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.


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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