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
&=(a_i-a_j)e_{ij}\\\left[\mathrm{ad}(y)\right](e_{ij})&=(f(a_i)-f(a_j))e_{ij}\end{aligned}](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Cbegin%7Baligned%7D%5Cleft%5B%5Cmathrm%7Bad%7D%28s%29%5Cright%5D%28e_%7Bij%7D%29%26%3D%28a_i-a_j%29e_%7Bij%7D%5C%5C%5Cleft%5B%5Cmathrm%7Bad%7D%28y%29%5Cright%5D%28e_%7Bij%7D%29%26%3D%28f%28a_i%29-f%28a_j%29%29e_%7Bij%7D%5Cend%7Baligned%7D&bg=e6e6e6&fg=333333&s=0&c=20201002)
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.
Like this:
Like Loading...
Related
August 31, 2012 -
Posted by John Armstrong |
Algebra, Lie Algebras, Linear Algebra
The Unapologetic Mathematician 
[…] 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 […]
Pingback by Cartan’s Criterion « The Unapologetic Mathematician | September 1, 2012 |