Roots of invertibly weighted shifts with finite defect

Abstract:

Let $T$ be a unilateral invertibly weighted shift; i.e., $T$ maps a square-summable vector sequence $\left \{ {{x_{0,}}{x_1}, \ldots } \right \}$ from a Hilbert space $H$ to the sequence $\left \{ {0,{A_0}{x_0},{A_1}{x_{1,}} \ldots } \right \}$, where $\left \{ {{A_n}} \right \}$ is a uniformly bounded sequence of invertible operators on $H$. If ${S_0}$ is the identity operator on $H$, and ${S_n} = {A_{n - 1}}{A_{n - 2}} \cdots {A_0}$ for $n \geqslant 1$, then $T$ is unitarily equivalent to multiplication by the variable $Z$ on the space ${H^2}\left ( T \right )$ consisting of formal series $\sum {x_n}{Z^n}$ having coefficients ${x_n} \in H$ which satisfy $\sum {\left \| {{S_n}{x_n}} \right \|^2} < + \infty$. The commutant of this multiplication consists of formal series $\sum {F_n}{Z^n}$ which define bounded operators on ${H^2}\left ( T \right )$—where each ${F_n}$ is an operator on $H$, and the action of such a series on an element of ${H^2}\left ( T \right )$ is given by the Cauchy product of the two series. Using these characterizations, it is shown that if $H$ has finite dimension $m \geqslant 2$, then $T$ has an $n$th root only if $n$ divides $m$. Examples are given of shifts $T$ with (a) $m = 2$, but $T$ has no square root, and (b) $m = 4$, $T$ has a square root, but no fourth root.