Square root of a compact normal operator

Question

Halmos expresses below problem in his book;

Problem: If $A$ is a normal operator and if $A^n$ is compact for some positive integer $n$, then $A$ is compact.

I have an example in my mind which I think is a counterexample for this problem.

Let $\{e_n\}$ be an orthonormal basis for Hilbert space $H$, and the bounded sequence $\{t_n\}$ in $\Bbb R$ such that $t_n\to 2$.

Define operator $A\in B(H)$ such that $$Ae_{2n} = t_n e_{2n+1},~~~Ae_{2n+1} = 0$$ for every $n$. Operator
$A$ is normal and non-compact while $A^2=0$ which is compact.

I'm confused about it. Where is my mistake? Please help me.

Answer 1

Unilateral shifts (including weighted shifts) are never normal.