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Spectral gap of the largest eigenvalue of the normalized graph Laplacian

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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 $\frac{n-1}2$.
arXiv:1910.14402v1 [math.SP] 31 Oct 2019
Spectral gap of the largest eigenvalue of the normalized graph
urgen JostRaffaella Mulas Florentin M¨unch
November 1, 2019
We prove that the maximal eigenvalue of the normalized graph Laplacian of a graph with nvertices
is at least n+1
n1provided 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 n1
1 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 λnn
n1which is only attained
if the graph is a complete graph [Chu97]. Here, λndenotes the largest eigenvalue and nis the number of
Naturally, the question arises what is the optimal a-priori estimate for λnfor non-complete graphs. In
this note, we show that for all non-complete graphs, one has
λnn+ 1
see Theorem 2.1. Equality is attained for each nby a complete graph with a single edge removed. For odd
n, it is also attained when two complete graphs of size n+1
2share a single vertex. These are the only cases
where equality is attained, see Theorem 3.1.
2 Eigenvalue estimate
A graph G= (V, E) consists of a finite non-empty vertex set Vand a symmetric edge relation EV×V
containing no diagonal elements (v, v). We write vwfor (v, w)E. Let G= (V, E) be a graph with n
vertices. The vertex degree is denoted by d(v) := |N(v)|where N(v) := {wV:wv}. We also denote
Nk(v) := {wV:d(v, w) = k}where d(v, w) := inf{n:v=x0... xn=w}is the combinatorial graph
distance. We say Gis connected if d(v, w)<for all v, w V. We write C(V) = RVand we denote the
positive semidefinite normalized Laplacian by L:C(V)C(V); it is given by
Lf(x) := 1
MPI MiS Leipzig,
MPI MiS Leipzig,
MPI MiS Leipzig,
We will always assume that d(x)1 for all xVas the Laplacian is not well defined otherwise. The inner
product is given by
hf, gi:= X
The operator Lis self-adjoint w.r.t. this inner product and the eigenvalues of Lare
0 = λ1λ2...λn.
In this paper, we are interested in estimating the largest eigenvalue λnwhich, by the min-max principle, can
be written as
λn= sup
hLf, f i
hf, f i.
It is well known that n
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 λnfor all non-
complete graphs.
Theorem 2.1. Let G= (V, E)be a non-complete graph with nvertices. Then,
λnn+ 1
Proof. We first assume that Gis connected. Since Gis not complete, there exists a vertex vwith d(v)n2.
As the graph is connected, we have N2(v)6=. Let wN2(v). Then, d(w)n2 as vand ware not
adjacent. Moreover, N(v)N(w)6=.
We write A:= |N(v)N(w)|. We aim to find a function fwith hf, Lf i ≥ n+1
n1hf, f i. To do so, it is
convenient to choose fin such a way that Lf =n+1
n1fin vand w. Particularly, let f:VRbe given by
f(x) :=
1 : xN(v)N(w),
0 : otherwise.
We observe
d(v)Lf(v) = d(v)f(v) + A=n+ 1
and thus, Lf (v) = n+1
n1f(v). Similarly, Lf(w) = n+1
n1f(w). We now claim that Lf (x)n+1
n1for all
xN(v)N(w). We observe A1(d(v) + d(w) + 2 n) where denotes the maximum, and we calculate
Lf(x) = d(x)− |N(x)N(v)N(w)|+f(v) + f(w)
d(x)1 + 1A+f(v) + f(w)
As f(v) + f(w)A, we can use d(x)n1 and continue
1A+f(v) + f(w)
Since d(v)n2 and d(w)n2, we see that the term in brackets is positive and thus,
n1[1 (d(v) + d(w) + 2 n)] 1
We write D:= (d(v) + d(w))/2, and by the harmonic-arithmetic mean estimate, we have 1
and thus,
n1[1 (2D+ 2 n)] 1
We aim to show that the latter term is at least 1
n1which, by multiplying with D(n1) and subtracting
D, is equivalent to
[1 (2D+ 2 n)] (n1D)D0.(5)
If Dn1
2, then the maximum equals 1 and the inequality follows immediately. If Dn1
2, then we
can discard the “1”, and so the left hand side becomes a concave quadratic polynomial in Dwith its zero
