The least eigenvalue of graphs whose complements are unicyclic Supported by National Natural Science Foundation of China (11071002), Program for New Century Excellent Talents in University (NCET-10-0001), Key Project of Chinese Ministry of Education (210091), Specialized Research Fund for the Doctoral Program of Higher Education (20103401110002), Science and Technological Fund of Anhui Province for Outstanding Youth (10040606Y33), Scientific Research Fund for Fostering Distinguished Young Scholars of Anhui University(KJJQ1001), Academic Innovation Team of Anhui University Project (KJTD001B).

# The least eigenvalue of graphs whose complements are unicyclic††thanks: Supported by National Natural Science Foundation of China (11071002), Program for New Century Excellent Talents in University (NCET-10-0001), Key Project of Chinese Ministry of Education (210091), Specialized Research Fund for the Doctoral Program of Higher Education (20103401110002), Science and Technological Fund of Anhui Province for Outstanding Youth (10040606Y33), Scientific Research Fund for Fostering Distinguished Young Scholars of Anhui University(KJJQ1001), Academic Innovation Team of Anhui University Project (KJTD001B).

Yi Wang, Yi-Zheng Fan , Xiao-Xin Li, Fei-Fei Zhang
School of Mathematical Sciences, Anhui University, Hefei 230039, P. R. China
Department of Mathematics and Computer Sciences, Chizhou University, Chizhou 247000, P.R. China
Corresponding author. E-mail addresses: wangy@ahu.edu.cn (Y. Wang), fanyz@ahu.edu.cn (Y.-Z. Fan), starlittlestar@sina.com (X.-X. Li), zhangfeifei2403@126.com (F.-F. Zhang).

Abstract: A graph in a certain graph class is called minimizing if the least eigenvalue of the adjacency matrix of the graph attains the minimum among all graphs in that class. Bell et al. have characterized the minimizing graphs in the class of connected graphs of order and size , whose complements are either disconnected or contain a clique of order at least . In this paper we discuss the minimizing graphs of a special class of graphs of order whose complements are connected and contains exactly one cycle (namely the the class of graphs whose complements are unicyclic), and characterize the unique minimizing graph in when .

Key words: Unicyclic graph; complement; adjacency matrix; least eigenvalue

2010 Mathematics Subject Classification:  05C50, 05D05, 15A18

## 1 Introduction

Let be a simple graph with vertex set and edge set . The adjacency matrix of is defined to be a matrix of order , where if is adjacent to , and otherwise. Since is real and symmetric, its eigenvalues are real and can be arranged as: . We simply call the eigenvalues of as the eigenvalues of . The eigenvalue is exactly the spectral radius of ; and there are many results in literatures concerning this eigenvalue; see e.g. [1, 4, 5].

The least eigenvalue is now denoted by , and the corresponding eigenvectors are called the first eigenvectors of . Relative to the spectral radius, the least eigenvalue has received less attention. In the past the main work on the least eigenvalue of a graph is focused on its bounds; see e.g. [6, 9]. Recently, the problem of minimizing the least eigenvalues of graphs subject to graph parameters has received much more attention, since two papers of Bell et al. [2, 3] and one paper of our group [7] appeared in the same issue of the journal Linear Algebra and Its Applications. Ye and Fan [15] discuss the connectivity and the least eigenvalue of a graph. Liu et al. [10] discuss the least eigenvalues of unicyclic graphs with given number of pendant vertices. Petrović et al. [11] discuss the least eigenvalues of bicyclic graphs. Wang et al. [13, 14] discuss the least eigenvalue and the number of cut vertices of a graph. Tan and Fan [12] discuss the least eigenvalue and the vertex/edge independence number, the vertex/edge cover number of a graph.

For convenience, a graph is called minimizing in a certain graph class if its least eigenvalue attains the minimum among all graphs in the class. Let denote the class of connected graphs of order and size . Bell et al. [2] have characterized the structure of the minimizing graphs in ; see Theorem 1.1.

