In [*Proof of a conjectured exponential formula*](http://www.tandfonline.com/doi/abs/10.1080/03081088608817715), R. C. Thompson (1986) [edit: apparently, assuming Horn's conjecture] proved that if $A$ and $B$ are Hermitian matrices, then there exist unitary matrices $U$ and $V$, such that


$$ e^{iA}e^{iB} = e^{i (UAU^\*+VBV^\*)}.$$


I was wondering whether a similar result holds without the $i$? More precisely, my question is:



> 
> Let $A$ and $B$ be Hermitian matrices. Do there exist unitary matrices $U$ and $V$ such that
>  $$e^{A/2}e^{B}e^{A/2} = e^{UAU^\* + VBV^\*}$$
> 
> 
> 




---


**EDIT** 


Denis Serre has pointed out an inconsistency in my statement above, in the sense that Thompson's result was based on an the hypothesis that Horn's conjecture is true. Now that this conjecture is indeed known to be true, Thompson's claim may officially be regarded as a theorem. 



Representing a product of matrix exponentials as the exponential of a sum
This follows from a result of Klyachko. Klyachko [proved](http://www.ams.org/mathscinet-getitem?mr=1799623):



> 
> Let $\alpha$, $\beta$ and $\gamma$ be three vectors in $\mathbb{R}^n$. Then the following are equivalent:
> 
> 
> (1) There exist Hermitian matrices $\mathfrak{a}$, $\mathfrak{b}$ and $\mathfrak{c}$ with $\mathfrak{a}+\mathfrak{b}= \mathfrak{c}$, and with eigenvalues $\alpha$, $\beta$ and $\gamma$ respectively.
> 
> 
> (2) There exist invertible matrices $A$, $B$ and $C$ with $AB=C$ and singular values $e^{\alpha}$, $e^{\beta}$ and $e^{\gamma}$ respectively. Here $e^{\alpha}$ etcetera mean termwise exponentiation.
> 
> 
> 


I will show that $(2) \implies (1)$ implies your statement (and is basically equivalent to it). I'll call your Hermitian matrices $X$ and $Y$, to leave the letters $(A,B,C)$ clear.


Let $A=e^{X/2}$, let $B=e^{Y/2}$ and let $AB=C$. Let $e^{\alpha}$, $e^{\beta}$ and $e^{\gamma}$ be the singular values of $A$, $B$ and $C$. So the eigenvalues of $C C^{\ast}$ are $e^{2 \gamma}$, and we note that $C C^{\ast} = e^{X/2} e^Y e^{X/2}$.


Using $(2) \implies (1)$, let's find Hermitian $\mathfrak{a}$, $\mathfrak{b}$ and $\mathfrak{c}$ with eigenvalues $2 \alpha$, $2 \beta$ and $2 \gamma$ and $\mathfrak{a}+\mathfrak{b} = \mathfrak{c}$. Then $C C^{\ast}$ and $e^{\mathfrak{c}}$ are Hermitian with the same eigenvalues and, conjugating by a unitary matrix, we can arrange that $C C^{\ast} = e^{\mathfrak{c}}$.


Now, $X$ and $\mathfrak{a}$ are both Hermitian with eigenvalues $2 \gamma$, so we can find unitary $U$ with $\mathfrak{a} = U X U^{\ast}$. Similarly, we can find unitary $V$ with $\mathfrak{b} = V Y V^{\ast}$. So $\mathfrak{c} = U X U^{\ast} + V Y V^{\ast}$ and $e^{X/2} e^Y e^{X/2} = e^{U X U^{\ast} + V Y V^{\ast}}$ as desired.


