# Algebraic Connectivity of Connected Graphs with Fixed Number of Pendant Vertices

**ABSTRACT** In this paper, we consider the following problem. Over the class of all simple connected graphs of order n with k pendant vertices (n, k being fixed), which graph maximizes (respectively, minimizes) the algebraic connectivity? We also discuss the algebraic connectivity

of unicyclic graphs.

KeywordsLaplacian matrix–Algebraic connectivity–Characteristic set–Perron component–Pendant vertex

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**ABSTRACT:**In this paper we characterize the unique graph whose algebraic connectivity is minimum among all connected graphs with given order and fixed matching number or edge covering number, and present two lower bounds for the algebraic connectivity in terms of the matching number or edge covering number.01/2014; - SourceAvailable from: Claudia Marcela Justel[Show abstract] [Hide abstract]

**ABSTRACT:**Several approaches for ordering graphs by spectral parameters are presented in the literature. We can find graph orderings either by the greatest eigenvalue (spectral radius or index) or by the sum of the absolute values of the eigenvalues (the energy of a graph) or by the second smallest eigenvalue of the Laplacian matrix (the algebraic connectivity), among others. By considering the fact that the algebraic connectivity is related to the connectivity and shape of the graphs, several structural properties of graphs relative to this parameter have been studied. Hence, a large number of papers about ordering graphs by algebraic connectivity, mainly about trees and graphs with few cycles, have been published. This paper surveys the significant results concerning these topics, trying to focus on possible points to be investigated in order to understand the difficulties to obtain partial orderings via algebraic connectivity.Linear Algebra and its Applications 01/2014; 458:429–453. · 0.97 Impact Factor

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arXiv:1003.4646v1 [math.CO] 24 Mar 2010

ALGEBRAIC CONNECTIVITY OF CONNECTED GRAPHS WITH

FIXED NUMBER OF PENDANT VERTICES

Arbind K. Lal1, Kamal L. Patra2and Binod K. Sahoo3

Abstract. In this paper we consider the following problem: Over the class of all simple

connected graphs of order n with k pendant vertices (n,k being fixed), which graph max-

imizes (respectively, minimizes) the algebraic connectivity? We also discuss the algebraic

connectivity of unicyclic graphs.

Keywords: Laplacian matrix; Algebraic connectivity; Characteristic set; Perron compo-

nent; Pendant vertex.

1. Introduction

All graphs considered here are finite, simple and undirected. Let G be a graph with

vertex set V = {v1,··· ,vn}. The number n of vertices is called the order of G. The

adjacency matrix A(G) of G is defined as A(G) = [aij], where aij is equal to one, if the

unordered pair {vi,vj} is an edge of G and zero, otherwise. Let D(G) be the diagonal

matrix of the vertex degrees of G. The Laplacian matrix L(G) of G is defined as L(G) =

D(G) − A(G). We refer to [13, 14] for a general overview on results related to Laplacians.

It is well known that L(G) is a symmetric positive semidefinite M-matrix. The smallest

eigenvalue of L(G) is zero with the vector of all ones as its eigenvector. It has multiplicity

one if and only if G is connected. In other words, the second smallest eigenvalue of L(G) is

positive if and only if G is connected. Viewing the second smallest eigenvalue as an algebraic

measure of connectivity, Fiedler termed this eigenvalue as the algebraic connectivity of G,

denoted µ(G). The following two lemmas are well known.

Lemma 1.1 ([4], p.223). Let G be a graph. Let?G be the graph obtained from G by adding

Lemma 1.2 ([5], 3.2, p.299). Let G be a non-complete graph. Let G′be the graph obtained

from G by joining two non-adjacent vertices of G with an edge. Then µ(G) ≤ µ(G′).

a pendant vertex to a vertex of G. Then µ(?G) ≤ µ(G).

Lemma 1.2 implies that, over all connected graphs, the maximum algebraic connectivity

occurs for the complete graph and the minimum algebraic connectivity occurs for a tree.

It is also known that over all trees of order n, the path (denoted by Pn) has the minimum

1Department of Mathematics and Statistics, Indian Institute of Technology, Kanpur–208016, India;

e-mail: arlal@iitk.ac.in

2School of Mathematical Sciences, National Institute of Science Education and Research, Sainik School

Post, Bhubaneswar–751005, India; e-mail: klpatra@niser.ac.in

3School of Mathematical Sciences, National Institute of Science Education and Research, Sainik School

Post, Bhubaneswar–751005, India; e-mail: bksahoo@niser.ac.in

1

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2A. K. LAL, K. L. PATRA AND B. K. SAHOO

algebraic connectivity and this minimum is given by 2(1 − cosπ

proved that, among all trees of order n > 2, the star K1,n−1uniquely attains the maximum

algebraic connectivity which is equal to 1 ([12], Corollary 2, p.118).

