most recent 30 from http://mathoverflow.net 2013-05-22T01:46:07Z http://mathoverflow.net/feeds/question/74388 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/74388/extensions-to-the-golden-thompson-inequality S. Sra 2011-09-02T18:49:13Z 2011-09-02T19:14:01Z

<p>Let $A$ and $B$ be two Hermitian matrices. The famous <a href="http://en.wikipedia.org/wiki/Golden-Thompson_inequality" rel="nofollow">Golden-Thompson inequality</a> states that </p> <p>$$\text{tr}(e^{A+B}) \le \text{tr}(e^Ae^B)$$</p> <p>However, for determinants we have equality</p> <p>$$\det(e^{A+B}) =\det(e^Ae^B)$$</p> <p>I was wondering if similar results can be shown, if instead of trace and determinant, we use any of the other <em>fundamental scalar functions</em> 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 $\phi_n$)</p> <p>PS: Please feel free to add more tags, if you deem it to be necessary.</p>

http://mathoverflow.net/questions/74388/extensions-to-the-golden-thompson-inequality/74390#74390 Gjergji Zaimi 2011-09-02T19:14:01Z 2011-09-02T19:14:01Z

<p>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.</p>