For any matrix $A_{n\times n}$ with $\mathrm{tr}(A)=0$ show that there exist two matrices $B$ and $C$ such that $$A=BC-CB.$$
I know to prove this: if $A=BC-CB$, then we have $\mathrm{tr}(A)=0$ because $$\mathrm{tr}(BC)=\mathrm{tr}(CB)$$ so $$\mathrm{tr}(A)=\mathrm{tr}(BC-CB)=\mathrm{tr}(BC)-\mathrm{tr}(CB)=0.$$ But my problem is that I can't prove it.
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$\begingroup$I have shown, in an answer to a related question, that over a field $\mathbb F=\mathbb R$ or $\mathbb C$, every traceless matrix is a commutator (i.e. $\operatorname{trace}(C)=0\Rightarrow C=AB-BA$ for some square matrices $A$ and $B$ over $\mathbb F$). For a general ground field, a reference is also given in the answer.
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