###### Theorem 1.1

[2] Let be a minimizing graph in . Then is either

(i) a bipartite graph, or

(ii) a join of two nested split graphs (not both totally disconnected).

We find that the complements of the minimizing graphs in are either disconnected or contain a clique of order at least . This motivates us to discuss the least eigenvalue of graphs whose complements are connected and contain clique of small size. In a recent work [8] we characterize the unique minimizing graph in the class of graphs of order whose complements are trees.

In this paper, we continue this work on the complements of unicyclic graphs, and determine the unique minimizing graph in for , where denotes the class of the complements of connected unicyclic graphs of order . It is easily seen that . However the minimizing graph in does not hold the conditions in Theorem 1.1.

## 2 Preliminaries

We begin with some definitions. Given a graph of order , a vector is called to be defined on , if there is a 1-1 map from to the entries of ; simply written for each . If is an eigenvector of , then it is naturally defined on , i.e. is the entry of corresponding to the vertex . One can find that

 XTAX=2∑uv∈E(G)XuXv,

and is an eigenvalue of corresponding to the eigenvector if and only if and

 λXv=∑u∈NG(v)Xu,~{} for each vertex ~{}v∈V(G).

where denotes the neighborhood of in the graph . The equation (2.2) is called -eigenequation of . In addition, for an arbitrary unit vector ,

 λmin(G)≤XTA(G)X,

with equality if and only if is a first eigenvector of .

In this paper all unicyclic graphs are assumed to be connected. Denote by the set of unicyclic graphs of order , and denote , where denotes the complement of the graph . Note that , where denote the all-ones matrix and the identity matrix both of suitable sizes, respectively. So for any vector ,

 XTA(Gc)X=XT(J−I)X−XTA(G)X.

A star of order , denoted by , is a tree of order with pendant edges attached to a fixed vertex. The vertex of degree in is called the center of . A cycle and a complete graph both of order are denoted by respectively. Denote by the graph obtained from by adding a new edge between two pendant vertices. Next, we introduce two special unicyclic graphs denoted by and respectively, where is obtained from two disjoint graphs and by adding a new edge between one pendant vertex of and one pendant vertex of , is obtained from two disjoint graphs and by adding a new edge between one pendant vertex of and one vertex of ; see Fig. 2.1.

Fig. 2.1. The graphs (left side) and (right side)

For a graph containing at least one edge, then with equality if and only if is a complete graph or a union of disjoint complete graphs at least one component of which is nontrivial (or contains more than one vertices). So, for a unicyclic graph other than , . In addition, if is disconnected, then contains a complete multipartite graph as a spanning subgraph, which implies is or . When , consists of an isolated vertex and a connected non-complete subgraph of order .

At the end of this section, we will discuss the least eigenvalues of and . Let be a first eigenvector of the graph with some vertices labeled in Fig. 2.1. By eigenequations (2.2), as , all the pendant vertices attached at have the same value as given by , say . Similarly all the pendant vertices attached at have the same value as , say ; two vertices of degree on the triangle have the same value as , say . Write for and simply. Then by eigenequations (2.2) on for , we have

 ⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩λ1X1=(p−2)X1+X3+X4+X5+2X6+(q−3)X7λ1X2=X4+X5+2X6+(q−3)X7λ1X3=(p−1)X1+X5+2X6+(q−3)X7λ1X4=(p−1)X1+X2+2X6+(q−3)X7λ1X5=(p−1)X1+X2+X3λ1X6=(p−1)X1+X2+X3+X4+(q−3)X7λ1X7=(p−1)X1+X2+X3+X4+2X6+(q−3)X7

Transform (2.5) into a matrix equality , where and the matrix of order is easily to seen. We have

 f(λ;p,q):=det(B−λI)=(−8+2p+2q)+(13−11p−7q+4pq)λ+(20−6q−4qp)λ2+(−1+11p+7q−7pq)λ3+(−20+12p+12q−2pq)λ4+(−16+6p+6q)λ5+(−6+p+q)λ6−λ7.

It is easily found that is the least root of the polynomial .