There is a good deal of work on the algebraic connectivity of graphs. We refer to [1]–

[7],[10],[17] for various works on the algebraic connectivity of connected graphs having

certain graph theoretic properties. For all graph theoretic terms used in this paper (but

not defined), the reader is advised to look at the book Graph Theory by Harary [9]. In this

paper, we consider the problem of extremizing the algebraic connectivity over all connected

graphs of order n with k pendant vertices for fixed n and k, 0 ≤ k ≤ n − 1.

The paper is arranged as follows: In Section 2, we recall some results from the literature

that will be used in the subsequent sections. In Section 3, we record some elementary

results and prove a result which gives conditions under which equality is attained in Lemma

1.2. We study the algebraic connectivity of connected graphs with and without pendant

vertices in Section 4 and Section 5, respectively. Finally, in Section 6, we discuss the

algebraic connectivity of unicyclic graphs.

n) ([5], p.304). Merris

2. Preliminaries

Let G be a connected graph. The distance d(u,v) between two vertices u and v of G is

the length of a shortest path from u to v. The diameter of G is defined by max

u,vd(u,v).

A pendant vertex of G is a vertex of degree one. For a set W of vertices of G, G − W

denotes the graph obtained from G by removing the vertices in W and all edges incident

with them. If W = {v} consists of one vertex only, we simply denote G − W by G − v

and refer to the connected components of G−v as the connected components of G at v. A

vertex v of G is called a cut-vertex if there are at least two connected components of G at

v.

An eigenvector of L(G) corresponding to µ(G) is called a Fiedler vector of G. Let Y

be a Fiedler vector of G. By Y (v), we mean the co-ordinate of Y corresponding to the

vertex v. A vertex v of G is called a characteristic vertex with respect to Y if Y (v) = 0 and

there exists a vertex w adjacent to v such that Y (w) ?= 0. An edge {u,w} of G is called a

characteristic edge with respect to Y if Y (u)Y (w) < 0. The characteristic set of G with

respect to Y , denoted C(G,Y ), is the set of all characteristic vertices and edges of G. By

a result of Fiedler [6], either C(G,Y ) consists of one vertex only or C(G,Y ) is contained

in a block of G. For a tree, it is known that C(G,Y ) is independent of Y and it contains

either a vertex or an edge. We refer to [1] for more general results on the size of C(G,Y ).

.............

..................

v1

v2

vd

l

k

Figure 1. The tree T(k,l,d)

Page 3

ALGEBRAIC CONNECTIVITY WITH PENDANT VERTICES3

For a fixed positive integer n, the path [v1v2···vn] on n vertices is denoted by Pn. For

positive integers k,l,d with n = k + l + d, let T(k,l,d) be the tree of order n obtained by

taking the path Pdand adding k pendant vertices adjacent to v1and l pendant vertices

adjacent to vd(see Figure 1). The path Note that T(1,1,d] is a path on d+2 vertices. The

next proposition determines the tree, up to isomorphism, which minimizes the algebraic

connectivity over all trees of order n with fixed diameter.

Proposition 2.1 ([2], Theorem 3.2, p.58). Among all trees of order n with fixed diameter

d + 1, the minimum algebraic connectivity is uniquely attained by T?⌈n−d

Now, consider a path v1v2···vdon d ≥ 3 vertices and add n−d pendant vertices adjacent

to the vertex vj, where j = ⌊d+1

(see Figure 2). Then Td

n−dis a tree of order n with diameter d − 1. The next proposition

determines the tree, up to isomorphism, which maximizes the algebraic connectivity over

all trees of order n with fixed diameter.

2⌉,⌊n−d

2⌋,d?.

2⌋. The new graph thus constructed is denoted by Td

n−d

.........

v2

.........

. . ...........

?

???

v1

vj

vd

vd−1

n − d

Figure 2. The tree Td

n−d

Proposition 2.2 ([2], p.62–65). Among all trees of order n with fixed diameter d+1, the

maximum algebraic connectivity is uniquely attained by Td+2

n−d−2.

Fix a vertex v of G and let l,k ≥ 1. We construct a new graph Gk,lfrom G by attaching

two new paths P : vv1v2...vkand Q : vu1u2...ulat v of lengths k and l, respectively,

where u1,u2,...,ul,v1,v2,...,vkare distinct new vertices. Let?Gk,lbe the graph obtained

We say that?Gk,lis obtained from Gk,lby grafting an edge. The next result compares the

a tree.

from Gk,lby removing the edge {vk−1,vk} and adding a new edge {ul,vk} (see Figure 3).

algebraic connectivity of Gk,lwith that of?Gk,l≃ Gk−1,l+1whenever the initial graph G is

Proposition 2.3 ([17], Theorem 2.4, p.861). Let G be a tree of order n ≥ 2 and v be a

vertex of G. For l,k ≥ 1, let Gk,land?Gk,l≃ Gk−1,l+1be the graphs as defined above with

Let v be a vertex of G and C1,C2,··· ,Ckbe the connected components of G at v. Note

that k ≥ 2 if and only if v is a cut-vertex of G. For each i = 1,··· ,k, we denote by?L(Ci)

M-matrix and has nullity 1, it follows that?L(Ci) is a non-singular M-matrix and hence

Theorem [15],?L(Ci)−1has a simple positive dominant eigenvalue, called the Perron value

respect to v. If l ≥ k, then µ(Gk−1,l+1) ≤ µ(Gk,l).

the principal sub-matrix of L(G) corresponding to the vertices of Ci. Since L(G) is an

?L(Ci)−1is a positive matrix, called the bottleneck matrix of Ci. By the Perron-Frobenius

Page 4

4A. K. LAL, K. L. PATRA AND B. K. SAHOO

GG

...........

