TITLE: Spectrum of a compact operator
QUESTION [11 upvotes]: If the spectrum of a compact operator is finite, I don't understand why $0$ has to be a member. I have proved that for all $\epsilon > 0$, there is only a finite number of eigenvectors which have eigenvalues $x$ satisfying $|x|>\epsilon$. I can see that when the spectrum is countable, the sequence of eigenvalue tends to zero so since the spectrum is closed, $0$ is a member but don't see why when the spectrum is finite. 
Also, since every operator on a finite dimensional space is compact, am I right in thinking this results is only valid for infinite dimensions?
Thanks

REPLY [2 votes]: Suppose $X$ is a Banach space and suppose that $T : X\rightarrow X$ is linear. Then the following are true:

1. If $X$ is finite-dimensional, then $T$ is compact, regardless of whether or not $0 \in \sigma(T)$.
2. If $X$ is infinite-dimensional and $T$ is compact, then $0 \in \sigma(T)$.