 Solution

We prove closure using complete reducibility of semisimple Lie algebra representations, rather than assuming that a Lie subalgebra is closed under polynomials in its elements. Write the matrix Jordan–Chevalley decomposition as $X=S+N$. On $\mathrm{End}(V)$, the commutator operator $\mathrm{ad}S$ is diagonalizable: if $S$ has eigenspaces $V_{\lambda }$, then its eigenvalue on $\mathrm{Hom}(V_{\lambda },V_{\mu })$ is $\mu -\lambda$. The operator $\mathrm{ad}N$ is nilpotent: if $N^r=0$, thenThe two operators commute because $[S,N]=0$. Uniqueness of the Jordan–Chevalley decomposition therefore says that $\mathrm{ad}S$ is the semisimple part of $\mathrm{ad}X$. Since a matrix's semisimple part is a polynomial in that matrix, and $\mathfrak{g}$ is invariant under $\mathrm{ad}X$, it follows thatNow regard $\mathrm{End}(V)$ as a Lie algebra representation of $\mathfrak{g}$ under commutators. By complete reducibility of semisimple Lie algebra representations, there is an invariant complement $M$ with $\mathrm{End}(V)=\mathfrak{g}\oplus M$. Write $S=Y+C$, with $Y\in \mathfrak{g}$ and $C\in M$. For $A\in \mathfrak{g}$, the normalization property gives $[C,A]=[S,A]-[Y,A]\in \mathfrak{g}$, while invariance of $M$ gives $[C,A]\in M$. Thus $[C,A]=0$: $C$ commutes with the whole Lie algebra.

Decompose $V={⨁}_j{V}_j$ into irreducible $\mathfrak{g}$-modules. Each $V_j$ is preserved by $X$, hence by its polynomial parts $S,N$, and also by $Y$; therefore it is preserved by $C$. By the Schur lemma, $C{|}_{V_j}=c_{j}I$. A semisimple Lie algebra is a perfect Lie algebra, so every element is a sum of Lie brackets and has trace zero on every finite-dimensional representation. Consequently $\mathrm{tr}(X{|}_{V_j})=\mathrm{tr}(Y{|}_{V_j})=0$. Also $N{|}_{V_j}$ is nilpotent, so it has trace zero; henceThus every $c_j$ is zero, $C=0$, and $S=Y\in \mathfrak{g}$. Finally $N=X-S\in \mathfrak{g}$. We have proved