Timeline for Maximum of $\sum_{n=1}^N z^T X(P_n X + I)^{-1}z$ over unit trace, positive semidefinite matrices?
Current License: CC BY-SA 4.0
24 events
| when toggle format | what | by | license | comment | |
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| S May 29 at 17:04 | history | bounty ended | CommunityBot | ||
| S May 29 at 17:04 | history | notice removed | CommunityBot | ||
| May 28 at 21:56 | answer | added | ResearchMath | timeline score: 1 | |
| May 28 at 19:52 | comment | added | Drew Brady | If you have the time, yes it would be great to see it. | |
| May 28 at 19:31 | comment | added | ResearchMath | It will take me about one hour to type the argument for you. The derivation is quite elegant. Let me know if you want it. | |
| May 28 at 19:15 | comment | added | ResearchMath | These are purely algebraic equations in terms of the unknown $L$. | |
| May 28 at 19:13 | comment | added | ResearchMath | No it doesn't look like that at all. | |
| May 28 at 19:05 | comment | added | Drew Brady | Does this condition look something like equating a trace inner product to a maximal eigenvalue? | |
| May 28 at 8:32 | comment | added | ResearchMath | I can give you necessary and sufficient conditions for a local maximum (without using Lagrange multipliers). They seem to simplify even further if I rewrite the functional as you suggested. Although they are not that complicated, I don't know how to solve them. Let me know if this is worth anything to you, or if you knew this already. | |
| May 27 at 21:11 | comment | added | ResearchMath | Ok thanks for explaining, I will have one final look at it tomorrow, using this rewrite you just gave me. | |
| May 27 at 20:56 | comment | added | Drew Brady | Well, if $X \succ 0$, we certainly have $X(PX + I)^{-1} = \sqrt{X} (\sqrt{X} P \sqrt{X} + I)^{-1} \sqrt{X}$. However, now replace $X = X + \tau I$ and take the limit $\tau \to 0^+$ to see it holds if $X \succeq 0$ is singular. | |
| May 27 at 20:45 | comment | added | ResearchMath | How do you mean, by continuity? Please explain. | |
| May 27 at 20:37 | comment | added | Drew Brady | One can also write by continuity $X(PX + I)^{-1} = \sqrt{X}(\sqrt{X} P \sqrt{X} + I)^{-1} \sqrt{X}$. So we can instead look at optimizing $z^TL(L^T P L + I)^{-1} L^T z$ identifying $L$ as the Cholesky factor; $X = LL^T$. | |
| May 27 at 20:33 | comment | added | Drew Brady | Indeed, I mean $\leq$ (sorry, I had originally written the operator inequality, because as I mentioned, it holds for all $z$). I came up with $X^\star_\mathcal{P}$ in the case $|\mathcal{P}| = 1$ with the intuition that $(P + X^{-1})^{-1}$ should "align" with $z$. I looked at rank-one products $vv^T$, and maximizing over these leads to the result. However, I can see from simulation already that in the case $\mathcal{P}$ the optimum could have rank larger than 1. | |
| May 27 at 20:14 | comment | added | ResearchMath | How did you come up with $X^\star_{\mathcal{P}}$? | |
| May 27 at 20:10 | comment | added | ResearchMath | Did you mean $\le$? | |
| May 27 at 19:57 | comment | added | Drew Brady | Well, as I mentioned above, the only case I could make progress on was the case of $|\mathcal{P}| = 1$. In that case, it is clear from the upper bound argument it has no hope of generalizing to cardinality larger than 1. (Simply because the inequality $z^T X (PX + I)^{-1} z \preceq z^T (P + I)^{-1}z $ is not very strong for $X \in \mathcal{X}$, in the sense that it holds for any direction $z$.) | |
| May 27 at 10:09 | comment | added | ResearchMath | This is an interesting problem, there are many ways to look at it. It might help if you say more on what you tried so far and why it failed. You may know things others don't know and vice versa. | |
| S May 21 at 15:37 | history | bounty started | Drew Brady | ||
| S May 21 at 15:37 | history | notice added | Drew Brady | Draw attention | |
| May 21 at 15:36 | history | edited | Drew Brady | CC BY-SA 4.0 |
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| May 18 at 15:14 | history | edited | Drew Brady | CC BY-SA 4.0 |
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| May 17 at 21:43 | comment | added | Drew Brady | I should add, even a solution in the case $|\mathcal{P}| = 2$ would be helpful. | |
| May 17 at 21:22 | history | asked | Drew Brady | CC BY-SA 4.0 |