- #1

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[tex]

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

[/tex]

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- Thread starter jostpuur
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- #1

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- 18

[tex]

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

[/tex]

- #2

Hurkyl

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

[tex]

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

[/tex]

- #3

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[tex]

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

[/tex]

implies

[tex]

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

[/tex]

- #4

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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]

then

[tex]\det (e^X) = 1[/tex]

but

[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.

- #5

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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.

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:

- #6

matt grime

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- #7

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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.

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- #8

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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

[tex]

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

[/tex]

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

[tex]

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

[/tex]

Last edited:

- #9

matt grime

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

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