 Solution

For the ordered matrix exponential product $\Phi (t)=e^{tA}{e}^{tB}$, differentiation givesThe matrix commutator here is $[X,Y]=XY-YX$. Put $E(t)=\Phi (t)-e^{t(A+B)}$. Then $E(0)=0$ andMultiplying on the left by $e^{-t(A+B)}$ gives an exact product derivative. Integration from zero to $t$, followed by multiplication by $e^{t(A+B)}$, proves the exponential product defect identity:No commutation of $A$ and $B$ was assumed. The ordered product is the usual Lie-Trotter splitting; the PDF calls this product the Beam-Warming splitting. If $[A,B]=0$, the commutator term vanishes and the product is exact. In general its small-step expansion has local error $t^2[A,B]/2+O(t^3)$.