Real vs. complex Jordan-Chevalley decomposition Let $A$ be a real square matrix. Then over ${\mathbb C}$, $A$ has a unique Jordan-Chevalley 
decomposition $A=D+N$ with $D$ diagonalizable, $N$ nilpotent 
and $DN=ND$.
My question is : must $D$ and $N$ be real ? 
If $A$ is already trigonalizable over $\mathbb R$, then it will also
have a Jordan-Chevalley decomposition over $\mathbb R$ and by
uniqueness $D$ and $N$ will be real.
I'm not sure about what happens when $A$ is not trigonalizable over $\mathbb R$
though.
 A: In general, if $F$ is a perfect field, then the Jordan-Chevallery decomposition exists, namely for every square matrix $A$ over $F$, there's a unique semisimple matrix $D$ (namely, diagonalizable over the algebraic closure) and nilpotent matrix $N$ such that $A=D+N=N+D$. See the wiki page.
I suppose that the theorem is proved if the eigenpolynomial splits in the ground field, whose proof is canonical. Suppose $A$ is a matrix over $F$, and $f$ is its eigenpolynomial. Consider the splitting field $K$ of $f$ over $F$. Since $K/F$ is normal and separable, it's (finitely) Galois, and denote $G=\operatorname{Gal}(K/F)$. Suppose $A=D+N=N+D$ is a Jordan-Chevallery decomposition of $A$ over the field $K$. We should prove that $D,N$ are in fact matrices over $F$. To prove this, note that $G$ acts naturally on the space of matrices over $K$. Take each $\sigma\in G$, we have $\sigma(A)=\sigma(D)+\sigma(N)=\sigma(N)+\sigma(D)$. Note that $\sigma(A)=A$ since $A$ is a matrix over $F$, and $\sigma(N)$ is nilpotent, therefore by uniqueness $\sigma(D)=D$ and $\sigma(N)=N$ for all $\sigma\in G$. Note that the fixed field $K^G=F$ by fixed field theorem, we have $N,D$ are matrices over $F$.
