On graphs with some normalized Laplacian eigenvalue of extremal multiplicity
aa r X i v : . [ m a t h . C O ] J u l On graphs with some normalized Laplacian eigenvalue ofextremal multiplicity
Fenglei Tian ∗ , Junqing Cai, Zuosong Liang, Xuntuan Su School of Management, Qufu Normal University, Rizhao, China.
Abstract:
Let G be a connected simple graph on n vertices. Let L ( G ) be the normalizedLaplacian matrix of G and ρ n − ( G ) be the second least eigenvalue of L ( G ). Denote by ν ( G ) the independence number of G . Recently, the paper [Characterization of graphs withsome normalized Laplacian eigenvalue of multiplicity n −
3, arXiv:1912.13227] discussed thegraphs with some normalized Laplacian eigenvalue of multiplicity n −
3. However, there is oneremaining case (graphs with ρ n − ( G ) = 1 and ν ( G ) = 2) not considered. In this paper, wefocus on cographs and graphs with diameter 3 to investigate the graphs with some normalizedLaplacian eigenvalue of multiplicity n − Keywords:
Normalized Laplacian eigenvalues; Normalized Laplacian matrix; Eigenvalue mul-tiplicity
AMS classification:
Throughout, only connected and simple graphs are considered here. Let G = ( V ( G ) , E ( G ))be a graph with vertex set V ( G ) and edge set E ( G ). Let N G ( u ) be the set of all the neighborsof the vertex u . Then d u = | N G ( u ) | is called the degree of u . For a subset S ⊂ V ( G ), S is called a set of twin points if N G ( u ) = N G ( v ) for any u, v ∈ S . By u ∼ v , we mean that u and v are adjacent. A subset S of V ( G ) is called an independent set of G , if the verticesof S induce an empty subgraph. The cardinality of the maximum independent set of G iscalled the independence number, denoted by ν ( G ). The rank of a matrix M is written as r ( M ). Let R v i be the row of M indexed by the vertex v i . Denote by G ( n, n −
3) the set of all n -vertex ( n ≥
5) connected graphs with some normalized Laplacian eigenvalue of multiplicity n −
3. Let A ( G ) and L ( G ) = D ( G ) − A ( G ) be the adjacency matrix and the Laplacian matrixof graph G , respectively. Then the normalized Laplacian matrix L ( G ) = [ l uv ] of graph G isdefined as L ( G ) = D − / ( G ) L ( G ) D − / ( G ) = I − D − / ( G ) A ( G ) D − / ( G ) , ∗ Corresponding author. E-mail address: tfl[email protected]. Supported by ” the Natural Science Founda-tion of Shandong Province (No. ZR2019BA016) ”. l uv = , if u = v ; − / √ d u d v , if u ∼ v ;0 , otherwise . For brevity, the normalized Laplacian eigenvalues are written as L -eigenvalues. It is wellknown that the least L -eigenvalue of a connected graph is 0 with multiplicity 1 (see [9]).Then let the L -eigenvalues of a graph G be ρ ( G ) ≥ ρ ( G ) ≥ · · · ≥ ρ n − ( G ) > ρ n ( G ) = 0 . The normalized Laplacian spectrum of graphs has been studied intensively (see [1–8]),because it reveals some structural properties and some relevant dynamical aspects (such asrandom walk) of graphs [9]. Recently, graphs with some eigenvalue of large multiplicity haveattracted much attention (see [10–12]). However, there are few results on the normalizedLaplacian eigenvalues. Van Dam and Omidi [1] determined the graphs with some normalizedLaplacian eigenvalue of multiplicity n − n −
2, respectively. Tian et al. [13] characterizedtwo families of graphs belonging to G ( n, n − ρ n − ( G ) = 1 and graphs with ρ n − ( G ) = 1 and ν ( G ) = 2, leaving the last case of ρ n − ( G ) = 1 and ν ( G ) = 2 not considered.Hence, in this paper we further discuss the remaining case and obtain the following conclusion.For convenience, denote by G ( n, n −
3) the graphs of G ( n, n −
3) with ρ n − ( G ) = 1 and ν ( G ) = 2. If a graph contains no induced path P , then it is called a cograph. Theorem 1.1.
