Is there a compact operator in $\ell^2$ satisying these properties?

Question

Question: Is there any operator $T:\ell^2\to\ell^2$ satisfying all these properties?

1. $T$ is compact,
2. $T$ admits an adjoint $T^*$ but $T^*\neq T$,
3. $T(\ell^2)$ is infinite dimensional,
4. $T$ does not have eigenvalues.

Comment: If you know of an operator with these characteristics in a book, paper, notes, etc, and you do not want to spend the time typing it you can just cite it and once I check it I will consider it as an answer.

Answer 1

Let us denote the standard basis vectors by $e_n$. Let $T$ be given by $Te_n=e_{n+1}/2^n$. Then $T$ is compact (quick proof: let $T_n$ agree with $T$ on $e_n$ but map every other basis vector to $0$, then $\sum T_n$ converges in the norm topology to $T$ and each partial sum has finite rank). We have $T\neq T^*$ (the adjoint will shift the basis vectors left instead of right), and the image of $T$ is obviously infinite dimensional.

Finally, I claim $T$ has no eigenvalues. Indeed, for any nonzero sequence $x\in\ell^2$, the first nonzero term of $x$ is one term earlier than the first nonzero term of $Tx$, so $Tx$ cannot be a scalar multiple of $x$.