Let be a first eigenvector of the graph with some vertices labeled in Fig. 2.1. By a similar discussion, all the pendant vertices attached at have the same values given by , say . Two vertices of degree 2 on the triangle have the same values, say . Write for and simply. Then by eigenequations on for ,

 ⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩λ′1Y1=(p−2)Y1+Y3+Y4+2Y5λ′1Y2=Y4+2Y5λ′1Y3=(p−1)Y1+Y4+2Y5λ′1Y4=(p−1)Y1+Y2λ′1Y5=(p−1)Y1+Y2+Y3

It is easily found that is the least root of the following polynomial:

 g(λ;p):=(−4+2p)+(3−5p)λ+(6−p)λ2+(1+4p)λ3+(−2+p)λ4−λ5.
###### Lemma 2.1

If , .

Proof: Write , simply. By above discussion, (respectively, ) is the least root of (respectively, ). Denote by

 ¯g(λ;n−4):=(λ+1)2g(λ;n−4).

Since , is also the least root of . From (2.8), , and consequently if . Furthermore, when , , which implies . Obverse that when ,

 ¯g(λ;n−4)−f(λ;n−5,3)=(−6+n)λ(1+λ)(−2+5λ+2λ2)>0.

In particular, , which implies . The result follows.

###### Lemma 2.2

Given a positive integer , for any positive integers such that and ,

 λmin(U(p,q)c)≥λmin(U(⌈(n−2)/2⌉,⌊(n−2)/2⌋)c),

with equality if and only if and .

Proof: Write simply. By (2.6), we have

 f(λ;p,q)−f(λ;p+1,q−1) =−λ(2+λ)(−1+2λ)[(p−q+1)(2+λ)+2], f(λ;p,q)−f(λ;p−1,q+1) =λ(2+λ)(−1+2λ)[(p−q−1)(2+λ)+2].

In addition, , which implies .

If , for we have

 f(λ;p,q)−f(λ;p+1,q−1)>0.

In particular, , which implies

 λmin(U(p+1,q−1)c)<λ1=λmin(U(p,q)c).

If , by(2.6), we have

 f(−3;p,q)=241−19p+23q−21pq=241−19(p−q)+(4−21p)q<0,

which implies . Observe that when . In particular, , which implies

 λmin(U(p−1,q+1)c)<λmin(U(p,q)c).

To complete the proof, we need to prove when . In this case, , , and

 f(λ;p,q)−f(λ;p−1,q+1)=λ(2+λ)(−1+2λ)(4+λ).

So it is enough to prove or

 f(−4;n2,n2−2)=2376+582n−36n2<0.

If the above inequality holds, and hence the result follows.

## 3 Main results

By rearranging the edges of graphs, we first give an maximization of the quadratic form among all trees or all unicyclic graphs of order , where is a non-negative or non-positive real vector defined on .

###### Lemma 3.1

Let be a tree of order , and let be a non-negative or non-positive real vector defined on whose entries are arranged as with respect to their moduli. Then

 ∑uv∈E(T)XuXv≤n∑i=2X1Xi=∑uv∈E(K1,n−1)XuXv,

where is defined on such that the center has value . If, in addition, is positive or negative, and , then the above equality holds only if .

Proof: We may assume is non-negative; otherwise we consider . Let be a vertex with value given by . If there exists a vertex not adjacent of , letting be the neighbor of on a path of connecting and , and deleting the edge and adding a new edge , we will arrive at a new graph (tree) , which holds

 ∑uv∈E(T)XuXv≤∑uv∈E(T′)XuXv.

Repeating the process on the tree for the non-neighbors of , and so on, we at last arrive at a star with as its center, and

 ∑uv∈E(T)XuXv≤∑uv∈E(K1,n−1)XuXv=n∑i=2X1Xi.

If is positive, , and is not adjacent to all other vertices in , then the inequality (3.1), and hence (3.2), cannot hold as an equality. The result follows.

