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$ L$ is semisimple Lie algebra in ${\rm gl}\ V$ where $V$ is a complex vector space. Then we have two Jordan forms $$ x=d+n,\ {\rm ad}_x = {\rm ad}_d + {\rm ad}_n\ (x\in L) $$

(cf. Theorem 9.15 in the Erdmann and Wildon's book. The following proof is in page 87.)

Sketch of Proof We have Jordan : $$ {\rm ad}_x =\sigma + \nu $$ where $\sigma$ is diagonalizable and $\nu$ is nilpotent.

Since $L$ is semisipmle we have $$\sigma={\rm ad}_d,\ \nu = {\rm ad}_n $$ where $d,\ n\in L$ (cf. proposition 9.14)

So we must prove that $$ x= d+n $$ is Jordan.

(Commutativeness) $$ {\rm ad}_{[d,n]} =[{\rm ad}_d,{\rm ad}_n]=0$$ Since $L$ is semisimple so $ {\rm ad} $ is an Lie algebra isomorphism and $ [d,n]=0$

(Nilpotency of $n$) Note that there exists $p$ s.t. $${\rm ad}_n = p({\rm ad}_x) = \sum_{i=0}^k c_i ({\rm ad}_x)^i$$ so that $$0=({\rm ad}_n)^K(x)= (c_0)^K (x),\ c_0=0$$

But i cannot proceed anymore

Question : I have two questions : I cannot prove that $n$ is nilpotent and $d$ is diagonalizable.

Thank you

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(Just a little more detail on your first step Commutativity) The nullity of $\mathrm{ad}_{[d,n]}$ implies that there exists some scalar $\lambda\in\mathbb{C}$ such that $[d,n]=\lambda\mathrm{id}$. But since $L$ is semisimple, we have $L=[L,L]\subset[\mathfrak{gl}(V),\mathfrak{gl}(V)]=\mathfrak{sl}(V)$ and thus $\lambda=0$. – Olivier Bégassat Dec 20 '13 at 2:47
What is your question? I don't know what you are asking. – Olivier Bégassat Dec 20 '13 at 2:50
Thank you for your reply. I added – HK Lee Dec 20 '13 at 4:37

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