It’s obvious that if is nilpotent then will be solvable. And Engel’s theorem 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 algebras.
If is a linear Lie algebra and for all and , then is solvable. We’d obviously like to use the trace criterion to show this, but we need a little massaging first.
The catch is that our consists of all the such that sends to . Clearly , but it may not be all of ; our hypothesis states that for all , but the criterion needs it to hold for all .
To get there, we use the following calculation, which is a useful lemma in its own right:
Now, if — so — and then
But since we know that , which means that the hypothesis kicks in: and so .
Then we know that all are nilpotent endomorphisms, which makes them ad-nilpotent. Engel’s theorem tells us that is nilpotent, which means is solvable.
We can also extend this out to abstract Lie algebras: if is any Lie algebra such that for all and , then is solvable. Indeed, we can apply the linear version to the image to see that this algebra is solvable. The kernel is just the center , which is abelian and thus automatically solvable. The image is thus the solvable quotient of by a solvable kernel, so we know that itself is solvable.