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 $n=x_n$ .

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August 28, 2012 - Posted by John Armstrong | Algebra, Linear Algebra

1 Comment »

Very clear. Thank you.

Comment by Ralph Dratman | August 29, 2012 | Reply

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