Timeline for answer to Matrix-tree theorem for inverse matrices by Abdelmalek Abdesselam
Current License: CC BY-SA 4.0
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7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 22, 2024 at 19:37 | comment | added | Federico Poloni | got it, thanks! | |
| Nov 22, 2024 at 18:21 | comment | added | Abdelmalek Abdesselam | Oops, I forgot my formula gives two trees if the column sums are zero. So you need to apply it to the transpose of your matrix. | |
| Nov 22, 2024 at 18:19 | comment | added | Abdelmalek Abdesselam | For the numerator, one first sums over the two possible pairings of an $i$ with a $j$, then the forest is made of two trees (because you impose zero sum rows) and each $i,j$ pair is inside one of these two trees. Within one of theses trees, the orientation is as in the above comment with backbone rooting etc. | |
| Nov 22, 2024 at 18:17 | comment | added | Abdelmalek Abdesselam | For your denominator there is only one $i$ and one $j$ so they are automatically paired and in just one tree. The rule is you look at the backbone, i.e., path connecting $i$ and $j$. There the arrows must flow from $j$ to $i$. The backbone treated as a single entity is the "root" from which all the other arrows are outflowing. | |
| Nov 22, 2024 at 16:07 | comment | added | Federico Poloni | Thanks! That seems on point, but tricky to process for me. One question for now: in the figure on page 7 (resp. 57), it seems that the edges point not only away from $j$, but also towards $i$. Is that an additional condition that needs to be verified? | |
| Nov 22, 2024 at 15:59 | vote | accept | Federico Poloni | ||
| Nov 21, 2024 at 15:13 | history | answered | Abdelmalek Abdesselam | CC BY-SA 4.0 |