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Roots of invertibly weighted shifts with finite defect

Author: Gerard E. Keough
Journal: Proc. Amer. Math. Soc. 91 (1984), 399-404
MSC: Primary 47B37
MathSciNet review: 744638
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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\Vert {{S_n}{x_n}} \right\Vert^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.

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Keywords: Invertibly weighted shift, commutant, root of an operator
Article copyright: © Copyright 1984 American Mathematical Society

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