Timeline for Solving a nonlinear matrix equation
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| S Jun 26, 2019 at 8:25 | history | suggested | Rodrigo de Azevedo |
added tag.
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| Jun 26, 2019 at 7:03 | review | Suggested edits | |||
| S Jun 26, 2019 at 8:25 | |||||
| Jun 26, 2019 at 0:43 | comment | added | ppp | Thanks, all! Especially, thanks, Igor, for the detail comments which actually hit the final point that resolved my problem! | |
| Jun 25, 2019 at 13:46 | comment | added | Igor Khavkine | Following up on Anthony's comment, letting $Y = (x_1^{-1}, \ldots, x_n^{-1})$ as a row vector, the resulting equation is $Y(\hat{B}-\hat{P}A)=0$, where $\hat{B}$ and $\hat{P}$ are now diagonal matrices with the elements of $B$ and $P$ on the diagonal, respectively. | |
| Jun 25, 2019 at 5:56 | comment | added | Anthony Quas | No. If you call $y_i=x_i^{-1}$, then both sides are linear expressions in the $y_i$ (try writing out the 2 sides). | |
| Jun 25, 2019 at 5:54 | comment | added | Mahdi - Free Palestine | If $X$ is a solution then $\alpha X$ also is a solution, for each $\alpha > 0$. So, there is no uniqueness condition, in general. | |
| Jun 25, 2019 at 5:39 | history | edited | YCor |
edited tags
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| Jun 25, 2019 at 5:18 | comment | added | ppp | Thanks, Anthony! But I think it would still be nonlinear equation in $x_i^{-1}$, no? | |
| Jun 25, 2019 at 4:33 | comment | added | Anthony Quas | If you multiply on the right by $X^{-1}$, then you obtain a linear equation in the $x_i^{-1}$. | |
| Jun 25, 2019 at 4:02 | review | First posts | |||
| Jun 25, 2019 at 6:13 | |||||
| Jun 25, 2019 at 4:00 | history | asked | ppp | CC BY-SA 4.0 |