# Compact operators on Hilbert Spaces

• Feb 15th 2009, 05:20 PM
matt33
Compact operators on Hilbert Spaces
Hi, here is my problem :

Let H be a Hilbert space with basis {u_k}_k>0.
Verify that the operator T defined by :
T(u_k) = 1/k . u_k+1
is compact but has no eigenvectors.

To show it is compact, I tried to use the definition : T is compact if any bounded sequence {f_k} in H has a subsequence {f_nk}_k>0 such that {T(f_nk)} converges.
I took a bounded sequence {f_k} in H and approximated the f_k's by linear combinations of the basis elements {u_k}. But I don't see where this is supposed to lead me.

For the eigenvectors, I am lost.

Any ideas ?
Thanks.
• Feb 16th 2009, 02:19 PM
phuriku
For the eigenvectors, assume that $\displaystyle Tx = \alpha x$ for some $\displaystyle \alpha \in \mathbb{C}$, and write $\displaystyle x = \sum^{\infty}_{i=1} \lambda_i \varphi_i$. Then the equation reads $\displaystyle T(\sum \lambda_i \varphi_i) = \sum \alpha \lambda_i \varphi_i$, which, carrying out the operation, brings us to$\displaystyle \sum_{i=1}^{\infty} \frac{\lambda_i}{i} \varphi_{i+1} = \sum^{\infty}_{i=1} \alpha \lambda_i \varphi_i$. Now solve for $\displaystyle \lambda_i$. What do you get?