Cospectral graphs. Presented by: Steve Butler. 21 April 2011

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1 ospectral graphs Presented by: Steve utler 21 April 2011 Two non-isomorphic graphs are said to be cospectral with respect to a given matrix if they have the same eigenvalues. ospectral graphs help to show the limitations that the spectrum of a particular matrix might have in distinguishing properties of a graph. There are several different matrices that are used in spectral graph theory, and these different matrices can reveal different information about a graph. So graphs may be cospectral with respect to some matrix but not cospectral with respect to another matrix (though there are graphs which are cospectral with respect to all matrices). Some of the common matrices that are studied include the adjacency matrix A, the combinatorial Laplacian L := D A, the signless Laplacian Q := D + A, the normalized Laplacian L := D 1/2 (D A)D 1/2 (with the convention that if a vertex is isolated then the corresponding entry id D 1/2 is 0), and the Seidel matrix S := J I 2A. The Seidel matrix is not as commonly studied and is defined by putting a 1 for each edge, a 1 for each non-edge, and a 0 on the diagonal entries. A comparison of the different matrices and some properties the spectrum can and cannot distinguish as well as an example of a cospectral pair of graphs for each is given below. Matrix # edges biparite # components # bipartite components A Yes Yes No No cospectral pair L Yes No Yes No Q Yes No No Yes L No Yes Yes Yes S No No No No One thing that jumps out is just how bad the Seidel matrix is at distinguishing these properties. This is because the spectrum of S for a graph is invariant under Seidel switching, 1

2 which for a vertex v flips all the adjacency relationships with other vertices. In other words we remove all of the edges incident to v and any edges that were previously not there are added in. So for example if we flip a vertex on a K n then the resulting graph is K n 1 K 1 (in general it is not hard to show that we can by a sequence of these Seidel switching moves we can always get to a graph that has an isolated vertex). Suppose that G is the original graph and Ĝ is the graph after we perform a Seidel switch. Then the fact that these graphs are cospectral follows by noting SĜ =.... O.. O 1 S G 1. O... O.. i.e., we multiply on the left and right by a diagonal matrix with 1s in every entry except for the entry corresponding to v which has a 1. (In terms of the Seidel matrix this says we flip the sign on the row/column corresponding to v.) This shows that SĜ and S G are similar and so have the same eigenvalues. In general Seidel switching will not produce cospectral graphs for the other matrices, but with some appropriate assumptions it will work. The most well known example of two such graphs which are cospectral for A and related by Seidel switching is shown below., The application of a generalization of Seidel switching to form cospectral graphs for the adjacency matrix was developed by hris Godsil and rendan Mckay (who wrote the definitive paper on the subject of forming cospectral graphs). This is sometimes known as Godsil-Mckay switching or GM switching. Theorem 1 (Godsil-Mckay). Let M, M be symmetric 0-1 matrices with ( ) ( ) N M = N T, and M N =. N T Where is a b b (symmetric) matrix with constant row sum, is a c c (symmetric) matrix, N is a b c matrix with column sums equal to either 0, b/2, or b, and N is a b c matrix which is found by copying the columns that are all 0s or all 1s from N and the columns that have half 0s and half 1s we flip the 0s and 1s. Then M and M are cospectral. In terms of graphs this says that we can split our graph into two parts, namely and. On the graph is regular, and on it is whatever we want it to be. Further the vertices of can be partitioned into three groups. In the vertices of the first group there is no edge 2

