By similar arguments as for the proof of the golden-thompson inequality (see "Log majorization and complementary Golden-Thompson type inequalities" by T.Ando and F.Hiai) we can show that for all A,B symmetric positive definite we have $$\|\log(A)+\log(B)\|_{tr}\leq \|\log(A^{1/2}BA^{1/2})\|_{tr},$$ where $\log$ is the matrix logarithm and $\|\cdot\|_{tr}$ is the trace norm, i.e. $\|A\|_{tr}=\sqrt{tr(AA^*)}$. My conjecture is that for each $n\in \mathbb{N}$ there exists a constant $c_n$ such that $$\|\log(A)+\log(B)\|_{tr}\geq c_n\|\log(A^{1/2}BA^{1/2})\|_{tr}$$ for all symmetric positive definite matrices $A,B$ of dimension $n$. However I have no idea how to prove it. The question is interesting because it would show that the two metrics $d_1$, $d_2$ defined by $$d_1(A,B)=\|\log(A)-\log(B)\| \text{ and } d_2(A,B)=\|\log(A^{-1/2}BA^{-1/2})\|$$ on the space of positive definite Matrices are strongly equivalent.
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
Check out:
Reverse inequality to Golden–Thompson type inequalities: Comparison of $e^{A+B}$ and $e^Ae^B$ Jean-Christophe Bourin, Yuki Seo (2007), Linear Algebra and its Applications Volume 426, Issues 2–3, 15 October 2007, Pages 312-316.
answered Nov 19, 2013 at 14:20
-
$\begingroup$ Thanks for this reference. But I dont understand why the inequalities in this reference should be sharp. $\endgroup$ Commented Nov 24, 2013 at 9:10
-
$\begingroup$ Now I found a simple counterexample to my conjecture for the case $n=2$: $$A_m=\exp\left(\begin{matrix}m & 0\\0 & 0\end{matrix}\right) \text{ and } B_m=\exp\left(\begin{matrix}m & 1\\1 & 0\end{matrix}\right)$$ $\endgroup$ Commented Dec 12, 2013 at 13:10