連通圖的拉普拉斯與無符號拉普拉斯譜半徑之研究 - 政大學術集成

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(1)國立政治大學應用數學系碩士學位論文. 連通圖的拉普拉斯與無符號拉普拉斯政治大譜半徑之研究立. ‧ 國. 學. ‧. On the Laplacian and the Signless Laplacian Spectral Radius of a Connected Graph n. er. io. sit. y. Nat. al. Ch. engchi. i Un. v. 碩士班學生：羅文隆撰指導教授：張宜武博士中華民國 104 年 6 月 30 日.

(2) 中文圖的譜半徑在數學方面以及其他領域有非常多的應用。在這篇論文裡，我們整理有關連通圖的拉普拉斯與無符號拉普拉斯譜半徑的論文。本文一開始探討一些圖的譜理論，並找出這些界限的關係。然後，我們將討論更精確的圖之拉普拉斯與無符號拉普拉斯譜半徑。最後，我們給一個例子，並使用前面所探討過的性質分析之。關鍵字：圖、鄰接矩陣、拉普拉斯矩陣、無符號拉普拉斯矩陣、譜半徑、拉普拉斯譜半徑、無符號拉普拉斯譜半徑. 立. 政治大. ‧. ‧ 國. 學. n. er. io. sit. y. Nat. al. Ch. engchi. i. i Un. v.

(3) Abstract The spectral radius of a graph has been applied in mathenatics and in diverse disciplines. In this thesis, we survey some papers about the Laplacian spectral radius and the signless Laplacian spectral radius of a connected graph. Initially, we dis-. 政治大 bounds. Then, we discuss 立 the upper bounds and lower bounds of the Laplacian and. cuss some properties about the spectral graphs and find the relations between these. ‧ 國. analyze it.. 學. signless Laplacian spectral radius of a graph. In the end, we give an example and. ‧. Keywords: grpah, adjacency matrix, Laplacian matrix, signless Laplacian ma-. n. al. er. io. sit. y. Nat. trix, spectral radius, Laplacian spectral radius, signless Laplacian spectral radius. Ch. engchi. ii. i Un. v.

(4) Contents 中文. i. Abstract. ii. Contents. 2. Preliminaries. ‧ 國. Introduction. ‧. 2.2. Some Basics in Matrix Theory . . . . . . . . . . . . . . . . . . . . . . . . . .. 2.3. Some Properties of Spectral Graphs . . . . . . . . . . . . . . . . . . . . . . .. 5. 6 8. er. n. al. 3. y. Definitions and Notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. sit. 2.1. io. 4. 1 3. Nat. 3. iii. 學. 1. 立. 政治大. Ch. Some Properties of the Spectral Radius of a Graph. engchi. i Un. v. 11. 3.1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 11. 3.2. More Connections to Matrix Theory . . . . . . . . . . . . . . . . . . . . . . .. 13. 3.3. Some Relations Among ρ( G ), λ( G ) and µ( G ) . . . . . . . . . . . . . . . . .. 14. 3.4. More Discussions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 16. Main Results. 23. 4.1. Sharp Upper Bounds For λ( G ) and µ( G ) . . . . . . . . . . . . . . . . . . . .. 23. 4.2. Sharp Lower Bounds For λ( G ) and µ( G ) . . . . . . . . . . . . . . . . . . . .. 27. 4.3. Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 30. Conclusion. 32 iii.

(5) Chapter 1 Introduction 政治大 Let G = G (V, E) be a simple undirected connected graph with n vertices and m edges, 立 We follow [1] for notation in graph theory.. ‧ 國. and | E( G )| = m.. 學. where V=V ( G ) is the vertex set and E = E( G ) is the edge set and hence we have |V ( G )| = n. ‧. Graph spectrum plays an important role in many fields. In [10], the authors mention that the graph spectrum is important, not only because of its relations to numerous graph invariants (eg:. y. Nat. io. sit. algebraic connectivity, spread of a graph, expanding property, isoperimetric number, maximum. n. al. er. cut, independence number, genus, diameter, mean distance and bandwidth-type parameters of a. i Un. v. graph), but also because of its applications in diverse disciplines.. Ch. engchi. In [11], the authors explain why the graph spectrum is important, and mention the spectral of graphs applied in graph theory and many diverse disciplines (eg: chemistry, physics, computer science, and so on). In [12], the authors list many bibliographies on some applications to particular branches of science, and give some simple results about graph spectrum. Therefore, we know that studying the spectral radius of a graph is an important topic in many fields of mathematics and diverse disciplines. In this thesis, we focus on the upper bounds and lower bounds of the Laplacian and signless Laplacian spectrum of a graph. For the completeness, we still mention some properties about the adjacency matrix of a graph. 1.

(6) This thesis is organized as follows: in chapter 2, we give some preliminaries about graph theory and matrix theory. In chapter 3, we discuss some properties of the graph spectrum, and use these properties to identify some simple upper and lower bounds. In chapter 4, we disciss the upper bounds and lower bounds of the Laplacian and signless Laplacian spectral radius of a graph, and give some examples to make our results clear. In chaper 5, we give some conclusions. In addition, the main results of this thesis are from [2], [3] and [4].. 立. 政治大. ‧. ‧ 國. 學. n. er. io. sit. y. Nat. al. Ch. engchi. 2. i Un. v.

(7) Chapter 2 Preliminaries 政治大 theory and matrix theory. For detail duscussions, please refer to [1], [6] and [11] . 立. To study the spectral radius of a graph, we have to recall some preliminaries about graph. ‧ 國. 學. 2.1. Definitions and Notations. ‧. In this section, we give some definitions about graph theory. For convenience, we define. y. Nat. n. al. We use Mn to denote the set of all n × n real matrices.. Ch. engchi. We denote the n × n identity matrix by In .. er. io. We use Mm×n to denote the set of all m × n real matrices.. sit. some notations which shall be used frequently in this thesis.. i Un. v. In this thesis, we use det( x, K ) to denote the characteristic polynomial of K, that is det( xIn − K ), where K ∈ Mn . Let B ∈ Mn , we denote σ ( B) to be the set of all eigenvalues of B. Let G = G (V, E) be a simple undirected connected graph with n vertices and m edges, where V = V ( G ) is the vertex set and E = E( G ) is the edge set and hence we have |V ( G )| = n and | E( G )| = m. Thus, we can easy to say the order of G is n. When u and v are the endpoints of an edge, they are adjacent. We use N (v) to denote the set of vertices adjacent to a vertex v in G. The completement G of G defined by uv ∈ E( G ) if and only if uv ∈ / E( G ). A subgraph of G is a graph H such that V ( H ) ⊆ V ( G ) and E( H ) ⊆ E( G ) and the assignment of endpoints 3.

(8) to edge in H is the same as in G. A spanning subgraph of G is a subgraph with vertex set V ( G ). Let G be a graph with vertex set V ( G ) = {v1 , v2 , ...vn } and edge set E( G ) = {e1 , e2 , ..., em }. The degree of vertex vi , written di , is the number of edges incident to vi . The maximum degree is ∆( G ) and the minimum degree is δ( G ). The adjacency matrix of G, written A( G ), is the n × n matrix in which entry ai,j is the number of edges in G with endpoints {vi , v j }. The incidence matrix M ( G ) is the n × m matrix in which entry mi,j is 1 if vi is an endpoint of e j and otherwise is 0. The diagonal matrix of vertex degree of G is the n × n diagonal matrix D ( G ) whose its diagonal entry is di . The Laplacian matrix of G is the n × n matrix given by L( G ) = D ( G ) − A( G ). The signless Laplacian matrix of G is the n × n matrix given by Q ( G ) = D ( G ) + A ( G ).. 政治大. Remark 2.1. Every adjacency matrix, Laplacian matrix and signless Laplacian matrix are sym-. 立. metric.. ‧ 國. 學. Note 2.2.. ‧. In this thesis, we use A, L and Q to represent A( G ), L( G ) and Q( G ), respectively.. y. sit. al. er. io. Definition 2.3.. Nat. Now, we give some definitions which are mentioned in [2] and [3].. iv n C U of degree of the vertices adjacent h eti , nwhich (1) The 2-degree of vertex vi is denoted by g cishthei sum n. Let G be a graph with vertex set V ( G ) = {v1 , v2 , ..., vn }.. to vi . (2) Let Tn,k be a tree with n vertices, which is obtained by adding paths P1 , P2 , ..., Pk of almost equal the number of its vertices to the pendant vertices of the star graph K1,k . Figure 2.1 is T12,5 . Obviously, Tn,k is a tree with k pendant vertices. (3) Let Gn,k be a graph with n vertices, which is obtained by adding paths P1 , P2 , ..., Pn−k of almost equal length to the vertices of the complete graph Kn−k . Figure 2.2 is G12,5 Clearly, Gn,k is a graph with n vertices and k cut vertices. (4) We say G is a semi-regular bipartite graph if all vertices in the same part of a bipartition have the same degree.. 4.

