Theorem: Let $L\subseteq \mathfrak{gl}(V)$ be finite dimensional semisimple Lie algebra. Then every $x\in L$ has Jordan decomposition within $L$ (semi-simple + nilpotent).
Part of proof:(Humphreys, p.29) Let $W\subseteq V$ be an $L$-submodule. Define $$L_W :=\{ y\in \mathfrak{gl}(V): y(W)\subseteq W \mbox{ and } Tr(y|_W)=0\}.$$ Since $L$ is semi-simple, $L=[L,L]$.
Thus, $L$ is contained in $L_W$. How?
Here $L=[L,L]$ implies that $L$ is sitting inside $\mathfrak{sl}(V)$, so every element of $L$ has trace $0$ on $V$; but how can it be zero also on any subspace of $V$ which is invariant under it?
$L\subset gl(W)=\{M\in gl(V): M(W)\subset W\}$ since $W$ is a submodule. Now use the fact that $[gl(V),gl(V)]=sl(V)=\{M: tr(M)=0\}$, this implies that $[L,L]\subset [gl(V),gl(V)]=sl(V)$, we deduce $L=[L,L]\subset sl(V)$ and $L=\subset sl(V)\cap gl(W)=L_W$.