# In which cases the spectrum of an operator contains only eigenvalues?

Let $X\neq \{0\}$ be a complex normed spaces (not necessarily finite-dimensional) and $T:D(T)\subset X\to X$ a closed linear operator (not necessarily bounded). I would like to know under what conditions can we conclude that every spectral value $\lambda\neq 0$ of $T$ is, is fact, an eigenvalue of $T$.

For example, a common condition is: $T$ compact and $D(T)=X$ (see Kreyszig, p. 420). Are there some other?

Thanks.

Here is another condition: $X$ is Banach and $T$ has compact resolvent (see Engel, p. 248).
Remark. $T$ having compact resolvent means that $(\lambda - T)^{-1}$ is a comapct operator for some $\lambda\in\rho(T)$, which is the case if and only if the embedding $(D(T),\|\cdot\|_{D(T)})\to |(X,\|\cdot\|_X)$ is compact (see Engel, p. 117).