Consider a linear map $\varphi : E \to E$ where $E$ is a linear space over the field $\mathbb C$ of the complex numbers. When $E$ is a finite dimensional vector space of dimension $n \ge 1$, the number of eigenvalues is finite. The eigenvalues are the roots of the characteristic polynomial $\chi_\varphi$ of $\varphi$. $\chi_\varphi$ is a complex polynomial of degree $n \ge 1$. Therefore the set of eigenvalues of $\varphi$ is non-empty and its cardinal is less than $n$.

Things are different when $E$ is an infinite dimensional space.

A linear map having all numbers as eigenvalue

Let’s consider the linear space $E=\mathcal C^\infty([0,1])$ of smooth complex functions having derivatives of all orders and defined on the segment $[0,1]$. $E$ is an infinite dimensional space: it contains all the polynomial maps.

On $E$, we define the linear map $$\begin{array}{l|rcl} \varphi : & \mathcal C^\infty([0,1]) & \longrightarrow & \mathcal C^\infty([0,1]) \\ & f & \longmapsto & f^\prime \end{array}$$

The set of eigenvalues of $\varphi$ is all $\mathbb C$. Indeed, for $\lambda \in \mathbb C$ the map $t \mapsto e^{\lambda t}$ is an eigenvector associated to the eigenvalue $\lambda$.

A linear map having no eigenvalue

On the same linear space $E=\mathcal C^\infty([0,1])$, we now consider the linear map $$\begin{array}{l|rcl} \psi : & \mathcal C^\infty([0,1]) & \longrightarrow & \mathcal C^\infty([0,1]) \\ & f & \longmapsto & x f \end{array}$$

Suppose that $\lambda \in \mathbb C$ is an eigenvalue of $\psi$ and $h \in E$ an eigenvector associated to $\lambda$. By hypothesis, there exists $x_0 \in [0,1]$ such that $h(x_0) \neq 0$. Even better, as $h$ is continuous, $h$ is non-vanishing on $J \cap [0,1]$ where $J$ is an open interval containing $x_0$. On $J \cap [0,1]$ we have the equality $$(\psi(h))(x) = x h(x) = \lambda h(x)$$ Hence $x=\lambda$ for all $x \in J \cap [0,1]$. A contradiction proving that $\psi$ has no eigenvalue.