Traceless matrices are spanned by commutators
= Traceless matrices are spanned by commutators
{title2=$\ker\operatorname{Tr}=\operatorname{span}\{AB-BA\}$}
Every <commutator> of square <matrices> is traceless by cyclicity of the <trace>. Conversely, off-diagonal <matrix units> and differences of diagonal units are themselves <commutators> and span all <traceless matrices>. This proves equality with the linear span of commutators, without needing a theorem about representation as one commutator.