Kernel of selfadjoint compact operator is trivial

Question

Let $H$ be separable Hilbert space and $K(H)$ the space of all compact operators acting in $H$. If $x\in K(H)$ and $x=x^*$ then is it true that $\ker(x)\neq \{0\}$?. Where $\ker(x)$ is the kernel of $x$. Thanks for any help.

Answer 1

The answer is no. An example of a self-adjoint compact with a trivial kernel is as follows: define $x:\ell^2 \to \ell^2$ by $$ x\left[(\xi_k)_{k = 1}^\infty\right] = (\xi_k/k)_{k=1}^\infty $$