# Decomposition of Matrices in Semisimple and Nilpotent Parts

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?

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Also $L$ is redundant. You may take $L=E$ (the identity matrix). –  Anton Klyachko Sep 29 '12 at 12:24
There is no uniqueness if you don't require that the two matrices in the decomposition commute. –  BS. Sep 30 '12 at 10:00

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 $K=F_2(t)$.

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We should demand that $R$ and $M$ commute. –  Anton Klyachko Sep 29 '12 at 15:58
The example in #3 (which readily adapts to any imperfect field $k$ using the $k$-linear multiplication by $a^{1/p}$ on $V = k(a^{1/p})$) isn't an entirely satisfactory counterexample because it is semisimple over $k$ (though not "geometrically semisimple"; i.e., not diagonalizable over $\overline{k}$). The Wikipedia entry has now been updated to give an example over any imperfect field $k$ in which the operator isn't a sum of two commuting $k$-linear operators that are respectively semisimple (just over $k$!) and nilpotent. –  grp Oct 1 '12 at 11:23