The Grone-Merris Conjecture

In spectral graph theory, Grone and Merris conjecture that the spectrum of the Laplacian matrix of a finite graph is majorized by the conjugate degree sequence of this graph. We give a complete proof for this conjecture.

The Laplacian of a simple graph G with n vertices is a positive semidefinite n × n matrix L(G) that mimics the geometric Laplacian of a Riemannian manifold; see §1 for definitions, and [2,14] for comprehensive bibliographies on the graph Laplacian.The spectrum sequence λ(G) of L(G) can be listed in non-increasing order as For two non-increasing real sequences x and y of length n, we say that x is majorized by y (denoted x y) if This notion was introduced because of the following fundamental theorem.
Theorem 1 (Schur-Horn Dominance Theorem [18,11]).There exists a Hermitian matrix H with diagonal entry sequence x and spectrum sequence y if and only if x y.
For a non-negative integral sequence d, we define its conjugate degree sequence as the sequence Another important majorization relation is the following.

The Laplacian matrix and the majorization relation
Let G = (V, E) be a simple finite graph with n = |V | vertices.We write i ∼ j when the i-th vertex is adjacent to the j-th vertex, and we let d i denote the degree of the i-th vertex.
The Laplacian matrix L(G) of the graph G is the n × n matrix defined by otherwise.We can also express the Laplacian as L(G) = D−A, where D is the diagonal matrix defined by the degree sequence, and A is the adjacency (0, 1)-matrix of the graph.
It is well-known that L(G) is positive semi-definite, since it corresponds to the quadratic form Let λ = (λ 1 , λ 2 , . . ., λ n ) T be the non-increasing spectrum sequence of the Laplacian matrix L(G).The smallest eigenvalue is λ n = 0, with eigenvector Given two vectors x = (x 1 , . . ., x n ) T and y = (y 1 , . . ., y n ) T in R n , rearrange their components in non-increasing order as We say that x is majorized by y, and write x y, if We will make use of the following majorization inequality.

Split graphs
A graph is split (also called semi-bipartite in [12]) if its vertices can be partitioned into a clique V 1 and a co-clique V 2 .This is equivalent to saying that the subgraph induced by V 1 is complete, and that the subgraph induced by V 2 is an independent set.See [5,20,15,10] for many characterizations and properties of split graphs.
Given a split graph G = (V, E), let N = |V 1 | be the size of the clique, and M = |V 2 | be the size of the co-clique.Let δ(G) be the maximum degree of vertices in V 2 .Clearly δ(G) ≤ N , and the Laplacian matrix of the split graph G is of the form where K N is the Laplacian matrix of the complete graph on N vertices, where D 1 and D 2 are diagonal matrices with diagonal entries the vertex degrees of V 1 , V 2 respectively, and where A is the adjacency matrix for edges between V 1 and V 2 .The Laplacian matrix is symmetric, and therefore Hermitian.
Theorem 4 (Courant-Fischer-Weyl [16]).Let the n × n matrix H be Hermitian, with eigenvalues where the max (resp.min) is taken over all k-dimensional (resp.
We first investigate the Laplacian spectrum of a split graph.
Proposition 5.If G is a split graph of clique size N , then Proof.To prove the inequalities involving λ N −1 (G) and λ N +1 (G) by the Courant-Fischer-Weyl Min-Max Principle, it suffices to find an (N − 1)dimensional (resp.M -dimensional) subspace for which the action of L(G) has a desirable lower (resp.upper) bound.There are natural candidates.Let P ⊂ R M +N be the (N − 1)-dimensional subspace consisting of all vectors of the form u 0 M with u ∈ R N and u ⊥ 1 N .Then for any unit This proves our first statement part that When λ N (G) ≥ N , we assert that the degree of any vertex in the clique V 1 is at least N .For this, suppose that our assertion is false, namely that there exists a vertex v 0 ∈ V 1 with degree less than N .Then this vertex v 0 is adjacent to none of the vertices of the co-clique V 2 .Consequently G can be regarded as a split graph with new clique V 1 \ {v 0 } and new co-clique V 2 ∪ {v 0 }.The size of the new clique is N = N − 1. Applying the first part of the proposition, we obtain that For a conjugating pair of non-negative integral sequences, the partial sum of one sequence can be computed in a different way as where χ is the characteristic function.The second part of the proposition now follows from the observation that The next lemma will play an essential role in our proof of the Grone-Merris Conjecture.Its proof is presented in the next section.