......

......

vv

v1

vk−1

vk

u1 u2

ul−1 ul

vk−1

vk−2

v1

u1

u2

ul−1ul

vk

Gk,l

˜Gk,l≃ Gk−1,l+1

Figure 3. Grafting an edge

of Ciand is denoted by ρ(?L(Ci)−1). A corresponding eigenvector of ρ(?L(Ci)−1) with all

G at v if its Perron value is maximum among that the Perron values of C1,··· ,Ck.

The next proposition gives the description of the entries of the bottleneck matrices of a

tree.

entries positive is called a Perron vector of Ci. We say that Ciis a Perron component of

Proposition 2.4 ([11], Proposition 1, p.313). Let T be a tree, v be a vertex of T and C

be a connected component of T at v. Then?L(C)−1= [mij], where mij is the number of

common edges between the paths Pivjoining i and v and Pjvjoining j and v.

The following proposition connecting Perron components, bottleneck matrices and alge-

braic connectivity recasts some of the results obtained in [2, 10]. Throughout the paper,

we denote by I the identity matrix and by J the matrix of all ones of appropriate orders.

For a symmetric matrix M, λ(M) denotes its largest eigenvalue.

Proposition 2.5. Let G be a connected graph with a cut-vertex v. Let C1,C2,··· ,Ckbe

the connected components of G at v with C1as a Perron component. Then the following

results hold.

(i) There is a unique non-negative number x such that

??L(C1)−1− xJ

(ii) If there exists a non-negative number x such that

??L(C1)−1− xJ

then α is an eigenvalue of L(G).

λ

?

= ρ

??L(C2)−1⊕ ··· ⊕?L(Ck)−1⊕ [0] + xJ

?

=

1

µ(G).

λ

?

= ρ

??L(C2)−1⊕ ··· ⊕?L(Ck)−1⊕ [0] + xJ

?

=1

α,

The next result is a special case of Proposition 2.5(i).

Proposition 2.6 ([10], Theorem 8, p.146). Let G be a connected graph and {vi,vj} be

an edge not on any cycle of G. Let Cibe the connected component of G at vj containing

vi and Cj be the connected component of G at vi containing vj. Then Ci is the Perron

Page 5

ALGEBRAIC CONNECTIVITY WITH PENDANT VERTICES5

component of G at vjand Cjis the Perron component of G at viif and only if there exists

a γ ∈ (0,1) such that

??L(Ci)−1− γJ

In that case, µ(G) is a simple eigenvalue of L(G) and the characteristic set of G is a

singleton consisting of the edge {vi,vj}.

ρ

?

= ρ

??L(Cj)−1− (1 − γ)J

?

=

1

µ(G).

Identification of Perron components at a vertex helps to determine the location of the

characteristic set in G. The next proposition is a result in that direction.

Proposition 2.7 ([10], Theorem 1 and Corollary 1.1). Let G be a connected graph. For

any vertex v of G which is not a characteristic vertex, the unique Perron component at v

contains the vertices in any characteristic set. A cut-vertex v is a characteristic vertex of

G if and only if there are at least two Perron components of G at v, and in that case

µ(G) =

1

ρ(?L(C)−1),

where C is a Perron component of G at v.

Remark 2.8. Recall that P2n+1 is a path on 2n + 1 vertices with the vertex vi being

adjacent to the vertex vi+1for i = 1,2,...,2n. Let C1and C2be the two components of

P2n+1−vn+1. Then using symmetry, it can be easily observed that the vertex labelled vn+1

is the characteristic vertex. Hence by Proposition 2.7

ρ(?L(C)−1) =

1

µ(P2n+1)=

1

2(1 − cos(

π

2n+1)).

For non-negative square matrices A and B (not necessarily of the same order), A ≪ B

means that there exists a permutation matrix P such that PAPTis entri-wise dominated

by a principal sub-matrix of B, with strict inequality in at least one position in the case

A and B have the same order. A useful fact from the Perron-Frobenius theory is that if B

is irreducible and A ≪ B, then ρ(A) < ρ(B). We use this fact together with Propositions

2.4 and 2.5 (mostly without mention) to get information about the algebraic connectivity

of a graph, and in particular, for a tree.

