every compact normal operator can be represented by product of unitary operator and compact positive operator Could anyone show that every compact normal operator can be represented by commutative product of unitary operator and compact positive operator? I mean if $T$ is a compact normal operator, then there exists a compact positive operator $A$ and unitary operator $B$ such that $T=AB=BA$. Are these operators uniquely determined?
 A: The key here that every compact normal operator $T$ on any complex Hilbert space $\mathcal H$ is diagonalizable (see e.g. "Introduction to Hilbert Space" by Berberian, 1976) so there exists an orthonormal system $(e_n)_{n\in\mathbb N}$ in $\mathcal H$ and a complex null sequence $(\tau_n)_{n\in\mathbb N}$ such that $|\tau_n|\geq |\tau_{n+1}|$ for all $n\in\mathbb N$ and
$$
T=\sum_{n=1}^\infty \tau_n\langle e_n,\cdot\rangle e_n.
$$
Now $A=\sum_{n=1}^\infty |\tau_n|\langle e_n,\cdot\rangle e_n=|T|$ is the positive-semidefinite (obvious) compact operator we are looking for. The compactness of $A$ can be shown e.g. by showing $A$ is the limit of the norm-convergent sequence of the finite-rank operators $(\sum_{n=1}^k|\tau_n|\langle e_n,\cdot\rangle e_n)_{k\in\mathbb N}$ or by using the following argumentation:
$$
|T|\text{ compact }\Leftrightarrow |T|^\dagger |T|=|T|^2=T^\dagger T\text{ compact }\Leftrightarrow T\text{ compact}
$$
The unitary operator $B$ now somehow has to transform the $\tau_n$ into their absolute value. As unitary operators are strongly connected to orthonormal bases, we extend $(e_n)_{n\in \mathbb N}$ to an orthonormal basis $(e_n)_{n\in J}$ of $\mathcal H$ with $\mathbb N\subseteq J$. Define $I:=\lbrace n\in\mathbb N\,|\,\tau_n\neq 0\rbrace$ and the operator $B$ via
$$
Be_n:=\begin{cases} \frac{\tau_n}{|\tau_n|}e_n & \text{if }n\in I \\ e_n & \text{if }n\in J\,\backslash \,I \end{cases}
$$
and its linear and continous extension onto all of $\mathcal H$. Observe that
$$
\Big\langle \frac{\tau_j}{|\tau_j|}e_j,\frac{\tau_k}{|\tau_k|}e_k\Big\rangle=\delta_{jk}
$$
for all $j,k\in I$ so $B$ maps $(e_n)_{n\in J}$ to an orthonormal basis of $\mathcal H$ and thus is unitary. With this it is easy to see that $T=BA$ and $BAB^\dagger =A$ (where the latter obviously equals $BA=AB$).
For your last question, note that $A$ and $B$ are not unique as the orthonormal system $(e_n)_{n\in\mathbb N}$ belonging to $T$ is not uniquely determined. For this one may consider $(e^{i\varphi}e_n)_{n\in\mathbb N}$ for any $\varphi\in(0,2\pi)$, this still is an orthonormal system and satisfies the above decomposition of $T$. For something more concrete one may choose $T=A=0$ so $B$ can be literally any unitary operator so satisfies the required properties.