3 to, in the vertices of the second group exactly half of the possible edges connect to, and in the vertices of the third group all of the possible edges connect to. The adjacency matrix of this graph is M. To form the adjacency matrix of M we now do everything the same except that the vertices in the second group which were adjacent to half the vertices of now flip the vertices that it was adjacent to. In the example above we have that the outer cycle on eight vertices is and the central vertex is, which is adjacent to exactly half of the vertices. We now flip the adjacency relationship on the central vertex to form the second graph. The proof is shorter than the statement of the theorem (at least if we skip the computations, which we will). Proof. onsider the matrix Q = ( 2 J ) b b I b O. O I c Then a computation shows that Q 2 = I and QMQ = M so M and M are similar, i.e., cospectral. This switching generates lots of cospectral graphs for the adjacency matrix, and for small graphs tends to be the leading way to form cospectral graphs. It should also be noticed that with additional assumptions the operation of switching can be shown to form cospectral pairs for other matrices as well. For constructing cospectral graphs for the normalized Laplacian, less is known. Some of the highlights include the following: If in the switching operation above the degrees of the vertices do not change, then the two graphs are cospectral. (This is a result of Michael avers in his Ph.D. dissertation.) K p,q is cospectral with K r,s if p + q = r + s. This is a trivial result but is actually a theorem in a paper showing that cospectral graphs do exist on n vertices for each n 4. There are known examples of graphs which are regular and cospectral for A, such graphs are automatically also cospectral for L, Q, L and S. There is a method to unfold bipartite graphs in two ways, and if the two parts of the bipartite graphs have an equal number of vertices then the resulting graphs are cospectral. An example of this is shown below G = G 1 = G 2 = 3

4 In each of these cases though we end up either looking at graphs which are bipartite or graphs which do not have differing degree sequences. These are highly restrictive elements and so a search for cospectral pairs for the normalized Laplacian for graphs which were nonbipartite and had differing degree sequences was undertaken. y examining small graphs ( 8 vertices) a new type of graph was discovered. Namely, fuzzy balls which for each partition generate a graph. So for example if we has b 1 + b b k = n then the corresponding graph would have n+k vertices composed of two parts. First a ball which was a K n and then k pieces of fuzz where in the fuzz the ith vertex connected to exactly b i (unique) vertices of K n. Examples are shown below Fact 1. The eigenvalues of the fuzzy ball graph for the normalized Laplacian depends only on the values of n and k, and not on the partition of n into k parts. In particular the two graphs shown above are cospectral, moreover they are clearly not bipartite and have differing degree sequences. So this gives an example of the desired class. Looking at the two graph we see that they both have a K 5 which is fixed and then if we remove the K 5 the two remaining graphs are K 1,3 K 1,1 K 1,1 and K 1,2 K 1,2 K 1,1 which by themselves are cospectral. So in particular what seems to be happening here is that we have a pair of bipartite cospectral graphs that we are attaching into another graph and the result is that the new graphs are still cospectral. This idea works more generally. Theorem 2 (utler-grout). Let P 1, P 2 be bipartite, L-cospectral graphs on with degree of vertices in all equal to k for both graphs. Let G 1 be a graph on the vertices A A where G 1 [A A ] is arbitrary; G 1 [] and G 1 [] have no edges; there are no edges going between A and, between A and, or between A and ; all possible edges are between A and ; and G 1 [ ] = P 1. The graph G 2 is the same except that G 2 [ ] = P 2. Then G 1 and G 2 are L-cospectral. If the dimension of the eigenspace for λ = 1 restricted to is the same for both P 1 and P 2, then the graphs H 1 and H 2 are also L-cospectral, where H 1 only differs from G 1 in that H 1 [] is the complete graph and similarly H 2 only differs from G 2 in that H 2 [] is the complete graph. 4