Let G be a graph of order n ≥ . Then(i) G ∈ G ( n, n − with diameter 3 if and only if G = G (see Fig. 1),(ii) G ∈ G ( n, n − and G is a cograph if and only if G = G (see Fig. 1). G G K a+ K b+ n- K v v v vv v a+ n Fig. 1: The graphs G ( a + b + 4 = n ) and G .Before showing the proof of Theorem 1.1, we first introduce some notations and lemmasin the next section. For a symmetric real matrix H of order n whose columns and rows are indexed by X = { , , · · · , n } , let { X , X , · · · , X t } be a partition of X . According to the partition of X , we2rite the block form of H as H = H · · · H t ... . . . ... H t · · · H tt , where H ji is the transpose of H ij . Denote by q ij the average row sum of H ij , then the matrix Q = ( q ij ) is called the quotient matrix of H . If the row sum of H ij is constant, then thepartition of X is equitable (see [14]). Lemma 2.1. [14]
Suppose that H is a real symmetric matrix with an equitable partition. Let Q be the corresponding quotient matrix of H . Then, each eigenvalue of Q is an eigenvalue of H . Lemma 2.2. [5, 6]
Let G be a graph with order n . Denote by { v , . . . , v p } a set of twin pointsof G , then is an L -eigenvalue of G with multiplicity at least p − . Lemma 2.3. [6]
Let G be a graph with n vertices. Let K = { v , . . . , v q } be a clique in G such that N G ( v i ) − K = N G ( v j ) − K (1 ≤ i, j ≤ q ) , then d vi is an L -eigenvalue of G withmultiplicity at least q − . Lemma 2.4. [13]
Let G ∈ G ( n, n − and θ be the L -eigenvalue of G with multiplicity n − .If ρ n − ( G ) = 1 , then θ = 1 . Lemma 2.5.
Let G ∈ G ( n, n − with an induced path P = v v v v . If there is a vertex u (resp., u ) such that u v v v (resp., v u v v ) is also an induced path, then d v = d u (resp., d v = d u ). Proof.
Let θ be the L -eigenvalue of G with multiplicity n −
3, then r ( L ( G ) − θI ) = 3. Since G ∈ G ( n, n − θ = 1. Denote by M the principal submatrixof L ( G ) − θI indexed by the vertices { v , v , v , v } , then M = − θ − √ d v d v − √ d v d v − θ − √ d v d v − √ d v d v − θ − √ d v d v − √ d v d v − θ . It is clear that the last three rows of M are linearly independent, which yields that the rows R v , R v , R v of L ( G ) − θI are linearly independent, and then R v can be written as a linearcombination of R v , R v , R v . Let R v = aR v + bR v + cR v , (1)3hen − a √ d v d v = 1 − θa (1 − θ ) − b √ d v d v = − √ d v d v − b √ d v d v + c (1 − θ ) = 0 − a √ d v d v + b (1 − θ ) − c √ d v d v = 0 . (2)From the first three equations of (2), we obtain that a = − (1 − θ ) p d v d v b = − (1 − θ ) d v p d v d v + q d v d v c = − θ ) √ d v d v − (1 − θ ) d v √ d v √ d v . (3)Taking (3) into the last equation of (2), we deduce that(1 − θ ) d v d v d v d v − ( d v d v + d v d v + d v d v )(1 − θ ) + 1 = 0 . (4)Analogously, for the induced path u v v v and v u v v , we can respectively get that(1 − θ ) d u d v d v d v − ( d u d v + d v d v + d u d v )(1 − θ ) + 1 = 0 (5)and (1 − θ ) d v d u d v d v − ( d v d u + d v d v + d v d v )(1 − θ ) + 1 = 0 . (6)Combining (4) and (5), it follows that(1 − θ ) = ( d v + d v )( d v − d u ) d v d v d v ( d v − d u ) . If d v = d u , then (1 − θ ) = ( d v + d v ) d v d v d v , and thus from (4) we have d v d v = 0, a contradiction. Hence, d v = d u . Combining (4) and(6), it follows that (1 − θ ) = d v − d u d v d v ( d v − d u ) . If d v = d u , then (1 − θ ) = 1 d v d v , and from (4) we have d v d v = 0, a contradiction. As a result, d v = d u . The proof is completed. (cid:3) Lemma 2.6.
Let G and G be the graphs in Fig. 1. Then their spectra (eigenvalues with ultiplicity) are respectively { , − n + √ n − n − , − n −√ n − n − , ( n − n − ) n − } , { , n − n +3+ √ n − n +42 n − n − n +2) , n − n +3 −√ n − n +42 n − n − n +2) , ( n − n − ) n − } . Proof.
