Operators with infinite rank and kernel

Question

Can a compact operator in $H=l^2$ have infinite rank and infinite kernel? I guess no because such an operator has more than countable eigenvalues.

Answer 1

I don't understand why do you say that such an operator should have more then countable eigenvalues. By the way, I think that it is possibile, take:

$T : H \to H$

$(Tx)_n = \begin{cases} 0 \quad \quad \,\,\,\text{if $n$ is odd} \\ n^{-1}x_n \quad \text{if $n$ is even} \end{cases} \quad \forall n \ \in \mathbb{N}$

Then

$Ker(T) = \{ x \in H : x_{2k} =0 \quad \forall k \in \mathbb{N}\}$

$Im(T) = \{x \in H : \exists k \in \mathbb{N} :x_{2k}\ne 0 \}$

which are both infinite dimensional, but $T$ is compact since it is the limit of finite rank operators.