5 What is going on is that we start with some arbitrary graph (this is the part on A A ) and we have two cospectral bipartite graphs P 1 and P 2 that we now glue on to this graph and the result is that after we glue on the graphs will still be cospectral. An illustration of what is going on is shown below. A A A A Also note that the fuzzy ball falls into the second part of the theorem where in this particular case we do not have anything in A A but simply throw in a complete graph on the vertices of. Examples of pairs of bipartite cospectral graphs that we can use for this operation include K l,p1 K l,p2 K l,pk and K l,q1 K l,q2 K l,qk with p 1 + +p k = q 1 + +q k as well as some other graphs such as the following two pairs: On a side note the graph on the left is 8 and it has a cospectral mate for the normalized Laplacian. This is surprising because it is well known for the matrices A and L that a cycle is uniquely determined by its spectrum, but clearly for the normalized Laplacian this need not be true. We will now work through the theorem. In order to do this we will find it convenient to work with harmonic eigenvectors which help remove the sting of the square roots involved in the definition of L. In particular if Lx = λx then the corresponding harmonic eigenvector is y = D 1/2 x. With a little bit of manipulation we now have Lx = λx (D A)y = λdy for each v V: y(u) = (1 λ)y(v)d(v). The other thing we will need are some basic properties of bipartite graphs. The following all follow from the above formulas, with the convention that two harmonic eigenvectors are orthogonal if and only if there corresponding eigenvectors are orthogonal. 5 u v

6 Lemma 1. Let P be a bipartite graph on with all the vertices in of degree k. If Lx = x then L(x ) = x and L(x ) = x. If Lx = λx then L ( x x ) = (2 λ) ( x x ). If y 1 = 1 +1,...,y i,...,y m = 1 1 are a set of orthonormal harmonic eigenvectors then for 1 < i < m we have y i 1 = 0, in particular, the sum of the entries of y i for the vertices of is 0. The first two properties are general properties for bipartite graphs. The third property follows by noting that y i will be orthogonal (as a harmonic eigenvector) to 1 2 (y 1+y m ) = 1. ut since the vertices in all have the same degree k then this means that they are actually orthogonal as vectors as well. We are now ready to step through the main ideas of the theorem. Proof. Let s be the degree of the vertices in for both G 1 and G 2. We will also use the notation in the lemma, i.e., m will be the sizes of P 1 and P 2 and y 1 and y m are defined as above. We will work through the first statement (the part where is empty). laim 1. For 1 < i < m, if y i = b i + c i (i.e., b i is the part of y i on the vertices of and similar for c i ) is a harmonic eigenvector of P 1 (respectively P 2 ) for the eigenvalue λ i, then ŷ i = b i + s c k i is a harmonic eigenvector of G 1 (respectively G 2 ) for the eigenvalue k 1 (1 λ s i). To verify this claim it suffices to show that for each vertex in our graph ŷ i (u) = u v k s (1 λ i)ŷ i (v)d(v). ( ) We will do this in a case-by-case basis: v A: If the vertex is in A then both sides of ( ) are trivially 0 and we are done. In the case that v A then the entry on the right hand side of ( ) is again 0, while for the sum on the left hand side the only nonzero terms would come from, but by definition v is adjacent to all of and the sum of the entries from is 0 by the above lemma. v : In this case we will work through the sum. The important thing to remember is that y i is an eigenvector for P 1 (respectively P 2 ) and so we can use our knowledge about those corresponding sums to simplify. So we have: ŷ i (u) = u v s k y i (u) = u v Showing that ( ) is satisfied. s k (1 λ i)y i (v)k = k s (1 λ i)ŷ i (v)s. 6