###### Lemma 3.2

Let be a unicyclic graph of order , and let be a non-negative or non-positive real vector defined on whose entries are arranged as with respect to their moduli. Then

 ∑uv∈E(U)XuXv≤n∑i=2X1Xi+X2X3=∑uv∈E(S3n)XuXv,

where is defined on such that the vertex with degree has value , and the other two vertices on the triangle have values respectively. If, in addition, is positive or negative, and , then the above equality holds only if .

Proof: We may assume is non-negative; otherwise we consider . Let be a vertex with value given by . By a similar discuss to the proof in Lemma 3.1, we have a graph of order , for which the vertex is adjacent to all other vertices, and

 ∑uv∈E(U)XuXv≤∑uv∈E(U′)XuXv=n∑i=2X1Xi+Xu′Xv′,

where is an edge of not incident to . Surely,

 Xu′Xv′≤X2X3.

So,

 ∑uv∈E(U)XuXv≤n∑i=2X1Xi+X2X3=∑uv∈E(S3n)XuXv.

If is positive, and , then the equality (3.5) holds only if (3.3) holds, which implies is adjacent to all other vertices and consequently . The result follows.

Next we give a sign property of the first eigenvector of graphs in for . Note that the first eigenvector of a connected graph with order at least contains both positive entries and negative entries.

###### Lemma 3.3

Let be a unicyclic graph of order , and let be a first eigenvector of . Then contains at least two positive entries and two negative entries.

Proof: Denote by the unique cycle in of length , and write simply. When , we know is connected or , the latter of which consists of an isolated vertex and a connected non-complete subgraph of order . So contains at least one positive entry and one negative entry.

Assume to the contrary that contains exactly one positive entry corresponding the vertex . For an arbitrary vertex , considering the eigenequation (2.2) on for the graph , we have

 0≤λmin(Uc)Xu=∑v∈NUc(u)Xv≤0.

So

 Xu=0,Xv=0,~{}for each~{}u∈NU(w),v∈NUc(u).

If a vertex , then , and by (3.6),

 λXr=∑v∈NUc(r)Xv=∑v∈NU(u)∖(NU(r)∪{r})Xv.

Similarly, if the vertex , then

 λXr=∑v∈NUc(r)Xv=∑v∈NU(u)∖NU(r)Xv.

We divide the discussion into two cases.

Case 1: . As is a unicycle graph of order , there exists a vertex such that . Let . Note that as so that , and consequently for by (3.6). By (3.7) we have

 λXw′=∑v∈NU(u)∖(NU(w′)∪{w′})Xv=∑v∈NU(u)∖{w′}Xv.

By (3.6) and (3.7), we also have

 λXw=∑v∈NU(u)∖(NU(w)∪{w})Xv=∑v∈NU(u)∖{w}Xv.

So

 (1+λ)Xw=∑v∈NU(u)Xv=(1+λ)Xw′,

which implies as , a contradiction.

Case 2: . We assert that has exactly one neighbor in . Otherwise, let be two vertices adjacent to in . As , . Applying (3.6) on the vertex , we have for each , and hence for each . Also by (3.6) and for each . So contains exactly one nonzero entry, i.e. ; a contradiction.

Let be the unique neighbor of in the graph . If there exists a neighbor of in other than , say , not lying on the cycle, then , and by (3.7),

 λXw′=∑v∈NU(u)∖{w′}Xv.

Combining this equality with (3.9), we have , a contradiction. So the vertex lies on the cycle and has exactly two neighbors except .

Now let be the neighbors of on the cycle of . By (3.6) and (3.7), we have

 λXw′=∑v∈NU(u)∖{w′,w′′}Xv=λXw′′,

which implies . As the graph has more than vertices, one of , say , has a neighbor not on the cycle which is also not adjacent to . By (3.6), , and by (3.8),

 0=λX¯w′=∑v∈NU(u)∖{w′}Xv.

Combining (3.10) with (3.11), we have , and hence as . So, we have for each . Also by (3.6) and for each . Therefore contains exactly one nonzero entry, i.e. ; a contradiction.

By above discussion, we get contains at least two positive entries. Similarly if considering , we also get contains at least two negative entries. The result follows.

