# Commuting operators and polar decomposition

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...
"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. –  Jonas Meyer Dec 7 '12 at 16:41
@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 $\ell^2$. –  Jonas Meyer Dec 7 '12 at 16:44
@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 $V^*V\neq VV^*$. –  Jonas Meyer Dec 11 '12 at 17:34