The Unapologetic Mathematician

Mathematics for the interested outsider

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 | Algebra, Linear Algebra

1 Comment »

  1. Very clear. Thank you.

    Comment by Ralph Dratman | August 29, 2012 | Reply


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