# 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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