Proposition 17

— Let $E$ be a finite dimensional vector space over a commutative field, and let $f$ be an endomorphism of $E$. Then the following conditions are equivalent: (i) $f$ has a Jordan decomposition and is an automorphism ; (ii) $f$ has a Jordan decomposition and $f_s$ is an automorphism ; (iii) there exists an absolutely semi-simple automorphism $a$ of $E$ and a unipotent endomorphism $u$ of $E$ such that $f = ua = au$. Moreover, under these conditions, in the notation of (iii), we must have $a = f_s$ and $u = 1 + f_s^{-1}f_n$. (i) $\Rightarrow$ (ii): this follows from Remark 1. (ii) $\Rightarrow$ (iii): take $a = f_s$ and $u = 1 + f_s^{-1}f_n$; then $f = ua = au$, while $a$ is an absolutely semi-simple automorphism, and $u$ is unipotent. (iii) $\Rightarrow$ (i): in the notation of (iii), take $n = a(u - 1) = (u - 1)a$. Then $an = na$ and $f = a + n$, and $n$ is nilpotent. It follows that $(a, n)$ is the Jordan decomposition of $f$. This implies (i) as well as the relations $a = f_s$ and $u = 1 + f_s^{-1}f_n$.

Put $f_r = f_s^{-1}f = ff_s^{-1} = 1 + f_s^{-1}f_n$, and call this the unipotent component of $f$. The pair $(f_s, f_r)$ is often called the multiplicative Jordan decomposition of the automorphism $f$.