The question is to show that if $T$ is normal, there exists a unitary operator $U$ such that $T^{*}=UT$. My guess is that we use the polar decomposition of $T$- into a product of a unitary and positive operator- in some way, but am not sure how to start.
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I should have mentioned that all operators are bounded linear operators on an arbitrary Hilbert space $H$. – Arundhathi Sep 10 '13 at 18:12
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I think I figured it out after all.
Since $T$ is normal, there exists a unitary operator $V$ and a positive operator $P$ such that $T=VP$, and $P$, $V$ and $T$ commute with each other. $T=VP \Longrightarrow T^{*}=PV^{*}$.
$\Longrightarrow T^{*}V^{2}=PV^{*}VV=PV=T$ $\Longrightarrow T^{*}=T(V^{*})^{2}=(V^{*})^{2}T$
The last equality follows since $VT=TV$ and $T$ normal, by Fuglede-Putnam-Rosenblum, $VT^{*}=T^{*}V \Longrightarrow V^{*}T=TV^{*}$
Let $U=(V^{*})^{2}$
Then $U$ is unitary and satisfies $T^{*}=UT$
Arundhathi
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2In the polar decomposition of $T$ as $T=UP$, with $U$ a unitary operator and $P$ a positive operator, $P$ is always uniquely determined as the positive square root of $T^{*}T$. So if $T$ is invertible clearly $U$ must be uniquely given by $U=TP^{-1}$. – Arundhathi Dec 09 '13 at 04:55