(9) (5) Let G be a graph and a be one of the vertex of G. We define adding a path to G is adds a path at a as one of the end points.. 政治大. Figure 2.1: The T12,5 graph. 立. ‧. ‧ 國. 學. n. Ch. engchi. er. io. al. sit. y. Nat Figure 2.2: The G12,5 graph. i Un. v. Example 2.4. Let G be a complete graph with 3 vertices, then [ ] [2 1 1] [0 1 1] 2 −1 −1 A( G ) = 1 0 1 , L( G ) = −1 2 −1 and Q( G ) = 1 2 1 . −1 −1 2. 110. 112. Lemma 2.5. Let G be a graph, then ∑in=1 ti = ∑in=1 d2i . Proof. We prove the statement by induction on | G |=n. Basis step:n=1 is true obviously . Induction step:If n=k is true. ′. For n=k+1. Let G be the new graph that add vertex vk+1 and vk+1 adjacent to v1 , ..., vl . Hence, 5.

(10) d k +1  = l.   di + 1 if i = 1, ..., l    ′ di = di if i = l + 1, ..., k      l if i = k + 1 ′2. ′2. ′2. ′2. ′2. ′2. Thus, ∑ik=+11 di = d1 + ... + dl + dl +1 + ... + dk + dk+1. = (d1 + 1)2 + ... + (dl + 1)2 + d2l +1 + ... + d2k + l 2 = (t1 + l ) + ... + (tl + l ) + tl +1 + ... + tk+1 + (d1 + ... + dl ) ′. = ∑ik=+11 ti . Now, we recall some basic properties about graph theory in [1] and [3] which we needed.. 政治大. Let G be a graph, we say that LG is the line graph of G if its vertices are the edge of G,. 立. with e f ∈ E( LG ) when e = uv and f = vw in G. If LG is a connected bipartite graph, then. ‧ 國. 學. ∆( G ) 6 2. Let G be a graph with n vertices, then G is Cn or Pn if and only if LG is Cn or Pn−1 . Let G be a graph, then Pn−1 is semiregular bipartite graph if and only if n 6 4. Let G be a. ‧. bipartite graph and LG be its line graph. It is easy to say that LG is a regular graph if and only. al Some Basics in Matrix Theory n. 2.2. Ch. engchi. er. io. sit. y. Nat. if G is either a regular graph or a semiregular graph.. i Un. v. We recall the definitions and results from [6] and [7]. Lemma 2.6. Let C ∈ Mm×n , then CC T and C T C have the same nonzero eigenvalues. Lemma 2.7. Let K be a nonegative symmetric matrix and v be a unit vector of Rn . If ρ = v T Kv , then Kv = ρv, where ρ is the spectral radius of K (will state later). Definition 2.8. Let S ∈ Mn . The graph g(S) of S is defined to be the directed graph on n vertices N1 , N2 , ..., Nn. 6.

(11) in which there is a directed edge leading from Ni to Nj if and only if sij ̸= 0, where sij is the element of A( g(S)). [0 1 0 0] 1011 0100 0100. Example 2.9. Consider S =. 立. 政治大. ‧ 國. 學 Figure 2.3: The g(S) graph. ‧ y. Nat. Definition 2.10.. , then we get g(S) in figure 2.3.. sequence of directed edges leading from Ni to Nj .. n. al. Note 2.11.. Ch. engchi. er. io. sit. We say that a graph is a strongly connected if for each pair of vertices (Ni , Nj ), there exists a. i Un. v. Clearly, if G is a connected graph, g( A( G )) is a strongly connected. Definition 2.12. Let R ∈ Mn . We say that R is irreducible if g( R) is strongly connected. Definition 2.13. Let R ∈ Mn , the spectral radius of R is defined by ρ( R) = maxi |ρi |, where ρi are eigenvalues of R (1 6 i 6 n) . Theorem 2.14. (Perron-Frobenius Theorem) If R ∈ Mn is nonnegative irreducible, then each of the following is true. 7.

(12) (1) ρ( R) is a simple eigenvalue of R. (2) There exists a unique unit positive eigenvector v corresponding to ρ( R). This vector is called the Perron vector. There are no nonnegative eigenvectors for R except for positive multiples of v, regardless of the eigenvalue. Theorem 2.15. If G is a graph and ρ is the spectral radius of A, then the following are equivalent: (1) G is a bipartite graph. (2) Any eigenvalue of A is symmetric about the origin. (3) −ρ is an eigenvalue of A.. 政治大 ≥ ... ≥立 ρ are eigenvalues of R. Then we have ρ. Theorem 2.16. (Rayleight Principle). 1. = maxx̸=0. x T Rx . xT x. 學. Note 2.17.. n. ‧ 國. If R ∈ Mn and ρ1 ≥ ρ2. Let B, C ∈ Mn be nonnegative. We say that B < C if B1 < C1 , where 1 is a n-dimention. ‧. vector which all entris are 1.. n. al. er. io. Some Properties of Spectral Graphs. Definition 2.18.. sit. y. Nat. 2.3. Ch. engchi. i Un. v. Let G be a graph, the spectral radius of G is defined by ρ( G ) = maxi |ρi |, where ρi are eigenvalues of the adjacency matrix A (1 6 i 6 n) . In example 2.4, we have discussed the graph G which is a complete graph with 3 vertices. To count ρ( G ), λ( G ) and µ( G ), we have to find the following matrices:. . . 0 1 1    A( G ) =  1 0 1     1 1 0. 8.

(13) . .  2 −1 −1    L( G ) =  − 1 2 − 1     −1 −1 2 . . 2 1 1    Q( G ) =  1 2 1     1 1 2 First, consider det( xI3 − A( G )) = 0, then we find x = −1, −1, 2. Thus, ρ( G ) = 2.. 政治大. Similarly, we consider det( xI3 − A( G )) = 0 and det( xI3 − A( G )) = 0. Thus, λ( G ) = 3 and µ( G ) = 4.. 立. ‧ 國. 學. Remark 2.19.. ‧. By theorem 2.14, we have:. (1) ρ( G ) is a simple eigenvalue of A and thus we can simply say the spectral radius of G is the. n. al. Definition 2.20.. Ch. engchi. er. io. (2) There is a Perron vector of A corresponding to ρ( G ).. sit. y. Nat. largest eigenvalue of A.. i Un. v. Let G be a graph, the Laplacian spectral radius of G is the largest eigenvalue of L, and we denoted it as λ( G ) Definition 2.21. Let G be a graph, the signless Laplacian spectral radius of G is the largest eigenvalue of Q, and we denoted it as µ( G ) From [16], we know that given any nonzero vector x = ( x1 , x2 , ..., xn ) T , x T Lx = ∑vi ∈V di xi2 − ∑vi ∈V ∑v j ̸=vi ,v j ∼vi xi x j. =. 1 2. ∑vi ∈V di xi2 + 12 ∑v j ∈V d j x2j − ∑vi ∈V ∑v j ∈V,v j ∼vi xi x j. =. 1 2. ∑vi ∈V ∑v j ∈V,v j ∼vi ( xi2 − 2xi x j + x2j ) 9.

