For a simple graph G, the following relationships hold: $$RR^T=\Delta+A$$ and $$R^TR=2I+A_{L(G)}$$ where R is the incidence matrix, A is the adjacency matrix, I is the identity matrix, $A_{L(G)}$ is the adjacency of the line graph of G, and $\Delta$ is the diagonal matrix of the vertex degrees; Q, $Q=\Delta+A$, is the signless Laplacian.

Since $RR^T$ and $R^TR$ have the same non-zero eigenvalues,is not the following true? $${\kappa}_l = \sum_{i=1}^n(2+\mu_i)^l$$ where $\kappa_l$ is the $l^{th}$ spectral moment of the signless Laplacian and $\mu_i$ are eigenvalues of the line graph?


$${\kappa}_l = \sum_{i=1}^m(2+\mu_i)^l$$ where m is the number of edges. Now I correctly reproduce the first three spectral moments by use of line graphs.

  • $\begingroup$ This is certainly true. It's not clear why you are asking. $\endgroup$ – Chris Godsil Jul 20 '13 at 13:33
  • $\begingroup$ Cvetkovic gives decomposition of the first three spectral moments via subgraph contributions. I have not seen decompositions of higher order moments (For the Laplacian proper, 4th and 5th moments are given by Preciado ea; their method can be applied for the signless Laplacuan also). On the other hand, spectral moments of line graphs can be easily accounted that way. Hence my confusion: why it happened that this obvious relationship has been ignored for handling the spectral moments of signless Laplacian, at least in chemical graph theory? Thus I assessed the beginning. $\endgroup$ – giorgi Jul 20 '13 at 18:22
  • $\begingroup$ There's a possibility that the chemists were not aware of the relation. Mathematically it's obvious. $\endgroup$ – Chris Godsil Jul 20 '13 at 18:36

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