For graph G , it is clear that L ( G ) has an equitable partition with respect to thevertex partition V ( G ) = { v } ∪ V ( K a +1 ) ∪ V ( K b +1 ) ∪ { v n } . Note that d v = a + 1, d v = d v a +3 = n − d v n = b + 1 and a + b + 4 = n in G . Denote thequotient matrix of L ( G ) by Q , then Q = − ( a +1) √ d v d v − √ d v d v − ad v − ( b +1) d v d va +3 − ( a +1) d v d va +3 − bd va +3 − √ d va +3 d vn − ( b +1) √ d va +3 d vn = − ( a +1) √ ( a +1)( n − − √ ( a +1)( n − n − − an − − ( n − a − n − − ( a +1) n − a +2 n − − √ ( n − a − n − − ( n − a − √ ( n − a − n − . By direct calculation, the eigenvalues of Q are { , n − n − , − n ± √ n − n − } . From Lemma 2.3, we see that n − n − is an L -eigenvalue of G with multiplicity at least n − L ( G ) and Lemma 2.1 that the last unknown eigenvalue is n − ( n − n − n − − − n − √ n − n − − − n + √ n − n − n − n − . Therefore, the multiplicity of n − n − is n − G , we obtain that n − n − is an L -eigenvalue of G with multiplicityat least n −
3. Divide the vertex set V ( G ) into three parts V ( G ) = { v } ∪ { v } ∪ { V ( G ) \ { v , v }} . Accordingly, L ( G ) has an equitable partition. Note that d v = 1, d v = n − V ( G ) \ { v , v } is n −
2. Then the quotient matrix of L ( G ) can be written5s Q = − √ n − − √ n − − n √ ( n − n − − √ ( n − n −
2) 1 n − , whose eigenvalues are { , n − n + 3 ± √ n − n + 42 n − n − n + 2) } . From Lemma 2.1, the eigenvalues of Q are also the eigenvalues of L ( G ), which implies thatthe multiplicity of n − n − is n − (cid:3) Suppose that G ∈ G ( n, n − i.e., G contains some L -eigenvalue of multiplicity n − ρ n − ( G ) = 1 and ν ( G ) = 2, then the diameter of G is not larger than 3. It is clear that thecomplete graph K n do not belong to G ( n, n − G is 2 or 3, andthe proof of Theorem 1.1 is divided into Theorems 3.1 and 3.3 based on the diameter. Theorem 3.1.
Let G be a connected graph of order n ≥ . Then G ∈ G ( n, n − with diam ( G ) = 3 if and only if G is the graph G in Fig. 1. Proof.
The sufficiency part is clear from Lemma 2.6. In the following, we present thenecessity part. Suppose that G ∈ G ( n, n −
3) and θ is the L -eigenvalue of multiplicity n − θ = 1 from Lemma 2.4. Denote by P = v v v v a diametrical path of G . Assumethat U is a subset of V ( P ) and S U = { u ∈ V ( G ) \ V ( P ) : N G ( u ) ∩ V ( P ) = U } . Since theindependence number ν ( G ) = 2, we obtain that any vertex out of V ( P ) must be adjacent toat least one of V ( P ) and S { v } = S { v } = S { v } = S { v } = S { v ,v } = S { v ,v } = S { v ,v } = ∅ . Recalling that diam ( G ) = 3, we only need to discuss the vertices of S { v ,v } , S { v ,v } , S { v ,v ,v } and S { v ,v ,v } . The following claims are useful. Claim 1. If S { v ,v } = ∅ , then S { v ,v ,v } = ∅ .Suppose that S { v ,v ,v } = ∅ and v ∈ S { v ,v } , v ∈ S { v ,v ,v } . Lemma 2.5 yields that d v = d v and d v = d v . Let M be the principal submatrix of L ( G ) − θI indexed by v i (1 ≤ i ≤ M = − θ − √ d v d v − √ d v d v − √ d v d v − θ − √ d v d v − √ d v d v − √ d v d v − √ d v d v − θ − √ d v d v − √ d