Commutator on a finite-dimensional $\mathcal{H}$ is scalar multiple of an identity

Question

Suppose $A$ and $B$ are operators on a finite-dimensional Hilbert space and suppose that $[A, B] = c I$ for some constant c. Show that $c = 0$.

I have tried approaching the problem using matrices $A, B$ so \begin{align} A_{ik}B_{kj}- B_{il}A_{lj} = c \delta_{ij} \end{align} and somehow show that it is the same as $[B,A]$, so the $c$ is forced to be 0.

Answer 1

Observe that, as $\operatorname{tr}(AB)=\operatorname{tr}(BA)$, the trace is linear, and $\operatorname{tr}(I)\neq0$,

$$c\operatorname{tr}(I)=\operatorname{tr}(cI)=\operatorname{tr}([A,B])=\operatorname{tr}(AB)-\operatorname{tr}(BA)=0,$$

implying that $c=0$.