Is the Trace of a Nilpotent Matrix Always Zero?

[jostpuur]

If $$X\in\textrm{End}(\mathbb{R}^n)$$ is some arbitrary nxn-matrix, is it true that

$$X^2 = 0\quad\implies\quad \textrm{Tr}(X)=0?$$

 

[Hurkyl]

If $$X\in\textrm{End}(\mathbb{R}^n)$$ is some arbitrary nxn-matrix, is it true that

$$X^2 = 0\quad\implies\quad \textrm{Tr}(X)=0?$$

Sounds plausible. Doesn't it follow immediately from putting X into a normal form, or from computing its generalized eigenvalues?

 

[jostpuur]

hmhm.. yes. If X is upper triangular, then

$$\sum_{k=1}^n X_{ik} X_{kj} = 0\quad\forall i,j$$

implies

$$0=\sum_{k=1}^n X_{ik} X_{ki} = X_{ii}^2\quad\forall i$$

 

[DavidWhitbeck]

It doesn't have to be upper triangular.

My hunch is that you could take advantage of higher powers of X being zero to say that

$$X^n = 0, n \geq 2 \Rightarrow$$
$$e^X = I + X$$
and then if you could argue that
$$\det (I + X) = 1$$
then
$$\det (e^X) = 1$$
but
$$\det (e^X) = e^{(\textup{tr} X)}$$
which would imply that
$$e^{(\textup{tr} X)} = 1 \Rightarrow \textup{tr} X = 0$$

but I'm missing the crucial step, so I guess it's no good. It's an interesting problem and I'll follow this thread to see if anyone posts the solution.

 

[jostpuur]

This got settled already. We don't need to assume that X is upper triangular in the beginning, because there is a theorem that says that for any matrix there is a coordinate transformation that transforms the matrix into the Jordan normal form, http://en.wikipedia.org/wiki/Jordan_normal_form (edit: hmhm... although it could be that the transform involves complex numbers...). The Jordan normal form is a special case of upper triangular matrices, so if the claim is true for them, its all done.

Actually the proof (the one I know) of the formula $$\textrm{det}(e^X) = e^{\textrm{Tr}(X)}$$ uses the fact that we can choose a basis so that X is upper triangular. I'm not fully sure what you were doing, but I believe that if you could make your idea work, it would probably still be using the same underlying facts.

 

[matt grime]

If X^2 is zero then either X is zero, or X^2 is the minimal poly of X, and in particular divides its characteristic polynomial, which implies that the char poly has 0 constant term, but the constant term is +/- the trace of X. No need to invoke a choice of basis.

 

[jostpuur]

If X^2 is zero then either X is zero, or X^2 is the minimal poly of X, and in particular divides its characteristic polynomial, which implies that the char poly has 0 constant term, but the constant term is +/- the trace of X. No need to invoke a choice of basis.

Okey, no need for change of basis if you know lot of linear algebra! I think I'll skip the proof of Cayley-Hamilton theorem for now, http://en.wikipedia.org/wiki/Cayley–Hamilton_theorem, because it looks too unpleasant.

 

[jostpuur]

wait a minute... why is the constant term of the characteristic polynomial plus or minus the trace?

For example $$\textrm{tr}(1_{2\times 2})=2$$, but

$$\textrm{det}(1_{2\times 2}-\lambda) = \lambda^2 - 2\lambda + 1$$

 

[matt grime]

Duh. Idiot. It is the coefficient of the second highest term that is the trace (or minus the trace). It is the determinant that is the constant coefficient. Sorry. Stick with the e-values all being zero, hence the sum being zero, and thus the trace is zero.

 

[DavidWhitbeck]

That was slick Matt. Any eigenvalue has to be zero (since any e-vector of X is an e-vector of X^2), so the characteristic polynomial is just $$\lambda^n$$ and both the trace and the determinant must vanish.

 

[trambolin]

The argument is easier if you consider a nilpotent matrix with index 2 (or 1 depending on the definition of nilpotency index). Proof is immediate.

http://en.wikipedia.org/wiki/Nilpotent_matrix