# A curious determinantal inequality I

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 $\det(A+B)^2$.

• What evidence or examples do you have that this is true; for example, with $2 \times 2$ matrices? Nov 18, 2017 at 23:30
• My evidence is weak, I could prove $\det(|AB+BA|)\leq \det(|AB|+|BA|)$... But I had run numerical experiments for the proposed inequality... Nov 18, 2017 at 23:40
• @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 $\det(A+B)^2$? Nov 19, 2017 at 20:45
• @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. Nov 19, 2017 at 20:50
• @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 $n=2$. Nov 20, 2017 at 3:25

• 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 $-58.97\dots$. Nov 20, 2017 at 20:03
• 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. Nov 20, 2017 at 20:10
• @ChrisRamsey As Suvrit remarked, what if $A, B$ are assumed to be positive definite? Nov 20, 2017 at 23:58