3. Initial Results

We start with the following observation. Let Knbe the complete graph of order n ≥ 2

and v be a vertex of Kn. Let C denote the only connected component of Knat v. We

have L(Kn) = nI − J, an n-by-n matrix. Let?L(C) be the principal sub-matrix of L(Kn)

can be verified that?L(C)−1=1

the Perron-Frobenius Theorem, the Perron value of C is one.

corresponding to the vertices of C. Then?L(C) = nI − J, an (n − 1)-by-(n − 1) matrix. It

vector of all ones is an eigenvector of?L(C)−1corresponding to the eigenvalue one. So, by

nI +1

nJ. Since the sum of each row of?L(C)−1is one, the

Page 6

6A. K. LAL, K. L. PATRA AND B. K. SAHOO

Lemma 3.1. Let v be a vertex of a connected graph G. For a connected component C of

G at v, let?Cvdenote the induced subgraph of G on the vertices of C together with v. Then

(i) If? Cvis complete, then the Perron value of C is one.

then µ(G) = 1.

the following results hold.

(ii) If v is a cut-vertex and?Cvis complete for each connected component C of G at v,

Proof. (i) The principal sub-matrix of L(G) and that of L(?Cv) corresponding to the vertices

this section, it follows that the Perron value of C is one.

(ii) By (i), ρ(?L(C)−1) = 1 for each connected component C of G at v as µ(G) = 1

A consequence of Lemma 3.1 is the following.

of C are the same. If?Cvis complete, then replacing Knby?Cvin the first paragraph of

follows from Proposition 2.7.

?

Corollary 3.2. Let G be a connected graph with a cut-vertex v. Then µ(G) ≤ 1.

Proof. Let?G be the graph obtained from G by making?Cv complete for each connected

pair (?G,v), we get that µ(?G) = 1. So µ(G) ≤ 1.

The statements of Lemma 3.1 and Corollary 3.2 were implicitly mentioned in ([2], Ex-

ample 1.5, p.51). We write them here with their proofs for the sake of completeness and

for later use in this paper. We also note that Corollary 3.2 follows from ([5], 4.1, p.303)

since the vertex connectivity of G is one.

component C of G at v. By Lemma 1.2, µ(G) ≤ µ(?G). Applying Lemma 3.1(ii) to the

?

Lemma 3.3. Let v be a vertex of a connected graph G. If v is not a cut-vertex of G, then

the Perron value of C = G − v (the only connected component of G at v) is at least one.

Proof. Let?G be the graph obtained from G by adding one pendant vertex adjacent to v.

of L(G) and that of L(?G) corresponding to the vertices of C1are the same. The Perron

component (use Proposition 2.5) and hence ρ(?L(C)−1) = ρ(?L(C1)−1) ≥ ρ(?L(C2)−1) =

The next theorem gives conditions under which the equality is attained in Lemma 1.2.

Let C1= C and C2be the two connected components of?G at v. The principal sub-matrix

value of C2is 1 and µ(?G) ≤ 1 (Corollary 3.2). Then the component C2cannot be a Perron

1.

?

Theorem 3.4. Let v be a vertex of a connected graph G such that v is not a cut-vertex of

G. Form a new graph?G from G by adding t ≥ 1 pendant vertices v1,··· ,vtadjacent to v.

Then µ(?G) = µ(?G).

Proof. Let G∗be the graph obtained from?G by addingt(t−1)

we assume that µ(?G) < 1 and prove that µ(G∗) ≤ µ(?G).

Let?G be the graph obtained from?G by adding k, 0 ≤ k ≤

t(t−1)

2

, edges among v1,··· ,vt.

2

edges among v1,··· ,vt. By

Lemma 1.2 and Corollary 3.2, µ(?G) ≤ µ(?G) ≤ µ(G∗) ≤ 1. If µ(?G) = 1, we are done. So,

Page 7

ALGEBRAIC CONNECTIVITY WITH PENDANT VERTICES7

Let C1,C2,··· ,Ct+1be the t+1 connected components of?G at v, where C1= G−v. For

the Perron value of C1is at least 1. So, by definition, C1is a Perron component of?G at v.

??L(C1)−1− θJ

Since µ(?G) < 1, the second equality implies that θ > 0.

matrix with Perron value

2 ≤ i ≤ t + 1, the Perron value of Ciis one. As v is not a cut-vertex of G, by Lemma 3.3,

By Proposition 2.5(i), there is a unique non-negative number θ such that

?

λ

= ρ

??L(C2)−1⊕ ··· ⊕?L(Ct+1)−1⊕ [0] + θJ

?

=

1

µ(?G).

Let M =?L(C2)−1⊕ ··· ⊕?L(Ct+1)−1⊕ [0] + θJ. Thus M is a (t + 1)-by-(t + 1) positive

take β = θ(y1+ ··· + yt+1). Then β > 0. A simple calculation shows that yi+ β =

1

µ(?G)yt+1. Since 0 < µ(?G) < 1, it follows that y1= ··· = yt. So

There are two connected components of G∗at v, say D1 = C1and D2. The Perron

value of D2is one by Lemma 3.1(i). So D1is a Perron component of G∗at v since it has

Perron value at least one. Let M∗=?L(D2)−1⊕ [0] + θJ. Using?L(D2)−1=

Perron-Frobenius Theorem, ρ(M∗) =

µ(? G). Thus,

??L(D2)−1⊕ [0] + θJ

Now, by Proposition 2.5(ii), µ(? G) is an eigenvalue of L(G∗) and hence µ(G∗) ≤ µ(? G)

proof.

