# $AB=BA$ from $e^{A+B} = e^A e^B$, given Hermitian matrices

Let $$A$$ and $$B$$ be Hermitian matrices.

• If $$AB=BA$$, we know that $$e^{A+B} = e^A e^B$$.
• In this paper, the author showed that $$\text{Tr } e^{A+B} = \text{Tr } e^A e^B$$ iff. $$AB=BA$$.

As such, $$e^{A+B} = e^A e^B$$ is equivalent to $$\text{Tr } e^{A+B} = \text{Tr } e^A e^B$$ in the context of Hermitian matrices.

My question is how we can derive the commutation relation between $$A$$ and $$B$$ directly from $$e^{A+B}=e^A e^B$$ without bringing in the Golden-Thompson inequality (as in the paper I linked). Since the condition $$e^{A+B} = e^A e^B$$ has a simpler form than that involving the trace, I think there should be some way.

Edit: rephrase the question

• I don't understand... $a \implies b$ and $b \implies c$ imply that $a \implies c$. So if the results you mention are true, so is the conclusion. Apr 10, 2021 at 11:59
• So the next question is how we can derive the commutation relation between $A$ and $B$ directly from $e^{A+B} = e^A e^B$ without bringing in the Golden-Thompson inequality (as in the paper I linked). The condition $e^{A+B} = e^A e^B$ has a simpler form than that involving the trace, so I think there should be some way. Apr 10, 2021 at 12:05
• I would suggest that you rephrase your question in that direction then. Apr 10, 2021 at 12:12
• I edited the question Apr 10, 2021 at 12:19
• Do you know that commuting hermitian matrices are simultaneously diagonalizable? Apr 10, 2021 at 13:05

The idea here is $$A+B$$ is Hermitian and the exponential map preserves Hermicity. Taking the conjugate transpose of each side, we have
$$e^Ae^B = e^{A+B} = \big(e^{A+B}\big)^*=\big(e^B\big)^*\big(e^A\big)^*=e^Be^A$$
so $$e^A$$ and $$e^B$$ commute.

Now call on a lemma twice:
for Hermitian $$X,Y$$
$$e^XY= Ye^X$$
iff $$XY=YX$$
proof sketch: the same unitary matrix $$U$$ that simultaneously diaogonalizes $$e^X$$ and $$Y$$ must diagonalize $$X$$ as well since all are Hermitian. And the same argument also runs backwards. (Underlying idea: the exponential map is injective on reals and Hermitian matrices are diagonalizable with real spectrum. So $$e^X \mathbf v = \sigma \cdot \mathbf v\implies X\mathbf v = \log(\sigma)\cdot \mathbf v$$ and of course $$X \mathbf v = \lambda \cdot \mathbf v\implies e^X\mathbf v = e^{\lambda}\cdot \mathbf v$$)

after applying the lemma once, with $$Y:=e^B$$, $$X=A$$, we know $$Ae^B = e^BA$$
and a 2nd application of the lemma, with $$Y:=A$$ and $$X:= B$$, tells us $$AB = BA$$

• Could you elaborate more on $e^X \mathbf v = \sigma \cdot \mathbf v\implies X\mathbf v = \log(\sigma)\cdot \mathbf v$? It's not clear to me. Apr 11, 2021 at 2:18
• $X$ is Hermitian so diagonalizable with d distinct eigenvalues thus $\mathbb C^n=W_1 \oplus ... \oplus W_d$, i.e. the vector space is a direct sum of Xs eigenspaces with $\mathbf w_i \in W_i$, compute the same thing two different ways $\mathbf 0 \neq \mathbf v=\sum_{i=1}^d \alpha_i \mathbf w_i$ and $\sigma \mathbf v =e^X\mathbf v=e^X\sum_{i=1}^d \alpha_i \mathbf w_i=\sum_{i=1}^d e^{\lambda_i} \alpha_i \mathbf w_i$ $\implies \mathbf 0=\sum_{i=1}^d (\sigma - e^{\lambda_i}) \alpha_i \mathbf w_i$ $\implies (\sigma - e^{\lambda_i}) \alpha_i=0$ for all i by linear independence. Apr 11, 2021 at 4:45
• But at most one $(\sigma - e^{\lambda_i}) \neq 0$ since $\lambda_i$ are distinct, real and the exponential function is injective on reals. Thus at least $d-1$ of the $\alpha_i=0$ and at most $d-1$ of the $\alpha_i=0$ because $\mathbf v\neq \mathbf 0$. $\implies\mathbf v\propto\mathbf w_k$ for some $k\in \big\{1,2,...,d\big\}$ (and of course $\lambda_k = \log(\sigma))$. Apr 11, 2021 at 4:45
• Thank you. I think it's correct, with just a small typo: at most one $(\sigma - e^{\lambda_i}) = 0$. Apr 11, 2021 at 8:23
• Yes indeed. Unfortunately a typo like this in the form of a comment can't be edited outside the original 5 minute window. I may end up dropping this explanation (sans typo) into the posting. Apr 11, 2021 at 17:10