The homotopy method
This section is devoted to proving Lemma 6.We adopt a homotopy method, following an idea of Katz [12] in his proof of the Grone-Merris Conjecture for 1-regular semi-bipartite graph. Let where J M ×N denotes the M × N matrix whose entries are all equal to 1.Note that L 1 = L(G) is the matrix we are interested in, and that L 0 is the Laplacian of a complete split graph.The spectrum of L 0 is well-understood: Lemma 7. The Laplacian spectrum of the complete split graph of clique size N and co-clique size M is where P (Q) denotes Q copies of the number P .The eigenspace corresponding to the eigenvalue N consists of all vectors of the form 0 N v , where v is M -dimensional and v ⊥ 1 M ; the eigenspace corresponding to the eigenvalue (M + N ) is spanned by the orthogonal vectors N for all 0 ≤ α < 1. Proof.We again make use of the Courant-Fischer-Weyl Min-Max Principle.Recall that the M -dimensional subspace Q ⊂ R M +N consists of all vectors of the form 0 N u with u ∈ R M .Then for any unit vector u, Therefore, the (N + 1)-th largest eigenvalue λ (α) N +1 is at most N .For the eigenvalue λ (α) N , let P be the N -dimensional subspace which is spanned by the eigenvectors of L 1 corresponding to the N largest eigenvalues.Clearly P ⊥ 1 M +N .For any unit vector v ∈ P , we know from Lemma Therefore, the N -th largest eigenvalue λ (α) N is at least N .We next proceed to show that the inequality on λ (α) N is strict, when 0 ≤ α < 1.We already know that λ N = N for some 0 < α < 1, then the above arguments show that necessarily The first condition λ N (G) = N implies that δ(G) < N , from our assumption on λ N (G); the third condition L 0 (v) = N v implies that v is a unit vector in Ker(L 0 − N ), thus in turn a unit vector of Q.Then We now consider all possible N -dimensional subspaces where V (α) is an M × N matrix.Here the notation of the subspace means that the subspace is spanned by the column vectors of the matrix Lemma 9.If the subspace ⊆ (1 M +N ) ⊥ is an invariant subspace of L α , then the matrix V (α) solves the quadratic matrix equation In terms of matrix entries, this means that where the non-negative integers d i , f j are the entries of the diagonal matrices Proof.It is easy to see that the orthogonal complement in R M +N of the subspace is an invariant subspace of L α , then so is its orthogonal complement, since L α is a symmetric matrix.
The L α -invariance property is equivalent to the existence of two square matrices X α and Y α such that By comparison of the corresponding four block matrices, we immediately obtain that T , together with a quadratic matrix equation for V (α) : This condition, in terms of matrices, is equivalent to α) , with which the above quadratic matrix equation can be simplified to The quadratic matrix equation is complicated, and is almost impossible to be solved explicitly.Fortunately, we do not have to do so.
From Lemma 8 and the assumption on λ N (G), we know that Thus the subspace spanned by the eigenvectors of L α corresponding to the N largest eigenvalues is unique.Assume that this subspace is given by α) , so that the matrix V (α) is well defined.
Lemma 10.The map According to the algebraic multiplicity of eigenvalues of L α , there exist positive integers l = l(α) and i 1 , . . ., i l (i 0 = 0 by convention) such that be an orthonormal basis consisting of the eigenvectors corresponding to the eigenvalues λ (β) i for any β ∈ [0, 1], and {Z αn k } l k=1 , {W α k } l k=1 denote two sequences of monotonic subspaces of R M +N given by Any vector u ∈ Z αn l can now be expressed as Assume that the maximum of |c k,s | is achieved at |c k 0 ,s 0 |.Due to the orthogonality of {u α i } i , the absolute value of the coefficient of is at most u .But when n is sufficiently large, it is at least Hence |c k 0 ,s 0 | ≤ 2 u .For any given vector v ∈ W α l+1 , we see that which goes to zero as n goes to infinity.
Without loss of generality, we can assume that this integer k is minimum over all counterexamples.Then we have Moreover, we can further assume that the number |E| of edges is minimum over all counterexamples with the same k.Under this assumption, we claim that Lemma 13.For any two vertices i, j in the graph G, if d i ≤ k and d j ≤ k, then they are not adjacent in G.
Proof.We will prove this by contradiction.Assuming that the lemma is false, namely there exists a pair of vertices such that Let G be the graph obtained from G by deleting the edge ij.Due to the minimum property of |E|, we must have Two Laplacian matrices are related via L(G) = L( G)+H, where H n×n is a positive semi-definite matrix whose only non-zero entries are H ii = H jj = 1 and H ij = H ji = −1.Applying Fan's Theorem 3, we see that This contradicts our assumption that G was a counterexample, and therefore concludes the proof.
Next, we add new edges to G to get a new graph G. Add to G a new edge ij for any pair of vertices i, j in G such that The new graph G so obtained is a split graph.
The clique of G consists of all vertices of G whose degree is at least k, so the size of the clique is equal to d ′ k (G).Let N = d ′ k (G) denote this size.The co-clique consists of all vertices of G whose degree is less than k, so the maximum degree of vertices in the co-clique is δ while λ i ( G) ≥ λ i (G) for all 1 ≤ i ≤ n, so these two inequalities are still valid for the new graph G, namely Let us discuss the relationship between N and k.
≤ N , which leads to a contradiction.The second inequality comes from Proposition 5.
If N = k, then G is a split graph of clique size N , with the property that This contradicts Lemma 6.So k < N .Note that G is a split graph of clique size N .In this graph G, the maximum degree of vertices in the co-clique is at most (k − 1), while the minimum degree of vertices in the clique is at least (N − 1).This means that According to Proposition 12, Therefore, G is a split graph of clique size M , with the additional property that This again contradicts Lemma 6.
All possible cases are eliminated, and the Grone-Merris Conjecture is proved.

Lemma 6 .
Assume that G is a split graph of clique size N .If either λ N (G) > N or λ N (G) = N > δ(G), then the N -th inequality of the Grone-Merris Conjecture holds, namely N i=1 Combining this with λ k+1 ( G) ≥ . . .≥ λ N −1 ( G) ≥ N from Proposition 5, we see immediately that the inequality Then we proceed to compare λ N ( G) with the clique size N .First consider the case where λN ( G) ≥ N .Because N = d ′ N −1 ( G) ≥ d ′ N ( G), the split graph G has clique size N , with the additional property that N i=1 λ i ( G) > N i=1 d ′ i ( G) and λ N ( G) ≥ N > δ( G).This again contradicts Lemma 6.In the other case, where λ N ( G) < N , we switch attention to the complement graph of G.This complement graph is another split graph G. Its clique size is M , andλ M ( G) = (N + M ) − λ N ( G) > M.