1

µ(? G). Let Y = [y1,··· ,yt,yt+1]Tbe a Perron vector of M and

1

µ(? G)yi

for 1 ≤ i ≤ t and β =

without loss of generality we may take Y = [1,··· ,1,y]Tfor some y > 0.

1

t+1I +

1

t+1J

(see the first paragraph of this section), it can be verified that M∗Y =

1

µ(? G)Y. So, by the

1

ρ

?

=

1

µ(?G)

= λ

??L(C1)−1− θJ

?

= λ

??L(D1)−1− θJ

?

.

as, by definition, µ(G∗) is the smallest nonzero eigenvalue of L(G∗). This completes the

?

4. With Pendant Vertices

Let Hn,kdenote the class of all connected graphs of order n with k pendant vertices.

We may assume that 1 ≤ k ≤ n − 1 and hence n ≥ 3. We first consider the question of

maximizing the algebraic connectivity over Hn,k. We now define a collection of graphs,

denoted Pk

n, which have exactly k pendant vertices and hence are members of Hn,k. For

k ?= n − 2, the graph Pk

nis obtained by adding k pendant vertices adjacent to a single

vertex of the complete graph Kn−k. There is no graph of order n = 3 with exactly one

pendant vertex. For n ≥ 4 and k = n − 2, the graph Pn−2

pendant vertices adjacent to a pendant vertex of the path P3.

n

is obtained by adding n − 3

Theorem 4.1. The graph Pk

Hn,kand this maximum value is equal to one.

n, k ?= n−2, attains the maximum algebraic connectivity over

Proof. Let G be a graph in Hn,k. Since k ≥ 1, G has a cut-vertex. By Corollary 3.2,

µ(G) ≤ 1. Also Pk

n, k ?= n−2, has a cut-vertex satisfying the hypothesis of Lemma 3.1(ii).

So µ(Pk

n).

n) = 1 and hence µ(G) ≤ µ(Pk

?

Page 8

8A. K. LAL, K. L. PATRA AND B. K. SAHOO

.....

?

n − 3

Pn−2

n

Kn−k

.....

?

k

Pk

n,k ?= n − 2

Figure 4. The graph Pk

n

Remark 4.2 (Uniqueness). For k = n − 1, Hn,k contains only one graph, namely,

K1,n−1≃ Pn−1

n

. For k = n − 3, let G be a graph in Hn,n−3. If G is a tree then clearly

µ(G) < 1. Otherwise G has a unique 3-cycle. By ([2], Theorem 4.14, p.73), it follows that

µ(G) = 1 if and only if G ≃ Pn−3

n

. However, for 1 ≤ k ≤ n−4, Pk

(up to isomorphism) in Hn,khaving algebraic connectivity one. We give an example below.

nis not the unique graph

Example 4.3. Let 1 ≤ k ≤ n − 4. Let v1,··· ,vn−k−1be n − k − 1 pendant vertices in

the star K1,n−1. Let G be the graph obtained from K1,n−1by adding new edges {vi,vi+1},

1 ≤ i ≤ n − k − 2. Then G is a member of Hn,k. Since n ≥ 5, G is not isomorphic to Pk

But by Theorem 3.4, µ(G) = µ(K1,n−1) = 1.

n.

Theorem 4.4. The graph Pn−2

n

uniquely maximizes the algebraic connectivity over Hn,n−2.

Proof. Any graph in Hn,n−2is a tree with diameter three. By Proposition 2.2, the maximum

algebraic connectivity is uniquely attained by the tree T4

n−4≃ Pn−2

n

over Hn,n−2.

?

We next consider the question of minimizing the algebraic connectivity over Hn,k. We

first prove the result for k = 1 (and so n ≥ 4). We denote by Cn−3

adding a path of order n − 3 to a vertex of K3(see Figure 5). Then Cn−3

Hn,1.

3

the graph obtained by

is a member of

3

...........

vn−4

v1

v2

vn−3

Figure 5. The graph Cn−3

3

Theorem 4.5. The graph Cn−3

Hn,1.

3

uniquely attains the minimum algebraic connectivity over

Proof. For n = 4, C1

adjacent to a vertex of a 4-cycle) is the only graph in H5,1, non-isomorphic to C2

least possible edges. The other graphs, non-isomorphic to C2

C1

4by joining some non-adjacent vertices. So, for any other graph G on 5 vertices and

with one pendant vertex,

3is the only graph in H4,1. For n = 5, C1

4(adding a pendant vertex

3, with

3, in H5,1can be obtained from

µ(G) ≥ µ(C1

4) ≈ 0.8299 > 0.5188 ≈ µ(C2

3).

