Identity for natural log of matrices

Question

Let $A$ and $B$ be square matrices such that $\ln (A)$ , $\ln (B)$ and $\ln (AB)$ are all defined. Is it true that $$\ln (AB)= \ln (A) + \ln(B)$$ only if $AB=BA$ ?

I appreciate any help, Thanks

Answer 1

If $A$ and $B$ commute, we can check that $\ln(AB)=\ln(A)+\ln(B)$ by considering the series involved. And if $A$ and $B$ do not commute, we can't have both $\ln(AB)=\ln(A)+\ln(B)$ and $\ln(BA)=\ln(A)+\ln(B)$, as that would imply $\ln(AB)=\ln(BA)$.

In all generality, we have the Baker-Campbell-Hausdorff to express $Z = \ln(AB)$ in terms of $X = \ln(A)$ and $Y = \ln(B)$ (whether $A$ and $B$ commute or not). The formula is rather complicated and involves nested commutators : if $C$ $D$ are matrices, we define the commutator of $C$ and $D$ to be $[C,D] = CD-DC$ (this quantity is $0$ if and only if $C$ and $D$ commute). With this notations, the formula can be written as

\begin{eqnarray*} Z &=& X + Y \\ &+& \frac{1}{2}[X,Y] \\ &+& \frac{1}{12}([X,[X,Y]]+[Y,[Y,X]]) \\ &-& \frac{1}{24} [Y,[X,[X,Y]]] \\ && \quad\vdots \end{eqnarray*}

Of course, if $A$ and $B$ commute, all commutators terms vanish and we simply get $Z = X + Y$, as expected.