v d v − √ d v d v − θ − √ d v d v − √ d v d v − √ d v d v − θ ∗ − √ d v d v − √ d v d v − √ d v d v ∗ − θ . R v , R v , R v of L ( G ) − θI are linearly independent from the observationof M . Suppose the equation (1) still holds. Then applying (1) to the columns of M , we get(2) and − a √ d v d v = − √ d v d v − a √ d v d v − b √ d v d v − c √ d v d v = 0 . (7)The first equations of (2) and (7) indicate that 1 − θ = − d v , which together with (4) yieldsthat d v ( d v − d v − d v ) = d v d v ( d v − d v ) . (8)The fourth equation of (2) and the second one of (7) imply that b √ d v = b √ d v d v . Further, as b = 0 (otherwise a = c = 0 from (2), a contradiction), then we get d v = d v . (9)Bringing (9) into (8), we derive d v ( d v − d v − d v ) = d v d v ( d v − d v ), that is,( d v − d v )( d v − d v ) = d v d v . (10)Note that each vertex of V ( G ) \ V ( P ) just belongs to S { v ,v } , S { v ,v } , S { v ,v ,v } or S { v ,v ,v } .Then it is easy to know that d v > d v and d v > d v . As a result, the left side of (10) isnegative, but the right side is positive, a contradiction. Claim 2. S { v ,v } = S { v ,v } = ∅ .By symmetry, it suffices to prove S { v ,v } = ∅ . Suppose on the contrary that S { v ,v } = ∅ and v ∈ S { v ,v } . Let M be the principal submatrix of L ( G ) − θI indexed by v i (1 ≤ i ≤ M is a principal submatrix of M and the equation (8) still holds. From Claim 1, we seethat S { v ,v ,v } = ∅ , which implies that d v = d v + 1. Further, if S { v ,v } = ∅ at this moment,then d v = 1. Reconsidering (8), we obtain that 2 d v = d v , contradicting with the fact that d v ≥ d v . On the other hand, suppose S { v ,v } = ∅ and v ∈ S { v ,v } , then S { v ,v ,v } = ∅ fromClaim 1 and d v = d v + 1. For the subgraph induced by { v , v , v , v , v } , similar deductionwith (8) leads to d v ( d v − d v − d v ) = d v d v ( d v − d v ) . (11)Applying d v = d v + 1 and d v = d v + 1 to (8) and (11), we derive that d v = d v and d v = d v , which imply that d v = d v = 1, a contradiction. Therefore, S { v ,v } = ∅ .Next, we show that G must be isomorphic to G in Fig. 1. From Claim 2, the vertices of V ( G ) \ V ( P ) belong to S { v ,v ,v } or S { v ,v ,v } . First, any two vertices of S { v ,v ,v } (resp., S { v ,v ,v } ) are adjacent. Otherwise, it is easy to see that ν ( G ) ≥
3, contradicting with ν ( G ) = 2. Further, suppose u ∈ S { v ,v ,v } (resp., S { v ,v ,v } ), then d u = d v (resp., d u = d v )by Lemma 2.5, which indicates that u is adjacent to each of S { v ,v ,v } (resp., S { v ,v ,v } ).Hence, G is isomorphic to G .The proof is completed. (cid:3) Now, we discuss the case of diam ( G ) = 2. In the following, we always let P = v v v be7 diametrical path of G . Assume that U is a subset of V ( P ) and S U = { u ∈ V ( G ) \ V ( P ) : N G ( u ) ∩ V ( P ) = U } . Lemma 3.2.
Let G be a cograph with diam ( G ) = 2 and G ∈ G ( n, n − . The diametricalpath P and the notation S U are stated as above. Then(i) S { v } = S { v } = S { v } = ∅ ;(ii) If S { v ,v } = ∅ , then S { v ,v } = S { v ,v } = S { v ,v ,v } = ∅ . Proof.
