Does the abstract Jordan decomposition agree with the usual Jordan decomposition in a semisimple Lie subalgebra of endomorphisms? Is it true that for every element $x$ of a semisimple Lie subalgebra of endomorphisms $L\subseteq \text{End}(V)$, where $V$ is a finite dimensional vector space over $\mathbb{C}$, the abstract Jordan decomposition of $x$ coincides with the usual Jordan decomposition of $x$ considered as a linear endomorphism $x : V \to V$? 
I guess it can be false, as I see no reason why the summands of the usual Jordan decomposition must belong to the semisimple Lie subalgebra $L$.
Am I right or wrong? Could you please write me a proof or a counterexample of this statement?
Thanks! :) 
 A: It is true as long as the Lie subalgebra is semisimple, and the field has characteristic zero. If not, then there are counterexamples, see  this MO question. We also may assume that the field is algebraically closed.
The theorem is as follows. Let $L$ be a linear, semisimple Lie algebra over an algebraically closed field of characteristic zero. Then $L$ contains the semisimple and nilpotent parts in $gl(V )$ of all of its elements. In particular, the abstract and usual Jordan decompositions in $L$ agree.
Proof: Let $x ∈ L$ be given. Let $x = s+n$ be the usual Jordan decomposition in $gl(V)$. The problem is to show $s,n ∈ L$. Since $ad( x)(L) ⊆ L$, we know that both $ad (s)(L) ⊆ L$ and $ad ( n)(L) ⊆ L$. Hence, $s, n ∈ N_{gl(V)}(L) $. In general it is not the case that $N = L$ so we will need to construct a smaller subalgebra containing both $s, n$ which can be shown to equal $L$. To this end we need to use Weyl’s theorem and it’s Lemma that $L ⊆ sl(V )$. Let $W$ be any $L$-submodule of $V$ ,and define $L_W =\lbrace y∈gl(V)|y(W)⊆W, tr(y|W)=0\rbrace$. For example $L_V = sl(V )$. We have $L ⊆ L_W$ , for all W. Define, $L′ = \cap_W (N ∩ L_W ) ⊆ N$. This is a subalgebra which contains $L$, and we will show it is equal to $L$. Before that, because $W$ is $L$-stable, it is in particular $x$-stable and again we have $W$ is stable under both $s, n$. Furthermore,
since $0=tr(x)=tr(s+n)$ and $tr(n)=0$ we have $tr(s)=0$ hence $s,n∈L_W$ for all $W$ and thus in $L′$.
Hence,it remains to show $L=L′$.From the fact $L′ ⊆N$ we have $[L,L′]⊆L$ so in particular
$L′$ is an $L$-module. As $L$ is a submodule in $L′$, Weyl’s Theorem allows us to write
$L′ = L ⊕ M$ for some $L$-module $M$. Since $[L, L′] ⊆ L$, we have $L$ acts trivially on $M$. Let $W$ be an irreducible $L$-submodule of $V$. If $y ∈ M$ then, because $[L,y] = 0$, i.e.
$y$ commutes with all of $L$, Schur’s Lemma implies that $y$ acts on $W$ as a scalar. On the other hand $y∈M ⊆L′ ⊆L_W$, so we have $tr(y|W)=0$. With the fact $charF = 0$ this implies
$y = 0$, hence, $M = 0$, because $V$ is a sum of irreducibles by Weyl’s Theorem, proving $L=L'$.
