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? $\endgroup$– David HandelmanNov 18, 2017 at 23:30
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$\begingroup$ 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. LinNov 18, 2017 at 23:40
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$\begingroup$ @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 $\det(A+B)^2$? $\endgroup$– Iosif PinelisNov 19, 2017 at 20:45
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4$\begingroup$ @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. LinNov 19, 2017 at 20:50
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1$\begingroup$ @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 $n=2$. $\endgroup$– Iosif PinelisNov 20, 2017 at 3:25
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1 Answer
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Consider the following matlab output:
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.
answered Nov 20, 2017 at 18:05
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3$\begingroup$ 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
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$\begingroup$ 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
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$\begingroup$ 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
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$\begingroup$ @ChrisRamsey As Suvrit remarked, what if $A, B$ are assumed to be positive definite? $\endgroup$– M. LinNov 20, 2017 at 23:58
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