Can an operator have spectrum consisting of just one point? 
Could it happen that an operator has a spectrum consisting of just one point? 

Sorry if this is really short. Was just pondering to myself random questions and this one came up. The good news is that I am reading a functional analysis book so my question is not completely random.
 A: A good example of an operator with one point in the spectrum is the integral operator
$$
                 Lf=\int_{0}^{t}f(u)\,du
$$
defined on $X=L^{2}[0,1]$, or defined on $X=C[0,1]$. In both cases $\sigma(L)=\{0\}$. This operator is also compact but it has no eigenvalues, which makes it a good counterexample to remember when studying compact operators.
A: It turns out that for any bounded linear operator on a Hilbert space, the spectrum of that operator is a compact subset of $\mathbb{C}$.
In turn, any compact subset, $K$, of $\mathbb{C}$ is the spectrum of some linear operator. In particular, if $\chi(z)$ is the indicator function of that set, then $M_{\chi}$ has spectrum $K$. Here $M_{\chi}$ is the multiplication operator over $L^2(\mathbb{C})$ (with respect to Lebesgue measure).
Since a singleton set in $\mathbb{C}$ is compact, it is the spectrum of some bounded operator.
A: Let me add an example of operators with spectrum consisting of a single eigenvalue. For example
$$
\begin{bmatrix}0&0\\1&0\end{bmatrix}
$$
has spectrum consisting of just its eigenvalue $0$. Similarly, 
$$
\begin{bmatrix}0&0&0\\1&0&0\\0&1&0\end{bmatrix},
$$
and we can construct examples in $n\times n$ for any $n$. 
