# Simple proof of Jordan-Chevalley decomposition for complex matrices

I am not sure what you would call this theorem for complex matrices - Jordan-Chevalley decomposition, Jordan normal form or maybe SN decomposition (although here it is really the multiplicative form)!

This is a problem in Lie Groups by Rossmann

Show that any invertible matrix can be uniquely written as $a=bc$ where $b$ is semisimple, $c$ is unipotent and $b$ and $c$ commute

(I"m assuming matrices are complex valued, although it is not stated as such)

The usual proof I know involves using a basis of generalised eigenvectors for $\mathbb{C}^n$, and is reasonably complex, at least for an exercise. (For example, see Appendix B of Lie groups, Lie algebras, and representations: an elementary introduction by Brian Hall)

This leads me to believe there may be a simpler proof, although it seems to be eluding me.

Any hints?

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Maybe this makes it a little bit simpler – user13838 Sep 23 '11 at 1:06
It isn't so hard to see if you recall the existence of a Jordan normal form, and a book on Lie groups should be able to assume something like that, no? You're way past basic linear algebra now! – Dylan Moreland Sep 23 '11 at 1:28
@Dylan - Yes, that kind of felt like a sledge hammer approach! But maybe that is what was intended. If I recall, Jordan normal form is usually covered in an undergraduate linear algebra course (it has been a long time...!) – Juan S Sep 23 '11 at 2:31

Let $a$ be an endomorphism of a finite dimensional vector space $V$. Assume that the eigenvalues $\lambda_1,\dots,\lambda_k$ of $a$ are in the ground field $K$. Then there is a unique pair $(s,n)$ of commuting endomorphisms of $V$ such that $s$ is diagonalizable, $n$ is nilpotent, and $a=s+n$.
The uniqueness is clear. Let's prove the existence. We have a $K[X]$-algebra isomorphism $K[a]\simeq K[X]/(f)$, where $f$ is the minimal polynomial of $a$. By the Chinese Remainder Theorem, there is a $K[X]$-algebra isomorphism $$\phi:\prod_{i=1}^k\ \frac{K[X]}{(X-\lambda_i)^{m_i}}\to K[a],$$ and it suffices to put $s:=\phi(\lambda_1,\dots,\lambda_k)$.