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The equality holds if and only ifG is a regular graph.

Proof. Let x = (x₁, x₂, ..., x_n)^T and y = (y₁, y₂, ..., y_n)^T be the Perron vector of A and Q, respectively.

From last chapter, we know that µ(G) = _{∑vi∼vi,i6j}(y_i +y_j)². Thus, ρ(G) = x^TAx = 2∑v_i∼v_i,i6jx_ixj6 ¹₂∑v_i∼v_i,i6j(x_i+x_j)²= ¹₂x^TQx6 ¹₂µ(G).

If the equality holds, all the inequalities above must be equalities.

Then, ρ(G) = 2∑v_i∼v_i,i6jx_ixj = ¹_{2∑vi∼vi,i6j}(x_i+x_j)² . Since 2x_ix_j 6 ^x2_i +x²_j. Hence, x_i = x_j whenv_iandv_j are adjacent. BecauseG is connected, x is the multiple of 1, where all entries of1 are 1. Since Ax = _ρ(G)x, we haveG is regular. Now, suppose G is regular, say k-regular graph. We haveρ(G) = k andµ(G) = 2k are easy to be found.

Example 3.13. LetG be the graph which is the same as example 3.4.

We find thatρ(G) =3.0766,λ(G) = 5.3028 andµ(G) =6.3014.

It is easy to seeρ(G) = 3.0766<3.1052= ¹_2µ(G).

Corollary 3.14.

IfG is a regular bipartite graph, thenρ(G) = ¹_2λ(G) = ¹_2µ(G)

By above discussion,we have the following important corollary which tells us the relation betweenρ(G),λ(G)andµ(G).

Corollary 3.15.

IfG is a graph, thenρ(G) < _λ(G) 6µ(G).

3.4 More Discussions

We introduced some results from [8] and extend these results to more cases.

Lemma 3.16.

LetG be a graph and a∈V(G). We construct a graphG₁by adding a vertex toG adjacent to

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Proof. Using the definition ofdet(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 toa. 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)). Lemma 3.18.

LetG be a graph and G₁be a proper spanning subgraph ofG.

Ifρ(G)6x, then det(x, A(G)) <det(x, A(G₁)). Proof. We prove the statement by induction on |G|=n.

Basis step:The result holds clearly for |G|=2.

Induction step:Assume |G|=|G₁|=n≥^{2. Suppose}V(G) =V(G₁) = {x₁, ..., x_n}^. are nonnegative and at least one item is positive.

Whenx≥ρ(G), we have _dx^d det(x, A(G₁))−_dx^{d det}(x, A(G)) >0 ...(3).

By (2) and (3), we’re done.

Using the same method, we get the following two corollaries.

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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 G_k,l denote the graph obtained fromG by adding path P_k+1 and P_l+1 at a as one of the end points. If 2 6 ^l 6 ^{k and} µ(G_k+1,l−1) 6 ^{x, then} det(x,µ(G_k,l)) < det(x,µ(G_k+_1,l−1)).

In particular, we haveµ(G_k+_1,l−1) <_µ(G_k,l).

Proof. Use lemma 3.17 and by the same method as lemma 3.21, we finish this corollary.

Theorem 3.24. right hand side of (1) < 0. Hence, we’re done.

Case 2: Whenk>l ≥2.

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Proof. Use corollary 3.19 and by the same method as theorem 3.24, we finish this corollary.

Corollary 3.26.

LetG be a graph and a and b be two adjacent vertices of G. Let G_k,l⁽¹⁾denote the graph obtained fromG by adding path P_k+₁andP_l+₁ata and b, respectively.

If26l 6k andµ(G⁽_k¹₊⁾_1,l−1)6x, then det(x,µ(G_k,l⁽¹⁾)) <det(x,µ(G_k⁽¹₊⁾_1,l−1)). In particular, we haveµ(G⁽_k¹₊⁾_1,l−1) <_µ(G_k,l⁽¹⁾).

Proof. Use corollary 3.20 and by the same method as theorem 3.24, we finish this corollary.

Using above theorem and corollaries, the following three corollaries are easy to be ckecked.

Corollary 3.27.