Since ν ( G ) = 2, then each vertex out of V ( P ) must be adjacent to v or v .Hence, S { v } = ∅ . Note that G is a cograph (i.e., containing no induced path P ), then S { v } = S { v } = ∅ . Therefore, the vertices out of V ( G ) \ V ( P ) belong to S { v ,v } , S { v ,v } , S { v ,v } or S { v ,v ,v } . The remaining proof can be completed by the following claims. Claim 1. If S { v ,v } = ∅ , then S { v ,v } = ∅ . In other words, G contains no induced subgraphisomorphic to H (see Fig. 2).Let v ∈ S { v ,v } . Suppose for a contradiction that S { v ,v } = ∅ and v ∈ S { v ,v } . Then v ∼ v , otherwise { v , v , v , v } induce a path P , a contradiction. Thus, the vertices v i (1 ≤ i ≤
5) induce a subgraph of G isomorphic to H in Fig. 2. Denote by M the principalsubmatrix of L ( G ) − θI indexed by v i (1 ≤ i ≤ M = − θ − √ d v d v − √ d v d v − √ d v d v − √ d v d v − θ − √ d v d v − √ d v d v − √ d v d v − θ − √ d v d v − √ d v d v − √ d v d v − θ − √ d v d v − √ d v d v − √ d v d v − √ d v d v − θ . Since the following minor D of M is nonzero, D = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − √ d v d v − √ d v d v − θ − √ d v d v − θ − √ d v d v (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = − − θ p d v d v ( 1 d v + 1 d v ) = 0 , then the second, third and fourth rows of M are linearly independent, and thus the rows R v , R v , R v of L ( G ) − θI are linearly independent. As r ( L ( G ) − θI ) = 3, any row of L ( G ) − θI can be represented as a linear combination of R v i (2 ≤ i ≤ R v = aR v + bR v + cR v . (12)8pplying (12) to the columns of M , we get − a √ d v d v − c √ d v d v = 1 − θa (1 − θ ) − b √ d v d v − c √ d v d v = − √ d v d v − a √ d v d v + b (1 − θ ) = 0 − a √ d v d v + c (1 − θ ) = − √ d v d v − b √ d v d v − c √ d v d v = − √ d v d v . (13)The second and the fifth equations of (13) imply that a = 0. Then b = 0 from the thirdequation of (13), which indicates that c = q d v d v by the fifth one of (13). Taking the values of a and c into the first and the fourth ones of (13), we obtain that1 − θ = − d v = − d v , (14)which yields that d v = d v . Moreover, let R v = sR v + tR v + kR v , and we have the following equations from the columns of M , − s √ d v d v − k √ d v d v = − √ d v d v s (1 − θ ) − t √ d v d v − k √ d v d v = 0 − s √ d v d v + t (1 − θ ) = − √ d v d v − s √ d v d v + k (1 − θ ) = − √ d v d v − t √ d v d v − k √ d v d v = 1 − θ. (15)It follows from the third and the fourth equations of (15) that t p d v = k p d v , which, togetherwith (14) and the fifth one of (15), implies that t = √ d v d v d v + d v k = d v √ d v √ d v ( d v + d v ) . Bringing the value of k into the first one of (15), we have s = q d v d v − d v √ d v d v d v ( d v + d v ) . Then nowthe second one of (15) can be simplified to be d v d v + d v d v + d v d v d v + d v d v d v = d v d v d v , (16)implying that d v > d v . Thus it follows from d v = d v that d v > d v , which indicatesthat S { v ,v } = ∅ . Let v ∈ S { v ,v } , then v ∼ v (otherwise { v , v , v , v } induce P ).9rom the symmetry of v and v , applying similar discussion to the subgraph induced by { v , v , v , v , v } , one can obtain that1 − θ = − d v = − d v . (17)Combining (14) and (17), d v = d v holds. Then the equation (16) can be rewritten as d v d v + d v d v + d v d v + d v d v = d v d v . (18)Recalling that any vertex out of V ( P ) must be adjacent to v or v , we can see d v + d v > d v ,i.e., d v + d v > d v as d v = d v , contradicting with (18). As a result, if S { v ,v } = ∅ , then S { v ,v } = ∅ . H v H v v v v v v v v v H v v v K a + K b + K c + H v v v K a + K c + H v v v K a + K b + Fig. 2: The graphs H , H , H ( a + b + c + 3 = n ), H ( a + c + 3 = n ) and H ( a + b + 3 = n ). Claim 2. If S { v ,v } = ∅ , then S { v ,v ,v } = ∅ . In other words, G contains no inducedsubgraph isomorphic to H (see Fig. 2).Suppose on the contrary that S { v ,v ,v } = ∅ and v ∈ S { v ,v ,v } . Also, let v ∈ S { v ,v } .It is easy to see that v ∼ v , otherwise { v , v , v , v } induce a path P , a contradiction.Then G contains H as an induced subgraph. Since S { v ,v } = ∅ from Claim 1, then all othervertices maybe spread in S { v ,v } , S { v ,v ,v } or S { v ,v } . Case 1.
