Timeline for Solving Matrix/Operator Equation $H P X + X P H + HQ = 0$
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 13, 2017 at 23:51 | vote | accept | Conner DiPaolo | ||
| Apr 12, 2017 at 16:58 | answer | added | Conner DiPaolo | timeline score: 0 | |
| Apr 12, 2017 at 2:59 | history | edited | Conner DiPaolo | CC BY-SA 3.0 |
some interesting updates!
|
| Apr 11, 2017 at 16:34 | history | edited | Conner DiPaolo | CC BY-SA 3.0 |
update question to be more meaningful
|
| Apr 11, 2017 at 16:30 | comment | added | Conner DiPaolo | Ah I think I see an issue. This question came from looking at the function $f(X) = X P X^* + XQ + R$ and trying to find an $X$ such that $f(X) \preceq_K f(Y)$ for all $Y\in\mathcal{L}(\mathcal{H})$ where $K = \mathcal{L}_+(\mathcal{H})$ is the positive cone. Since the $XQ$ term in general won't be positive, we would likely need to restrict the domain $X$ to be positive and require $Q$ to be positive as well. The resulting equation would force $Q$ to commute with all positive $H$ and would look more like the Lyapunov equation. | |
| Apr 11, 2017 at 12:16 | comment | added | J. E. Pascoe | It's not totally clear how you used $\mathcal{H} = \mathbb{C}^n,$ so the fact should probably hold for any Hilbert space. I guess my question was, is there some way the question needs to be changed to get a meaningful answer? | |
| Apr 11, 2017 at 7:57 | comment | added | Conner DiPaolo | Okay in general this can't hold. Consider $\mathcal{H}=\mathbb{C}^n$ and $H = iI$. Then $HQ - QH^* = 2iQ = 0$ so $Q=0$. I guess that does provoke an interesting question of whether $Q$ must be $0$ for any Hilbert space $\mathcal{H}$. | |
| Apr 11, 2017 at 7:43 | comment | added | Conner DiPaolo | If $Q$ is necessarily zero I guess the problem becomes trivial since $X=0$ would satisfy the equation. I can see that $HQ$ must be self-adjoint for all $H$ (specifically $Q$ must be self-adjoint) which is basically what you said. I guess this sub-question is equivalent to asking whether there exists a nonzero left-ideal on the multiplicative set of self-adjoint operators on a Hilbert space. I agree with your intuition that this can't always be the case, but I'll look into it tomorrow. | |
| Apr 11, 2017 at 6:02 | comment | added | J. E. Pascoe | I have a question about the formulation of the problem: Shouldn't the equation imply that $HQ - QH^* = 0$ for all $H \in \mathcal{L}(\mathcal{H})?$ (Say, take the imaginary part of the equation.) I would guess this implies that $Q$ is zero. | |
| Apr 11, 2017 at 5:32 | history | edited | Conner DiPaolo | CC BY-SA 3.0 |
clarify problem
|
| Apr 11, 2017 at 5:26 | history | edited | Conner DiPaolo | CC BY-SA 3.0 |
fixed grammar
|
| Apr 11, 2017 at 5:20 | review | First posts | |||
| Apr 11, 2017 at 6:22 | |||||
| Apr 11, 2017 at 5:17 | history | asked | Conner DiPaolo | CC BY-SA 3.0 |