Scalar square of a traceless two-by-two matrix ($M^2=-(\det M)I$)

A two-by-two matrix $M$ with zero trace satisfies $M^2=-(\det M)I$ by the Cayley-Hamilton theorem. Over the complex numbers, either $\det M\ne 0$ and its two distinct eigenvalues make it diagonalizable, or $M^2=0$. A matrix commutator has zero trace, so this alternative applies to it. Diagonalization over the real numbers need not follow when $-\det M<0$.