Assume that S { v ,v } = ∅ .We point out that all vertices of S { v ,v } (resp., S { v ,v } ) induce a clique of G , otherwise onecan see ν ( G ) ≥
3, a contradiction. Further, each vertex of S { v ,v } (resp., S { v ,v } ) is adjacentto each one of S { v ,v ,v } , otherwise one can easily obtain an induced path P , a contradiction.Additionally, all vertices of S { v ,v ,v } also induce a clique. If not, let u, w ∈ S { v ,v ,v } and u ≁ w , then { v , v , v , u, w } induce a subgraph isomorphic to H in Fig. 2, contradictingwith Claim 1. Let | S { v ,v } | = a ≥ | S { v ,v } | = b ≥ | S { v ,v ,v } | = c ≥
1, then G is isomorphic to H ( a + b + c + 3 = n ) in Fig. 2. The remaining proof is divided into thefollowing cases. • Suppose that a = b and a + b ≤ c in H . Then we declare that a = b = 1. Otherwise, a = b ≥
2, and then c ≥ a + b ≥
4. From Lemma 2.3, 1 + d v = 1 + d v (resp., 1 + d v ) isan L -eigenvalue with multiplicity at least 4. Noting that d v = d v = d v , then G contains noeigenvalue of multiplicity n −
3, a contradiction. Hence, a = b = 1 and c ≥
2. By Lemma 2.3again, the multiplicity of 1 + d v (resp., 1 + d v ) is at least 2. As a result, θ = 1 + 1 d v = 1 + 1 n − θ = 1 + 1 d v = 1 + 1 n − . θ = 1 + n − , then the multiplicity of 1 + n − is equal to 2. Then from the trace of L ( G ), n = ( n − n − n − n + 2 n − > n, a contradiction. If θ = 1 + n − , then the multiplicity of 1 + n − is equal to 2. Then n = ( n − n − n − n + 4 n − n + 3 > n, a contradiction. • Suppose that a = b and a + b > c in H . Then we say that c = 1. Otherwise, c ≥ a + b ≥
4. By Lemma 2.3, the multiplicity of 1 + d v (resp., 1 + d v ) as an L -eigenvalue isat least 4 (resp., at least 2). Thus, θ = 1 + d v and the multiplicity of 1 + d v is equal to 2,implying that c = 2. Note that n = a + b + c + 3 > L ( G ), n = ( n − d v ) + 2(1 + d v )= ( n − n +1 ) + 2(1 + n − )= n + n − n +1 + n − > n a contradiction. Thus, c = 1 and a = b = n − . Clearly, L ( G ) has an equitable partitionaccording to V ( G ) = { V ( K a +1 ) , V ( K b +1 ) , V ( K c +1 ) } . Let the quotient matrix of L ( G ) be Q ,then from d v = d v = n and d v = n − Q = − n − n − √ √ n ( n − − n √ n ( n − − n − − n √ n ( n − − √ √ n ( n − − n − n . By direct calculation, the eigenvalues of Q are { , n , n +2 n − n − n } . By Lemma 2.3, 1 + d v = n +2 n and 1 + d v = nn − are two distinct L -eigenvalues of G . Then from Lemma 2.1, we see that G has 5 distinct L -eigenvalues, contradicting with G ∈ G ( n, n − • Suppose that a = b and a > b without loss of generality. From b ≥ c ≥
1, then a ≥ d v ( with multiplicity at least 2), 1 + d v and 1 + d v are three distinct L -eigenvaluesof G by Lemma 2.3. Thus, one can derive that θ = 1 + d v , which yields that c = b = 1.Therefore, the trace of L ( G ) is n = ( n − d v ) + (1 + d v ) + (1 + d v )= ( n − n − ) + (1 + n − ) + (1 + )= n + n − + > n, a contradiction. From above three subcases for S { v ,v } = ∅ , we can always obtain contradic-tions. Case 2.