(14) = ∑ v i ∼v i ( x i − x j )2 > 0 . Similarly, we can obtain the result x T Qx = ∑vi ∼vi ( xi + x j )2 > 0 . By above, we can easily say L and Q are positive semidefinite matrix. Hence every eigenvalue of L and Q are nonnegative.. Remark 2.22. Let G be a graph, we can place each eigenvalue of A, L and Q in nonincreasing order as follows: (1) ρ( G ) = ρ1 ( G ) > ρ2 ( G ) > ... > ρn ( G ) (2) λ( G ) = λ1 ( G ) > λ2 ( G ) > ... > λn ( G ) > 0 (3) µ( G ) = µ1 ( G ) > µ2 ( G ) > ... > µn ( G ) > 0. 治政大 Now, we use this principal to In last section, we have introduced the Rayleight principal. 立 define ρ( G ), λ( G ) and µ( G ) . ‧ 國. 學. Definition 2.23.. ‧. Let G be a graph, x be a nonzero unit vector.. y. Nat. (1) ρ( G ) = max x T Ax .. n. al. er. io. (3) µ( G ) = max x T Qx .. sit. (2) λ( G ) = max x T Lx .. Ch. engchi. 10. i Un. v.

(15) Chapter 3 Some Properties of the Spectral Radius of a Graph 立. 政治大. ‧ 國. 學. In this chapter, we introduce some properties of the graph spectrum and find the relations among ρ( G ), λ( G ) and µ( G ). Besides, use these properties to identify simple upper and lower. sit. y. Nat. Introduction. io. n. al. er. 3.1. ‧. bound for ρ( G ), λ( G ) and µ( G ) .. i Un. v. In this section, we survey [17], [18] and [19] and find the relation between the spectral radius of a graph and it’s subgraph.. Ch. engchi. Lemma 3.1. Let G be a graph and H be a proper subgraph of G. (1) ρ( H ) < ρ( G ) . (2) λ( H ) 6 λ( G ) . (3) µ( H ) < µ( G ) .. Proof. (1) Let x be the Perron vector of ρ( H ), y be any nonzero unit vector. Then. 11.

(16) ρ( H ) = x T A( H ) x 6 x T A( G ) x 6 max y T A( G )y = ρ( G ). Since A( H ) < A( G ), the above inequality must be strict. (2) Let x be the unit eigenvector of λ( H ), y be any nonzero unit vector. λ( H ) = xT L( H ) x. = ∑vi ∼vi ,vi vi ∈E( H ) ( xi − x j )2 6 ∑vi ∼vi ,vi vi ∈E(G) ( xi − x j )2 6 max ∑vi ∼vi ,vi vi ∈E( H ) (yi − y j )2 = λ ( G ). (3) Similar to (1). In [13], the author tell us in some cases, which will state below, λ( H ) = λ( G ), where H is a subgraph of G.. 政治大. 立. Remark 3.2.. ‧ 國. 學. Let G be a graph, v ∈ V ( G ), and v1 , v2 , ..., vn are pendant vertices of G which are adjacent to s ( s −1) 2 ). among v1 , v2 , ..., vn .. ‧. v. If we add any t edges (1 6 t 6. Then their Laplacian spectral radius are equal.. sit. n. al. er. io. Let G be a graph.. y. Nat. Lemma 3.3.. (1) δ( G ) 6 ρ( G ) 6 ∆( G ) . (2) 1 + ∆( G ) 6 λ( G ) 6 n .. Ch. engchi. i Un. v. (3) 2δ( G ) 6 µ( G ) 6 2∆( G ) .. Proof. (1) Let x = ( x1 , x2 , ..., xn ) T be the Perron vector of A( G ). Then, A( G ) x = ρ( G ) x. n. For convenience, we assume ∑ xi = 1. Now, x T A( G ) = ( A( G ) x) T = (ρ( G ) x) T = ρ( G ) x T n. ρ( G ) = ρ( G )( ∑ xi ) = i =1. n. i =1. ∑ ρ ( G ) xi =. i =1. Similarly, we have δ( G ) 6 ρ( G ).. n. ∑ (xT A(G))i =. i =1. n. n. i =1. i =1. ∑ xi di 6 ∆ ( G ) ∑ xi = ∆ ( G ). (2) Clearly, any graph with n vertices is a subgraph of Kn , hence, we have λ( G ) 6 n. Since any graph must contain a subgraph K1,∆ , and λ(K1,∆ ) = ∆ + 1, we get the lower bound. 12.

(17) (3) Similar as (1) . Example 3.4. Let G be the graph which is shown in figure 3.1. We find that ρ( G ) = 3.0766, λ( G ) = 5.3028 and µ( G ) = 6.3014. It is easy to see δ( G ) = 2 < ρ( G ) < 4 = ∆( G ), 1 + ∆( G ) = 5 < λ( G ) < 8, and 2δ( G ) = 4 < µ( G ) < 8 = 2∆( G ) .. 立. ‧. ‧ 國. 學. Figure 3.1: The graph G. More Connections to Matrix Theory. sit. y. Nat. 3.2. 政治大. al. er. io. Using matrix can help us study spectral graph theory conveniencely. More detail related to. n. matrix and graph, please refer to [5]. Lemma 3.5.. Ch. engchi. i Un. v. Let G be a graph and A L be the adjacency matrix of LG . (1) Q = MM T . (2) M T M = 2Im + A L . Proof. (1) Since M is the incidence matrix of G, we can easily find that M is a 0-1 matrix. Consider MM T , the diagonal entries of MM T represent the degree of vi , that is, di . On the other hand, any nondiagonal entries of MM T represent the number of edges adjacent to vi and v j . In this thesis, we just consider simple graph, so we have the nondiagonal entries of MM T is 1 if vi adjacent to v j , otherwise are zero. Hence, MM T = D + A = Q. (2) Consider M T M, we have the diagonal entries of M T M is 2, other elements can be found by 13.

(18) the same reasons as in (1) and we have the nondiagonal entries of M T M are 1 if ei adjacent to e j and 0 otherwise. Since M T M ∈ Mm×m , we have to consider another m × m matrix. Consider A( LG ), we find that any enry of A( LG ) is 1 if ei adjacent to e j and 0 otherwise. Hence, by above discussion, we have M T M = 2Im + A L . Lemma 3.6. Let G be a graph, then µ( G ) = 2 + ρ( LG ) , where LG be the line graph of G. Proof. By lemma 3.5, we know that Q = MM T and M T M = 2Im + A L . Because MM T and M T M have the same nonzero eigenvalues and σ ( M T M ) = {2 + ρi | ρi are eigenvalues o f A( LG )}, the result holds.. Some Relations 立 Among ρ( G ), λ( G ) and µ( G ). 學. ‧ 國. 3.3. 政治大. In this section, we will find the relation between ρ( G ), λ( G ) and µ( G ). For detail discus-. y. Nat. Lemma 3.7.. ‧. sions, please refer to [9] and [14].. n. Proof. Since ρ( G ) 6 ∆( G ) < ∆( G ) + 1 6 λ( G ) .. Ch. engchi. er. io. al. sit. Let G be a graph, then ρ( G ) < λ( G ).. i Un. v. Example 3.8. Let G be the graph which is the same as example 3.4. We find that ρ( G ) = 3.0766, λ( G ) = 5.3028 and µ( G ) = 6.3014. It is easy to see ρ( G ) = 3.0766 < 5.3028 = λ( G ) .. Lemma 3.9. [9] If G be a graph, then λ( G ) 6 µ( G ). The equality holds if and only if G is a bipartite graph. Proof. Let x = ( x1 , x2 , ..., xn )T be unit eigenvector of L beloning to λ( G ) and y = (y1 , y2 , ..., yn ) T be the Perron vector of Q. Let X = (| x1 |, | x2 |, ..., | xn |) T . 14.

