Traceless matrices are spanned by commutators ($\ker \mathrm{Tr}=\mathrm{span}\{AB-BA\}$)

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.