= Solution
Work over $\mathbb C$. Decompose $V$ as the <direct sum> of the <generalized eigenspaces> $V_\lambda$ of $X$. Define $X_s$ to be $\lambda I$ on $V_\lambda$ and set $X_n=X-X_s$. Then $X_s$ is a <diagonalisable endomorphism>, $X_n$ is a <nilpotent endomorphism>, and they commute. For uniqueness, any commuting decomposition $X=S+N$ has $S$ and $N$ commuting with $X$, hence preserving each $V_\lambda$. Decompose $V_\lambda$ further into <eigenspaces> of $S$. On a nonzero such space with <eigenvalue> $\sigma$, the operator $X=\sigma I+N$ has only the <eigenvalue> $\sigma$, so $\sigma=\lambda$. Since $S$ is diagonalizable, $S=\lambda I$ throughout $V_\lambda$. Thus $S=X_s$ and $N=X_n$. This is the <Additive Jordan decomposition>.
The <Chinese remainder theorem> for the pairwise coprime polynomials $(t-\lambda)^{r_\lambda}$ also gives a polynomial $p$ with $p(t)\equiv\lambda\pmod{(t-\lambda)^{r_\lambda}}$, where $r_\lambda$ is the largest <Jordan block> size. Consequently $X_s=p(X)$ and $X_n=X-p(X)$. This polynomial description shows that both parts preserve every $X$-invariant subspace.
On $\operatorname{Hom}(V_\lambda,V_\mu)$ the operator $\operatorname{ad}X_s$ is the scalar $\mu-\lambda$, so it is diagonalizable. The <nilpotence of commutation by a nilpotent endomorphism> makes $\operatorname{ad}X_n$ nilpotent. They commute, since $[\operatorname{ad}X_s,\operatorname{ad}X_n]=\operatorname{ad}[X_s,X_n]=0$. Uniqueness of the <Additive Jordan decomposition> therefore gives the <adjoint compatibility of additive Jordan decomposition>:
$$
\boxed{(\operatorname{ad}X)_s=\operatorname{ad}X_s,\qquad(\operatorname{ad}X)_n=\operatorname{ad}X_n.}
$$
Now let $\mathfrak g\subseteq\mathfrak{gl}(V)$ be a complex <semisimple Lie algebra> and $X\in\mathfrak g$. Since the semisimple part of $\operatorname{ad}X$ is a polynomial in $\operatorname{ad}X$, it preserves $\mathfrak g$. Thus $[X_s,\mathfrak g]\subseteq\mathfrak g$. By the <Weyl complete reducibility theorem>, the <Adjoint representation> of $\mathfrak g$ on $\operatorname{End}(V)$ has a decomposition $\operatorname{End}(V)=\mathfrak g\oplus M$ into invariant subspaces. Write $X_s=y+z$ with $y\in\mathfrak g$ and $z\in M$. For $a\in\mathfrak g$, the vector $[z,a]=[X_s,a]-[y,a]$ lies in $\mathfrak g$, and invariance of $M$ puts it in $M$ as well. Hence $[z,a]=0$.
Decompose $V=\bigoplus_j U_j$ into <Irreducible Lie algebra representations>. Each $U_j$ is preserved by $X_s=p(X)$ and by $y$, hence by $z$. The <Schur lemma> makes $z|_{U_j}=c_jI$. A <semisimple Lie algebra> is a <perfect Lie algebra>, so every representing element of $\mathfrak g$ has zero <trace> on every $U_j$. Also $\operatorname{tr}(X_s|_{U_j})=\operatorname{tr}(X|_{U_j})$, because $X_n$ is nilpotent there. It follows that $c_j\dim U_j=\operatorname{tr}(z|_{U_j})=0$. In <characteristic zero>, $c_j=0$, so $z=0$. Therefore $X_s\in\mathfrak g$ and $X_n=X-X_s\in\mathfrak g$. This proves that <semisimple matrix Lie algebras are closed under additive Jordan decomposition>, including representations with repeated isomorphic irreducible summands.
Back to article page
