# Representing a product of matrix exponentials as the exponential of a sum

In Proof of a conjectured exponential formula, 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.

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This follows from a result of Klyachko. Klyachko proved:

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.

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Thanks for the clean argument; W. So also proved his claim by using Klyachko's result. – Suvrit Jan 26 '12 at 8:57
Oops! I totally forgot about accepting your answer back in 2012, until this question got linked to today (3rd Oct, 2013); sorry!!!!!! – Suvrit Oct 3 '13 at 22:30

Today serendipity prevailed and I chanced upon the following paper: Products of exponentials of hermitian and complex symmetric matrices, by W. So and R. C. Thompson (Linear and Multilinear Algebra, 1991). Apparently, it was this little-known paper, in which So and Thompson introduced the problem mentioned in my question; moreover, they also solved it for the special case where either $A$ or $B$ is restricted to be rank one.

Then, I searched for papers that cite the above one, to see if progress had been made on the general question. The answer was found in:

"The high road to an exponential formula" by W. So (Linear Algebra and its Applications, 379 (2004), 69--75.

This paper answers my (and now in retrospect, actually So and Thompson's) question affirmatively.

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You should look at:

MR0866114 (88i:17007) Newman, Morris(1-UCSB); Thompson, Robert C.(1-UCSB) Numerical values of Goldberg's coefficients in the series for log(exey).

and

MR0960151 (89i:15020) Thompson, Robert C.(1-UCSB) Special cases of a matrix exponential formula. Proceedings of the Victoria Conference on Combinatorial Matrix Analysis (Victoria, BC, 1987). Linear Algebra Appl. 107 (1988), 283–292.

The second gives partial results on your question, the first is quite entertaining and has related results.

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Thanks Igor---I should have heeded your word, and taken the hint to at least skim Thompson's work more carefully. Today by random chance I came upon a paper of his (with W. So), which actually led to the answer!! – Suvrit Jan 25 '12 at 21:43

Not really an answer, but more than a comment.

It seems to me that Thompson's paper was an if-theorem, in the following sense. It relied upon the announcement by Lidskii (Spectral polyhedron of a sum of two Hermitian matrices, published in Funktsional'nyi Analiz i Ego Prilozheniya 16 (1982), pp 76-77) of the proof of A. Horn's conjecture. What did happen with that? Something like Fermat's last theorem? Eventually Lidskii never published a proof until his death in 2004. But the conjecture was eventually proved in the early 2000s, by Klyachko and Knutson & Tao. Therefore, Thomson's result is now a theorem.

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Dear Denis, thanks a lot for pointing this out this dependence (though, Thompson does note in his abstract that his proof depends on the alleged claim of Lidskii, which apparently, actually never saw the light of day). – Suvrit Dec 9 '11 at 19:47