Let S { v ,v } = ∅ . 11n this case, there are only S { v ,v } and S { v ,v ,v } nonempty. Similar as Case 1, all verticesof S { v ,v } (resp., S { v ,v ,v } ) induce a clique of G and each vertex of S { v ,v } is adjacent toeach one of S { v ,v ,v } . Let | S { v ,v } | = a ≥ | S { v ,v ,v } | = c ≥
1, then G is isomorphic to H in Fig. 2. • Suppose that a ≥ c , then we claim that c ≤ c = 2, then 1 + d v and1 + d v are two distinct L -eigenvalues of G with multiplicity at least 2 by Lemma 2.3. Thus, θ = 1 + d v or θ = 1 + d v . If θ = 1 + d v , then the multiplicity of 1 + d v is 2 and from thetrace of L ( G ), n = ( n − d v ) + 2(1 + d v )= ( n − n − ) + 2(1 + n − )= n + n − n − n − > n, a contradiction. If θ = 1 + d v , then the multiplicity of 1 + d v is 2, which yields that a = 2(as a ≥ c = 2). Then G is a graph isomorphic to H with order 7 and by direct check G / ∈ G ( n, n − c = 1, then a = n −
4. According to V ( G ) = { V ( K a +1 ) , V ( K c +1 ) , v } ,there is an equitable partition for L ( G ). Let Q be the corresponding quotient matrix of L ( G ), then by d v = n − d v = n − d v = 2, Q = − n − n − − √ ( n − n − − n √ ( n − n − − n − − √ n − − √ n − . By calculation, the eigenvalues of Q are { , n − n + 4 ± √ n − n + 56 n − n − n + 2) } , which are also the eigenvalues of L ( G ) from Lemma 2.1. Furthermore, 1 + d v = n − n − and1 + d v = nn − are two distinct L -eigenvalues of G by Lemma 2.3. Thus G has 5 distinct L -eigenvalues, contradicting with G ∈ G ( n, n − • Suppose that a < c , then a ≤
2. Otherwise, a ≥ d v and1 + d v as two distinct L -eigenvalues of G are at least 3 by Lemma 2.3, contradicting with G ∈ G ( n, n − a = 2, then c ≥ θ = 1 + d v clearly, which impliesthat the multiplicity of 1 + d v must be 2. Thus, from the trace of L ( G ), we obtain n = 2(1 + d v ) + ( n − d v )= 2(1 + n − ) + ( n − n − )= n + n − n − > n, a contradiction. Now, assume that a = 1, then c = n −
4. It is clear that L ( G ) contains anequitable partition with respect to V ( G ) = { V ( K a +1 ) , V ( K c +1 ) , v } . Note that d v = n − v = n − d v = n −
3, then the corresponding quotient matrix of L ( G ) is Q = − n − − n √ ( n − n − − √ ( n − n − − n − n − − √ ( n − n − − n √ ( n − n − . Further, by calculating, the eigenvalues of Q are { , n − n − ± √ n − n + 332( n − n + 2) } , which are also the eigenvalues of L ( G ) by Lemma 2.1. Recalling that 1 + d v = n − n − and1 + d v = nn − are two distinct L -eigenvalues of G , then we see that G has 5 distinct L -eigenvalues, contradicting with G ∈ G ( n, n − S { v ,v } = ∅ ,we can also obtain contradictions.Consequently, we conclude that if S { v ,v } = ∅ , then S { v ,v ,v } = ∅ from Cases 1 and 2. Claim 3. If S { v ,v } = ∅ , then S { v ,v } = ∅ .Suppose for a contradiction that S { v ,v } = ∅ when S { v ,v } = ∅ . Then the vertices of S { v ,v } (resp., S { v ,v } ) induce a clique, otherwise ν ( G ) ≥
3, a contradiction. Further, eachvertex of S { v ,v } is not adjacent to any of S { v ,v } . If not, one can easily obtain an induced P , a contradiction. Let | S { v ,v } = a | and | S { v ,v } = b | , then G is isomorphic to H in Fig.2. • Assume that a = b = n − , then d v = d v = n − and d v = n −
1. According to the partition V ( G ) = { V ( K a +1 ) , v , V ( K b +1 ) } , L ( G ) has an equitable partition and the correspondingquotient matrix is Q = n − −√ n − − √ − √ −√ n − n − . From calculation, the eigenvalues of Q are { , n − , n +1 n − } , which are also the eigenvalues of L ( G ). Moreover, the multiplicity of 1+ d v = n +1 n − is at least n − L -eigenvalue is n − ( n − n + 1 n − − n − n + 1 n − . As a result, the multiplicity of n +1 n − is n −
2, contradicting with G ∈ G ( n, n − • Assume that a = b and a < b without loss of generality. Then we say that a ≤
2, otherwise a ≥ b ≥ G contains two distinct L -eigenvalue with multiplicity at least 3 from Lemma2.3, a contradiction. If a = 2, then the multiplicity of 1 + d v is at least 2 by Lemma 2.3.Since b = n − >
2, then θ = 1 + d v clearly, which yields that the multiplicity of 1 + d v is13. It follows from the trace of L ( G ) that n = 2(1 + d v ) + ( n − d v = 2(1 + ) + ( n − n − )= n + + n − a contradiction. If a = 1, with respect to the partition V ( G ) = { V ( K a +1 ) , v , V ( K b +1 ) } , then L ( G ) has an equitable partition and the corresponding quotient matrix is Q = − − √ n − − √ n − − n ( n − n − − n − n − − n − n − , By computing, the eigenvalues of Q are { , n − ± q n +13 n − ∗ n +2394( n − n − } , which are also L -eigenvalues of G from Lemma 2.1. Noting that n ≥ d v = and 1 + d v = n − n − are two distinct L -eigenvalues of G from Lemma 2.3. Thus,one can observe that G contains 5 distinct L -eigenvalues, a contradiction.Combining above discussion, we conclude that S { v ,v } = ∅ when S { v ,v } = ∅ .All the proofs are completed. (cid:3) Theorem 3.3.
