Timeline for How to prove that for the real stable characteristic polynomial $P=\Phi_T$ of a tree $T$, $P_iP_j-PP_{ij}=(\Phi_{T-[v_i,v_j]})^2$?
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| Mar 5, 2022 at 17:56 | comment | added | CHUAKS | For a general graph $G$ and $P=\det(I_x-A_G)$, we still have $\Delta_{ij}(P)=(\sum_{p_{ij} \in [v_i,v_j]} \Phi_{G-p_{ij}}(x))^2$, where $p_{ij}$ range over all simple paths form $v_i$ to $v_j$. Here simple means all vertices along the path are distinct. | |
| Jun 9, 2021 at 3:55 | comment | added | CHUAKS | One get a more interesting result if one replace determinant by permanent. If we define $\Psi_G=permanent(I_x-A)$, then for a tree $T$, we have $\Delta_{ij}(\Psi_T)=(-1)^{len([v_i,v_j])} (\Psi_{T-[v_i,v_j]})^2$ where $len([v_i,v_j])$ is the length of the unique path $[v_i,v_j]$. | |
| Jul 8, 2020 at 19:25 | vote | accept | Chua KS | ||
| Jun 19, 2020 at 7:28 | history | edited | Chua KS | CC BY-SA 4.0 |
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| Jun 11, 2020 at 16:42 | vote | accept | Chua KS | ||
| Jun 19, 2020 at 17:41 | |||||
| Jun 11, 2020 at 15:29 | answer | added | Abdelmalek Abdesselam | timeline score: 1 | |
| Jun 11, 2020 at 7:36 | history | edited | Chua KS | CC BY-SA 4.0 |
added 125 characters in body; edited tags; edited title
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| Jun 10, 2020 at 20:10 | history | asked | Chua KS | CC BY-SA 4.0 |