Remark 4

If $u$ is semi-simple but not absolutely semi-simple, then it has no Jordan decomposition.

An endomorphism $u$ of a vector space $V$ over a commutative field is said to be unipotent if the endomorphism $u - \mathrm{Id}$, is nilpotent, that is if there exists an integer $r$ such that $(u - \mathrm{Id},)' = 0$; then $u$ is an automorphism of $V$, since if $u = \mathrm{Id}, - n$ with $n' = 0$, then

If $V$ has finite dimension $m$, then $u$ is unipotent if and only if $\chi_u(X) = (X - 1)^m$ (VII, p. 33, Cor. 3 to Prop. 5).