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Operators in the commutant of a reductive algebra


Author: Robert L. Moore
Journal: Proc. Amer. Math. Soc. 66 (1977), 99-104
MSC: Primary 47C05; Secondary 47A65
DOI: https://doi.org/10.1090/S0002-9939-1977-0454730-3
MathSciNet review: 0454730
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Abstract: Let $ \mathcal{A}$ be a reductive algebra. It is shown that there is a subspace $ \mathcal{M}$ that reduces $ \mathcal{A}$ and such that the commutant of $ \mathcal{A}\vert\mathcal{M}$ is selfadjoint and the commutant of $ \mathcal{A}\vert{\mathcal{M}^ \bot }$ consists of hyporeductive operators. It is then shown that under a variety of conditions, if an operator T is in $ \mathcal{A}'$, then $ {T^ \ast }$ is in $ \mathcal{A}'$.


References [Enhancements On Off] (What's this?)

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Additional Information

DOI: https://doi.org/10.1090/S0002-9939-1977-0454730-3
Keywords: Reductive algebra, hyperinvariant subspace, hyporeductive operator
Article copyright: © Copyright 1977 American Mathematical Society

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