3
$\begingroup$

If we take polar decomposition of $T$, where $T$ is compact operator, that is $T=U|T|$, where $U$ is partial isometry and $|T|$ is positive part, then does it imply the partial isometry $U$ belong to $\mathcal{K}(H)$, the space of compact operators?? The motivation for asking the question is in general for von Neumann algebra the partial isometry lies in itself, but for what if von Neumann algebra is replaced by noncommutative $C$*-algebra.

$\endgroup$

1 Answer 1

4
$\begingroup$

The answer is negative. Take a compact diagonal operator $T:\ell^2 \to \ell^2$ given by $T(e_n) = a_n \, e_n$, for some sequence $a_n$ with $a_n \to 0^+$ and assume $a_n \neq 0$. Then $|U|$ is the multiplication operator associated with $$\mathrm{sgn}(a_n) = \frac{a_n}{|a_n|},$$ which is not compact (it lays in the multiplier algebra of $K(\ell^2)$

$\endgroup$
2
  • $\begingroup$ Thank you Adrian $\endgroup$
    – user548061
    May 14, 2018 at 16:07
  • 4
    $\begingroup$ Even more simply, If $T$ is compact and positive, then $U$ is the identity operator. $\endgroup$
    – Aweygan
    May 14, 2018 at 22:07

You must log in to answer this question.