= Solution
The <trace> of a square <matrix> is $\operatorname{tr}A=\sum_iA_{ii}$. Expanding a product and interchanging its finite sums gives
$$
\boxed{\operatorname{tr}(AB)=\sum_{i,j}A_{ij}B_{ji}=\sum_{j,i}B_{ji}A_{ij}=\operatorname{tr}(BA).}
$$
Hence every <commutator> has trace zero. Taking the trace of $AB-BA=cI$ gives $0=cn$. Over the complex numbers, with $n\geq1$, this forces
$$
\boxed{c=0.}
$$
For that value, choose $A=B=0$, or any commuting pair. For every nonzero $c$ the trace contradiction proves nonexistence. The finite <matrix> size is essential to this trace argument.
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