Page 9

ALGEBRAIC CONNECTIVITY WITH PENDANT VERTICES9

Let n ≥ 6. Any graph in Hn,1has at least one cycle. Let G be a graph in Hn,1which is

not isomorphic to Cn−3

3

. Delete some of the edges from the cycles of G to get a tree G1.

Further, we delete these edges in such a way that G1is neither isomorphic to the path Pn

nor to the graph G2(G is not isomorphic to Cn−3

3

) as shown in Figure 6. This is possible

............

vn−1

vn

vn−2

v2

v1

Figure 6. The tree G2

since n ≥ 6. By Lemma 1.2, µ(G1) ≤ µ(G). Starting with G1, by a finite sequence of

graph operations consisting of grafting of edges, we can get the tree G2from G1. Also note

that we must get the graph?G2as shown in Figure 7 in the penultimate step of getting

G2. Since G2≃ T(2,1,n − 3), µ(G2) < µ(?G2) by Proposition 2.1 and µ(?G2) ≤ µ(G1) by

vn

...........

v1

v2

vn−3

vn−2

vn−1

vn−4

Figure 7. The tree?G2

3

is obtained from G2by adding the edge {vn−1,vn}.

) = µ(G2) and it follows that µ?Cn−3

We now consider the case k ≥ 2. Recall that the tree T?⌈k

k pendant vertices and diameter n − k + 1.

Theorem 4.6. The tree T?⌈k

nectivity over Hn,k, k ≥ 2.

Proposition 2.3. Now, the graph Cn−3

By Theorem 3.4, µ(Cn−3

the proof.

33

?< µ(G). This completes

2⌋,n − k?of order n has

?

2⌉,⌊k

2⌉,⌊k

2⌋,n − k?uniquely attains the minimum algebraic con-

Proof. The result is obvious for k = 2. Let k ≥ 3 and G be a graph in Hn,kwhich is not

isomorphic to T?⌈k

So, let us assume that G is not a tree. Now, carefully remove some of the edges from the

cycles of G, to get a tree?G which is non-isomorphic to T?⌈k

less than n−k+1, then form a new tree?G of diameter n−k+1 from?G by grafting of edges

2.1 and 2.3 complete the proof.

2⌉,⌊k

2⌋,n − k?. If G is a tree then the result follows from Proposition 2.1.

2⌉,⌊k

2⌋,n − k?. Since?G has at

least k pendant vertices, it has diameter at most n−k +1. If the diameter of?G is strictly

such that?G is not isomorphic to T?⌈k

Let Tn,kdenote the subclass of Hn,kconsisting of all trees of order n with k, 2 ≤ k ≤ n−1,

pendant vertices. By Theorem 4.6, the tree T(⌈k

2⌉,⌊k

2⌋,n − k?. Now, Lemma 1.2 and Propositions

?

2⌉,⌊k

2⌋,n − k) has the minimum algebraic

Page 10

10A. K. LAL, K. L. PATRA AND B. K. SAHOO

connectivity over Tn,k. Now the following natural question arises: Which tree attains the

maximum algebraic connectivity over Tn,k? We answer this question in Theorem 4.7 below.

Let q = ⌊n−1

k⌋ and n − 1 = kq + r, 0 ≤ r ≤ k − 1. Let Tn,kbe the tree obtained by

adding r paths each of length q and k − r paths each of length q − 1 to a given vertex.

Then Tn,kis a member of Tn,k.

...........

..........

...........

..........

...........

u2

...........

...........

..........

r

k − r

u1

uq−1

uq

v1

v2

vq

vq+1

......

.......

Figure 8. The tree Tn,k

Theorem 4.7. The tree Tn,k uniquely attains the maximum algebraic connectivity over

Tn,k.

Proof. For k = 2, Tn,2≃ Pnis the only element in Tn,2and the result is obvious. Assume

that k ≥ 3. Let v be the vertex of Tn,kof degree k. There are k connected components of

Tn,kat v, and r of them are paths of order q + 1 and the rest k − r components are paths

of order q. Let dkbe the diameter of Tn,k. Then

Any tree in Tn,khas diameter at least dk, and equality holds if and only if the tree is

isomorphic to Tn,k.

If r = 0 then, by Proposition 2.7, each connected component of Tn,k at v is a Per-

ron component consisting of paths of order q and therefore by Remark 2.8, µ(Tn,k) =

21 − cos

2q+1

. If r ≥ 1, then from Proposition 2.4 together with the Perron-Frobenius

theory, it follows that the Perron components of Tn,kat v are the ones with q +1 vertices.

So µ(Tn,k) ≥ 2 1 − cos

2q+3

with strict inequality if r = 1. Note that if T is any tree of

?

for any tree T in Tn,kwith diameter β ≥ dk+ 1, we have

?

β + 1

dk=

2q

2q + 1

2q + 2

if r = 0

if r = 1

if 2 ≤ r ≤ k − 1

.

