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?
2026-04-01 22:16:56.1775081816
If $\mathrm{Tr} (X)=0$ then there exists $A, B\in \mathcal {M}_n (\mathbb {C}) $ s.t $ X=A\cdot B-B\cdot A $.
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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: