The resolvent of a matrix $M$ is defined as $$R_M(z)=(z-M)^{-1}.$$ The spectrum of $M$ can then be defined as the set of points $\lambda$ for which $R_M(\lambda)$ is singular. Since all the eigenvalues of a Hermitian matrix are real, and all elements of the spectrum of a matrix are eigenvalues of that matrix, it follows that all elements of the spectrum of a Hermitian matrix are real.
This is great, but I'm interested in the generalization to the operator case. Consider a Hermitian operator $H$ whose resolvent is $$R_H(z)=(z-H)^{-1}$$ and which has a spectrum defined by the singularities of $R_H(z)$. All the eigenvalues of $H$ must be real due to the hermiticity of $H$. However, my understanding is that there can be elements of the spectrum of an operator which are not eigenvalues of that operator. Can a Hermitian operator have elements of its spectrum which are outside of the set of eigenvalues, and can these eigenvalues be complex?
If a Hermitian operator has an eigenvalue $\lambda$, then it must be real because of the standard argument involving $Hx=\lambda x$ for some $x\ne 0$: $$ \lambda\|x\|^2=\langle \lambda x,x\rangle=\langle Hx,x\rangle = \langle x,Hx\rangle=\langle x,\lambda x\rangle=\overline{\lambda}\|x\|^2. $$ Any $\lambda$ in the spectrum of $H$ must be real; if such $\lambda$ is not an eigenvalue, then it is an approximate eigenvalue, meaning that there is a sequence of unit vectors $\{ x_n \}$ such that $\|(H-\lambda I)x_n\|\rightarrow 0$ as $n\rightarrow\infty$.