Timeline for A curious eigenvalue inequality
Current License: CC BY-SA 3.0
33 events
| when toggle format | what | by | license | comment | |
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| Feb 21, 2017 at 13:42 | comment | added | rucarden | If you consider the iteration proposed involving $T(U)$, and let $U_0=I$, then the eigenvalues of $AB$ are log majorized by the eigenvalues of $AU_1B$. This is because the $AB$ and $AU_1B$ are similar. Unfortunately, I don't see how this can be used to prove the desired inequality. | |
| Jun 2, 2016 at 7:38 | comment | added | M. Lin | @rucarden: Frankly speaking, I did not think of this before. | |
| May 24, 2016 at 12:15 | comment | added | rucarden | Was the motivation for the inequality the operator $T(U)=AU^*B$. This inequality must have something to do with doing the power method with this operator and normalizing using a polar factorization. $$T(U_i)=U_{i+1}^*P_{i+1}$$ | |
| May 15, 2016 at 8:51 | comment | added | M. Lin | @rucarden This is also my conjecture,^_^ | |
| May 13, 2016 at 17:06 | comment | added | rucarden | It seems a stronger result is true. That is the eigenvalues of $U^*AU+B$ are majorized by the eigenvalues of $A+B$. | |
| S May 11, 2016 at 3:48 | history | bounty ended | CommunityBot | ||
| S May 11, 2016 at 3:48 | history | notice removed | CommunityBot | ||
| May 5, 2016 at 23:29 | comment | added | M. Lin | @jjcale: yes, you are right. Part of the motivation for the question comes from this observation. | |
| May 5, 2016 at 18:17 | comment | added | jjcale | If the condition would be that $A^{1/2} U B^{1/2}$ is Hermitian then it would not be difficult to proof the inequality. | |
| May 3, 2016 at 15:08 | answer | added | rucarden | timeline score: 2 | |
| S May 3, 2016 at 2:12 | history | bounty started | M. Lin | ||
| S May 3, 2016 at 2:12 | history | notice added | M. Lin | Draw attention | |
| Apr 26, 2016 at 14:09 | comment | added | rucarden | I am generating $U$ in the manner suggested by the discussion above. We want $$AUB=BU^*A$$ so assume that $AU=BH$ where $H$ is a Hermitian matrix. Then $A^2=BH^2B$, and I get $H$ from taking the positive square root of $B^{-1}A^2B^{-1}$, and then $U=A^{-1}BH$. As the discussion above suggests, different roots could be used for generating $H$. And having thought about this a little more, I would say the resulting iterations seems more like a modified Duggal iteration. | |
| Apr 25, 2016 at 0:25 | comment | added | M. Lin | @jjcale I knew a proof of $tr((U^*AU+B)^2) \le tr((A+B)^2)$, my proof relies on THEOREM 4.3 in the paper 'Inequalities related to $2\times 2$ block PPT matrices, Oper. & Matrices, 9 (2015) 917-924.' I will email you the proof details if you want. | |
| Apr 24, 2016 at 15:24 | comment | added | jjcale | With the notation I used in my previous comment the conjectured inequality $tr((U^*AU+B)^2) \le tr((A+B)^2)$ is then equivalent to $tr((R B^2 R)^{1/2} B) \le tr((B R^2 B)^{1/2} B)$ . | |
| Apr 24, 2016 at 13:17 | comment | added | jjcale | If we set $R=B^{-1} A U$ then $R$ is Hermitian and we have to show : $\rho((R B^2 R)^{1/2} + B)\le \rho((B R^2 B)^{1/2} + B)$ . | |
| Apr 23, 2016 at 15:58 | comment | added | jjcale | Could there exist a completely positive map that maps $A+B$ to $U^* A U + B$ and the identity to itself ? If yes this would be the solution. | |
| Apr 23, 2016 at 2:24 | comment | added | M. Lin | @rucarden Not sure if I understand you correctly, but your way constructing $U$ does not guarantee $AUB$ to be Hermitian? | |
| Apr 22, 2016 at 14:34 | comment | added | rucarden | I don't have an answer nor do I have the reputation to comment on the previous comments, but it occurred to me to look at what happens when you iterate this operation. That is given A and B construct $U$, let $A=U^*AU$ and repeat. It seems that going in the forward direction, eventually the matrices $A$ and $B$ commute with the eigenvalues of $A$ matched with the eigenvalues of $B$ so that largest is paired with largest. If you run the iteration in reverse, again they eventually commute but at the end the eigenvalues are paired sorted so that the largest of $A$ is paired with smallest of $B$ a | |
| Apr 12, 2016 at 17:14 | comment | added | Iosif Pinelis | @FedorPetrov : That's right; there can be an extra factor on the right, which is a diagonal matrix with diagonal entries of modulus $1$. | |
| Apr 12, 2016 at 15:03 | comment | added | Fedor Petrov | @IosifPinelis it may be non-positive square root! | |
| Apr 11, 2016 at 18:47 | comment | added | Iosif Pinelis | Some easy suggestions: (i) note that $U=BC(CB^2C)^{-1/2}$, where $C$ is as in the comments by Fedor Petrov and M. Lin. (ii) Let $L:=C+B$ and $R:=UCU^*+B$. Then it suffices to show that $\text{tr}(L^k)\le\text{tr}(R^k)$ for all (large enough) natural $k$. This inequality trivially holds for $k=1$, and numerical experiments suggests it holds for all natural $k$. An advantage of the latter inequality is that it is polynomial in the elements of the matrices $L$ and $R$. | |
| Apr 11, 2016 at 13:01 | comment | added | Iosif Pinelis | Thank you Minghua for your email. My simulations, too, so far support your conjecture. | |
| Apr 11, 2016 at 11:34 | comment | added | M. Lin | @IosifPinelis I sent via email the background. | |
| Apr 11, 2016 at 4:09 | comment | added | Iosif Pinelis | Can you say what your reasons are to believe that the inequality is true? | |
| Apr 11, 2016 at 2:10 | history | edited | M. Lin |
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| S Apr 10, 2016 at 13:29 | history | suggested | Amir Sagiv |
I don't think this has anything to do with functional analysis. The question is solely about matrices.
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| Apr 10, 2016 at 13:10 | review | Suggested edits | |||
| S Apr 10, 2016 at 13:29 | |||||
| Apr 9, 2016 at 23:00 | comment | added | M. Lin | @FedorPetrov: thanks, at least I could see an equivalent statement of the question: If $B, C$ are positive definite and $U$ is unitary such that $UCB$ is Hermitian, then $\rho(C+B)\le \rho(UCU^*+B)$. | |
| Apr 9, 2016 at 22:14 | comment | added | Fedor Petrov | @ChristianRemling in generic situation yes, but if $CB^2C$ has multiple eigenvalue, there are infinitely many Hermitian matrices $R$ satisfying $R^2=BC^2B$, for each of them $U^*:=CBR^{-1}$ is unitary. | |
| Apr 9, 2016 at 18:19 | comment | added | Christian Remling | Another observation (though somewhat beside the point) is that $U(n)$ has (real) dimension $n^2$, and you're imposing exactly that many conditions, so typically there will probably only be finitely many $U$'s satisfying your assumption for given $A,B$. | |
| Apr 9, 2016 at 9:32 | comment | added | Fedor Petrov | Denoting $C=U^*AU$ we get $UCB=AUB=R$ is Hermitian, so $CB=U^*R$. That is, $U$ comes from the polar decomposition of $CB$, if $R$ is additionally assumed to be positive definite. | |
| Apr 9, 2016 at 4:20 | history | asked | M. Lin | CC BY-SA 3.0 |