(19) Since λ( G ) = x T Lx = ∑vi ∼vi ,i6 j ( xi − x j )2 and µ( G ) = y T Qy = ∑vi ∼vi ,i6 j (yi + y j )2 . We have λ( G ) = ∑vi ∼vi ,i6 j ( xi − x j )2 6 ∑vi ∼vi ,i6 j (| xi | + | x j |)2 = | x| T Q | x| 6 µ( G ). If λ( G ) = µ( G ), then all inequalities above must be equal. Thus, by lemma 2.7 and above equality, we have | x| T Q | x| = µ( G ) and | x| be eigenvector of Q beloning to µ( G ). Hence,. | x| = ± X. By theorem 2.14, we have y > 0, X = y, and | x1 |, | x2 |, ..., | xn | > 0. Since ∑vi ∼vi ,i6 j ( xi − x j )2 = ∑vi ∼vi ,i6 j (| xi | + | x j |)2 and ( xi − x j )2 6 (| xi | + | x j |)2 . We have ( xi − x j )2 = (| xi | + | x j |)2 when vi ∼ vi . Thus, xi x j < 0 if vi ∼ vi . That is, if vi ∼ vi , then xi x j < 0. That is ,if xi x j > 0, then vi vi . Now, we consider V1 = {vi | xi > 0} and V2 = {v j | x j < 0}. For each edge vi v j , we have xi x j < 0, and one of the vertices of edge vi v j is in V1 , the other is in V2 . Thus, G is a bipartite graph. Conversely,. 政治大. if G is a bipartite graph, then we assume V ( G ) = V1 ∪ V2 , where V1 = {v1 , v2 , ..., vs } and. 立. V2 = {vs+1 , vs+2 , ..., vn }. Let P ∈ Mn×n and. ‧ 國. . 學. . ‧. − I O P=  O I. sit. y. Nat. Then obviously, P is nonsingular and P = P−1 . For some simple operations, we have. n. al. er. io. PLP−1 = Q. Namely, L and Q are similar. Thus, σ ( L) = σ ( Q). Hence, λ( G ) = µ( G ).. i Un. v. Example 3.10. Let G be the graph which is the same as example 3.4.. Ch. engchi. We find that ρ( G ) = 3.0766, λ( G ) = 5.3028 and µ( G ) = 6.3014. It is easy to see λ( G ) = 5.3028 < 6.3014 = µ( G ) .. Corollary 3.11. If G is a graph, then λ( G ) 6 2 + ρ( LG ) , where LG is the line graph of G. The equality holds if and only if G is a bipartite graph. Proof. By lemma 3.6 and lemma 3.9, we finish. Lemma 3.12. [14] If G be a graph, then ρ( G ) 6 12 µ( G ) . 15.

(20) The equality holds if and only if G is a regular graph. Proof. Let x = ( x1 , x2 , ..., xn )T and y = (y1 , y2 , ..., yn ) T be the Perron vector of A and Q, respectively. From last chapter, we know that µ( G ) = ∑vi ∼vi ,i6 j (yi + y j )2 . Thus, ρ( G ) = x T Ax = 2 ∑vi ∼vi ,i6 j xi xj 6. 1 2. ∑vi ∼vi ,i6 j ( xi + x j )2 = 12 x T Qx 6 12 µ( G ) .. If the equality holds, all the inequalities above must be equalities. Then, ρ( G ) = 2 ∑vi ∼vi ,i6 j xi xj =. 1 2. ∑vi ∼vi ,i6 j ( xi + x j )2 . Since 2xi x j 6 xi2 + x2j . Hence,. xi = x j when vi and v j are adjacent. Because G is connected, x is the multiple of 1, where all entries of 1 are 1. Since Ax = ρ( G ) x, we have G is regular. Now, suppose G is regular, say k-regular graph. We have ρ( G ) = k and µ( G ) = 2k are easy to be found.. 政治大 Example 3.13. Let G be the graph 立which is the same as example 3.4.. ‧ 國. 學. We find that ρ( G ) = 3.0766, λ( G ) = 5.3028 and µ( G ) = 6.3014. It is easy to see ρ( G ) = 3.0766 < 3.1052 = 21 µ( G ) .. ‧. n. al. er. io. If G is a regular bipartite graph, then ρ( G ) = 12 λ( G ) = 12 µ( G ). sit. y. Nat. Corollary 3.14.. i Un. v. By above discussion,we have the following important corollary which tells us the relation between ρ( G ), λ( G ) and µ( G ).. Ch. engchi. Corollary 3.15. If G is a graph, then ρ( G ) < λ( G ) 6 µ( G ).. 3.4. More Discussions. We introduced some results from [8] and extend these results to more cases. Lemma 3.16. Let G be a graph and a ∈ V ( G ). We construct a graph G1 by adding a vertex to G adjacent to. 16.

(21) a. Then we have det( x, A( G1 )) = x det( x, A( G )) − det( x, A( G − a)) . Proof. Using the definition of det( x, A( G1 )) and we can easily find the results. Using the same method, we get the following corollary. Corollary 3.17. Let G be a graph and a ∈ V ( G ). We construct a new graph G1 by adding a new vertex and adjacent the new vertex to a. Then we have (1) det( x, L( G1 )) = ( x − 1) det( x, L( G )) − det( x, L( G − a)) and (2) det( x, Q( G1 )) = ( x − 1) det( x, Q( G )) − det( x, Q( G − a)) .. 立. Lemma 3.18.. 政治大. ‧ 國. 學. Let G be a graph and G1 be a proper spanning subgraph of G. If ρ( G ) 6 x, then det( x, A( G )) < det( x, A( G1 )).. ‧. Proof. We prove the statement by induction on |G|=n.. y. Nat. Basis step:The result holds clearly for |G|=2.. d dx. det( x, A( G1 )) −. al. er. det( x, A( G )) = ∑in=1 det( x, A( Gi )), where Gi = G − vi .. n. So,. d dx. io. Then,. sit. Induction step:Assume |G|=|G1 |=n ≥ 2. Suppose V ( G ) = V ( G1 ) = { x1 , ..., xn }.. Ch. i Un. v. det( x, A( G )) = ∑in=1 (det( x, A( G1i )) − det( x, A( Gi ))) ...(1) .. d dx. engchi. Because ρ( G ) > ρ( G1 ) and det( x, A( G1 )) is monic,. det( x, A( G1 )) − det( x, A( G )) = det( x, A( G1 )) > 0 ...(2) . Since G1 is a proper spanning subgraph of G, all G1 − vi are subgraph of G − vi (i=1,..., n) and when n ≥ 3, there is i such that G1 − vi is a proper spanning subgraph of G − vi . By induction, when x ≥ ρ( G ) > ρ( G − vi ), we can easy say that every n item of the right hand side of (1) are nonnegative and at least one item is positive. When x ≥ ρ( G ), we have. d dx. det( x, A( G1 )) −. d dx. det( x, A( G )) > 0 ...(3).. By (2) and (3), we’re done. Using the same method, we get the following two corollaries.. 17.

