# Matrix X^2=0 -> Tr(X)=0?

1. May 21, 2008

### jostpuur

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

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

2. May 21, 2008

### Hurkyl

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

3. May 21, 2008

### jostpuur

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

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

implies

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

4. May 21, 2008

### 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.

5. May 21, 2008

### 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.

Last edited: May 21, 2008
6. May 21, 2008

### 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.

7. May 21, 2008

### jostpuur

Okey, no need for change of basis if you know lot of linear algebra! :surprised 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.

Last edited: May 21, 2008
8. May 21, 2008

### 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

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

Last edited: May 21, 2008
9. May 21, 2008

### 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.

10. May 21, 2008

### 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.

11. May 22, 2008