Definition 1

*Let $E$ be a vector space over a commutative field $K$, and $u$ an endomorphism of $E$. An element $x$ of $E$ is said to be an eigenvector of $u$ if there exists $a \in K$ such that $u . x = \alpha x$; if $x \neq 0$ then the scalar $a$ is called the eigenvalue of $u$ corresponding to the eigenvector $x$. For every scalar $a$, the vector subspace $V_a$ consisting of $x \in E$ such that $u . x = \alpha x$ is called the eigenspace of $E$ corresponding to $a$.

The geometric multiplicity of the eigenvalue $a$ is the cardinal $\dim V_a$.

Suppose $E$ is finite dimensional. The eigenvalues of $u$ are those elements $a$ of $K$ such that the endomorphism $a . 1 - u$ of $E$ is not injective, in other words (III, p. 524, Prop. 3) such that $\det(a . 1 - u) = 0$. But, by the definition of the characteristic polynomial $\chi_u$ of $u$ (III, p.541, Def. 3), we have $\det(a . 1 - u) = \chi_u(\alpha)$. Consequently :