Jordan–Chevalley decomposition ($X=X_s+X_n,[X_s,X_n]=0$)

Over an algebraically closed field, a linear operator $X$ has a unique decomposition into a commuting diagonalisable endomorphism $X_s$ and nilpotent endomorphism $X_n$. On its generalized eigenspace for $\lambda$, set $X_s=\lambda I$ and $X_n=X-\lambda I$. The Chinese remainder theorem makes both parts polynomials in $X$, so they preserve every $X$-invariant subspace. Uniqueness follows by restricting any other commuting decomposition to those generalized eigenspaces. Over a perfect field the semisimple part need only become diagonalizable after scalar extension.

The commutator action of $X_s$ is diagonalizable on endomorphisms: on $\mathrm{Hom}(V_{\lambda },V_{\mu })$ it acts by $\mu -\lambda$. The action of $X_n$ is nilpotent by nilpotence of commutation by a nilpotent endomorphism. The two actions commute, so uniqueness of the Additive Jordan decomposition identifies them as the parts of $\mathrm{ad}X$.

For a complex semisimple Lie algebra $\mathfrak{g}\subseteq \mathrm{End}(V)$ and $X\in \mathfrak{g}$, both parts of its Additive Jordan decomposition lie in $\mathfrak{g}$. The polynomial semisimple part of $\mathrm{ad}X$ shows $X_s$ normalizes $\mathfrak{g}$. Split $\mathrm{End}(V)=\mathfrak{g}\oplus M$ by the Weyl complete reducibility theorem; the $M$-component of $X_s$ centralizes $\mathfrak{g}$. On each irreducible summand of $V$ it is scalar by the Schur lemma. Its trace is zero because $\mathfrak{g}$ is a perfect Lie algebra and the nilpotent part has zero trace, so those scalars vanish. Nonsemisimple subalgebras may fail this property: $\mathbb{C}(I+N)$ with $N\ne 0$ nilpotent contains neither Jordan part of $I+N$.