Timeline for Determinant of the oriented adjacency matrix of a tree
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 6, 2018 at 5:14 | comment | added | Ranveer Singh | @Allen Knutson: I have added the proof now. | |
| Nov 23, 2014 at 15:21 | comment | added | Chris Godsil | @darij grinberg: you're right, it will not be zero. | |
| Nov 23, 2014 at 4:28 | comment | added | darij grinberg | Forget the sentence where I said to label the vertices increasingly. This is not necessary. Whatever way they are labelled, as long as the edges are labelled accordingly, the matrix will still be unitriangular for an appropriate choice of labels for both rows and columns (and the choice for rows is the same as the choice for columns, so different choices do not force different signs), and so the determinant will still be $1$. | |
| Nov 23, 2014 at 3:48 | comment | added | Allen Knutson | I added an example to show a cut vertex in action; darij is right. | |
| Nov 23, 2014 at 3:34 | comment | added | darij grinberg | @AllenKnutson: I'm not sure how explicit you want it. Algorithmically, it is simple: Re-orient all edges away from the vertex $v$, thus making $v$ the root of the tree. Remove $v$, thus obtaining a forest and some dangling edges with only a target but no source. Label the vertices increasingly (i.e., for every edge $a \to b$, we must have $a < b$). Label the edges such that the label of every edge is that of its target (this is possibly because every vertex is now the target of exactly one edge). Then, the determinant is $1$ since the matrix is unitriangular. | |
| Nov 23, 2014 at 3:33 | comment | added | darij grinberg | @ChrisGodsil: I don't think it will be $0$. | |
| Nov 23, 2014 at 2:57 | history | edited | Chris Godsil | CC BY-SA 3.0 |
fixed typo
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| Nov 23, 2014 at 2:46 | comment | added | Allen Knutson | I want to know whether it's $1$ or $-1$. | |
| Nov 23, 2014 at 2:37 | history | answered | Chris Godsil | CC BY-SA 3.0 |