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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


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?

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I don't know. Here's what I'd try. Find a real matrix whose Jordan form is $$\pmatrix{i&1&0&0\cr0&i&0&0\cr0&0&-i&1\cr0&0&0&-i}$$ and see whather you can find your matrices $L$, $R$, and $M$. – Gerry Myerson Sep 29 '12 at 6:29
crossposted at… – m_t_ Sep 29 '12 at 12:21

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