The Jordan-Chevalley Decomposition (proof)

We now give the proof of the Jordan-Chevalley decomposition. We let $x$ have distinct eigenvalues $\{a_i\}_{i=1}^k$ with multiplicities $\{m_i\}_{i=1}^k$ , so the characteristic polynomial of $x$ is

$\displaystyle\prod\limits_{i=1}^k(T-a_i)^{m_i}$

We set $V_i=\mathrm{Ker}\left((x-a_iI)^{m_i}\right)$ so that $V$ is the direct sum of these subspaces, each of which is fixed by $x$ .

On the subspace $V_i$ , $x$ has the characteristic polynomial $(T-a_i)^{m_i}$ . What we want is a single polynomial $p(T)$ such that

$\displaystyle\begin{aligned}p(T)&\equiv a_i\mod (T-a_i)^{m_i}\\p(T)&\equiv0\mod T\end{aligned}$

That is, $p(T)$ has no constant term, and for each $i$ there is some $k_i(T)$ such that

$\displaystyle p(T)=(T-a_i)^{m_i}k_i(T)+a_i$

Thus, if we evaluate $p(x)$ on the $V_i$ block we get $a_i$ .

To do this, we will make use of a result that usually comes up in number theory called the Chinese remainder theorem. Unfortunately, I didn’t have the foresight to cover number theory before Lie algebras, so I’ll just give the statement: any system of congruences — like the one above — where the moduli are relatively prime — as they are above, unless $0$ is an eigenvalue in which case just leave out the last congruence since we don’t need it — has a common solution, which is unique modulo the product of the separate moduli. For example, the system

$\displaystyle\begin{aligned}x&\equiv2\mod3\\x&\equiv3\mod4\\x&\equiv1\mod5\end{aligned}$

has the solution $11$ , which is unique modulo $3\cdot4\cdot5=60$ . This is pretty straightforward to understand for integers, but it works as stated over any principal ideal domain — like $\mathbb{F}[T]$ — and, suitably generalized, over any commutative ring.

So anyway, such a $p$ exists, and it’s the $p$ we need to get the semisimple part of $x$ . Indeed, on any block $V_i$ $x_s=p(x)$ differs from $x$ by stripping off any off-diagonal elements. Then we can just set $q(T)=T-p(T)$ and find $x_n=q(x)$ . Any two polynomials in $x$ must commute — indeed we can simply calculate

$\displaystyle\begin{aligned}x_sx_n&=p(x)q(x)\\&=q(x)p(x)\\&=x_nx_s\end{aligned}$

Finally, if $x:B\to A$ then so must any polynomial in $x$ , so the last assertion of the decomposition holds.

The only thing left is the uniqueness of the decomposition. Let’s say that $x=s+n$ is a different decomposition into a semisimple and a nilpotent part which commute with each other. Then we have $x_s-s=n-x_n$ , and all four of these endomorphisms commute with each other. But the left-hand side is semisimple — diagonalizable — but the right hand side is nilpotent, which means its only possible eigenvalue is zero. Thus $s=x_s$ and $n=x_n$ .

1 Comment »

Very clear. Thank you.

Comment by Ralph Dratman | August 29, 2012 | Reply

About this weblog

This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).

I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.

RSS Feeds

RSS - Posts
RSS - Comments
Feedback

Got something to say? Anonymous questions, comments, and suggestions at Formspring.me!
Subjects
Subjects
Archives

August 2012

M T W T F S S

« Jul Sep »

1 2 3 4 5

6 7 8 9 10 11 12

13 14 15 16 17 18 19

20 21 22 23 24 25 26

27 28 29 30 31
Search for:

The Unapologetic Mathematician

Mathematics for the interested outsider