Counterexamples for "every linear map on an infinite dimensional complex vector space has an eigenvalue" Every linear map on a finite dimensional complex vector space has an eigenvalue. Not so in the infinite case.
I'm interested in nice counterexamples anyone might have. 
Here's one:
Consider the vector space $\mathbb C^\infty$ of sequences and the right shift map $R$ defined by
$$R(a_1, a_2, a_3, ...) = (0, a_1, a_2, a_3, ...)$$
$R$ has no eigenvalue (using the usual convention that there must be a non-trivial eigenvector).

Let $C[0,1]$ be the linear space of continuous functions on $[0,1]$; this linear space is a complete normed linear space (i.e., a Banach Space) when given the max norm $\|f\|=\max_{t \in [0,1]}|f(t)|$. The Volterra operator
$$
                   (Vf)(x) = \int_{0}^{x}f(t)dt
$$
maps continuous functions to continuous functions. $V$ has trivial null space because $(Vf)(x)=0$ for all $x$ implies $f(x)=0$ for all $x$ by the Fundamental Theorem of Calculus. So $V$ does not have $0$ as an eigenvalue. If $Vf=\lambda f$ for a non-zero $\lambda$, then $f=\frac{1}{\lambda}Vf$ is continuously differentiable with $f'=\frac{1}{\lambda} f$ and $f(0)=0$, which has only the trivial solution $f\equiv 0$.
