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Let $M$ be the adjacency matrix of a directed graph $G$. Is there any known relation between $\det(\textrm{id}-M)$ and the cycles of $G$?

It is easy to see that if $G$ is acyclic then this determinant is $1$ (because we can take $M$ to be strictly upper triangular). What does this measure in general?

Somewhat related: How to tell if a directed graph is acyclic from the adjacency matrix


Using sage I computed these determinants for all the 9608 directed graphs on $5$ vertices and I found that we have $\det(I-M)=1$ iff $G$ is acyclic. Moreover this is the list of possible determinants together with the number of graphs with such determinant:

[(-48, 1), (-40, 1), (-36, 2), (-32, 6), (-30, 3), (-28, 9), (-27, 1),
(-26, 4), (-25, 4), (-24, 36), (-23, 4), (-22, 18), (-21, 9), (-20, 49),
(-19, 12), (-18, 75), (-17, 23), (-16, 144), (-15, 76), (-14, 124),
(-13, 69), (-12, 361), (-11, 116), (-10, 339), (-9, 290), (-8, 676),
(-7, 294), (-6, 917), (-5, 500), (-4, 1195), (-3, 889), (-2, 1144), (-1,
749), (0, 1166), (1, 302)]
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  • $\begingroup$ Note that, up to an overall sign, this determinant is the same as the characteristic polynomial $\det{(M-\lambda \text{ id})}$ evaluated at $\lambda=1$. Consequently this determinant is the same as $\prod_{k}(1-\lambda_k)$. Interestingly, it vanishes if 1 is an eigenvalue. $\endgroup$ – Semiclassical Aug 15 '14 at 20:31
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According to my calculations in sage, there are 11 graphs (out 156) on six vertices such that $\det(I-M)=1$ and exactly two of these eleven are trees. This shows that is is going to be very difficult to determine information about cycles from the value of $\det(I-M)$. (There is nothing special about the value '1' here, for example example there are 35 graphs with $\det(I-M)=0$ and two of these are trees.)

Note that, trees aside, no useful combinatorial interpretation of $\det(M)$ is known, and there seems to be little reason to expect $\det(I-M)$ to work better.

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  • $\begingroup$ I was working with directed graphs... If you have an DAG, then you can label the vertices using a topological order and the matrix $I-M$ becomes upper triangular with $1$ in the diagonal. So, it seems that the analysis is a little bit different with directed graphs. $\endgroup$ – Quimey Aug 18 '14 at 12:35
  • $\begingroup$ Are you trying to decide if a directed graph is acyclic using the value of $\det(I-M)$? That's a different question. $\endgroup$ – Chris Godsil Aug 18 '14 at 15:47
  • $\begingroup$ That's what's written in the question ;) $\endgroup$ – Quimey Aug 18 '14 at 15:49
  • $\begingroup$ It seemed to me that there was a relation between that determinant and the cycles of the graph. I wanted to know if that was written somewhere I could read. $\endgroup$ – Quimey Aug 18 '14 at 15:51
  • $\begingroup$ After thinking a little on the problem and in the examples you give here I see that there are non-acyclic graphs for which the determinant is 1 (take any non directed non trivial graph and double the edges), so I am accepting this answer. $\endgroup$ – Quimey Aug 25 '14 at 20:11
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If you have a graph with disjoint pairs of vertices such that you have just two edges that connect each pair of vertices in opposite directions so that you just have a bunch of disjoint cycles with 2 vertices each, then the determinant will be zero, regardless of how many cycles you have. Similarly, if you an arbitrary graph where two vertices are disconnected from all others and form a cycle among themselves, the determinant will similarly be zero regardless of the structure of the rest of the graph. However I'm not sure if the determinant will always be zero if you have arbitrary cycles, and if it's non-zero, then what it means in terms of the cycles.

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