7 v : We again work through the sum. So we have: u vŷ i (u) = u v Showing that ( ) is satisfied. y i (u) = (1 λ i )y i (v)d(v) = k s (1 λ i)ŷ i (v)d(v). laim 2. AnyharmoniceigenvectoronG 1 (respectivelyg 2 ) whichis orthogonalto ŷ 2,...,ŷ m 1 must be constant on and constant on. This claim simply follows by noting that with the addition of 1 and 1 that we have a full set of orthogonal harmonic eigenvectors (by the above lemma) for P 1 (respectively P 2 ). Therefore any harmonic eigenvector orthogonal to one of these y i for 1 < i < m must be some linear combination of 1 and 1. laim 3. Any harmonic eigenvector on G 1 orthogonal to ŷ 2,...,ŷ m 1 is also a harmonic eigenvector on G 2. From this claim the proof of the theorem now follows because we have first shown that we have generated m 2 eigenvalues for both G 1 and G 2 that must match from the eigenvalues of P 1 and P 2 (here is where we are using that P 1 and P 2 are cospectral). Further this claim says that any remaining eigenvalue of G 1 will also be an eigenvalues of G 2 and therefore all the eigenvalues match and the graphs are cospectral. This follows again by verifying that u vy(u) = (1 λ)y(v)d(v) works for both graphs. For vertices in A,A, both sides are unchanged. The only tricky part is what happens for a vertex in since in we might change the degree of a vertex. ut in this case note that we have for u v with v that y(u) = (1 λ)y(v)d(v) y(u )d(v) = (1 λ)y(v)d(v) y(u ) = (1 λ)y(v), u v in particular the degree term drops out and the relationship will still be satisfied. In the case that we want to have a complete graph, i.e., H 1 and H 2 in the statement of the theorem, then we go through the exact same process as before, only we have to do a little more bookkeeping, so we will omit most of the details and only sketch what to do in the first part. So suppose that λ i = 1 for P 1 then by the lemma we can assume that the eigenvector either has the form b i or c i. Suppose that y i = c i then we let ŷ i = c i and the new eigenvalue for G 1 will be 1. On the other hand if the harmonic eigenvector y i = b i then we let ŷ i = b i and the new eigenvalue for G 1 will be 1+ 1 s. Finally, if λ i 1 and y i = b i +c i then we can generate the new eigenvalues for G 1 of λ i = 2s+1± 1+4(1 λ i ) 2 sk 2s 7

8 which come from the harmonic eigenvectors ( ) 1± 1+4(1 λi ) ŷ i = b i + 2 sk c i. 2k(1 λ i ) A cursory glance might make us think we are now generating too many new eigenvalues and eigenvectors. ut using the lemma we have a way to naturally pair up eigenvalues. In particular if we had used b i c i and (2 λ i ) we would have generated the same eigenvalue/eigenvector pair, so everything works out. This construction will only generally work for the normalized Laplacian, so a natural question is to look and see at what point was the normalized Laplacian critical in the above computations. It boils down to handling the degree change of a vertex in. The normalized Laplacian had the degree built into each side and so could cancel making it easy to work with. ut for a matrix such as the adjacency matrix this is not present and so the technique fails. So this also indicates how to modify our result to work for other matrices and we have the following theorem. Theorem 3 (utler-grout). Let P 1, P 2 be (k,l)-biregular, bipartite, A-cospectral graphs on. Further, the dimension of the eigenspace for λ = 0 restricted to is the same for both P 1 and P 2. Let G 1 be a graph on the vertices A A where G 1 [A A ] is arbitrary; G 1 [] and G 1 [] have no edges; there are no edges going between A and, between A and, or between A and ; all possible edges are between A and ; and G 1 [ ] = P 1. The graph G 2 is the same except that G 2 [ ] = P 2. Then G 1 and G 2 are cospectral with respect to A+tD (t arbitrary). The graphs H 1 and H 2 are also cospectral with respect to A+tD (t arbitrary), where H 1 only differs from G 1 in that H 1 [] is the complete graph and similarly H 2 only differs from G 2 in that H 2 [] is the complete graph. The proof is similar to before (with more bookkeeping). However to make sure this is not a vacuous statement we need to find examples of (k, l)-biregular, bipartite, A-cospectral graphs on. Well that is not too hard, an example of this is shown below. These are clearly cospectral and more than biregular, they are regular. Of course swapping one out for the other is not particularly interesting since this does not change the graph! 8

