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

  1. May 21, 2008 #1
    If [tex]X\in\textrm{End}(\mathbb{R}^n)[/tex] is some arbitrary nxn-matrix, is it true that

    X^2 = 0\quad\implies\quad \textrm{Tr}(X)=0?
  2. jcsd
  3. May 21, 2008 #2


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    Sounds plausible. Doesn't it follow immediately from putting X into a normal form, or from computing its generalized eigenvalues?
  4. May 21, 2008 #3
    hmhm.. yes. If X is upper triangular, then

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


    0=\sum_{k=1}^n X_{ik} X_{ki} = X_{ii}^2\quad\forall i
  5. May 21, 2008 #4
    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

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

    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.
  6. May 21, 2008 #5
    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 [tex]\textrm{det}(e^X) = e^{\textrm{Tr}(X)}[/tex] 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
  7. May 21, 2008 #6

    matt grime

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    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.
  8. May 21, 2008 #7
    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
  9. May 21, 2008 #8
    wait a minute... why is the constant term of the characteristic polynomial plus or minus the trace?

    For example [tex]\textrm{tr}(1_{2\times 2})=2[/tex], but

    \textrm{det}(1_{2\times 2}-\lambda) = \lambda^2 - 2\lambda + 1
    Last edited: May 21, 2008
  10. May 21, 2008 #9

    matt grime

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    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.
  11. May 21, 2008 #10
    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 [tex]\lambda^n[/tex] and both the trace and the determinant must vanish.
  12. May 22, 2008 #11
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