Let $D=[d_{ij}]_{i,j=1,\ldots,4}$ be a $4 \times 4$ “density matrix”, that is, a Hermitian (possibly symmetric) positive definite matrix having trace 1—that is, the (nonnegative) diagonal entries sum to 1. Let $D_1$ and $D_2$ be the two diagonal $2 \times 2$ blocks of $D$. Further, $p= \sqrt{d_{11} d_{44}/(d_{22} d_{33})}$ and $q$ equal the ratio of the two singular values of $D_2^{1/2} D_1^{-1/2}$. Show that $\min (p,1/p) \geq \min (q,1/q)$.

  • $\begingroup$ As a brute force, one can show that in diagonal case the inequality turns into the equality, and introducing two parameters for $d_{12}$ and $d_{34}$ just analyze the behavior of singular values ratio as a function of two variables. It may be more convenient then to do a similarity transform to $D_1^{-1/2} D_2 D_1^{-1/2}$ and work with $D_1^{-1} D_2$ $\endgroup$ – VorKir Feb 27 '17 at 19:13
  • $\begingroup$ I verified the (equality) assertion of VorKir in the diagonal case. Further, "the behavior of the singular values ratio as a function of two variables" that he requests is $\epsilon=\exp \left(-\cosh ^{-1}\left(\frac{2 d_{12} d_{34} \mu -\mu ^2-1}{2 \sqrt{d_{12}^2-1} \sqrt{d_{34}^2-1} \mu }\right)\right)$. I can give some interesting background (shortly to be an arXiv posting) on this result, if requested. $\endgroup$ – Paul B. Slater Feb 28 '17 at 23:57
  • $\begingroup$ What is $\mu$ in your formula? In general, it would be interesting to know the background, just out of curiosity. I think there should be a nice and elegant argument working to proof the inequality, but cannot suggest it:) $\endgroup$ – VorKir Mar 1 '17 at 0:09
  • $\begingroup$ My apologies--I mixed some notation. Actually, $\mu \equiv p$. See Lemma 5 in arxiv.org/pdf/1610.01410.pdf for what I used to get the formula. In that paper, $\epsilon$ is the ratio of singular values, denoted $q$ in my original question. (I didn't fully understand your earlier point about a similarity transform.) $\endgroup$ – Paul B. Slater Mar 1 '17 at 2:11
  • $\begingroup$ OK, let me try to get this right using the notation of the problem I put (rather than the one I had been working with). Now, $q=\exp \left(-\cosh ^{-1}\left(\frac{\frac{2 d_{12} d_{34} p}{\sqrt{d_{11} d_{22}} \sqrt{d_{33} d_{44}}}-p^2-1}{2 \sqrt{\frac{d_{12}^2}{d_{11} d_{22}}-1} \sqrt{\frac{d_{34}^2}{d_{33} d_{44}}-1} p}\right)\right)$. $\endgroup$ – Paul B. Slater Mar 1 '17 at 18:05

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