The Unapologetic Mathematician

Mathematics for the interested outsider

A Trace Criterion for Nilpotence

We’re going to need another way of identifying nilpotent endomorphisms. Let A\subseteq B\subseteq\mathfrak{gl}(V) be two subspaces of endomorphisms on a finite-dimensional space V, and let M be the collection of x\in\mathfrak{gl}(V) such that \mathrm{ad}(x) sends B into A. If x\in M satisfies \mathrm{Tr}(xy)=0 for all y\in M then x is nilpotent.

The first thing we do is take the Jordan-Chevalley decomposition of xx=s+n — and fix a basis that diagonalizes x with eigenvalues a_i. We define E to be the \mathbb{Q}-subspace of \mathbb{F} spanned by the eigenvalues. If we can prove that this space is trivial, then all the eigenvalues of s must be zero, and thus s itself must be zero.

We proceed by showing that any linear functional f:E\to\mathbb{Q} must be zero. Taking one, we define y\in\mathfrak{gl}(V) to be the endomorphism whose matrix with respect to our fixed basis is diagonal: f(a_i)\delta_{ij}. If \{e_{ij}\} is the corresponding basis of \mathfrak{gl}(V) we can calculate that

\displaystyle\begin{aligned}\left[\mathrm{ad}(s)\right](e_{ij})&=(a_i-a_j)e_{ij}\\\left[\mathrm{ad}(y)\right](e_{ij})&=(f(a_i)-f(a_j))e_{ij}\end{aligned}

Now we can find some polynomial r(T) such that r(a_i-a_j)=f(a_i)-f(a_j); there is no ambiguity here since if a_i-a_j=a_k-a_l then the linearity of f implies that

\displaystyle\begin{aligned}f(a_i)-f(a_j)&=f(a_i-a_j)\\&=f(a_k-a_l)\\&=f(a_k)-f(a_l)\end{aligned}

Further, picking i=j we can see that r(0)=0, so r has no constant term. It should be apparent that \mathrm{ad}(y)=r\left(\mathrm{ad}(s)\right).

Now, we know that \mathrm{ad}(s) is the semisimple part of \mathrm{ad}(x), so the Jordan-Chevalley decomposition lets us write it as a polynomial in \mathrm{ad}(x) with no constant term. But then we can write \mathrm{ad}(y)=r\left(p\left(\mathrm{ad}(x)\right)\right). Since \mathrm{ad}(x) maps B into A, so does \mathrm{ad}(y), and our hypothesis tells us that

\displaystyle\mathrm{Tr}(xy)=\sum\limits_{i=1}^{\dim V}a_if(a_i)=0

Hitting this with f we find that the sum of the squares of the f(a_i) is also zero, but since these are rational numbers they must all be zero.

Thus, as we asserted, the only possible \mathbb{Q}-linear functional on E is zero, meaning that E is trivial, all the eigenvalues of s are zero, and x is nipotent, as asserted.

August 31, 2012 Posted by | Algebra, Lie Algebras, Linear Algebra | 1 Comment

   

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