Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?
share|cite|improve this question
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.