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?