## A Trace Criterion for Nilpotence

We’re going to need another way of identifying nilpotent endomorphisms. Let be two subspaces of endomorphisms on a finite-dimensional space , and let be the collection of such that sends into . If satisfies for all then is nilpotent.

The first thing we do is take the Jordan-Chevalley decomposition of — — and fix a basis that diagonalizes with eigenvalues . We define to be the -subspace of spanned by the eigenvalues. If we can prove that this space is trivial, then all the eigenvalues of must be zero, and thus itself must be zero.

We proceed by showing that any linear functional must be zero. Taking one, we define to be the endomorphism whose matrix with respect to our fixed basis is diagonal: . If is the corresponding basis of we can calculate that

Now we can find some polynomial such that ; there is no ambiguity here since if then the linearity of implies that

Further, picking we can see that , so has no constant term. It should be apparent that .

Now, we know that is the semisimple part of , so the Jordan-Chevalley decomposition lets us write it as a polynomial in with no constant term. But then we can write . Since maps into , so does , and our hypothesis tells us that

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

Thus, as we asserted, the only possible -linear functional on is zero, meaning that is trivial, all the eigenvalues of are zero, and is nipotent, as asserted.

[…] tells us that if each is ad-nilpotent, then is itself nilpotent. We can now combine this with our trace criterion to get a convenient way of identifying solvable Lie […]

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