# On distance spectral radius of graphs

## On distance spectral radius of graphs

Linear Algebra and its Applications 438 (2013) 3490–3503 Contents lists available at SciVerse ScienceDirect Linear Algebra and its Applications jour...

Linear Algebra and its Applications 438 (2013) 3490–3503

Contents lists available at SciVerse ScienceDirect

Linear Algebra and its Applications journal homepage: w w w . e l s e v i e r . c o m / l o c a t e / l a a

On distance spectral radius of graphs Yanna Wang, Bo Zhou ∗ Department of Mathematics, South China Normal University, Guangzhou 510631, PR China

ARTICLE INFO

ABSTRACT

Article history: Received 26 September 2012 Accepted 17 December 2012 Available online 4 February 2013

We determine the unique tree of given domination number with minimum distance spectral radius and the unique connected graph (tree) of given domination number with maximum distance spectral radius. We also determine the unique tree of given bipartition size with minimum distance spectral radius and the unique connected bipartite graph (tree) of given bipartition size with maximum distance spectral radius. Finally, we determine the unique trees with the second and the third minimum (maximum, respectively) distance spectral radii among the trees on n  6 vertices. © 2013 Elsevier Inc. All rights reserved.

Submitted by Shaun Fallat AMS classification: 05C50 05C35 05C12 15A18 Keywords: Distance matrix Distance spectral radius Domination number Trees Bipartition size

1. Introduction We consider only simple graphs. Let G be a connected graph with vertex set V (G) and edge set E (G). For u, v ∈ V (G), the distance between u and v in G is the length of a shortest path connecting them, denoted by duv (G) or duv . The distance matrix of G is defined as D(G) = (duv )u,v∈V (G) . The eigenvalues of D(G) are called the distance eigenvalues of G. Since D(G) is real and symmetric, the distance eigenvalues of G are real. The distance spectral radius of G, denoted by ρ(G), is the largest distance eigenvalue of G. Let |V (G)|  2. Since D(G) is irreducible, by the Perron–Frobenius theorem, ρ(G) is positive, simple and there is a unique positive unit eigenvector x(G) corresponding to ρ(G), which is called the distance Perron vector of G. The study of distance eigenvalues of graphs dates back to the classical work of Graham and Pollack [7], Graham and Lovász [6] and Edelberg et al. [5]. Balaban et al. [1] proposed the use of the distance ∗ Corresponding author. E-mail address: [email protected] (B. Zhou). 0024-3795/\$ - see front matter © 2013 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.laa.2012.12.024

Y. Wang, B. Zhou / Linear Algebra and its Applications 438 (2013) 3490–3503

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Then ρ(G )

> ρ(G).

3

A non-pendent edge in a graph G is an edge of G not incident with a pendent vertex.

3

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Y. Wang, B. Zhou / Linear Algebra and its Applications 438 (2013) 3490–3503

Lemma 2.3. Let G be a connected graph and uv be a non-pendent cut edge of G. Let G be the graph obtained from G by contracting uv to a vertex u and attaching a pendent vertex v to u. Then ρ(G ) < ρ(G). The above lemma appeared in [17] without proof. A proof is given here for completeness. Let xz be the component of x(G ) corresponding to the vertex z ∈ V (G ). Let G1 (G2 , respectively) be the component of G − uv containing u (v, respectively). Let u1 be a neighbor of u in G1 . It is easily seen that

ρ(G )(s(G1 ) − xv )  ρ(G )xu + ρ(G )xu1 − ρ(G )xv = 3xv − xu1 +



w∈V (G )\{u,u1 ,v}

(du1 w (G ) − 1)xw

> xv − xu1 . Since s(G1 )

> xu1 , we have s(G1 ) > xv , and thus

ρ(G) − ρ(G )  x(G ) D(G)x(G ) − x(G ) D(G )x(G ) = 2s(G2 − v)(s(G1 ) − xv ) > 0, implying that ρ(G ) < ρ(G). Let D(n, a, b) be the tree obtained from the path Pn−a−b by attaching a and b pendent vertices to the two end vertices, respectively, where a  b  1 and a + b  n − 1. Lemma 2.4 [13]. If k

 1 and t  2, then ρ(D(n, k + t , k)) < ρ(D(n, k + t − 1, k + 1)).