(22) Corollary 3.19. Let G be a graph and G1 be a proper spanning subgraph of G. If λ( G ) 6 x, then det( x, L( G )) < det( x, L( G1 )). Corollary 3.20. Let G be a graph and G1 be a proper spanning subgraph of G. If µ( G ) 6 x, then det( x, Q( G )) < det( x, Q( G1 )). Theorem 3.21. Let G be a graph and a ∈ V ( G ). Let Gk,l denote the graph obtained from G by adding path Pk+1 and Pl +1 at a as one of the end points. If 2 6 l 6 k and ρ( Gk+1,l −1 ) 6 x, then. 政治大立) < ρ(G ).. det( x, ρ( Gk,l )) < det( x, ρ( Gk+1,l −1 )). In particular, we have ρ( Gk+1,l −1. k,l. ‧ 國. 學. Proof. Suppose |G| ≥ 2.. If l ≥ 3, by lemma 3.16, we have. ‧. det( x, A( Gk,l )) − det( x, A( Gk+1,l −1 )) = det( x, A( Gk−l +1,1 )) − det( x, A( Gk−l +2,0 )).. sit. y. Nat. Using lemma 3.16 again, we have. er. io. det( x, A( Gk−l +1,1 )) = x det( x, A( Gk−l +1,0 )) − det( x, A(( Gk−l +1,0 ) − a)). det( x, A( Gk−l +2,0 )) = x det( x, A( Gk−l +1,0 )) − det( x, A( Gk−l,0 )).. n. al. Ch. i Un. v. In fact, Gk−l +1,0 − a is a proper spanning subgraph of Gk−l,0 , we have det( x, A( Gk,l )) − det( x, A( Gk+1,l −1 )). engchi. = det( x, A( Gk−l,0 )) − det( x, A(( Gk−l +1,0 ) − a)) ...(*) . Since Gk−l +1,0 − a is a proper spanning subgraph of Gk−l,0 and Gk−l,0 − a is a proper spanning subgraph of Gk+1,l −1 , we have ρ( Gk+1,l −1 ) > ρ( Gk−l,0 ). When x ≥ ρ( Gk+1,l −1 ), the right hand side of (*) < 0. We have det( x, A( Gk,l )) − det( x, A( Gk+1,l −1 )) < 0. Namely, det( x, A( Gk,l )) < det( x, A( Gk+1,l −1 )). In particular, we have ρ( Gk+1,l −1 ) < ρ( Gk,l ). Corollary 3.22. Let G be a graph and a ∈ V ( G ). Let Gk,l denote the graph obtained from G by adding path Pk+1 and Pl +1 at a as one of the end points. If 2 6 l 6 k and λ( Gk+1,l −1 ) 6 x, then 18.

(23) det( x, λ( Gk,l )) < det( x, λ( Gk+1,l −1 )). In particular, we have λ( Gk+1,l −1 ) < λ( Gk,l ). Proof. Use lemma 3.17 and by the same method as theorem 3.21, we finish this corollary. Corollary 3.23. Let G be a graph and a ∈ V ( G ). Let Gk,l denote the graph obtained from G by adding path Pk+1 and Pl +1 at a as one of the end points. If 2 6 l 6 k and µ( Gk+1,l −1 ) 6 x, then det( x, µ( Gk,l )) < det( x, µ( Gk+1,l −1 )). In particular, we have µ( Gk+1,l −1 ) < µ( Gk,l ). Proof. Use lemma 3.17 and by the same method as lemma 3.21, we finish this corollary.. 政治大. 立. Theorem 3.24.. (1). ‧ 國. 學. Let G be a graph and a and b be two adjacent vertices of G. Let Gk,l denote the graph obtained from G by adding path Pk+1 and Pl +1 at a and b, respectively. If 2 6 l 6 k and (1). (1). (1). (1). ‧. ρ( Gk+1,l −1 ) 6 x, then det( x, ρ( Gk,l )) < det( x, ρ( Gk+1,l −1 )). (1). In particular, we have ρ( Gk+1,l −1 ) < ρ( Gk,l ).. sit. y. Nat. io. (1). (1). det( x, A( Gk,l )) − det( x, A( Gk+1,l −1 )). n. a. i Un. l GCk−l+1,0 − b))) ...(1) . = det( x, A( Gk−l,0 )) − det( x, A(( (1). hengchi. Case 1: When k = l ≥ 2. (1). (1). (1). er. Proof. By lemma 3.16, we have. v. (1). (1). Clearly, ( Gk−l +1,0 − b) = G1,0 − b. In fact, G1,0 − b is a proper spanning subgraph of Gk−l,0 = (1). (1). (1). G0,0 = G and G is a proper subgraph of Gk+1,l −1 . By lemma 3.18, when x ≥ ρ( Gk+1,l −1 ), the right hand side of (1) < 0. Hence, we’re done. Case 2: When k > l ≥ 2. (1). (1). (1). Obviously, Gk−l,1 − b is a proper subgraph of Gk−l,0 . By lemma 3.18, when x ≥ ρ( Gk−l,0 ) the right hand side of (1) satisfies (1). (1). det( x, A( Gk−l,0 )) − det( x, A( Gk−l +1,0 )) (1). (1). < det( x, A( Gk−l,1 − b)) − det( x, A( Gk−l +1,0 − b)) ...(2) . (1). (1). For Gk−l,1 − b, using lemma 3.18 can know that when x ≥ ρ( Gk−l,1 − b) we have 19.

(24) (1). (1). det( x, A( Gk−l,1 − b)) − det( x, A( Gk−l +1,0 − b)) ...(3) . (1). (1). Since Gk−l,0 is a proper subgraph of Gk+1,l −1 , we have (1). (1). (1). (1). ρ( Gk+1,l −1 ) > ρ( Gk−l,0 ) > ρ( Gk−l,1 − b) > ρ( Gk−l +1,0 − b) ...(4) . (1). By (1), (2), (3), (4) and x ≥ ρ( Gk+1,l −1 ), we finished. Corollary 3.25. (1). Let G be a graph and a and b be two adjacent vertices of G. Let Gk,l denote the graph obtained from G by adding path Pk+1 and Pl +1 at a and b, respectively. (1). (1). (1). If 2 6 l 6 k and λ( Gk+1,l −1 ) 6 x, then det( x, λ( Gk,l )) < det( x, λ( Gk+1,l −1 )). (1). (1). In particular, we have λ( Gk+1,l −1 ) < λ( Gk,l ).. 政治大. Proof. Use corollary 3.19 and by the same method as theorem 3.24, we finish this corollary.. 立. Corollary 3.26.. ‧ 國. 學. (1). Let G be a graph and a and b be two adjacent vertices of G. Let Gk,l denote the graph obtained from G by adding path Pk+1 and Pl +1 at a and b, respectively.. ‧. (1). (1). (1). If 2 6 l 6 k and µ( Gk+1,l −1 ) 6 x, then det( x, µ( Gk,l )) < det( x, µ( Gk+1,l −1 )). (1). (1). y. sit. Nat. In particular, we have µ( Gk+1,l −1 ) < µ( Gk,l ).. n. al. er. io. Proof. Use corollary 3.20 and by the same method as theorem 3.24, we finish this corollary.. Ch. i Un. v. Using above theorem and corollaries, the following three corollaries are easy to be ckecked.. engchi. Corollary 3.27. Let G be a graph and a and b be two vertices of G which has a path of lengh m adjacent from (m). a to b. Let Gk,l denote the graph obtained from G by adding path Pk+1 and Pl +1 at a and b, (m). respectively. If 2 6 m 6 k − l, 2 6 l, and ρ( Gk+1,l −1 ) 6 x, then (m). (m). det( x, ρ( Gk,l )) < det( x, ρ( Gk+1,l −1 )). (m). (m). In particular, we have ρ( Gk+1,l −1 ) < ρ( Gk,l ). Corollary 3.28. Let G be a graph and a and b be two vertices of G which has a path of lengh m adjacent from (m). a to b. Let Gk,l denote the graph obtained from G by adding path Pk+1 and Pl +1 at a and b, 20.