We now arrive at the main result of this paper.

###### Theorem 3.4

Let be a unicyclic graph of order . Then

 λmin(Uc)≥λmin(U(⌈(n−2)/2⌉,⌊(n−2)/2⌋)c),

with equality if and only if .

Proof: Let be the first eigenvector of with unit length. Denote , , both being nonempty and containing at least elements by Lemma 3.3. Denote by (respectively, ) the subgraph of induced by (respectively, ), the set of edges between and in . Since is connected, . Obviously,

 ∑vv′∈E(U)XvXv′=∑vv′∈E(U+)XvXv′+∑vv′∈E(U−)XvXv′+∑vv′∈E′XvXv′.

First assume . The cycle of may contain the edges of , or is contained in one of . Without loss of generality, we assume that the cycle of is not contained in ; otherwise we consider the vector instead. Let be a graph obtained from by possibly adding some edges within and , such that the subgraph of induced by , denoted by , is a tree, and the subgraph of induced by , denoted by , is a unicyclic graph.

In the tree , choose a vertex, say , with maximum modulus among all vertices of . By Lemma 3.1, we will have a star, say centered at , where , which holds

 ∑vv′∈E(U+)XvXv′≤∑vv′∈E(U∗+)XvXv′≤∑vv′∈E(K1,p)XvXv′.

In the unicyclic graph , choosing a vertex, say , with maximum modulus. By Lemma 3.2, we have a unicyclic graph , where and the vertex joins all other vertices of , which holds

 ∑vv′∈E(U−)XvXv′≤∑vv′∈E(U∗−)XvXv′≤∑vv′∈E(S3q+1)XvXv′.

Let be the vertices of with minimum modulus among all vertices of , respectively. Then

 ∑vv′∈E′XvXv′≤Xu′Xw′,

Now by (3.12-3.15), we have

 ∑vv′∈E(U)XvXv′≤∑vv′∈E(K1,p)XvXv′+∑vv′∈E(S3q+1)XvXv′+Xu′Xw′

Since , the vertex can be chosen within the pendent vertices of by Lemma 3.1. If , can be chosen within the pendent vertices of by Lemma 3.2, then from (3.16) we have

 12XTA(U)X=∑vv′∈E(U)XvXv′≤∑vv′∈E(U(p,q))XvXv′=12XTA(U(p,q))X.

and consequently

 λmin(Uc)=XTA(Uc)X=XT(J−I)X−XTA(U)X≥XT(J−I)X−XTA(U(p,q))X=XTA(U(p,q)c)X≥λmin(U(p,q)c).

If , that is, , by a similar discussion, we have . By Lemma 2.1, .

Next we consider the case when . In this case the cycle of cannot lies in . We form a graph from possibly by adding some edges within and , such that the subgraph of induced by is a unicyclic graph, and the subgraph of induced by is exactly . Also similar to the discussion for (3.13-3.18), we have . By Lemma 2.2 and the above discussion,

 λmin(Uc)≥λmin(U(p,q)c)≥λmin(U(⌈(n−2)/2⌉,⌊(n−2)/2⌋)c).

Finally we prove the necessity for the equality in the theorem holding. Assume the equalities in (3.19) hold. Then , and consequently only the case of occurs. Note that is also a first eigenvector of . Let have some vertices labeled as in Fig. 2.1, where .

Assertion 1: contains no zero entries; the vertices and are respectively the unique ones in with maximum and minimum modulus, and are respectively the unique ones in with maximum and minimum modulus. As is a first eigenvector of , from the assumption at the beginning, for and for . By (2.5),

 λ1(X4−X7)=−X3−X4<0, λ1(X6−X7)=−2X6, λ1(X5−X6)=−X4−(q−3)X7,

which implies that . If , deleting the edges , and adding two new edges , we get a graph with same quadratic as both associated with , which implies , a contradiction by Lemma 2.2. So . Also by (2.5),

 λ1(X1−X2)>0, λ1(X3−X1)=X1−X3−X4>X1−X3,

which implies .

Assertion 2: