the set of all vectors of the form $p(A)f$, where $p$ varies over all polynomials, is dense in $\mathbf{H}$.

A unilateral shift of multiplicity $p$ (where $p$ is a cardinal number, $1 \leq p \leq \aleph_0$) is the direct sum of $p$ copies of the simple unilateral shift $U$ on $l^{2}$, i.e., the operator $\left\langle \xi_0^{(1)}, \xi_1^{(1)}, \ldots \right\rangle \oplus \left\langle \xi_0^{(2)}, \xi_1^{(2)}, \ldots \right\rangle \oplus \cdots$ acting coordinatewise by the simple shift on each summand.

It is convenient to say that if an operator has a cyclic vector, then it is a cyclic operator.

If $\mu$ is a Borel measure with compact support in the complex plane, then the position operator $A$ on $L^{2}\left(\mu \right)$ is defined by $(Af)(z) = zf(z)$.

A matrix $\left\langle \alpha_{ij} \right\rangle$ is triangular (specifically, upper triangular) if every entry below the main diagonal is zero (that is, $\alpha_{ij} = 0$ whenever $i > j$). A matrix is triangular $+1$ if every entry more than one step below the main diagonal is zero (that is, $\alpha_{ij} = 0$ whenever $i > j+1$).

The orbit of $f$ is the set of all vectors of the form $A^{n}f$, $n = 0, 1, 2, \ldots$, with no scalar multiples and no sums allowed.