 Solution

No. On the complex l2 sequence space, let $L(x_n)=(x_n/n)$. This is a bounded self-adjoint operator. At $\lambda =0$, $\ker L=\{0\}$, so zero is not an eigenvalue (an eigenvector must be nonzero). But $y=(1/n)\in {\ell }^2$ has no preimage in ${\ell }^2$: its only formal preimage is the constant-one sequence. Thus $L$ is not surjective andAlternatively $\Vert L{e}_n\Vert =1/n\to 0$ excludes a bounded inverse. This example separates the spectrum of a bounded operator from its point spectrum.