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Let $F$ be a field and $n\in \mathbb N^*$.

The (additive) Jordan-Chevalley decomposition theorem states that for any matrix $M\in \mathcal M_n(F)$ whose characteristic polynomial splits into linear factors there exists a unique decomposition $M=D+N$ where $D$ is diagonalisable and $N$ is nilpotent and commutes with $D$.

This proof of the multiplicative Jordan-Chevalley decomposition theorem uses the additional (given, I think) fact that if $M$ is invertible, then $D$ (from the additive decomposition) is also invertible, but I can't see why.

James Well
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    See https://math.stackexchange.com/questions/119904/units-and-nilpotents. – Cave Johnson Aug 25 '18 at 22:10
  • I'm under the impression the statement I'm trying to prove is the converse, am I wrong ? He assumes $u$ (i.e. my $D$) to be a unit, I myself am trying to prove that while knowing that $D+N$ (or $u+a$ in the post you mention) is a unit. – James Well Aug 25 '18 at 23:05
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    $N$ is nilpotent, then so is $-N$. Thus $D=(D+N)+(-N)$ is a unit. – Cave Johnson Aug 26 '18 at 01:58
  • PS : I think I remember reading somewhere that $D$ even has the same eigenvalues as $M$ but I'm not sure ? Can you confirm this ? If so I'll post later on for a proof if needed. – James Well Aug 26 '18 at 14:20
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    The argument is identical: $D-\lambda I$ is not a unit if and only if $D-\lambda I +N$ is not a unit. – Cave Johnson Aug 26 '18 at 14:23

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