(25) (m). respectively. If 2 6 m 6 k − l, 2 6 l, and λ( Gk+1,l −1 ) 6 x, then (m). (m). det( x, λ( Gk,i )) < det( x, λ( Gk+1,l −1 )). (m). (m). In particular, we have λ( Gk+1,l −1 ) < λ( Gk,l ). Corollary 3.29. Let G be a graph and a and b be two vertices of G which has a path of lengh m adjacent from (m). a to b. Let Gk,l denote the graph obtained from G by adding path Pk+1 and Pl +1 at a and b, (m). respectively. If 2 6 m 6 k − l, 2 6 l, and µ( Gk+1,l −1 ) 6 x, then (m). (m). det( x, µ( Gk,l )) < det( x, µ( Gk+1,l −1 )). (m). (m). In particular, we have µ( Gk+1,l −1 ) < µ( Gk,l ).. 政治大. From above discussions and lemma 3.1, we have the following results.. 立. Corollary 3.30.. ‧ 國. 學. Let Pn be a path with n vertices, Tn be a tree with n vettices, and K1,n−1 be a star graph with n vertices. Then we have. ‧. (1) ρ( Pn ) 6 ρ( Tn ) 6 ρ(K1,n−1 ). sit. y. Nat. (2) λ( Pn ) 6 λ( Tn ) 6 λ(K1,n−1 ). n. al. er. io. (3) µ( Pn ) 6 µ( Tn ) 6 µ(K1,n−1 ). Corollary 3.31.. Ch. engchi. i Un. v. If G is a graph with n (3 6 n) vertices and Pn is a path with n vertices, then (1) ρ( Pn ) 6 ρ( G ) (2) λ( Pn ) 6 λ( G ) (3) µ( Pn ) 6 µ( G ). Example 3.32. Consider G and G1 which are shown in figure 3.2 and 3.3, respectively. Then we can find λ( G ) = 5.0922 > 4.2208 = λ( G1 ) and µ( G ) = 5.3900 > 4.6582 = µ( G1 ). It satisfies corollary 3.22 and 3.23, respectively.. 21.

(26) Figure 3.2: The graph G. 立. 政治大. ‧. ‧ 國. 學. n. er. io. sit. y. Nat. al. Ch. engchi. i Un. Figure 3.3: The graph G1. 22. v.

(27) Chapter 4 Main Results 政治大 of the results are from [2], [3], [4],and [15]. 立. Now, we are going to give some sharp upper and lower bounds for λ( G ) and µ( G ). Most. ‧ 國. 學. 4.1. Sharp Upper Bounds For λ( G ) and µ( G ). ‧. When a graph with n vertices is given, we can use some results in last chapter to find its. y. Nat. n. al. er. io. given graph.. sit. spectral radius. To obtain precisely bounds, we have to consider more properties about the. In the begining, we consider trees. Lemma 4.1. [3]. Ch. engchi. i Un. v. Let u, v ∈ V ( G ), v1 , x2 , ..., vs ∈ N (v) − ( N (u) ∪ u) and x = ( x1 , x2 , ..., xn ) T be the Perron vector of Q, where xi corresponds to the vetex vi . Let G1 be the graph obtained from G by deleting the edges vvi and adding edges uvi . If xu 6 xv , then µ( G ) < µ( G1 ) Proof. Let V ( G ) = { v1 , v2 , ..., vs , vs+1 = vu , vs+2 = vv , ..., vn }. Let v∗u , v∗v ∈ G1 , then d∗u = du + s and d∗v = dv − s. Then x T Q( G1 ) x − x T Q( G ) x = x T ( Q( G1 ) − Q( G )) x. = 2 ∑is=1 xi ( xu − xv ) + s( xu2 − xv2 ) 23.

(28) ≥ 0 since xu ≥ xv . Thus, x T Q( G1 ) x ≥ x T Q( G ) x. Hence, we have µ( G1 ) = max∥y∥=1 y T Q( G1 )y ≥ x T Q( G1 ) x ≥ x T Q( G ) x = µ( G ). If the equality hold, then the above inequalities must be equal. Hence, we have µ( G1 ) = x T Q( G1 ) x = x T Q( G ) x = µ( G ). Namely, µ( G1 ) x = Q( G1 ) x and µ( G ) x = Q( G ) x by lemma 2.7. That is, µ( G1 ) xv = ( Q1 x)v = d∗v xv + ∑vi ∈ NG(1) (v) xi = (dv − s) xv + ∑vi ∈ NG(1) (v) xi . Similarly, we have µ( G ) xv = dv xv + ∑vi ∈ NG (v) xi = dv xv + ∑vi ∈ NG(1) (v) xi + ∑is=1 xi . Because x is the Perron vector of Q, xi ≥ 0 (1 6 i 6 n) , µ( G1 ) xv < µ( G ) xv . Then, µ( G1 ) < µ( G ), a contradiction. Hence, the inequality must be strict. Therefore, we have µ( G ) < µ( G1 ). Theorem 4.2. [3]. 政治大. 立. Let T be a tree with n vertices and k pendant vertices, then. ‧ 國. 學. λ( T ) 6 λ( Tn,k ) and µ( T ) 6 µ( Tn,k ).. ‧. Proof. Given a tree T with n vertices and k pendant vertices.. Let t be the number of vertices whose degree are no less than 3 and such a vertex is called a. y. sit. n. al. er. io. Case 1: t = 0.. Nat. branch vertex.. i Un. v. In this case, T must be a path and we can express it by Tn,2 . We have λ( T ) = λ( Tn,k ). Case 2: t = 1.. Ch. engchi. Consider the line graph L T of T, it is easy to see that the edges which are incident to the branch vertex would form a clique in L T . Let k be the degree of the branch vertex. Then L T contains a Kk . Hence, we see that L T can be obtained by adding paths P1 , ... , Pn−k to each vertex of Kk . Clearly, T and Tn,k are bipartite. We use corollary 3.22 and 3.25 repeatedly if necessary. We have λ( T ) = 2 + ρ( L T ) < 2 + ρ( L Tn,k ) = λ( Tn,k ). By our construct above, we know that λ( T ) = λ( Tn,k ) if and only if T ∼ = Tn,k . Case 3: t > 1. Let x = ( x1 , ..., xn ) T be the Perron vector of Q and each xi corresponds to vi (1 6 i 6 n). Suppose u, v are two branch vertices of T and xu ≥ xv . T is a tree, so there exists a unique path 24.

(29) P between u and v. Let w be the neighbor of v which is along P. Consider {v1 , ..., vdv −2 } ( N (v) − w. We delete the edges vvi and add edges uvi (1 6 i 6 dv − 2), then we obtain a new graph T1∗ . By corollary 3.22, 3.25 and the property about tree, we have λ( T ) < λ( T1∗ ) and the number of the branch vertices decreases to t-1. If t-1 > 1, then we take T1∗ and repeat this construction until branch vertices become 1. So, we have the increasing sequence λ( T1∗ ) < λ( T2∗ ) < ... < λ( Tt∗−1 ). Referring to Case 2, we have λ( Tn∗−1 ) 6 λ( Tn,k ). Thus λ( T ) < λ( Tn,k ) and apply Case 2, we’re done. Since T is a tree, we can easily get µ( T ) 6 µ( Tn,k ). Example 4.3. Let G be a tree with 3 pendant vertices which is shown in figure 4.1. Then we can find µ( T ) = λ( T ) = 4.2143 < 4.3028 = λ( T6,3 ) = µ( T6,3 ). It satisfies theorem 4.2 .. 立. 政治大. ‧. ‧ 國. 學. n. er. io. sit. y. Nat. al. Ch. i Un. v. Figure 4.1: Graph with 6 vertices and 3 pendant vertices.. engchi. 25.

