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Let $A, B$ be Hermitian matrices. Does the following hold?

$$\det(A^{2}+B^{2}+|AB+BA|)\leq \det(A^{2}+B^{2}+|AB|+|BA|)$$

As usual, $|X|=(X^*X)^{1/2}$. Clearly, quantities on both sides are no less than $\det(A+B)^2$.

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  • $\begingroup$ What evidence or examples do you have that this is true; for example, with $2 \times 2$ matrices? $\endgroup$ Nov 18, 2017 at 23:30
  • $\begingroup$ My evidence is weak, I could prove $\det(|AB+BA|)\leq \det(|AB|+|BA|)$... But I had run numerical experiments for the proposed inequality... $\endgroup$
    – M. Lin
    Nov 18, 2017 at 23:40
  • $\begingroup$ @M.Lin : Can we see the proof of $\det(|AB+BA|)\le\det(|AB|+|BA|)$? Interestingly, inequality $|AB+BA|\le |AB|+|BA|$ does not hold in general, whereas inequality $A^{2}+B^{2}+|AB+BA|\ge0$ seems to hold. Also, how do you show that both sides of your proposed inequality are no less than $\det(A+B)^2$? $\endgroup$ Nov 19, 2017 at 20:45
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    $\begingroup$ @IosifPinelis As $\left(\begin{array}{cc} |X| & X^{*} \\ X & |X^{*}|\\ \end{array} \right) $ is positive semidefinite (psd) for any $X$, it follows that $\left( \begin{array}{cc} |AB| & BA \\ AB & |BA| \\ \end{array} \right)$ and $\left( \begin{array}{cc} |BA| & AB \\ BA & |AB| \\ \end{array} \right)$ are psd. Adding them gives the positivity of $\left( \begin{array}{cc} |AB|+|BA| & AB+BA \\ AB+BA & |AB|+|BA| \\ \end{array} \right)$... $|\det (AB+BA)|=\det(|AB+BA|)\le\det(|AB|+|BA|)$ follows. $\endgroup$
    – M. Lin
    Nov 19, 2017 at 20:50
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    $\begingroup$ @M.Lin : Nice! It appears that the following stronger inequality holds for $n=2$ -- but not for $n=3$ (!): $\begin{bmatrix}|AB|+|BA|& |AB+BA|\\ |AB+BA|&|AB|+|BA|\end{bmatrix}\ge0$. Of course, for any Hermitian $M\ge0$, this inequality (when it holds) implies $\begin{bmatrix}M+|AB|+|BA|& M+|AB+BA|\\ M+|AB+BA|&M+|AB|+|BA|\end{bmatrix}\ge0$ and hence the inequality in question -- but, again, this would work only for $n=2$. $\endgroup$ Nov 20, 2017 at 3:25

1 Answer 1

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Consider the following matlab output:

enter image description here

The square root function produces a very small imaginary error in the second case so I've just stripped that away in the final calculation. Therefore, the proposed determinantal inequality does not hold.

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    $\begingroup$ Exact calculations with Mathematica confirm that the proposed inequality does not hold for your $A$ and $B$. The difference between the right-hand side of that inequality and its left hand side is $-58.97\dots$. $\endgroup$ Nov 20, 2017 at 20:03
  • $\begingroup$ It appears now that the inequality will likely hold only for $2\times2$ matrices. So, the proof for that case should be quite dimension-specific. $\endgroup$ Nov 20, 2017 at 20:10
  • $\begingroup$ It took me a fair amount of tries before I got this pair which is why it becomes too hard to calculate by hand. That being said, I'm sure someone could find an easier pair. $\endgroup$ Nov 20, 2017 at 20:40
  • $\begingroup$ @ChrisRamsey As Suvrit remarked, what if $A, B$ are assumed to be positive definite? $\endgroup$
    – M. Lin
    Nov 20, 2017 at 23:58
  • $\begingroup$ @ChrisRamsey thank you for your solution. $\endgroup$
    – M. Lin
    Nov 21, 2017 at 0:02

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