LetG be a graph and a and b be two vertices of G which has a path of lengh m adjacent from a to b. Let G⁽_k,l^m) denote the graph obtained from G by adding path P_k+₁ andP_l+₁at a and b, respectively. If26m6k−l, 26l, andρ(G⁽_k^m₊⁾_1,l−1)6x, then

det(x,ρ(G_k,l^(m))) <det(x,ρ(G⁽_k^m₊⁾_1,l−1)). In particular, we haveρ(G_k^(m₊⁾_1,l−1) < _ρ(G_k,l^(m)). Corollary 3.28.

LetG be a graph and a and b be two vertices of G which has a path of lengh m adjacent from

(_m)

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respectively. If26m6k−l, 26l, andλ(G⁽_k^m₊⁾_1,l−1)6x, then det(x,λ(G⁽_k,i^m))) <det(x,λ(G_k^(m₊⁾_1,l−1)).

In particular, we haveλ(G_k^(m₊⁾_1,l−1) < _λ(G⁽_k,l^m)). Corollary 3.29.

LetG be a graph and a and b be two vertices of G which has a path of lengh m adjacent from a to b. Let G⁽_k,l^m) denote the graph obtained from G by adding path P_k+₁ andP_l+₁at a and b, respectively. If26^m6^k−^{l, 2}6^{l, and}µ(G_k^(m₊⁾_1,l−1) 6^{x, then}

det(x,µ(G_k,l^(m))) <det(x,µ(G⁽_k^m₊⁾_1,l−1)). In particular, we haveµ(G⁽_k^m₊⁾_1,l−1) <_µ(G_k,l^(m)).

From above discussions and lemma 3.1, we have the following results.

Corollary 3.30.

LetP_nbe a path with n vertices, T_n be a tree with n vettices, andK_1,n−1be a star graph with n vertices. Then we have

(1)ρ(P_n) 6ρ(T_n) 6ρ(K_1,n−1) (2)λ(P_n) 6λ(T_n) 6λ(K_1,n−1) (3)µ(P_n)6µ(T_n) 6µ(K_1,n−1). Corollary 3.31.

IfG is a graph with n (36n) vertices and Pn is a path with n vertices, then (1)ρ(P_n) 6ρ(G)

(2)λ(P_n) 6λ(G) (3)µ(P_n)6µ(G).

Example 3.32. Consider G and G₁which are shown in figure 3.2 and 3.3, respectively. Then we can find λ(G) = 5.0922 > 4.2208 = _λ(G₁)andµ(G) = 5.3900 > 4.6582 = _µ(G₁). It satisfies corollary 3.22 and 3.23, respectively.

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Figure 3.2: The graphG

Figure 3.3: The graphG₁

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Chapter 4 Main Results

Now, we are going to give some sharp upper and lower bounds forλ(G) andµ(G). Most of the results are from [2], [3], [4],and [15].

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 spectral radius. To obtain precisely bounds, we have to consider more properties about the given graph.

In the begining, we consider trees.

Lemma 4.1. [3]

Letu, v ∈ ^V(G), v₁, x₂, ..., v_s ∈ ^N(v)− (^N(u)∪^u)andx = (x₁, x₂, ..., x_n)^T be the Perron vector ofQ, where x_icorresponds to the vetexv_i.

LetG₁be the graph obtained fromG by deleting the edges vv_iand adding edgesuv_i. Ifx_u 6x_v, thenµ(G) <_µ(G₁)

Proof. LetV(G) ={^v1, v₂, ..., v_s, v_s+₁ =v_u, v_s+₂ =v_v, ..., v_n }^. Letv^∗_u, v^∗_v ∈ ^G1, thend^∗_u =d_u+s and d^∗_v =d_v−^s.

Thenx^TQ(G₁)x−x^TQ(G)x= x^T(Q(G₁)−Q(G))x

=2∑_i^s=₁x_i(x_u−x_v) +s(x²_u−x²_v)

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If the equality hold, then the above inequalities must be equal. Hence, we have µ(G₁) = x^TQ(G₁)x =x^TQ(G)x =_µ(G). contradiction. Hence, the inequality must be strict. Therefore, we haveµ(G) <_µ(G₁).

Theorem 4.2. [3]

LetT be a tree with n vertices and k pendant vertices, then λ(T) 6λ(T_n,k)andµ(T) 6µ(T_n,k).

Proof. Given a treeT 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 branch vertex.

Case 1: t=0.

In this case,T must be a path and we can express it by T_n,2. We haveλ(T) = _λ(T_n,k). Case 2: t=1.