(30) We say that v is a cut vertex of G when G − v has more connected components than G. Next, we consider this condition and discuss the upper bound for the spectral radius. For more detail materials, please refer to [15]. Theorem 4.4. [15] Let G be a graph with n vertices and k cut vertices, then λ( G ) 6 λ( Gn,k ). Proof. Let G be a graph with n vertices and k cut vertices. Without loss of generality, we assume that each cut vertex of G connects exactly two blocks and that both of them are cliques. Suppose that all the blocks of G are Ka1 , Ka2 , ..., Kak+1 and a1 ≥ a2 ≥ ... ≥ ak+1 ≥ 2. If k=0, then λ( G ) 6 n. The equality hold if and only if G = Kn = G0 . If k=n-1, then. 政治大 + 1 since λ( G立 ) 6 ∆( G ) + 1. If a = n − k, then a. G ∼ = Kn . Assume that 1 6 k 6 n − 2. Then a1 = n − ( a2 + ... + ak+1 ) + k 6 n − k. Thus, λ( G ) 6 a1. 1. 2. = a3 = ... = ak+1 = 2.. ‧ 國. 學. (Since a1 ≥ a2 ≥ ... ≥ ak+1 ≥ 2, a1 = n − k, and a1 + a2 + ... + ak+1 = n + k. ) Because a2 ≥ ... ≥ ak+1 ≥ 2, a2 = ... = ak+1 = 2. Hence, G = Kn−k v1 P1 v2 P2 · · · vn−k Pn−k , ∩. V (Kn−k ) = vi , and. ‧. where P1 , ..., Pn−l are disjoint path, Pi is a path of length li , V ( Pi ). y. Nat. ∑in=−1k li = n. Now, by corollary 3.22 repeatedly, we’re done.. io. sit. If a1 6 n − k − 1, then λ( G ) 6 n − k − 1 + 1 = n − k < λ( Gn,k ). ( Since λ( Gn,k ) > n. ). n. al. er. In this case, the inequality must be strict. Theorem 4.5. [15]. Ch. engchi. i Un. v. Let G be a graph with n vertices and k cut vertices, then µ( G ) 6 µ( Gn,k ). Proof. Consider the properties about µ( G ), use corollary 3.23 and use the same method as theorem 4.4. Example 4.6. Let G be the graph with 2 pendant vertices which is shown in figure 4.2. Then we can find λ( G ) = 4.17.1 < 4.3028 = λ( Gn,k ) and µ( G ) = 4.6412 < 4.9354 = µ( Gn,k ). It satisfies theorem 4.4 and 4.5, respectively.. 26.

(31) Figure 4.2: Graph with 5 vertices and 2 pendant vertices.. 4.2. Sharp Lower Bounds For λ( G ) and µ( G ). Let G be a graph. We have shown that ρ( G ) < λ( G ) and µ( G ). Hence, we consider a sharp lower bound of ρ( G ) in the beginning. Lemma 4.7. [2], [3]. ‧ 國. 1 n. ∑in=1 d2i .. 學. Let G be a graph, then ρ( G ) ≥. 立 √. 政治大. The equality holds if and only if G is a regular graph or a semigegular bipartite graph.. ‧. sit. y. Nat. Proof. Let x = ( x1 , x2 , ..., xn ) T be the Perron vector of A. Take c = n1 (1, ..., 1) T . √ √ √ √ √ Then ρ( G ) = ρ( A) = ρ( A2 ) = xA2 x ≥ cA2 c = ( A2 c) · ( A2 c) T = n1 ∑in=1 d2i .. er. io. If the equality holds, then c T A2 c = x T A2 x. Thus, we can say that c is a positive eigenvector. al. of A2 corresponds to ρ( A2 ). If the multiplicity of ρ( A2 ) is one, then by lemma 2.7, c is an. n. iv n C eigenvector of A corresponds to ρ( G ). h Hence, x = c. That e n g c h i Uis, Ac = ρ(G)c. Therefore, G is a 2 regular graph. If the multiplicity of ρ( A ) is two, then −ρ( G ) must be eigenvalue of A. Hence G is a bipartite graph.. [ ] B Without loss of generality,we assume A = BOT O , where B ∈ Ms×(n−s) . ] [ T O . Thus, BB T c1 = ρ( A2 )c1 and B T Bc2 = ρ( A2 )c2 , Then A2 = BB O BT B where c1 = n1 (1, ..., 1) T ∈ Rs and c2 = n1 (1, ..., 1) T ∈ Rn−s . Let x = ( x1 , x2 ), where x1 ∈ Rs and x2 ∈ Rn−s . Then BB T x1 = ρ( A2 ) x1 and B T Bx2 = ρ( A2 ) x2 . Since BB T and B T B have the same nonzero eigenvalues, BB T and B T B have eigenvalue ρ( A2 ) with multiplicity one, respectively. By theorem 2.14, we have x1 = ac1 and x2 = bc2 , for some nonzero constants a and b. Thus, we can obviously say G is a semireg-. 27.

(32) ular bipartite graph. Conversely, if G is a regular graph, say k-regular, then ρ( G ) = k = √ √ 1 n 1 n 2 2 n ∑ i =1 k = n ∑i =1 di . On the other hand, if G is a semiregular bipartite graph, then we [ ] B assume A = BOT O , where B ∈ Ms(n−s) Ni = p1 ti . Thus, Nj = p2 t j , where 1 6 i 6 s and s + 1 6 j 6 n. Let c1 =. 1 T n (1, ..., 1). ∈ Rs and c2 =. 1 T n (1, ..., 1). operations, we have A2 c = p1 p2 c. Then ρ( A2 ) = p1 p2 = c T A2 c = √ ρ( G ) = n1 ∑in=1 d2i .. ∈ Rn−s . Via some 1 n. ∑in=1 d2i , we have. Example 4.8. Let G be the graph is shown in figure 4.3. √ Then we can find ρ( G ) = 2.1701 > 2.1213 = 14 ∑4i=1 d2i . It satisfies lemma 4.7.. 立. 政治大. ‧. ‧ 國. 學. n. Remark 4.9.. Ch. i Un. engchi. By above lemma, we have µ( G ) ≥ λ( G ) ≥. √. 1 n. er. io. al. sit. y. Nat Figure 4.3: The graph G.. v. ∑in=1 ti .. Let G be a graph. We know that G is bipartite if and only if λ( G ) = µ( G ). Now, we consider a bipartite graph G. Theorem 4.10. [2], [3] Let G be a bipartite graph, then µ( G ) = λ( G ) ≥ 2. √. 1 n. ∑in=1 ti .. The equality holds if and only if G is a regular bipartite graph. Proof. Since G is a bipartite graph, µ( G ) = λ( G ). √ √ 1 n 2 By lemma 4.7, λ( G ) = ρ( D + A) ≥ n ∑i=1 (2di ) = 2 n1 ∑in=1 d2i . 28.

(33) By lemma 4.7 again, we have G is a regular bipartite graph if and only if λ( G ) = ρ( D + A) = √ 1 n 2 n ∑i =1 (2di ) . Example 4.11. Let G be the bipartite graph is shown in figure 4.4. √ Then we can find µ( G ) = λ( G ) = 4 > 3.4641 = 2 14 ∑4i=1 ti . It satisfies theorem 4.9.. 政治大. Figure 4.4: The bipartite graph G. ∑vi v j ∈ E(G),i6 j (di + d j − 2)2 .. y. 1 m. sit. √. ‧. Nat. Let G be a bipartite graph, then µ( G ) = λ( G ) ≥ 2 +. 學. Theorem 4.12. [3]. ‧ 國. 立. n. al. er. io. The equality holds if and only if G is a regular bipartite graph or a semiregular bipartite graph or G is an even cycle or G is a copy of P4 .. Ch. engchi. i Un. v. Proof. Since G is a bipartite graph, λ( G ) = 2 + ρ( LG ), where LG is the line graph of G. √ By lemma 4.7 and corollary 3.11, we have λ( G ) ≥ 2 + m1 ∑vi ∼v j ,i< j (di + d j − 2)2 . If the √ equality hold, then λ( G ) = 2 + m1 ∑vi ∼v j ,i< j (di + d j − 2)2 . Since G is bipartite and by √ 1 2 lemma 4.7, we have ρ( LG ) = m ∑vi ∼v j ,i < j ( di + d j − 2) if and only if L G is regular or semiregular. If LG is regular, then if and only if G is either regular or semiregular. On the other hand, if LG is semiregular, then ∆( G ) 6 2, then G is either Cn or Pn . Case 1: G is Cn . Since G is bipartite, n must be even. Thus G is an even cycle. Case 2: G is Pn . Since G is Pn if and only if LG is Pn−1 . 29.