The first part of the following lemma was proved in [13], and the second part follows from Lemma 2.3. Lemma 2.5. Let T be a tree on n  4 vertices with matching number m, where 1  m   2n . Then ρ(T )             ρ D n, n+2 1 − m, n+2 1 − m with equality if and only if T ∼ = D n, n+2 1 − m, n+2 1 − m .             1 − m1 , n+2 1 − m1 < ρ D n, n+2 1 − m2 , n+2 1 − m2 . Also, for m1 < m2 , ρ D n, n+ 2 Let Bn, be the tree obtained from the path Pn−+1 by attaching  terminal vertex, where 2    n − 1.

− 1 pendent vertices to a

Lemma 2.6 [16]. Let T be a tree on n  4 vertices with maximum degree , where 2    n − 1. Then ρ(T )  ρ(Bn, ) with equality if and only if T ∼ = Bn, . Also, for 1 > 2 , ρ(Bn,1 ) < ρ(Bn,2 ). Let G be a connected graph with v ∈ V (G). For k, l  1, let G(v, k, l) be the graph obtained from G by attaching a terminal vertex in each of Pk and Pl to v , and G(v, k, 0) be the graph obtained from G by attaching a terminal vertex in Pk to v. Lemma 2.7 [16]. Let G be a connected graph with v < ρ(G(v, k + 1, l − 1)).

∈ V (G). If k  l  1, then ρ(G(v, k, l))

For a graph G with u ∈ V (G) and vw ∈ E (G), G − u denotes the graph obtained from G by deleting the vertex u (and its incident edges) and G − vw denotes the graph obtained from G by deleting the edge vw. If zw is an edge of the complement of a graph G, then G + zw denotes the graph obtained from G by adding the edge zw.

Y. Wang, B. Zhou / Linear Algebra and its Applications 438 (2013) 3490–3503

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3. Distance spectral radius of graphs with given domination number In this section, we determine the unique tree of given domination number with minimum distance spectral radius, and the unique connected graph (tree) of given domination number with maximum distance spectral radius. Recall that if G is a graph on n vertices without isolated vertices, then γ (G)   2n  [14, p. 207]. Bollobás and Cockayne [2] showed that a graph G without isolated vertices has a γ (G)-set S such that for each vertex u ∈ S, there exists a vertex in V (G) \ S that is adjacent to u but no other vertices in S. Thus m(G)  γ (G). Theorem 3.1. Let T be a tree on n vertices with domination number ρ(T )  ρ(A(n, γ )) with equality if and only if T ∼ = A(n, γ ).

γ , where 1  γ   2n . Then

Proof. Note that m(T )  γ and γ (A(n, γ )) = γ . By Lemma 2.1, we have ρ(T )  ρ(A(n, m(T )))  ρ(A(n, γ )) with equalities if and only if T ∼ = A(n, m(T )) and m(T ) = γ , i.e., T ∼ = A(n, γ ).  Let G be a connected graph on n vertices with domination number  n3 . Note that γ (Pn ) =  3n

and that Pn is the unique graph among the connected graphs on n vertices with maximum distance spectral radius [15]. Thus ρ(G)  ρ (Pn ) with equality if and only if G ∼ = Pn . In the rest of this section, we determine the unique connected graph (tree) on n vertices of domination number γ with maximum distance spectral radius for 1γ < n3 and  n3 <γ  2n , separately. A quasi-pendent vertex of a graph G is a vertex adjacent to some pendent vertex. Theorem 3.2. Let T be a tree on n vertices with domination number γ , where 1  γ <  3n . Then ρ(T )            n−3γ +2 , n−32γ +2 . with equality if and only if T ∼ ρ D n, n−32γ +2 , n−32γ +2 = D n, 2



