MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Two graphs are isospectral if the combinatorial Laplacian on them has the same spectrum, equivalently, the adjacency matrix has the same the set of eigenvalues (including multiplicities). Two graphs have the same matroid if they are 2-isomorphic, that means there exists a bijection between their edge sets that preserves cycles. There are plenty of examples of isospectral graphs (that are not isomorphic) and it is fairly easy to draw examples of 2-isomorphic graphs but can both of these phenomena occur at the same time?

In general I don't see any reason why no example of this should exist. I know examples of isospectral graphs can be constructed with Seidel switching but this results in graphs with relatively large average degree and usually high connectivity. On the other hand, 3-connected graphs are uniquely determined by their matroid so any potential example cannot be 3-connected.

I am interested in connected non-weighted graphs, I don't care whether they are simple (ie have no loops or multiple edges). Thanks

share|cite|improve this question
By the way, graphs can be adjacency isospectral but not Laplacian isospectral, and vice versa. – Felix Goldberg Dec 13 '12 at 18:46
up vote 3 down vote accepted

Choose a graph $X$ with vertices $u$ and $v$ such that $X\backslash u$ and $X\backslash v$ are cospectral. (In this case I say that $u$ and $v$ are cospectral vertices.) Assume that there is no automorphism of $X$ that swaps $u$ and $v$. Now form the graph $Y$ from two copies of $X$ by identifying vertex $u$ in the first copy with vertex $u$ in the second, and vertex $v$ in the first with $v$ in the second. Next form $Z$ from two copies of $X$ by identifying $u$'s with $v$'s.

If $u$ and $v$ are adjacent, then then will be a double edge, so we assume that $u$ and $v$ are not adjacent. This means that $Y$ and $Z$ are related by a Whitney flip and hence they have the same cycle matroid.

The graphs $Y$ and $Z$ are cospectral. This follows from Corollary 4.3.2 in my book "Algebraic Combinatorics". Since we only need examples we do not need the proof though, we can just carry out the construction and check.

As example, use Schwenk's tree, constructed as follows. Start with the path on 8 vertices 0,1,...,7 and add a vertex 8 adjacent to 5. Vertices 3 and 6 are cospectral. The graphs we get from the construction are not cospectral. (I just checked using sage.)

For more examples, take any strongly regular graph with trivial automorphism group and take $u$ and $v$ to be two nonadjacent vertices. These examples will be 2-connected.

share|cite|improve this answer

There exist co-spectral trees. As there are no cycles to preserve, they fit the bill.

As far as 2-connected graphs are concerned, I would try finding a 2-connected graph $\Gamma$ which can be obtained from disjoint graphs $\Gamma_1$ and $\Gamma_2$ by identifying a pair of vertices $u_1,v_1\in V\Gamma_1$ with $u_2,v_2\in V\Gamma_2$, so that the Whitney twist, i.e. identification of $u_1,v_1$ with $v_2,u_2$, gives a non-isomorphic graph, but preserves the characteristic polynomial of the Laplacian matrix of $\Gamma$.

share|cite|improve this answer
That was quick, thanks. In these examples the matroid is trivial so the better question would be: Are there 2-connected graphs that are isospectral and have the same matroid. – Ralf Rueckriemen Dec 13 '12 at 9:24
I expanded my answer to suggest a possible approach to this. For terminology, look at – Dima Pasechnik Dec 13 '12 at 10:18

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.