hilbert spaces - Operators with infinite rank and kernel

can compact operator in $h=l^2$ have infinite rank , infinite kernel? guess no because such operator has more countable eigenvalues.

i don't understand why such operator should have more countable eigenvalues. way, think possibile, take:

$t : h \to h$

$(tx)_n = \begin{cases} 0 \quad \quad \,\,\,\text{if $n$ odd} \\ n^{-1}x_n \quad \text{if $n$ 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 both infinite dimensional, $t$ compact since limit of finite rank operators.

Duplicate