points in D=n2 and D=n1
2. Thus, the inequality (5) holds true for all Dbetween the zero points.
Moreover by assumption, Dhas to be between the zero points which proves the claim that Lf(x)n+1
for all xN(v)N(w). Particularly, this shows that f Lf n+1
n1f2. Integrating proves the claim of the
theorem for all connected graphs. For non-connected graphs, the smallest connected component has at most
2vertices. By the standard estimate λnn
n1applied to the smallest connected component, and the fact
that the right hand side of the estimate is a decreasing function of n, we get
n2>n+ 1
which proves the theorem for non-connected graphs.
3 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
K7with one edge removed. The graph on the right is made by two copies of the complete graph K3, joined
by the black vertex in the middle.
Theorem 3.1. Let G= (V, E)be a graph with nvertices. T.f.a.e.:
(i) λn=n+1
(ii) The complement graph of G, after removing isolated vertices, is a single edge or a complete bipartite
graph with both parts of size n1
Proof. We first prove (i)(ii). We first note that Gis 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 v6∼ wwith
N(v)N(w)6=. 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 n1. By equality in (3), we obtain
|N(v)N(w)|= (d(v) + d(w) + 2 n)1.
By equality in (4), we obtain d(v) = d(w) = D. Finally by equality in (5), we see that
2, n 2.
We first show that if D=n2, then the complement graph is a single edge. If d(v) = d(w) = D=n2,
then we get |N(v)N(w)|=n2. Since all vertices within N(v)N(w) are adjacent, we see that the only
missing edge is the one from vto wwhich shows that the complement graph is a single edge.
Now we assume D=n1
2. Then, A=|N(v)N(w)|= 1 and we can write N(v)N(w) = {x}. We recall
that d(x) = n1. We now specify the parts of the bipartite graph which we want to be the complement
graph. One part is
Pv:= {v} ∪ N(v)\ {x}
and similarly, Pw:= {w} ∪ N(w)\ {x}. Let evPvand ewPw. Then d(v, ew) = d(w, ev) = 2 as xis adjacent
to every other vertex. By applying the above arguments to the pair (v, ew), we see that N(v)N(ew) = {x}.
Particularly, ev6∼ ew. Moreover, we have d(ew) = d(v) = n1
2and similarly, d(ev) = n1
2. By a counting
argument, this shows that every evPvis adjacent to every vertex not belonging to Pw. An analogous
statements holds for all ewPw. Putting everything together, we see that the complement graph of Gis
precisely the complete bipartite graph with the parts Pvand Pw. 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 Pand Q. Then, φ:= 1P1Qis eigenfunction to the eigenvalue 2
n1and every
function orthogonal to φand 1 is eigenfunction to the eigenvalue n+1
We end with the case that the complement graph is a single edge (v , w). Then, φ= 1v1wis eigenfunction
to eigenvalue 1, and ψ=2 + (n+ 1)(1v+ 1w) is eigenfunction to eigenvalue n+1
n1. 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 λnhas multiplicity larger
than 1. With the notation of the proof of that theorem, we can take any vertex vPvand any vertex
wPwand a function that is 1 at vand 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 zand assumes the values ±1 on the two other vertices of each of the two triangles, to produce other
eigenfunctions with eigenvalue 3
[Chu97] Fan Chung. Spectral graph theory. Number 92. American Mathematical Soc., 1997.
[LGS14] Jianxi Li, Ji-Ming Guo, and Wai Chee Shiu. Bounds on normalized Laplacian eigenvalues of graphs.
Journal of Inequalities and Applications, 2014(1):316, 2014.
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Full-text available
Let G be a simple connected graph of order n, where n≥2. Its normalized Laplacian eigenvalues are 0=λ1≤λ2≤⋯≤λn≤2. In this paper, some new upper and lower bounds on λn are obtained, respectively. Moreover, connected graphs with λ2=1 (or λn−1=1) are also characterized. MSC: 05C50, 15A48.