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Every commutator $[A,B]=AB-BA$ has zero trace, by the cyclic identity. Hence its linear span $W$ is contained in $\ker \mathrm{Tr}$. For the reverse inclusion, the matrix units satisfyThe off-diagonal matrix units and the differences $E_{ii}-E_{nn}$ for $i<n$ span all traceless matrices: a traceless diagonal part is ${\sum }_{i<n}{a}_{ii}(E_{ii}-E_{nn})$. Thus every element of the kernel is a linear combination of commutators, provingThis establishes that traceless matrices are spanned by commutators. For $n=1$, both vector subspaces are zero. Notice that the argument needs a span of commutators; it does not assume that every such sum has already been written as a single commutator.