Timeline for Uniqueness of solution for a tridiagonal system
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 11, 2022 at 8:13 | history | edited | Rócherz | CC BY-SA 4.0 | Corrected spelling, improved formating |
| May 27, 2019 at 6:37 | history | edited | Rodrigo de Azevedo | edited tags | |
| Jun 23, 2016 at 2:33 | vote | accept | Hinazuki Kayo | ||
| Jun 22, 2016 at 19:06 | answer | added | Rodrigo de Azevedo | timeline score: 0 | |
| Jun 22, 2016 at 15:08 | answer | added | Jack D'Aurizio | timeline score: 3 | |
| Jun 22, 2016 at 14:41 | answer | added | Rodrigo de Azevedo | timeline score: 2 | |
| Jun 22, 2016 at 14:32 | comment | added | Ian | No, because an LU factorization doesn't guarantee uniqueness. In particular, if $A$ is not invertible then neither is $U$. LU factorization only reduces us to showing that $U$ is invertible. But if, say, you replace the $2$s in the top left and bottom right corner with $1$s, the result is not invertible (constant vectors get mapped to zero; this is a feature shared by all graph Laplacians), yet your logic goes through the same way. | |
| Jun 22, 2016 at 14:27 | comment | added | Hinazuki Kayo | Thanks guys, I see now. Could I also do it like this, since $A$ is tridiagonal, an $LU$ factorization exists which can be computed by Thomas algorithm. Since $A\mathbf{x}=\mathbf{b}$ iff $(LU)\mathbf{x}=\mathbf{b}$ We can solve the lower triangular system $L\mathbf{y} = \mathbf{b}$ which yields a unique solution and we can also solve the upper triangular system $U\mathbf{x} = \mathbf{y}$ resulting in $\mathbf{x}$ being unique. Also I cannot accept comment answers unfortunately! | |
| Jun 22, 2016 at 14:27 | comment | added | user265759 | This tridiagonal system has a unique solution according to the theorem here | |
| Jun 22, 2016 at 14:23 | comment | added | Ian | math.stackexchange.com/questions/388829/… This question is not quite a duplicate but the desired result follows immediately from the result in this question. | |
| Jun 22, 2016 at 14:13 | history | edited | Rodrigo de Azevedo | CC BY-SA 3.0 | Added tags to this question |
| Jun 22, 2016 at 14:11 | comment | added | Dark | Can you compute $\det A$ ? If $A$ is invertible, then your system has an unique solution. | |
| Jun 22, 2016 at 14:11 | history | edited | Riccardo.Alestra | CC BY-SA 3.0 | edited title |
| Jun 22, 2016 at 14:05 | history | asked | Hinazuki Kayo | CC BY-SA 3.0 |