Let $X\in \mathcal {M}_n (\mathbb {C}) $ with $\mathrm{Tr}(X)=0$. Then there exists $A, B\in \mathcal {M}_n (\mathbb {C}) $ s.t $X=A\cdot B-B\cdot A $.
It's there a solution that uses Normal Jordan Form?
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3Yes, there is. ${}{}{}$ – Pedro Oct 20 '17 at 17:13
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1Duplicate of https://math.stackexchange.com/q/95537/490083 – Learnmore Oct 20 '17 at 17:40
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1Also see this https://math.stackexchange.com/q/756131/490083 – Learnmore Oct 20 '17 at 17:42
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I need a proof by Jordan Normal Form – Problemsolving Oct 20 '17 at 18:00
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@PedroTamaroff - Are you familiar with a simpler proof than the one I mentioned below? It would be interesting to see, since Bhatia's seems like a bit of an overkill. – Nathaniel Bubis Oct 22 '17 at 09:54
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Rajendra Bhatia gives a sketch of such a proof in his book "Matrix Analysis" (pp 190).
Note that in his notation, $Z(A)$ is the set of all commuting matrices, $Z(A) = \{X\ |\ [A,X] = 0\}$, and $T_AO_A$ is the tangent orbit, $T_AO_A =\{[A,X]\ \forall \ X\in M_n \}$.
Screenshot from Google Books:
Nathaniel Bubis
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