Timeline for A (linear) optimization problem subject to (linear) matrix inequality constraints
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 4, 2019 at 15:12 | comment | added | Fabian Wirth | This answer assumes that $A$ is symmetric. But this is not assumed in the original question. For $A$ symmetric the answer is correct, but this is not the interesting case. | |
| Sep 29, 2018 at 16:31 | comment | added | Ludwig | If $A$ is diagonalizable and has strictly negative eigenvalues, then it is not true in general that $A+A^\top \le 0$. Take for instance $A=\begin{bmatrix}-\frac{1}{2} & 2\\ 0 & -\frac{1}{2}\end{bmatrix}$, $A+A^\top$ has eigenvalues $\{-3,+1\}$. | |
| Sep 29, 2018 at 16:28 | comment | added | hänsel | but $U(A+A^*)U^* = UAU^*+(UAU^*)^*=UAU^* + UA^* U^* \le 0$ | |
| Sep 29, 2018 at 16:24 | comment | added | Ludwig | Your proof is correct when $A+A^\top\le 0$. However when $A+A^\top\not\le 0$, taking $X=\frac{1}{2}I$ violates the constraint $AX+XA^\top\le 0$ and so it does not represent an admissible solution. | |
| Sep 29, 2018 at 16:21 | comment | added | hänsel | my proof is correct, so always $X=1/2$ is the solution. search the flaw in your comment. | |
| Sep 29, 2018 at 16:11 | comment | added | Ludwig | Ok, so your claim is that the solution is invariant under unitary similarity transformations. I agree. Further, I would say that the optimum is $X=\frac{1}{2}I$ in the case $A+A^\top\le 0$ (not just $A$ diagonal). Otherwise, the explicit form of the optimum $X$ seems tricky (cf. the explicit example in my edited OP). | |
| Sep 29, 2018 at 16:08 | comment | added | hänsel | Indeed, then optimum is $U X U^*$ as explained in my answer. $X=1/2$ for $A$ diagonal. | |
| Sep 29, 2018 at 16:05 | comment | added | Ludwig | Then I don't think your claim is true. See the edit in my OP. | |
| Sep 29, 2018 at 15:58 | comment | added | hänsel | yes. strangly you could even always take exactly this $X$. | |
| Sep 29, 2018 at 15:55 | comment | added | Ludwig | Thanks for your answer. However, I think I'm missing something. Are you claiming that the optimal $X$ is always of the form $X=\frac{1}{2}I$? | |
| Sep 29, 2018 at 15:52 | history | edited | hänsel | CC BY-SA 4.0 |
added 14 characters in body
|
| Sep 29, 2018 at 15:07 | history | answered | hänsel | CC BY-SA 4.0 |