Let $H$ be a Hilbert-space and $K:H\rightarrow H$ a compact operator. I know that if $K$ is self adjoint, then it has one eigenvalue $\mu$ such that $|\mu|=||K||$. Can some one give me a nice counter example of a not self adjoint compact operator such that such a eigenvalue does not exist?
The standard example is the Volterra operator, e.g., on $L^2[0,1]$, given by $Vf(x)=\int_0^x f(t)\,dt$. It is compact, but not self-adjoint, and has no eigenvalues.