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.