Consider the line graphL_TofT, it is easy to see that the edges which are incident to the branch vertex would form a clique inL_T. Let k be the degree of the branch vertex. Then L_T contains a K_k. Hence, we see that L_T can be obtained by adding paths P₁, ... , P_n−k to each vertex of K_k. Clearly, T and T_n,k are bipartite. We use corollary 3.22 and 3.25 repeatedly if necessary.

We haveλ(T) =2+_ρ(L_T) <2+_ρ(L_Tn,k) = _λ(T_n,k). By our construct above, we know that λ(T) = _λ(T_n,k)if and only if T∼= T_n,k.

Case 3: t>1.

Let x = (x₁, ..., x_n)^T be the Perron vector of Q and each x_i corresponds to v_i (1 6ⁱ 6 ^n).

Suppose u, v are two branch vertices ofT and x ≥^{x .} T is a tree, so there exists a unique path

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P between u and v. Let w be the neighbor of v which is along P.

Consider {v₁, ..., v_dv−2} ( N(v)−w. We delete the edges vv_i and add edgesuv_i (16ⁱ 6 d_v−2), then we obtain a new graph T₁^∗. By corollary 3.22, 3.25 and the property about tree, we haveλ(T) <_λ(T₁^∗)and the number of the branch vertices decreases to t-1. If t-1 > 1, then we takeT₁^∗ and repeat this construction until branch vertices become 1. So, we have the increasing sequenceλ(T₁^∗) <λ(T₂^∗) < ... <λ(T_t^∗₋₁). Referring to Case 2, we haveλ(T_n^∗₋₁) 6 λ(T_n,k). Thus λ(T) < _λ(T_n,k) and apply Case 2, we’re done. Since T is a tree, we can easily get µ(T)6µ(T_n,k).

Example 4.3. LetG be a tree with 3 pendant vertices which is shown in figure 4.1.

Then we can findµ(T) = _λ(T) =4.2143 <4.3028 =_λ(T_6,3) =_µ(T_6,3). It satisfies theorem 4.2 .

Figure 4.1: Graph with 6 vertices and 3 pendant vertices.

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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]

LetG be a graph with n vertices and k cut vertices, thenλ(G) 6λ(G_n,k).

Proof. Let G be a graph with n vertices and k cut vertices. Without loss of generality, we as-sume that each cut vertex ofG connects exactly two blocks and that both of them are cliques.

Suppose that all the blocks of G are K_a1, K_a2, ..., K_ak+1 and a₁ ≥ ^a2 ≥ ^... ≥ ^ak+₁ ≥ ^{2. If} k=0, then λ(G) 6 n. The equality hold if and only if G = K_n = G₀. If k=n-1, then G ∼= K_n. Assume that 1 6 k 6 n−2. Then a₁ = n− (a₂+...+a_k+₁) +k 6 n−k.

Thus,λ(G) 6a₁+1 sinceλ(G) 6∆(G) +1. If a₁ =n−k, then a₂ =a₃ =...= a_k+₁=2.

(Sincea₁ ≥a₂ ≥...≥a_k+₁ ≥2, a₁ =n−k, and a₁+a₂+...+a_k+₁=n+k. )

Becausea₂ ≥... ≥a_k+₁ ≥2, a₂ =... =a_k+₁ =2. Hence, G=K_n−kv₁P₁v₂P₂· · ·v_n−kP_n−k, where P₁, ..., P_n−l are disjoint path, P_i is a path of length l_i, V(P_i)^∩V(K_n−k) = v_i, and

∑ⁿ_i=⁻₁^kl_i =n. Now, by corollary 3.22 repeatedly, we’re done.

Ifa₁ 6ⁿ−^k−^{1, then}λ(G) 6ⁿ−^k−¹+1=n−^k <_λ(G_n,k). ( Sinceλ(G_n,k) >n. ) In this case, the inequality must be strict.

Theorem 4.5. [15]

LetG be a graph with n vertices and k cut vertices, thenµ(G)6µ(G_n,k).

Proof. Consider the properties about µ(G), use corollary 3.23 and use the same method as theorem 4.4.

Example 4.6. LetG 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(G_n,k) andµ(G) = 4.6412 < 4.9354 = µ(G_n,k). It satisfies theorem 4.4 and 4.5, respectively.

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Figure 4.2: Graph with 5 vertices and 2 pendant vertices.

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