 Solution

Put $A=\lambda I-T$. Since $T$ is a self-adjoint operator, $A^{\ast }=\overline{\lambda }I-T$. Applying the preceding image-kernel orthogonality for an adjoint identities to $A^{\ast }$ givesIf $\lambda$ is real, $A^{\ast }=A$ and $\ker A=\{0\}$ because $\lambda$ is not an eigenvalue. If $\lambda$ is nonreal and $A^{\ast }x=0$, then $Tx=\overline{\lambda }x$. The scalar $⟨Tx,x⟩$ is real by self-adjointness, while it equals $\overline{\lambda }\Vert x{\Vert }^2$. Its imaginary part forces $x=0$. Thus $\ker A^{\ast }=\{0\}$ in both cases, andThe conjugation on $\lambda$ in the adjoint operator is essential when $\lambda$ is nonreal.