= Solution
Every <commutator> $[A,B]=AB-BA$ has zero <trace>, by the cyclic identity. Hence its linear span $W$ is contained in $\ker\operatorname{Tr}$. For the reverse inclusion, the <matrix units> satisfy
$$
E_{ij}=[E_{ii},E_{ij}]\quad(i\ne j),\qquad E_{ii}-E_{jj}=[E_{ij},E_{ji}].
$$
The 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>, proving
$$
\boxed{W=\ker\operatorname{Tr}.}
$$
This 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.
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