I asked this question in http://math.stackexchange.com/posts/204115/edit ​​but remains unanswered.

For any matrix $A\in M_n(\mathbb F)$, where $\mathbb F$ is an algebraically closed field, there is a matrix $S\in M_n(\mathbb F)$ such that 

$$SAS^{-1}=D+N,$$
where $D$ is diagonal and $N$ nilpotent. Moreover, this decomposition is unique.




 Suppose now that $A\in M_n(\mathbb K)$, but $\mathbb K$ is not necessarily algebraically closed. It is also true that  there is a matrix $L\in M_n(\mathbb K)$ such that 

$$LAL^{-1}=R+M,$$

 where $M$ is nilpotent and $R$ is diagonalizable in the  algebraic closure of $\mathbb K$? Moreover when we consider the decomposition in $\mathbb K$ and in the algebraic closure of $\mathbb K$ the nilpotent part is the same?