(34) Since we suppose LG is semiregular and we know that LG is a semiregular if and only if n 6 4. If n=4, then G is a copy of P4 . If n=3, then G is P3 . Hence, G is semiregular. If n=2, then G is P2 . Hence, G is regular. Conversely, if G is a copy of P4 , then we’re done. Now, suppose G is regular, semiregular, or even cycle. Since G is regular, semiregular, or even cycle if and √ only if LG is regular if and only if ρ( LG ) = m1 ∑vi ∼v j ,i< j (di + d j − 2)2 by lemma 4.7. G is bipartite, so we finish this theorem. Example 4.13. Let G be the bipartite graph which is the same as example 4.11. √ Then we can find µ( G ) = λ( G ) = 4 = 2 + 13 ∑vi v j ∈E(G),i6 j (di + d j − 2)2 . It satisfies theorem 4.12.. 政治大 If G is an even cycle, then 2ρ( G立 ) = λ( G ) = µ( G ) = 4. Corollary 4.14.. ‧ 國. 學. 4.3. Examples. ‧. In this section, we use all tools which we discussed before to two specific graphs.. sit. y. Nat. io. er. Example 4.15. Let T be the tree is shown in figure 4.5. Then λ( T ) = µ( G ) = 7.0435. In begin with, we use lemma 3.3, we have. n. al. Ch. 1 + ∆( T ) = 7 < 7.0435 = λ( T ) = µ( T ) < 12.. engchi. i Un. v. By crollary 3.30, we have λ( P34 ) = µ( P34 ) < 4 < 7.0435 = λ( T ) = µ( T ) < 34 = λ(K1,34 ) = µ(K1,34 ). On the other hand, by theorem 4.2, λ( T ) = µ( T ) = 7.0435 < 19.0581 = λ( T34,18 ) = µ( T34,18 ). Hence, we find that lemma 3.3 give us a sharp bound.. Example 4.16. Let G be the graph is shown in figure 4.6. Then λ( T ) = 7.2649 and µ( G ) = 10.2329. Clearly, the tree graph T in above example is a subgraph of G. 30.

(35) Figure 4.5: The tree graph T with 34 vertices and 18 pendant vertices. Hence, λ( G ) > 7.0435 and µ( G ) > 7.0435.. 政治大. In begin with, we use lemma 3.3, we have. 立. 1 + ∆( T ) = 7 < 7.2649 = λ( G ) < 34 and 2δ( G ) = 4 < 10.2329 = µ( G ) < 12.. ‧ 國. 學. By crollary 3.30, we have λ( P34 ) < 4 < 7.2649 = λ( G ) < 34 = λ(K1,34 ) and µ( P34 ) < 4 < 10.2329 = µ( G ) < 34 = µ(K1,34 ) On the other hand, by theorem 4.2,. ‧. λ( G ) = 7.2649 < 23.0454 = λ( G34,18 ) and. y. Nat. µ( G ) = 10.2329 < 42.5703 = λ( G34,18 ). Hence, we have. io. sit. 7.0435 < λ( T ) = 7.2649 < 23.0454 and. n. al. er. 7.0435 < µ( G ) = 10.2329 < 12.. Ch. engchi. i Un. v. Figure 4.6: The graph G with 34 vertices and 12 cut vertices.. 31.

(36) Chapter 5 Conclusion 政治大 We recall some preliminaries in the beginning. In chapter 3, we discuss some simple bounds 立 In this thesis, we discuss the spectral radius of a connected graph.. ‧ 國. 學. about the spectral radius of graphs and find the relations among ρ( G ), λ( G ) and µ( G ). In. adition, we give a simple result about the spectral radius of a connected graph. In chapter four,. ‧. we discuss the upper bounds and lower bounds of λ( G ) and µ( G ). Also, we fnd for some kinds. y. Nat. of graph, their upper bound must less than Tn,k or Gn,k . In section 4.3, we give two examples. io. sit. and find that using simple properties of the spectral radius of a graph can help us get more sharp. n. al. er. bounds. Therefore, a more complex bound maynot ensure that we can get a sharp bound.. i Un. v. In the future, we can discuss other kinds of bounds and find more applications. We would. Ch. engchi. like to find bounds of λ( G ) and µ( G ) for other types of graphs.. 32.

(37) Bibliography [1] Douglas B.West, Introduction to graph theory, Prentice Hall, 1996. [2] A.M.Yu, M.Lu, F.Tian, On the spectral radius of graphs, Linear Algebra Appl, 387, 2004, 41-49.. 政治大 [3] Yuan Hong, Xiao-Dong Zhang, Sharp upper and lower bounds for largest eigenvalue of the 立. ‧ 國. 學. Laplacian matrices of trees, Discrete Mathematics, 296, 2005, 187-197. [4] Tomohiro Kawasaki, A sharp upper bound for the largest eigenvalue of the Laplacian matrix. ‧. of a tree, Portland State University M.S. in Mathematical Sciences, 296, 2011, 187-197.. y. Nat. n. al. er. io. 1995.. sit. [5] N.Biggs, Algebraic graph theory, second ed, Cambridge University Press, Cambridge,. Ch. i Un. v. [6] Meyer, C. D. (Carl Dean), Matrix analysis and applied linear algebra, Society for Industrial and Applied, 2000.. engchi. [7] G. Chris, R. Golden, Algebraic graph theory, Springer-Verlag, New York, Inc, 2001. [8] Q. Li, K. Feng, On the largest eigenvalue of a graph, Acta Math, Appl. Sinica 2 (in Chinese): 167-175, 1979 . [9] J. Shu, Y. Hong, K. Wnren, A sharp upper bound on the largest eigenvalue of the Laplacian matrix of a graph, Linear Algebra Appl, 347, 2002, 123-129 . [10] Jianxi Li, Wai Chee Shiu, Wai Hong Chan, The Laplacian spectral radius of some graphs, Linear Algebra Appl, 431, 2009, 99–103. 33.

(38) [11] Dragos M. Cvetkovic, Michael Doob, Horst Sachs, Spectra of graphs : theory and application , Academic Press, 1979. [12] Cvetkovic D., Applications of Graph Spectra: An introduction to the literature, Applications of Graph Spectra, Zbornik radova 13(21), ed. D.Cvetkovi c, I.Gutman, Mathematical Institute SANU, Belgrade, 2009, 7-31. [13] Ji-Ming Guo, The effect on the Laplacian spectral radius of a graph by adding or grafting edges, Linear Algebra Appl, 413, 2006, 59–71. [14] Lihua Feng, Qiao Li and Xiao-Dong Zhang, Some Sharp Upper Bounds on the Spectral Radius of Graphs, TAIWANESE JOURNAL OF MATHEMATICS, 2007. [15] Bao-Xuan Zhu,. 政治大 On the signless Laplacian spectral radius of graphs with cut vertices, 立. ‧ 國. 學. Linear Algebra Appl, 433, 2010, 928–933.. [16] JIAQI JIANG, Introduction To Spectral Graph Theory, 2012.. ‧. [17] Zdenek Dvorak, Bojan Mohar, Spectral radius of finite and infinite planar graphs and of. y. Nat. n. al. er. io. 0907.1591.. sit. graphs of bounded genus, Journal-ref: J. Combin. Theory Ser. B 100 (2010) 729-739, arXiv:. Ch. i Un. v. [18] M. N. Ellingham, Xiaoya Zha, The spectral radius of graphs on surfaces, Journal of. engchi. Combinatorial Theory, Series B, 78, 2000, 45–56.. [19] Xiao-Dong Zhang, The Laplacian eigenvalues of graphs: a survey, Linear Algebra Research Advances, Editor: Gerald D. Ling, pp. 201-228,2007, arXiv:1111.2897v1 .. 34.

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連通圖的拉普拉斯與無符號拉普拉斯 譜半徑之研究 - 政大學術集成

連通圖的拉普拉斯與無符號拉普拉斯譜半徑之研究 - 政大學術集成