# Polar decomposition of right shift operator

Let $T: H\to H$ be a bounded operator on Hilbert space $H$. $T(e_n) = a_n e_{n+1}$ where $\{e_n\}$ is orthonormal basis and $\{a_n\}$ is bounded sequence.

1. What is the polar decomposition of $T$?
2. For what sequences $T$ is Fredholm?
3. For what sequences $T$ compact?
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What have you tried? Where did you get stuck? Do you have any thoughts on the problem? – Jonas Meyer Dec 17 '12 at 18:58
I know T = |T| U but cant seem to compute U ( U partial isometry ) – user48931 Dec 17 '12 at 19:21
Write out the 'matrix' for $T$. The partial isometry should be fairly clear. – copper.hat Dec 17 '12 at 20:19
cant seem to get this hint – user48931 Dec 17 '12 at 20:34
$\begin{bmatrix}0 & 0 & \cdots \\ \text{sgn}\, a_1 & 0 & \cdots \\ 0 & \text{sgn}\, a_2 & \cdots \\ \vdots & &\end{bmatrix}$ (or something like $\frac{a_n}{|a_n|}$ if complex). – copper.hat Dec 17 '12 at 20:49

It is easy to check that $T^*Te_n=|a_n|^2\,e_n$ for all $n$. So $|T|\,e_n=|a_n|\,e_n$ for all $n$. Now, if $T=|T|\,U$, then $$\langle Te_j,e_k\rangle=\langle |T|Ue_j,e_k\rangle=\langle Ue_j,|T|e_k\rangle=|a_k|\,\langle Ue_j,e_k\rangle.$$ This shows that $U$ is the operator $Ue_n=\frac1{|a_n|}\,Te_n=\arg(a_n)\,e_{n+1}$.
As $T$ is compact if and only if $|T|$ is, the fact that $|T|$ is diagona and $a_1,a_2,\ldots$ are its eigenvalues imply that $T$ is compact if and only if $a_n\to0$.
As for Fredholm, if $a_n\to0$, then $T$ is compact so it cannot be Fredholm. If $a_n$ does not converge to zero, then the sequence is eventually bounded away from zero. This allows one to mimic what happens in the case of the shift (i.e. when $a_n=1$ for all $n$) to conclude that there exists $S$ such that $TS-I$ and $ST-I$ are compact. In conclusion $T$ is Fredholm precisely when it is not compact, i.e. when $a_n$ does not converge to zero.