Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2025 / Paper 1 / 6 / b / Solution

The cyclic property of the trace impliesTaking the matrix trace of the asserted equality therefore gives $nc=0$. For a positive integer $n$ over the complex numbers this requires $\boxed{c=0}$. It is also sufficient: choose $M=N=0$, or any two commuting matrices. Thus a finite-dimensional operator commutator cannot be a nonzero scalar multiple of the identity matrix.