Timeline for Explicit solution to a Rayleigh quotient equation
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
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| May 10, 2016 at 5:23 | comment | added | Hayabusananji | Yes, $\mathbf{x}=W(\lambda)\mathbf{1}$ is established form. But this equation forces numeric calculation to get $\lambda$ because of higher degree equation. On the other hand, main equation is also a non-linear simultaneous cubic equation. Theoretically, this is solvable. Just, this is my theme question. | |
| May 10, 2016 at 4:30 | comment | added | Igor Khavkine | $\mathbf{x} = W(\lambda) \mathbf{1}$ is already of that form. Of course, you also need to solve a separate equation for $\lambda$. But I doubt that the solution could be made much simpler, since you need to solve a generic $n$-degree polynomial equation for $\lambda$. | |
| May 10, 2016 at 3:07 | comment | added | Hayabusananji | Exactly, your statement is correct. On the contrary, the origin is that this consideration and constraint equations generate main equation eventually. Therefore, main equation must satisfy these conditions absolutely. Nevertheless, is it possible that we solve like $\mathbf{x}=\cdots$ ? | |
| May 10, 2016 at 2:43 | history | answered | Igor Khavkine | CC BY-SA 3.0 |