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For three symmetric positive semidefinite matrices $A, B,C$, I am trying to figure out if the following inequality holds, at least in some cases:

$$ \operatorname{tr} \left( A e^{B+C} \right) \leq \operatorname{tr} \left( A e^B e^C \right) $$

Note that if $A=I$ then this is the Golden-Thompson inequality:

$$\operatorname{tr} \left( e^{B+C} \right) \leq \operatorname{tr} \left( e^B e^C \right)$$

I wonder if the first inequality is true or if adding certain assumptions (such as $A≼I$ i.e. $I−A$ is positive semidefinite) will suffice to prove it. I have tried searching for similar inequalities with no luck.

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  • $\begingroup$ How is this related to non-negative matrices? $\endgroup$ May 13, 2023 at 6:37

2 Answers 2

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The difficulties with generalizations of the Golden-Thompson inequality to three matrices arise because the trace of a product of three positive symmetric matrices is in general not positive; unlike the trace of the product of two positive symmetric matrices, which is positive: ${\rm tr}\,e^B e^C={\rm tr}\,XX^t=\sum_{n,m}X_{nm}^2>0$, with $X=e^{B/2}e^{C/2}$ and $X^t$ the transpose of $X$.

So while ${\rm tr}\,Ae^{B+C}$ is a positive number, ${\rm tr}\,Ae^Be^C$ can be negative.

There do exist three-matrix generalizations of the Golden-Thompson inequality, but they take an entirely different form, see A survey of certain trace inequalities, equation 21.

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The Golden-Thompson inequality holds not only for the trace but in fact also for all unitarily invariant norms, for example the Schatten norms. That is, it holds that

$$\| e^{B+C} \|_p \leq \| e^{B} e^C \|_p$$

for any $p \geq 1$. (See, for example, Theorem 9.3.7 in ``R. Bhatia, Matrix analysis. Graduate Texts in Mathematics, 169. Springer Verlag, New York, 1997'') If we apply this to $p=\infty$ we get something similar to what you want, albeit including a maximization:

$$ \mathrm{tr}( A e^{B+C} ) \leq \max_{A' : \| A' \|_1 = 1} \mathrm{tr} \big( A' e^{B} e^{C} \big) .$$

This statement holds for any $0 \leq A \leq 1$ with $\mathrm{tr}(A) = 1$, or in fact any $A$ with $\| A\|_1 \leq 1$.

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