Timeline for Testing if two 12x12 matrices have the same determinant
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 22, 2013 at 13:36 | comment | added | hardmath♦ | @Jernej: If you believe something I stated is incorrect, I've created a chat room based on this Question where the discussion can be threaded without unnecessary commenting here. | |
| Jun 21, 2013 at 11:57 | comment | added | hardmath♦ | The off-diagonal entries of $Q$ will then generally contain 0 as well as -1. The $LDL^T$ decomposition suggested by Victor takes advantage of symmetry and reduces the leading term in operation count from $\frac{2}{3}n^3$ to $\frac{1}{3}n^3$. There is an exact integer approach, but that is probably not needed for your modest size matrix and entries. If I understand the construction, $12I-Q-J$ is positive definite for the same reason $Q$ is. | |
| Jun 21, 2013 at 8:08 | comment | added | Jernej | Hm.. $Q$ is in fact $D-A$ where $A$ is the adjacency matrix of $G$ so I think that this result may not be correct. In particular it would imply that the number of spanning trees of a graph $G$ is determined by its degree sequence which does not hold. | |
| Jun 20, 2013 at 19:48 | history | edited | hardmath♦ | CC BY-SA 3.0 |
Added a note computing det(Q) for a special case.
|
| Jun 20, 2013 at 12:54 | history | edited | hardmath♦ | CC BY-SA 3.0 |
Revised description of what Armadillo's det does with slow=false
|
| Jun 19, 2013 at 21:42 | history | answered | hardmath♦ | CC BY-SA 3.0 |