# The Unapologetic Mathematician

## Uses of the Jordan-Chevalley Decomposition

Now that we’ve given the proof, we want to mention a few uses of the Jordan-Chevalley decomposition.

First, we let $A$ be any finite-dimensional $\mathbb{F}$-algebra — associative, Lie, whatever — and remember that $\mathrm{End}_\mathbb{F}(A)$ contains the Lie algebra of derivations $\mathrm{Der}(A)$. I say that if $\delta\in\mathrm{Der}(A)$ then so are its semisimple part $\sigma$ and its nilpotent part $\nu$; it’s enough to show that $\sigma$ is.

Just like we decomposed $V$ in the proof of the Jordan-Chevalley decomposition, we can break $A$ down into the eigenspaces of $\delta$ — or, equivalently, of $\sigma$. But this time we will index them by the eigenvalue: $A_a$ consists of those $x\in A$ such that $\left[\delta-aI\right]^k(x)=0$ for sufficiently large $k$.

Now we have the identity:

$\displaystyle\left[\delta-(a+b)I\right]^n(xy)=\sum\limits_{i=0}^n\binom{n}{i}\left[\delta-aI\right]^{n-i}(x)\left[\delta-bI\right]^i(y)$

which is easily verified. If a sufficiently large power of $\delta-aI$ applied to $x$ and a sufficiently large power of $\delta-bI$ applied to $y$ are both zero, then for sufficiently large $n$ one or the other factor in each term will be zero, and so the entire sum is zero. Thus we verify that $A_aA_b\subseteq A_{a+b}$.

If we take $x\in A_a$ and $y\in A_b$ then $xy\in A_{a+b}$, and thus $\sigma(xy)=(a+b)xy$. On the other hand,

\displaystyle\begin{aligned}\sigma(x)y+x\sigma(y)&=axy+bxy\\&=(a+b)xy\end{aligned}

And thus $\sigma$ satisfies the derivation property

$\displaystyle\sigma(xy)=\sigma(x)y+x\sigma(y)$

so $\sigma$ and $\nu$ are both in $\mathrm{Der}(A)$.

For the other side we note that, just as the adjoint of a nilpotent endomorphism is nilpotent, the adjoint of a semisimple endomorphism is semisimple. Indeed, if $\{v_i\}_{i=0}^n$ is a basis of $V$ such that the matrix of $x$ is diagonal with eigenvalues $\{a_i\}$, then we let $e_{ij}$ be the standard basis element of $\mathfrak{gl}(n,\mathbb{F})$, which is isomorphic to $\mathfrak{gl}(V)$ using the basis $\{v_i\}$. It’s a straightforward calculation to verify that

$\displaystyle\left[\mathrm{ad}(x)\right](e_{ij})=(a_i-a_j)e_{ij}$

and thus $\mathrm{ad}(x)$ is diagonal with respect to this basis.

So now if $x=x_s+x_n$ is the Jordan-Chevalley decomposition of $x$, then $\mathrm{ad}(x_s)$ is semisimple and $\mathrm{ad}(x_n)$ is nilpotent. They commute, since

\displaystyle\begin{aligned}\left[\mathrm{ad}(x_s),\mathrm{ad}(x_n)\right]&=\mathrm{ad}\left([x_s,x_n]\right)\\&=\mathrm{ad}(0)=0\end{aligned}

Since $\mathrm{ad}(x)=\mathrm{ad}(x_s)+\mathrm{ad}(x_n)$ is the decomposition of $\mathrm{ad}(x)$ into a semisimple and a nilpotent part which commute with each other, it is the Jordan-Chevalley decomposition of $\mathrm{ad}(x)$.

August 30, 2012 - Posted by | Algebra, Lie Algebras, Linear Algebra

1. Could you elaborate a bit (or just give me a hint) on the first “identity: … which is easily verified”? I’m not seeing it…

Thanks.

Comment by Joe English | August 30, 2012 | Reply

2. Run an induction on $n$; it works out sort of like the binomial theorem does.

Comment by John Armstrong | August 30, 2012 | Reply

3. […] we know that is the semisimple part of , so the Jordan-Chevalley decomposition lets us write it as a polynomial […]

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