Compact diagonal operator

Question

Suppose $A : H \to H$, where $H$ is a Hilbert space, is bounded. Also, $A$ is a diagonal operator with diagonal $\{a_n\}$. Show: If $A$ is compact, then $a_n \to 0$ as $n \to \infty$.

Should I prove this by contrapositive?

Answer 1

Proof by contrapositive certainly works.

If $a_{n} \not \to 0$, there is an $\epsilon > 0$ and subsequence $a_{n_k}$ such that $|a_{n_k}| > \epsilon$ for each $k$. Then, show that the sequence $\{e_{n_k}\}$ is mapped to a sequence with no convergent subsequence.