Timeline for About the upper bound on the roots of the matching polynomial
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Mar 24, 2015 at 18:57 | history | edited | Chris Godsil | CC BY-SA 3.0 |
removed ambiguity, fixed typo
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| Mar 24, 2015 at 18:57 | comment | added | Chris Godsil | @user6818: I have edited my answer. I wrote "gives", which is ambiguous, now it's "divides". | |
| Mar 24, 2015 at 18:41 | comment | added | user6818 | But the statement I quoted is exactly from your own answer! Thats what was surprising to me. So is there an error in the answer you typed? | |
| Mar 23, 2015 at 23:37 | comment | added | Chris Godsil | The reason your "following statement" is not in the paper is that it is not true. The paper shows that given a graph $G$ thrre is a tree $T$ such that the matching polynomial of $G$ divides the matching polynomial of $T$. But the matching polynomial of a tree is equal to its characteristic polynomial and the graph and the tree have the same maximum valency, so standard eigenvalue bounds on the tree give the bound on the zeros of matching polynomial of $G$. | |
| Mar 23, 2015 at 21:46 | comment | added | user6818 | (this is what the abstract says though your theorem 2.5 is stated with only one vertex $v$!) | |
| Mar 23, 2015 at 21:28 | comment | added | user6818 | All I see there is that for any graph $G$ there, you construct the "path tree" $T$ such that any pair of vertices $v \in V.G$ and $w \in V.T$ you have that $\frac{ \alpha (G /\ v)}{ \alpha(G)} = \frac{\alpha (T /\ v) }{\alpha(T) }$ | |
| Mar 23, 2015 at 21:26 | comment | added | user6818 | I am not able to find the following statement in your paper, "if $G$ is a graph with maximum valency $k$, then there is a tree $T$ with maximum valency $k$ such that that the matching polynomial of $G$ gives that of $T$" - could you kindly point out what there implies this? | |
| Mar 19, 2015 at 3:45 | history | answered | Chris Godsil | CC BY-SA 3.0 |