Proof. Let T be a tree with maximum distance spectral radius among the trees on n vertices with domination number at most γ . We claim that T has exactly 2 quasi-pendent vertices. Otherwise, there is a vertex u in T such that T − u has at least 3 components, and at least two of them are nontrivial. Let Tv be the component of T − u containing v for v ∈ NT (u). Let S be a γ (T )-set of T. There are two cases. Case 1. u ∈ S. There are x, y in NT (u) such that Tx and Ty are nontrivial. Suppose without loss of generality that s(Tx )  s(Ty ). Since Ty is a nontrivial tree, V (Ty ) ∩ S = ∅. Let w ∈ V (Ty ) ∩ S. For z ∈ NT (u) with z = x, y, T  = T − uz + wz is a tree. Since w ∈ S, γ (T  )  γ (T )  γ . By Lemma 2.2, ρ(T  ) > ρ(T ), a contradiction. / S. Then there exists a vertex x ∈ NT (u) ∩ S. Let {y, z} ⊆ NT (u) \ {x}. Let T  = T − uz + yz Case 2. u ∈ if s(Tx )  s(Ty ), and T  = T − uz + xz if s(Tx ) < s(Ty ). Then T  is a tree with γ (T  )  γ (T )  γ . By Lemma 2.2, ρ(T  ) > ρ(T ), a contradiction. By combining Cases 1 and 2, we find that T has exactly 2 quasi-pendent vertices, i.e., T ∼ = D(n, a, b) for some a and b with a  b  1. If a = 1, then T ∼ = Pn and thus γ (T ) =  3n > γ , a contradiction. Thus a  2. Suppose that γ (T ) < γ . Let T  = D(n, a − 1, b) for a − 1  b and T  = D(n, a, b − 1) for a = b. Then T  is a tree with γ (T  )  γ (T ) + 1  γ , and we have by Lemma 2.3 that ρ(T ) < ρ(T  ), a contradiction. Thus γ (T ) = γ . Since T ∼ = D(n, a, b) and γ (T ) = γ , we have a + b = n − 3γ + 2, n − 3γ + 3, or n − 3γ + 4. For a + b = n − 3γ + 3 or n − 3γ + 4. Let T  = D(n, a − 1, b) for a − 1  b and T  = D(n, a, b − 1)  for a = b. Then T  is a tree with γ (T  ) = γ (T ) = γ , and we have byLemma 2.3  that ρ(T ) < ρ(T ), a contradiction. Thus a + b

= n − 3γ + 2. By Lemma 2.4, T ∼ = D n,

n−3γ +2 2

,

n−3γ +2 2

. 

Corollary 3.1. Let G be a connected graph on n vertices with domination number γ , where 1  γ <  3n .          n − 3γ + 2 n − 3γ + 2 Then ρ(G)  ρ D n, with equality if and only if G ∼ , n − 32γ + 2 , = D n, 2 2   n−3γ +2 . 2

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Proof. Since G has a spanning tree T with the same domination number as G, see, e.g., [8], and by the Perron–Frobenius theorem, ρ(G) < ρ(G − e) for e ∈ E (G) such that G − e is also connected, we have ρ(G)  ρ(T ) with equality if and only if G ∼ = T. The result follows from Theorem 3.2.  Lemma 3.1. Let G be a connected graph with u, v ∈ V (G). Let u and v be pendent neighbors of u and v, respectively. Let xz be the component of x(G) corresponding to the vertex z ∈ V (G). Then (ρ(G) + 2)(xu − xv ) = ρ(G)(xu − xv ).  Proof. Let s = z∈V (G) xz . Then ρ(G)(xu − xu ) s − 2xv . Now the result follows easily. 

= (s − xu )− xu = s − 2xu . Similarly, ρ(G)(xv − xv ) =

Let C (n, a, b) be the tree obtained by attaching a pendent vertex vi to vi of the path Pn−a−b = v1 v2 · · · vn−a−b for i = 1, 2, . . . , a, n − a − 2b + 1, n − a − 2b + 2, . . . , n − a − b, where b  a  1 and 2(a + b)  n. For i = 1, 2, . . . , n − a − b. Let si = xvi if vi has no pendent neighbor, and si = xvi + xv otherwise. i For a tree T with u ∈ V (T ), let δT (u) or δ(u) be the degree of u in T. Lemma 3.2. If b

 a + 2 and 2(a + b) < n, then ρ(C (n, a + 1, b − 1)) > ρ(C (n, a, b)).

Proof. Let T = C (n, a, b). Let xz be the component of x(T ) corresponding to the vertex z ∈ V (T ) and dxy be the distance between the vertices x and y in T. Let p =  n−2a−b  and p1 =  n−2a−b . Claim 1. xvi+1

− xvn−a−b−i > xvi − xvn−a−b+1−i > 0 for 1  i  a.

n−a−b p First we prove that xvp > xvp1 +1 . Suppose that j=1 sj  j=p1 +1 sj . We prove that xvp−i by induction on i for 0  i  p − 1. For i = 0, we have ⎛ ⎞ p n− a−b    ρ(T ) xvp − xvp1 +1 = (p1 + 1 − p) ⎝ sj − sj ⎠  0, j=p1 +1

j=1

and then xvp  xvp1 +1 . Suppose that i  1 and xvp−j  xvp1 +1+j for 0  j  i then sp−j = xvp−j  xvp1 +1+j  sp1 +1+j . If δ(vp−j ) = 3, then by Lemma 3.1, xv sp−j