?

π

?

?

π

?

order n with diameter d, then µ(T) ≤ 21 − cos

π

d+1

?

([8], Corollary 4.4, p.234). Thus,

µ(T) ≤ 21 − cos

π

?

≤ 2

?

1 − cos

π

dk+ 1

?

≤ µ(Tn,k).

Since the second inequality is strict if r = 0 and r ≥ 2, and the last inequality is strict if

r = 1, the uniqueness of Tn,khaving the maximum algebraic connectivity over Tn,kfollows.

This completes the proof.

?

Page 11

ALGEBRAIC CONNECTIVITY WITH PENDANT VERTICES 11

5. Without Pendant Vertex

Let Fn,n ≥ 3, be the class of all connected graphs of order n without any pendant

vertex. The complete graph Kn is a member of Fn and it has the maximum algebraic

connectivity, namely µ(Kn) = n. Here, our aim is to find the graph that has the minimum

algebraic connectivity over Fn. By Lemma 1.2, we consider only those graphs in Fnwith

minimum possible edges (if we delete any edge from such a graph, then the new graph is

either disconnected or has atleast one pendant vertex). Note that any two cycles in such

a graph are edge disjoint. For n = 3,4, the cycle Cnof order n is the only such graph. In

[5], Fiedler showed that µ(Cm) = 2?1 − cos2π

there are two such graphs G1and G2in F5, up to isomorphism, where G1is C5and G2

is the graph with two 3-cycles having one common vertex. By Lemma 3.1(ii), µ(G2) = 1

and hence,

?

m

?. So µ(C3) = 3 and µ(C4) = 2. For n = 5,

µ(G1) = 21 − cos2π

5

?

> 1 = µ(G2).

We next consider when n ≥ 6. Let C⌊n

C⌈n+1

the following.

2⌋,⌈n

2⌉be the graph with two cycles C⌊n+1

2⌉is a member of Fn. We first prove

2⌋and

2⌉with exactly one common vertex. Then C⌊n

2⌋,⌈n

........ ........ . ......... . . . . ........

C⌊n+1

2⌋

C⌈n+1

2⌉

Figure 9. The graph C⌊n

2⌋,⌈n

2⌉

Lemma 5.1. For n ≥ 6, µ

?

C⌊n

2⌋,⌈n

2⌉

?

< µ(Cn).

Proof. Let v be the cut-vertex of C⌊n

of C⌊n

2⌋,⌈n

2⌋,⌈n

2⌉. Let C1and C2be the two connected components

2⌋ and ⌈n−1

0···00−1

2⌉at v of orders ⌊n−1

2⌉, respectively. For i = 1,2, observe that

?L(Ci) =

2−1

2

−1

...

···

0 ···

0

−1

...

−1

00

0−1

0

...

0

−1

2

...

0

···

···

...

2

...

−1

2

?

.

It is well known that the smallest eigenvalue of?L(C1) is 2

?L(C2) is 2

1 − cos

π

2⌋

⌊n+1

?

and that of

?

1 − cos

π

2⌉

⌈n+1

?

. If n is odd, then ρ(?L(C1)−1) = ρ(?L(C2)−1) and hence C1and

Page 12

12A. K. LAL, K. L. PATRA AND B. K. SAHOO

C2both are Perron components of C⌊n

?

If n is even, then C2is the only Perron component of C⌊n

?

n + 2

This completes the proof.

2⌋,⌈n

2⌉at v. So by Proposition 2.7,

2π

n + 1

µC⌊n

2⌋,⌈n

2⌉

?

= 2

?

1 − cos

?

< 2

?

1 − cos2π

n

?

= µ(Cn).

2⌋,⌈n

2⌉at v and it follows that

1 − cos2π

n

21 − cos

2π

?

< µ

?

C⌊n

2⌋,⌈n

2⌉

?

< 2

?

?

= µ(Cn).

?

.....

v2

Cn−6

3,3

C0

3,3

v1

vn−6

Figure 10. The graph Cn−6

3,3

Now, take a path of order n − 6 and add a 3-cycle to each of its two pendant vertices.

The new graph thus obtained is a member of Fn. We denote it by Cn−6

3,3.

Theorem 5.2. For n ≥ 6, the graph Cn−6

nectivity over Fn.

3,3

uniquely attains the minimum algebraic con-

Proof. In Fn, Cnis the only graph whose maximum vertex degree is two. Since we are

minimizing the algebraic connectivity over Fn, by Lemma 5.1 we may consider graphs with

maximum vertex degree at least three. Let G1be such a graph in Fnnot isomorphic to

Cn−6

3,3. Then G1has at least two cycles. Form a tree G2by deleting one edge from each

cycle of G1in such a way that G2has diameter at most n − 3 and if equality holds, then

G2is not isomorphic to T(2,2,n − 4) (see Figure 11). By Lemma 1.2, µ(G2) ≤ µ(G1). If

.......

v1

v2

v3

v4

Figure 11. The tree T(2,2,n − 4)

the diameter of G2is less than n − 3, then form a new tree G3from G2, by grafting of

edges, such that the diameter of G3is n − 3 and G3is not isomorphic to T(2,2,n − 4).