Let G be a connected graph of order n ≥ . Then G ∈ G ( n, n − and G isa cograph if and only if G is the graph G in Fig. 1. Proof. If G = G , then clearly G ∈ G ( n, n −
3) from Lemma 2.6 and G is a cograph.Now suppose that G ∈ G ( n, n −
3) and G is a cograph. Also, let θ be the L -eigenvalue ofmultiplicity n −
3, then θ = 1 from Lemma 2.4. In the following, we will show that G must be G . Clearly, G cannot be the complete graph K n and then the diameter diam ( G ) = 2. Let P = v v v be a diametrical path of G and for a subset U of V ( P ), S U = { u ∈ V ( G ) \ V ( P ) : N G ( u ) ∩ V ( P ) = U } . From Lemma 3.2, we see that S { v } = S { v } = S { v } = ∅ , and suppose that S { v ,v } = ∅ , then S { v ,v } = S { v ,v } = S { v ,v ,v } = ∅ . Thus all the vertices out of V ( P ) belong to S { v ,v } .Since ν ( G ) = 2, the vertices of S { v ,v } induce a clique of G , that is, G is isomorphic to G .Next, suppose that S { v ,v } = S { v ,v } = ∅ . Then all the vertices out of V ( P ) spread in S { v ,v } or S { v ,v ,v } . In the following, we will show that this cannot hold.First, assume that S { v ,v } = ∅ , then we claim that | S { v ,v } | = 1. Otherwise, | S { v ,v } | ≥ v , v ∈ S { v ,v } . As ν ( G ) = 2, then v ∼ v , and thus the vertices v i (1 ≤ i ≤ H in Fig. 2, contradicting with Claim 1 of Lemma 3.2. So, | S { v ,v } | = 1 holds and let v ∈ | S { v ,v } | = 1. Since the order n ≥ G , then S { v ,v ,v } = ∅ .14urther, v is adjacent to each of S { v ,v ,v } . If not, one can also obtain an induced subgraphisomorphic to H , a contradiction. Now, we can see that v and v (resp., v and v ) are twinpoints, then the multiplicity of 1 as an L -eigenvalue is at least 2 from Lemma 2.2. As a result,either ρ n − ( G ) = 1 or θ = 1, a contradiction.Second, assume that S { v ,v } = ∅ , then all the vertices of V ( G ) \ V ( P ) belong to S { v ,v ,v } .If there are two vertices, say v and v , of S { v ,v ,v } are not adjacent, then we claim that allother vertices of S { v ,v ,v } are adjacent to both v and v . If not, there exists a vertex, say v , adjacent to exactly one of v and v (noting that ν ( G ) = 2). As a result, the vertices v i (2 ≤ i ≤
6) induce an subgraph isomorphic to H , contradicting with Claim 2 of Lemma3.2. Hence, v and v are twin points. Noting that v and v are also twin points, then themultiplicity of 1 as an L -eigenvalue is at least 2, a contradiction. Consequently, any two of S { v ,v ,v } are adjacent, i.e., G = K n − e . However, ρ n − ( K n − e ) = 1 from [13], a contradiction.Combining above discussion, we observe that G must be G . (cid:3) Remark 3.4.
To complete the characterization of graphs with some normalized Laplacianeigenvalue of multiplicity n − , one need to consider the remaining case, that is, graphs with ρ n − ( G ) = 1 and ν ( G ) = 2 and diam ( G ) = 2 . In this case, the edges of graphs are dense. So,it is a challenge to distinguish the edges. However, we conjecture that there is no graphs inthis case. Acknowledgements
The authors thank the anonymous referees for their valuable comments of this paper. Thiswork is supported by the Natural Science Foundation of Shandong Province (No. ZR2019BA016).
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