I asked this question in <https://math.stackexchange.com/questions/204115/decomposition-of-matrices-in-semisimple-and-nilpotent-parts> ​​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? 



Decomposition of Matrices in Semisimple and Nilpotent Parts
1. You don't need to conjugate if you want $R$ to be diagonalizable (as opposed to diagonal).
2. I assume you want $R$ and $M$ to have coefficients in $K$, otherwise just work in the algebraic closure.
3. The statement is then true if $K$ is perfect and possibly false otherwise, as you can see by taking $A=[[0, 1], [t, 0]]$ in $K=F\_2(t)$.


EDIT : see <http://en.wikipedia.org/wiki/Jordan%E2%80%93Chevalley_decomposition>