By Proposition 2.3, µ(G3) ≤ µ(G2). By Proposition 2.1, µ(T(2,2,n − 4)) < µ(G3). Now,

Cn−6

3,3

is isomorphic to the graph obtained from T(2,2,n − 4) by adding two new edges

{v1,v2} and {v3,v4}. Since µ(T(2,2,n − 4)) = µ

?

?

Cn−6

3,3

?

by Theorem 3.4, it follows that

µCn−6

3,3

?

< µ(G1). This completes the proof.

?

Page 13

ALGEBRAIC CONNECTIVITY WITH PENDANT VERTICES 13

6. Unicyclic Graphs

The algebraic connectivity of unicyclic graphs of order n with fixed girth has been

extensively studied by Fallat et al. [2, 3, 4]. Let Un, n ≥ 3, denote the class of all unicyclic

graphs of order n. Here we discuss the question of extremizing the algebraic connectivity

over Un. For n = 3, Uncontains only one graph. So we assume that n ≥ 4.

First consider the maximizing case. For the cycle Cnof order n, we have µ(Cn) > 1 if

and only if n ≤ 5, and µ(Cn) = 1 if and only if n = 6. Also, a graph in Un, which is not

isomorphic to Cn, has at least one cut-vertex and hence it has algebraic connectivity at

most one (Corollary 3.2). Again note that the graph Pn−3

Unand µ(Pn−3

n

) = 1 (Lemma 3.1(ii)). Therefore, if n ≤ 6, we have the following:

for n ≤ 5, clearly Cnis the only graph maximizing the algebraic connectivity over Unand

for n = 6, C6and P3

6are two non-isomorphic graphs in Unhaving algebraic connectivity

one.

Now, suppose that n ≥ 6 and let G be a graph in Unwith at least one cut-vertex which is

not isomorphic to Pn−3

n

. Let g be the girth of G. If g = 3, then µ(G) < 1 by ([2], Theorem

4.14, p.73). If g = 4,5 or 6 and G1is the graph in Ug+1obtained by adding one pendant

vertex adjacent to a vertex of Cg. Then µ(G1) ≈ 0.8299 if g = 4, µ(G1) ≈ 0.6972 if g = 5

and µ(G1) ≈ 0.5858 if g = 6. Since G is obtained from G1by a finite sequence of addition

of pendant vertices, by Lemma 1.1, µ(G) ≤ µ(G1) < 1. For g ≥ 7, G can be obtained from

Cgby a finite sequence of graph operations consisting of addition of pendant vertices. So,

by Lemma 1.1, µ(G) ≤ µ(Cg) < 1. Thus, we have the following result.

n

(see Figure 4) is a member of

Theorem 6.1. The maximum algebraic connectivity over Unis uniquely attained by Cnif

n ≤ 5 and uniquely attained by Pn−3

n

if n > 6. When n = 6, C6and P3

graphs, up to isomorphism, having the maximum algebraic connectivity over U6.

6are the only two

We next consider the minimizing case and prove the following.

Theorem 6.2. The graph Cn−3

connectivity over Un.

3

(see Figure 5) uniquely attains the minimum algebraic

Proof. If n = 4, then C4and C1

µ(C1

3) = 1 < µ(C4). If n = 5, then U5contains precisely five graphs (up to isomorphism).

They are C2

4(adding a pendant vertex to a vertex of C4) and H (adding

two pendant vertices to two distinct vertices of C3). We have µ(C2

0.6972, µ(C1

connectivity.

Now, let n ≥ 6. Let G be a graph in Unwith at least one pendant vertex which is not

isomorphic to Cn−3

3

. By the same argument used in the proof of Theorem 4.5, we can form

a graph G2 (see Figure 6) from G such that µ(Cn−3

need to show that µ(G2) < µ(Cn). In G2, it can be verified that

characteristic edge ([16], Lemma 2.2, p.386). So, in G2−v⌊n

3are the only two graphs, up to isomorphism, in U4with

3,C5,P2

5, C1

3) ≈ 0.5188, µ(H) ≈

3has the minimum algebraic

4) ≈ 0.8299, µ(P2

5) = 1 and µ(C5) > 1. Thus C2

3

) = µ(G2) < µ(G). Thus, we only

?

2⌋, the component C, consisting

v⌊n

2⌋,v⌊n

2⌋+1

?

is the

Page 14

14A. K. LAL, K. L. PATRA AND B. K. SAHOO

of the path on ⌊n

2⌋ − 1 vertices is not a Perron component. Therefore,

?

2(⌊n

2⌋ − 1) + 1

Thus, we have the required result.

µ(G2) < ρ(?(C)−1) = 21 − cos

π

?

< 2

?

1 − cos2π

n

?

= µ(Cn).

?

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