finite spectrum eigenvalue Let $T:X \to X$ be a linear bounded operator where X is Banach space ,and $\sigma (T)$ is a finite set.Then does the spectrum consist of eigenvalues only?
Any hint or counterexample is ok. thanks in advance.
 A: Any one-to-one quasinilpotent operator will do (quasinilpotent are the operators $T$ with $\sigma(T)=\{0\}$); the Volterra operator from user3808066's answer is one example. 
Any finite set $K\subset\mathbb C$ can be realized this way: if $K=\{k_1,\ldots,k_n\}$ and $T$ is the Volterra operator, then
$$
\bigoplus_{j=1}^nT+k_jI
$$
has spectrum $k_1,\ldots,k_n$ and none of them is an eigenvalue. 
For an example in $\ell^2(\mathbb N)$, let $T$ be the weighted unilateral shift given by $Te_j=\frac1{j+1}\,e_{j+1}$, where $\{e_j\}$ is the canonical basis. This operator $T$ is compact, so the nonzero points in its spectrum can only be eigenvalues, and it is easy to check that it has no eigenvalues. As it is compact, it is not invertible, so $\sigma(T)=\{0\}$.
A: A counter example is the Volterra operator, it is a bounded linear operator between Hilbert spaces, with no eigenvalues while the spectrum is $\{0\}$. It is defined as 
$$
 V:L^2(0,1)\to L^2(0,1), f\mapsto \left(t\mapsto \int_0^t f(x)\mathrm{d}x\right).
$$
