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Suppose that $V$ is an isometry and $X$ an arbitrary operator on a Hilbert space $H$. Let $X=U|X|$ be the polar decomposition for $X$.

If $VX=XV$, can I conclude that $VU=UV$?

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Edit: No, it does not hold in general. For some reason I always tended to use that $V$ commuted with $X^*$, which in general doesn't have to be true. A counterexample: Let $\mathcal{H}=\ell^2(\mathbb{N})$. Let $V$ be the right shift on $\mathcal{H}$, i.e. $V\delta_n=\delta_{n+1}$ where $\delta_n(m)=\left\{\begin{matrix} 1 & \mbox{if }n=m\\0 & \mbox{if }n \neq m\end{matrix}\right.$. Let $X=2+V$ (so $X\delta_n=2\delta_n+\delta_{n+1}$) then $XV=VX$. However, you can check that $X^*V \neq VX^*$. Now $X$ is invertible, so $U$ is unitary. But then $UV=VU \Leftrightarrow X^*V=VX^*$, which is false.

So the following only holds if $V$ commutes with $X^*$: Let $\mathcal{X}$ be the von Neumann algebra generated by $X$. One can show that in this case $U\in \mathcal{X}$. As clearly $V$ is an element of the commutant of $\mathcal{X}$ and $U \in \mathcal{X}$, we find that $VU=UV$. So I don't think you need the fact that $V$ is an isometry...

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    "As clearly $V$ is an element of the commutant of $\mathcal X$." I disagree. $V$ need not commute with $X^*$. If $X$ were normal or $V$ were unitary, this would be valid.2012-12-07
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    @JonasMeyer : the isometry V need not be unitary ????2012-12-07
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    @Ewan: Right, in the finite dimensional case $V$ is automatically unitary and it follows that $V$ is in the commutant of $\{X,X^*\}$. So the main problem is determining what can happen when $V$ is a nonunitary isometry, like the unilateral shift on $\ell^2$.2012-12-07
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    I corrected my answer. Now it should hold...2012-12-10
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    I'm sorry, for some reason I kept thinking in algebras, thus using that $V$ commutes with $X^*$ which doesn't have to be the case. I changed my answer and found a counterexample...2012-12-11
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    In your newest edit, I don't see why "$UV = VU \iff X^*V = VX^*$" holds. It seems to assume that $V$ commutes with $|X|$. Or am I missing something?2012-12-11
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    @Manny: Suppose to reach a contradiction that $UV=VU$. Then from $VU|X|=U|X|V$ we have $UV|X|=U|X|V$, which upon canceling $U$ yields $V|X|=|X|V$, which upon multiplying both sides by $|X|$ yields $VX^*X=X^*XV$, then again using $XV=VX$ yields $VX^*X=X^*VX$, and finally canceling $X$ yields $VX^*=X^*V$, which is not true because $V^*V\neq VV^*$.2012-12-11
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    Jan: Nice example.2012-12-11
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    @JonasMeyer, indeed I see now, thanks! And yes, Jan, nice example!2012-12-11
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    @ Jonas Meyer: Thanks, glad I found something :) it really kept bugging me...2012-12-12