Let $T$ be a compact operator on an infinite dimensional Hilbert space. Let $|T|=(T^*T)^{0.5}$. By the polar decomposition theorem there is a partial isometry $S$ of the closure of Im$(|T|)$ such that $T=S|T|$. Let $(e_{n})$ be the orthonormal sequence of eigenvectors of $|T|$ with corresponding eigenvalues ${\lambda}_{n}$. Prove that the sequence $(x_{n})$ defined by $x_{n}=Se_{n}$ is orthonormal.
My attempt: If ${\lambda}_{n}$ is non-zero, then $e_{n}$ is in $\mathrm{Im}|T|$ so since $S$ is an isometry on there, $x_{n}$ has norm 1. But if ${\lambda}_{n}$=0, then $e_{n}$ is in the kernel of $|T|$, which is the orthogonal complement of the closure of
$\mathrm{Im}(|T|)$ so $Se_{n}$=0 which would contradict $x_{n}$ having norm 1. So I am missing something here. Also, why would $\langle x_{n},x_{m}\rangle=\langle Se_{n},Se_{m}\rangle=0$ for distinct $n,m$? Thanks
As you said, some of the $x_n$ might be zero: note that all $x_n$ belong to the image of $T^*$, which could perfectly be finite-dimensional.
For your second question, $S^*Se_n=e_n$, as $S^*S$ is the projection onto the image of $|T|$.