Adjoint compatibility of additive Jordan decomposition ($(\mathrm{ad}X{)}_s=\mathrm{ad}{X}_s,(\mathrm{ad}X{)}_n=\mathrm{ad}{X}_n$)

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 $I+N$.