Lemma: if $\operatorname{tr}(Z)=0$,then there exist $X$, $Y$, $|Y|\not=0$ satisfy $Z=XY-YX$.
It's an old problem but I can't prove it.
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See also: http://math.stackexchange.com/questions/1045360/solving-ab-ba-c – Martin Sleziak Jan 13 '16 at 16:13
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It is easy to see that the vector space of trace zero matrices is equal to the space of commutators $[X,Y]=XY-YX$, using commutators of matrices $E_{ij}$ having entry $1$ at position $(i,j)$ and otherwise only zero entries. We have, for $j\neq k,m$, \begin{align*} [E_{jk},E_{kj}] & =E_{jj}-E_{kk},\\ [E_{jm},E_{mk}] & = E_{jk}. \end{align*} Here $[X,Y]=XY-YX$ is the commutator. This proof can be found in texts on Lie algebras, showing that $[\mathfrak{sl}(n),\mathfrak{sl}(n)]=\mathfrak{sl}(n)$.
Edit: To show that already every $A\in \mathfrak{sl}(n)$ is a commutator needs a further argument, see here.
Dietrich Burde
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+1, but "easily"? could you pleas add few lines to complete the answer? I don't see how to conclude from your observations without other ingredient, maybe I'm missing something and I would like to know a simple proof of this fact. – user126154 Apr 07 '15 at 15:42
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@user126154 Think about the basis of the vector space of $n\times n$ matrices with zero trace. You will see they are of the form of the given brackets. If I were you I would have accepted this answer. – Math137 Apr 07 '15 at 15:55
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A proof is given in this paper : On matrices of trace zero by A. A. Albert and Benjamin Muckenhoupt. Link to the paper. You could also have a look at this discussion (which is related) : Solutions to the matrix equation $\mathbf{AB-BA=I}$ over general fields.
pitchounet
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