Is there any way to count cycles of given size by that graph's spectrum? for example for $k=3$ the number of triangles in $G$ is $1/6\cdot\sum_{\lambda \in \mathrm{Spectrum}(G)} \lambda^3$ is there a way for $k>3$?
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$\sum_\lambda \lambda^n$, i.e. the trace of the n-th power of the adjacency matrix, is the number of closed paths of length $n$. – user8268 Aug 26 '13 at 20:52
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1@user8268 but doesn't that include backtracking as well? – draks ... Aug 26 '13 at 20:53
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It does, i.e. it doesn't answer the question – user8268 Aug 26 '13 at 20:56
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The complete bipartite graph $K_{1,4}$ and the 4-cycle together with an isolated vertex both have characteristic polynomial $t^5-4t^3$. This indicates there will be difficulties in reading off information about cycles from the spectrum. The number of shortest odd cycles can be determined from the spectrum. Let $g$ be the odd girth of the graph, the length of the shortest odd cycle and look at $\mathrm{tr}(A^g)$.
Chris Godsil
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nice, btw I still thinking about this relation to the Chebyshev polynomials you once mentioned in the link of my answer. It would be great if you could have another look (especially at my comment). Thx... – draks ... Aug 26 '13 at 22:37
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Though, according to some result in the back of my mind it is possible to count the number of $4$-cycles of a graph given its spectrum and degree sequence. – Jernej Aug 28 '13 at 19:19
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That's right, if you know the degrees, you can compute the number of closed walks of length four that do not go round a 4-cycle (and hence get the number of 4-cycles). – Chris Godsil Aug 28 '13 at 19:47
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Chris Godsil once gave me a nice answer how to calculate the number of Returning Paths on Cubic Graphs Without Backtracking:
If $r\ge3$ we have the recurrence $$ p_{r+1}(A) = Ap_r(A)- (\Delta-I) p_{r-1}(A). $$ where $\Delta$ is the diagonal matrix of valencies of the graph (not necessarily cubic).
When you deal with cycles of prime length, the sum of eigenvalues of $p_r(A)/r$, i.e. its by-$r$-scaled trace, should give you the number of such cycles.
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If $G$ is got by merging a vertex of $K_3$ with a vertex of $K_4$, there is a non-backtracking walk of length 7, but no 7-cycle. I believe it should be possible to get the number of cycles of odd length up to $2g-1$, where $g$ is the odd girth. (If I recall correctly, for regular graphs this can be done using the usual spectrum.) – Chris Godsil Aug 26 '13 at 23:41
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@Chris, would it help (my answer) to restrict it to at least $2$-vertex-connected graphs? – draks ... Aug 27 '13 at 19:44