Timeline for A curious determinantal inequality
Current License: CC BY-SA 3.0
16 events
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| Jun 27, 2015 at 22:38 | comment | added | M. Lin | @ChristianRemling: The partial order $A^{\frac{1}{2}}(A+B)A^{\frac{1}{2}}+B^{\frac{1}{2}}(A+B)B^{\frac{1}{2}}\ge (A+B)^2$ is definitely false (as their traces are equal), I meant there would be a majorization relation for eigenvalues between these matrices en.wikipedia.org/wiki/Majorization | |
| Jun 27, 2015 at 21:37 | comment | added | Christian Remling | @M.Lin: The stronger conjecture (without $\det$) is false, there are easy $n=2$ counterexamples (unless I miscalculated). | |
| Jun 27, 2015 at 20:52 | comment | added | M. Lin | I am surprised (and feel quite honored) that the question has attracted Terry's attention. Actually, I had tried his argument. As Terry mentioned majorization. Indeed, my original guess is that the eigenvalues of $A^{\frac{1}{2}}(A+B)A^{\frac{1}{2}}+B^{\frac{1}{2}}(A+B)B^{\frac{1}{2}}$ are majorized by those of $(A+B)^2$ (numerical experiments support this guess, I've also verified special cases). The determinantal inequality asked is a necessary condition for this conjectured majorization relation. I believe determinantal inequalities would attract more readers in MO... so the present form. | |
| Jun 27, 2015 at 18:51 | comment | added | GH from MO | Thank you! In fact your treatment suggested me to look at the same, namely to replace $D,E,F$ by $A^{1/2},B^{1/2},A+B$, but then I was lazy to work out the details. More precisely, I was not familiar with the Schur complement before your original post, so I decided to learn about it instead. | |
| Jun 27, 2015 at 17:14 | comment | added | Terry Tao | Apply Fischer's inequality to $\begin{pmatrix} A^{1/2} & B^{1/2} \\ -B^{1/2} & A^{1/2} \end{pmatrix} \begin{pmatrix} A + B & 0 \\ 0 & A+B \end{pmatrix} \begin{pmatrix} A^{1/2} & -B^{1/2} \\ B^{1/2} & A^{1/2} \end{pmatrix}$. By Schur complement, the first and last matrix have determinant $\det( A + A^{1/2} B^{1/2} A^{-1/2} B^{1/2} )$, giving $\det( A^{1/2} (A+B) A^{1/2} + B^{1/2} (A+B) B^{1/2} ) \geq \det(A+B) \det( A + A^{1/2} B^{1/2} A^{-1/2} B^{1/2} )$ (I had some typos in the previous inequality as I had changed notation in my computations by squaring $A,B$). | |
| Jun 27, 2015 at 17:10 | comment | added | GH from MO | @TerryTao: I am interested in the weaker inequality, too! If you have the time and mood, please share it with us. | |
| Jun 27, 2015 at 17:04 | comment | added | Terry Tao | You are right, of course; I had mistakenly identified a vector space with its dual when thinking about the problem, which translated into the sign error here when converted back into matrices. I can establish the weaker inequality $\det(A^{1/2} (A^2+B^2) A^{1/2} + B^{1/2} (A^2+B^2) B^{1/2}) \geq \det(A^2+B^2) \det(A^2 + A B A^{-1} B)$ with this approach, but it does not appear strong enough to recover the full inequality. | |
| Jun 27, 2015 at 17:01 | history | edited | Terry Tao | CC BY-SA 3.0 |
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| Jun 27, 2015 at 16:50 | comment | added | Fedor Petrov | @Semiclassical: it would not be true either. | |
| Jun 27, 2015 at 15:21 | comment | added | Semiclassical | I'm similarly perplexed by the statement of $D^2+E^2=1$; it would be manifestly true if one had $D=C^{-1/4}AC^{1/4}$ instead of $C^{-1/4}A C^{-1/4}$ and similarly for $E$, but as stated it seems invalid. | |
| Jun 27, 2015 at 13:47 | comment | added | Denis Serre | I tried with the choice $A^{1/2}={\rm diag}(2,1)$ and $B^{1/2}=\begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}$, which fails to pass the test $D^2+E^2=1$. | |
| Jun 27, 2015 at 13:29 | comment | added | Denis Serre | I doubt too that $D^2+E^2$ equals $1$. 14 votes pro without a verification ? | |
| Jun 27, 2015 at 10:52 | comment | added | GH from MO | Dear Terry, I don't see that $D^2+E^2=1$, can you please give more detail? | |
| Jun 27, 2015 at 9:54 | comment | added | GH from MO | In fact the last inequality is a special case of Fischer's inequality which preceded the work of Schur (see archive.org/stream/archivdermathem37unkngoog#page/n51/mode/2up). | |
| Jun 27, 2015 at 4:42 | history | edited | Terry Tao | CC BY-SA 3.0 |
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| Jun 27, 2015 at 4:36 | history | answered | Terry Tao | CC BY-SA 3.0 |