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Let $B, R\in M_{n}(\mathbb{C})$ hermitian and $B$ positive semidefinite. Let $s,t \in \mathbb{R}$ and $s,t \ge 0$ .

Does then hold $Tr[B^s (B R^2 B)^t] \ge Tr[B^s (R B^2 R)^t]$ ?

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  • $\begingroup$ Can you clarify what it means to raise a matrix to a real power? $\endgroup$
    – Rob Davis
    Aug 11 '17 at 18:59
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    $\begingroup$ $B$, $B R^2 B$ and $R B^2 R$ are all positive semidefinite. Therefore the positive real power is well defined for these matices . $\endgroup$
    – jjcale
    Aug 11 '17 at 19:09
  • $\begingroup$ @FrancoisZiegler Are you sure? I think I can prove it in the t=1 case. $\endgroup$
    – MTyson
    Aug 11 '17 at 21:09
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I think the inequality is false. Consider for instance the choices

\begin{equation*} B = \begin{bmatrix} 5& 6& -2\\ 6 & 13 & 2\\ -2 & 2 & 5\end{bmatrix},\quad R = \begin{bmatrix} -8 & 4 & 4\\ 4 & -2 & -1\\ 4 & -1 & 0 \end{bmatrix},\quad s=5,\ t=3. \end{equation*} Then, we have (computed using Mathematica)

lhs = Tr[MatrixPower[b, s].MatrixPower[b.r.r.b, t]]

which yields 415274500333934, whereas

rhs = Tr[MatrixPower[b, s].MatrixPower[r.b.b.r, t]]

yields 450223588494254, so that lhs-rhs = -34949088160320.

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  • $\begingroup$ Do you have also a counterexample for $t \le 2$ ? $\endgroup$
    – jjcale
    Aug 12 '17 at 22:41
  • $\begingroup$ Could you verify that $B=\{\{12,-4, 0\},\{-4, 2, 2\},\{0, 2, 8\}\}$ and $R=\{\{-2, -3 ,-2\},\{-3 ,-7,-3\},\{-2, -3, 3\}\}$ for $s=9$ and $t=7/4$ yields a counterexample. $\endgroup$
    – Suvrit
    Aug 12 '17 at 22:59
  • $\begingroup$ Yes, this is also a counterexample. Next interesting case is $t = 1/2$ . $\endgroup$
    – jjcale
    Aug 13 '17 at 11:41
  • $\begingroup$ seems like $0\le t \le 1/2$ could hold, though not yet fully sure. Perhaps the following paper: arxiv.org/abs/1507.00853 sparks some ideas.... $\endgroup$
    – Suvrit
    Aug 13 '17 at 16:08

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