Spectral gap of the largest eigenvalue of the normalized graph Laplacian

We prove that the maximal eigenvalue of the normalized graph Laplacian of a graph with $n$ vertices is at least $\frac{n+1}{n-1}$ provided the graph is not complete. Equality is attained if and only if the complement graph is a single edge or a complete bipartite graph with both parts of size $\frac{n-1}2$.


Introduction
Spectral graph theory investigates the fundamental relationships between geometric properties of a graph and the eigenvalues of the corresponding linear operator. A general overview is given in [Chu97]. In terms of the largest eigenvalue, the normalized graph Laplacian is particularly interesting as it measures how close a graph is to a bipartite graph. In this paper, we are interested in the reverse question, i.e., how far from a bipartite graph a graph can be. This translates to giving lower bounds on the largest eigenvalue. Several bounds are given in [LGS14]. However, the best known estimate so far is λ n ≥ n n−1 which is only attained if the graph is a complete graph [Chu97]. Here, λ n denotes the largest eigenvalue and n is the number of vertices.
Naturally, the question arises what is the optimal a-priori estimate for λ n for non-complete graphs. In this note, we show that for all non-complete graphs, one has λ n ≥ n + 1 n − 1 , see Theorem 2.1. Equality is attained for each n by a complete graph with a single edge removed. For odd n, it is also attained when two complete graphs of size n+1 2 share a single vertex. These are the only cases where equality is attained, see Theorem 3.1.

Eigenvalue estimate
A graph G = (V, E) consists of a finite non-empty vertex set V and a symmetric edge relation We write C(V ) = R V and we denote the positive semidefinite normalized Laplacian by L : C(V ) → C(V ); it is given by We will always assume that d(x) ≥ 1 for all x ∈ V as the Laplacian is not well defined otherwise. The inner product is given by The operator L is self-adjoint w.r.t. this inner product and the eigenvalues of L are In this paper, we are interested in estimating the largest eigenvalue λ n which, by the min-max principle, can be written as Lf, f f, f .

It is well known that
where the first inequality is an equality only for the complete graph, and the latter inequality is an equality only for bipartite graphs. The following theorem gives the optimal a-priori lower bound on λ n for all noncomplete graphs.
Theorem 2.1. Let G = (V, E) be a non-complete graph with n vertices. Then, n−1 f in v and w. Particularly, let f : V → R be given by (1) Since d(v) ≤ n − 2 and d(w) ≤ n − 2, we see that the term in brackets is positive and thus, We write D := (d(v) + d(w))/2, and by the harmonic-arithmetic mean estimate, we have 1 2d(v) + 1 2d(w) ≥ 1 D and thus, We aim to show that the latter term is at least 1 n−1 which, by multiplying with D(n − 1) and subtracting D, is equivalent to If D ≤ n−1 2 , then the maximum equals 1 and the inequality follows immediately. If D ≥ n−1 2 , then we can discard the "1∨", and so the left hand side becomes a concave quadratic polynomial in D with its zero points in D = n − 2 and D = n−1 2 . Thus, the inequality (5) holds true for all D between the zero points. Moreover by assumption, D has to be between the zero points which proves the claim that −Lf (x) ≥ n+1 n−1 for all x ∈ N (v) ∩ N (w). Particularly, this shows that f Lf ≥ n+1 n−1 f 2 . Integrating proves the claim of the theorem for all connected graphs. For non-connected graphs, the smallest connected component has at most n 2 vertices. By the standard estimate λ n ≥ n n−1 applied to the smallest connected component, and the fact that the right hand side of the estimate is a decreasing function of n, we get λ n ≥ n/2 n/2 − 1 = n n − 2 > n + 1 n − 1 which proves the theorem for non-connected graphs.

Rigidity
We now prove that Theorem 2.1 gives the optimal bound and we characterize equality in the eigenvalue estimate which can be attained only for two different graphs (Figure 1). One of the graphs is the complete graph with only one edge removed. The other graph is surprisingly significantly different. It can be seen as two copies of a complete graph which are joined by a single vertex.
(a) (b) Figure 1: For n = 7, these are the two graphs in Theorem 3.1. The graph on the left is the complete graph K 7 with one edge removed. The graph on the right is made by two copies of the complete graph K 3 , joined by the black vertex in the middle. Proof. We first prove (i) ⇒ (ii). We first note that G is non-complete but connected by the proof of Theorem 2.1. Thus, all inequalities in the proof of Theorem 2.1 must be equalities. Let v ∼ w with N (v) ∩ N (w) = ∅. By equality in (1), all vertices within N (v) ∩ N (w) must be adjacent. By equality in (2), all vertices of N (v) ∩ N (w) must have degree n − 1. By equality in (3), we obtain By equality in (4), we obtain d(v) = d(w) = D. Finally by equality in (5), we see that We first show that if D = n − 2, then the complement graph is a single edge. If d(v) = d(w) = D = n − 2, then we get |N (v) ∩ N (w)| = n − 2. Since all vertices within N (v) ∩ N (w) are adjacent, we see that the only missing edge is the one from v to w which shows that the complement graph is a single edge. Now we assume D = n−1 2 . Then, A = |N (v) ∩ N (w)| = 1 and we can write N (v) ∩ N (w) = {x}. We recall that d(x) = n − 1. We now specify the parts of the bipartite graph which we want to be the complement graph. One part is and similarly, P w := {w} ∪ N (w) \ {x}. Let v ∈ P v and w ∈ P w . Then d(v, w) = d(w, v) = 2 as x is adjacent to every other vertex. By applying the above arguments to the pair (v, w), we see that and similarly, d( v) = n−1 2 . By a counting argument, this shows that every v ∈ P v is adjacent to every vertex not belonging to P w . An analogous statements holds for all w ∈ P w . Putting everything together, we see that the complement graph of G is precisely the complete bipartite graph with the parts P v and P w . This finishes the case distinction and thus, the proof of the implication (i) ⇒ (ii) is complete.
We finally prove (ii) ⇒ (i). We start with the case that the complement graph is the complete bipartite graph. Let the parts be P and Q. Then, φ := 1 P − 1 Q is eigenfunction to the eigenvalue 2 n−1 and every function orthogonal to φ and 1 is eigenfunction to the eigenvalue n+1 n−1 . We end with the case that the complement graph is a single edge (v, w). Then, φ = 1 v −1 w is eigenfunction to eigenvalue 1, and ψ = −2 + (n + 1)(1 v + 1 w ) is eigenfunction to eigenvalue n+1 n−1 . Every function orthogonal to φ, ψ and 1 is eigenfunction to the eigenvalue n n+1 . This finishes the proof of (ii) ⇒ (i) and thus, the proof of the theorem is complete.
Remark. In the second equality case in Theorem 3.1, for n > 3, the eigenvalue λ n has multiplicity larger than 1. With the notation of the proof of that theorem, we can take any vertex v ′ ∈ P v and any vertex w ′ ∈ P w and a function that is 1 at v ′ and w ′ , −1 at their single joint neighbor z, and 0 everywhere else. For n = 5, that is when we have two triangles sharing a single vertex z. We can also take a function that is 0 at z and assumes the values ±1 on the two other vertices of each of the two triangles, to produce other eigenfunctions with eigenvalue 3 2 .