Proposition 4

Assume that $k$ is perfect. Let V be a finite dimensional vector space, $u$ an element of End(V)$,u_s, u_n$ the semi-simple and nilpotent components of $u($Algebra, Chap. VII, §5, no. 8).

(i) For all $\lambda \in k,V^{\lambda}(u) = V^{\lambda}(u_s) = V_{\lambda}(u_s)$.

(ii) If V has an algebra structure and if $u$ is a derivation of V$,u_s$ and $u_n$ are derivations of V.

(iii) If V has an algebra structure and if $u$ is an automorphism of V, then $u_s$ and $1 +u^{-1}_su_n$ are automorphisms of V.

In view of Prop. 1, we can assume that $k$ is algebraically closed, so

The semi-simple component of $u|V^{\lambda}(u)$ is the homothety with ratio $\lambda$ in $V^{\lambda}(u)$. This proves (i).

Assume from now on that V has an algebra structure. Let $x\in V^{\lambda}(u)$, $y\in V^\mu(u)$.

If $u$ is a derivation of V, then $xy\in V^{\lambda+\mu}(u)$ (Prop. 2 (ii)), so

This proves that $u_s$ is a derivation of V. Then $u_n=u-u_s$ is a derivation of V.

If $u$ is an automorphism of V, Ker($u_s$) $= V^0(u) = 0$, so $u_s$ is bijective. On the other hand, $xy\in V^{\lambda \mu}(u)$ (Prop. 2 (iii)), so

This proves that $u_s$ is an automorphism of V; but then so is