Journal of Operator Theory

Volume 51, Issue 2, Spring 2004 pp. 303-319.

On the polar decomposition of the Aluthge transformation and related results

Authors: Masatoshi Ito, (1) Takeaki Yamazaki, (2) and Masahiro Yanagida (3)
Author institution: (1) Department of Mathematical Information Science, Faculty of Science, Tokyo University of Science, Tokyo 162--8601, Japan
(2) Department of Mathematics, Kanagawa University, Yokohama 221--8686, Japan
(3) Department of Mathematical Information Science, Faculty of Science, Tokyo University of Science, Tokyo 162--8601, Japan

Summary: Let

$T=U|T|$ be the polar decomposition of a bounded linear operator

$T$ on a Hilbert space. The transformation

$\widetilde{T}=|T|^{\frac{1}{2}}U|T|^{\frac{1}{2}}$ is called the Aluthge transformation and

$\widetilde{T}_{n}$ means the

$n$ -th Aluthge transformation. In this paper, firstly, we show that

$\widetilde{T}=VU|\widetilde{T}|$ is the polar decomposition of

$\widetilde{T}$ , where

$|T|^{\frac{1}{2}}|T^{*}|^{\frac{1}{2}}= V\big| |T|^{\frac{1}{2}}|T^{*}|^{\frac{1}{2}}\big|$ is the polar decomposition. Secondly, we show that

$\widetilde{T}=U|\widetilde{T}|$ if and only if

$T$ is binormal, i.e.,

$[|T|,|T^{*}|]=0$ , where

$[A,B]=AB-BA$ for any operators

$A$ and

$B$ . Lastly, we show that

$\widetilde{T}_{n}$ is binormal for all non-negative integer

$n$ if and only if

$T$ is centered, i.e.,

$\{T^{n}(T^{n})^{*},\ (T^{m})^{*}T^{m}: \mbox{ $n$ and $m$ are natural numbers}\}$ is commutative.

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