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Let $A$ and $B$ be positive-definite matrices of the same size. For any $t \ge 0$, define $$ u(t) := \mbox{trace}(A^{-1/2}e^{-tB} (AB+BA) e^{-tB}A^{-1/2}). $$

Question. What are necessary and sufficient conditions on $A$ and $B$, such that $u(t) \ge 0$ for all $t \ge 0$ ?

Note. A sufficient condition is that $A$ and $B$ commute, or more generally, that $AB+BA$ is positive-semidefinite.

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    $\begingroup$ Could you clarify what you mean by positive definite? If $A$ and $B$ are Hermitian and positive definite in the sense $v^\dagger A v > 0$, then one can construct positive definite $\sqrt{A}$. $AB = \sqrt{A} \sqrt{A} B$ has the same eigenvalues as $\sqrt{A} B \sqrt{A}$ which is manifestly positive definite. Sums of positive definite matrices are also positive definite. Thus $AB + BA$ is always positive definite. $\endgroup$
    – user196574
    Commented Sep 14, 2022 at 4:38
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    $\begingroup$ @user196574 : counterexample to your claim : $A= \biggl(\begin{matrix} 1 & 0 \\ 0 & 2 \end{matrix} \biggr)$ , $B= \biggl(\begin{matrix} 1.01 & 1 \\ 1 & 1.01 \end{matrix} \biggr)$ $\endgroup$
    – jjcale
    Commented Sep 14, 2022 at 8:27
  • $\begingroup$ @jjcale Thanks, I see where I went wrong. It's true $AB$ and $BA$ have the same eigenvalues as $\sqrt{A} B \sqrt{A}$, which are indeed positive. However, these matrices don't obey the $v^\dagger A v > 0$ condition, which would be sufficient for their sum $AB+BA$ to still be positive. That is, as in jjcale's example, $AB+BA$ need not have only positive eigenvalues. I hadn't appreciated this before that the Hermitian part of a matrix with positive eigenvalues need not have positive eigenvalues, but I see it's true for actually many matrices! $\endgroup$
    – user196574
    Commented Sep 15, 2022 at 2:33

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