Compact linear operator on Hilbert space with no eigenvectors

Question

I have this question given to me as a Homework problem. I suspect there must be a typo somewhere. Can anyone confirm this for me?

Suppose $H$ is a Hilbert space with a countable basis $\{\phi_k\}$. Let $T:H\rightarrow H$ be a linear map defined via $$ T(\phi_k)= \frac{1}{k} \phi_k, \forall k.$$ Show that $T$ is compact, but it has no eigenvectors.

I can show compactness no problem. However, based on the statement, I think the $\phi_k$ are eigenvectors with eigenvalue $\frac{1}{k}$. Therefore, something is wrong with the question. Am I correct? If not, where is the flaw in my reasoning?

Answer 1

You are correct. They were likely trying to write $$ T\phi_k=\tfrac1k\,\phi_{k+1} $$ or something like that.