9 So let us fix that. In particular take the graph on the right and let us flip it over giving the following pair of graphs. In particular, we are still gluing in the same bipartite graph, but now we are changing which side we are gluing in. For the bipartite graph shown we can distinguish the two sides (for example we have marked two vertices on the right graph in which have the same set of neighbors, but on the left graph there are no two such vertices in ). In particular changing the way we glue these in will yield a different graph. As a simple example, we can also just throw in a complete graph on the vertices of and the resulting pair will still be cospectral. This gives the following pictures. These graphs are cospectral with respect to A + td, and so by varying t we have that they are cospectral with respect to A (for t = 0), L (for t = 1), Q (for t = 1) and also we still satisfy the original theorem and so are cospectral with respect to L. On a side note these graphs are also complements of one another, and so are examples of graphs which are cospectral with their complement. (Another example is 5 but that is not as surprising!) We can use the above ideas to now form not just a pair of cospectral graphs but more generally we can form large families of cospectral graphs. Theorem 4. For n large there is a family of 2 n/7 graphs on n vertices which are nonisomorphic and mutually cospectral with respect to the normalized Laplacian. Proof. We simply construct such a family. Start by choosing a graph G on n/7 vertices thathasatrivialautomorphismgroup. Foreachv V(G)wenowattachoneofthefollowing two widgets. or v v 9

10 The choice of widget does not change the spectrum because the two are cospectral by an application of the theorem for the normalized Laplacian. Further, given such a graph we can recover G (i.e., G will be the only part with degree 4) and we can then determine which choice was made in the attachment for v V(G), therefore the graphs are non-isomorphic. Finally each such graph has 7 n/7 n vertices, and we can always toss in some isolated vertices to get up to n vertices. So in particular we have a family of size n (a modification of this argument gives a family of size 282 n/ n ). Further, for each choice of G we have a different family and the number of such choices of G grows super exponentially. So we have super exponentially many families of exponential size of cospectral graphs. In other words there are a lot of cospectral graphs. ut there are also a lot of graphs. So a classic question in spectral graph theory is an you hear the shape of a graph? Or in other words, if you know the spectrum of a graph can you find a unique graph which produced it? We have shown that the answer is sometimes not, but generically it is believed that you can, though no proof is known. There is a partial result due to Allen Schwenk. Theorem 5 (Schwenk). Almost all trees are cospectral. In particular, with probability going to 1 as the number of vertices goes to a randomly chosen tree will have a cospectral mate. The basic idea behind this is there there are two twigs onninevertices which ifyouswaponeoutfortheotherdoesnotchangethespectrum of the tree, and then with probability tending to 1 a random tree will contain such a twig and hence a cospectral tree. For reference, below is a comparison of the number of graphs which have a cospectral mate for various matrices we have discussed (for S this is the number of graphs and so we will omit that matrix). n #graphs A L Q L It is tempting to try to generalize the trends seen in this table, but this is dangerous since the number of vertices is very small. For example small graphs tend to have non-trivial automorphism groups while larger graphs do not and so it is hard to pick out general trends by only looking at small graphs. 10

11 Also we should note that while we have mentioned several ways to construct cospectral graphs, we have by no means been exhaustive there are other techniques and also other graphs which are cospectral which have not yet been explained by some general tool. elow are some more examples of graphs which are cospectral for the normalized Laplacian. The pair on the left again show that the normalized Laplacian cannot determine if a graph is regular, and it also cannot tell the number of edges. ecause the normalized Laplacian cannot tell how many edges are in a graph we are led to the interesting possibility that a graph might be cospectral with a subgraph, an example of when this happens is shown for the pair on the right. Further reading S. utler, ospectral graphs for both the adjacency and normalized Laplacian matrices, Linear and Multilinear Algebra 58 (2010), S. utler and Jason Grout, A construction of cospectral graphs for the normalized Laplacian, preprint. Edwin van Dam and Willem Haemers, Which graphs are determined by their spectrum?, Linear Algebra and its Applications 373 (2003), hris Godsil, Are almost all graphs cospectral?, slides from a 2007 talk. hris Godsil and rendan Mckay, onstructing cospectral graphs, Aequationes Mathematicae 25 (1982), Willem Haemers and Edward Spence, Enumeration of cospectral graphs, European Journal of ombinatorics 25 (2004), Allen Schwenk, Almost all trees are cospectral, New Directions in the Theory of Graphs, Academic Press, 1973,