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Let $A$ and $B$ be two Hermitian matrices. The famous Golden-Thompson inequality states that

$$\text{tr}(e^{A+B}) \le \text{tr}(e^Ae^B)$$

However, for determinants we have equality

$$\det(e^{A+B}) =\det(e^Ae^B)$$

I was wondering if similar results can be shown, if instead of trace and determinant, we use any of the other fundamental scalar functions of a matrix (e.g., trace is $\phi_1(X) :=\sum_i \lambda_i(X)$; $\phi_2(X)=\sum_{i \neq j} \lambda_i(X)\lambda_j(X)$, determinant is $\phi_n$)

PS: Please feel free to add more tags, if you deem it to be necessary.

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1 Answer 1

up vote 8 down vote accepted

This is theorem IX.3.5 in "Matrix analysis" by R. Bhatia (Graduate Texts in Mathematics, 169). See also corollary IX.3.6 and theorem IX.3.7. The Golden-Thompson inequality holds when $Tr$ is replaced with a function $f$ which satisfies $f(XY)=f(YX)$ and $|f(X^{2m})|\le f(|XX^{\ast}|^m)$ for all $m\geq 1$. Such functions can be the elementary symmetric functions in the eigenvalues as in your question, the product of the $k$ largest eigenvalues (in absolute value) etc.

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