− 1. If δ(vp−j ) = 2,  xvp +1+j , and thus p−j

= xvp−j + xvp −j  xvp1 +1+j + xvp

1 +1+j



= 2⎝ ⎛

= 2⎝  0,

n− a−b 



j=p1 +1+i n− a−b 

sj

j=p1 +1

sj

p−i 

1

= sp1 +1+j . In either case, sp−j  sp1 +1+j . Then

ρ(T ) xvp−i − xvp1 +1+i − ρ(T ) xvp+1−i − xvp1 +i ⎛

 xvp1 +1+i

sj ⎠

j=1

p  j=1

sj ⎠ − 2

i−1   sp1 +1+j j=0

− sp−j



and thus xvp−i − xvp1 +1+i  xvp+1−i − xvp1 +i  0. Then xvp−i  xvp1 +1+i for 0  i  p − 1. It follows that sp−i  sp1 +1+i for 0  i  p − 1. Since b  a + 2, there exists an i with 0  i  p − 1 such that δ(vp−i ) = 2 and δ(vp1 +1+i ) = 3, and thus sp−i = xvp−i  xvp1 +1+i < sp1 +1+i . Thus p n−a−b n−a−b p j=p1 +1 sj , a contradiction. Then j=p1 +1 sj , and thus j=1 sj < j=1 sj < ⎛ ⎞ p n − a − b    ⎝ ρ(T ) xvp − xvp1 +1 = (p1 + 1 − p) sj − sj ⎠ > 0, j=p1 +1

from which we have xvp

> xvp1 +1 .

j=1

Y. Wang, B. Zhou / Linear Algebra and its Applications 438 (2013) 3490–3503

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 xvn −a−b . We prove that xvi  xvn−a−b+1−i by induction on i for 1  i  p. For  i  p and xvj  xvn−a−b+1−j for

Suppose that xv

1

i = 1, we have by Lemma 3.1 that xv1  xvn−a−b . Suppose that 2 1  j  i − 1. As above, we have sj  sn−a−b+1−j . It follows that 



ρ(T ) xvi − xvn−a−b+1−i − ρ(T ) xvi−1 − xvn−a−b+2−i = 2

i−1  j=1

(sj − sn−a−b+1−j )  0,

and then xvi − xvn−a−b+1−i  xvi−1 − xvn−a−b+2−i  0. Thus xvi  xvn−a−b+1−i for 1  i  p. In particular, , and as above, we have xvi+1 − xvn−a−b−i > xvp  xvp1 +1 , a contradiction. Therefore xv > xv n−a−b

1

xvi

− xvn−a−b+1−i > 0 for 1  i  a. Claim 1 follows.

Claim 2. xv

n−a−2b+1

< xvn−a−2b+2 + xvn −a−2b+2 .

It may be easily seen that for v ∈ V1 dvv − dvv  0. Then n−a−2b+2

= V (T ) \ {vn −a−2b+1 , vn−a−2b+2 , vn −a−2b+2 }, dvvn−a−2b+2 +

n−a−2b+1



ρ(T ) xvn−a−2b+2 + xvn −a−2b+2 − xvn −a−2b+1

= −xvn−a−2b+2 − 2xvn −a−2b+2 + 5xvn −a−2b+1 +

  v∈V1

dvvn−a−2b+2 +dvv

n−a−2b+2

− dvvn −a−2b+1 xv

 −xvn−a−2b+2 − 2xvn −a−2b+2 + 5xvn −a−2b+1 . Thus (ρ(T )



+ 2) xvn−a−2b+2 + xvn −a−2b+2 − xvn −a−2b+1  xvn−a−2b+2 + 3xvn −a−2b+1 > 0, from which

Claim 2 follows. Claim 3. xva+1

− xvn−a−2b+1 > xva+1+i − xvn−a−2b+1−i > 0 for 0 < i   n+2 1  − a − b − 1. − xn−2b+2−i > xvi+1 − xvn−2b+1−i 1 and t1 =  n+ . 2p n−a−b of Claim 1, j=1 sj < j=p1 +1 sj , and since t n−a−b j=n−b+2−t sj .

It suffices to prove that xvi Let t

=

 n+2 1 

By the proof

n−a−b

j=t1 −b+1 sj

=

We prove that xvi

> 0 with a + 1  i <  n+2 1  − b. − b  p1 − 1, we have

ρ(T ) xvt−b − xvn−b+2−t = (t1 + 1 − t ) ⎝

n− a−b  j=n−b+2−t

sj

t −b

> 0, and thus xvi

<

j=1

j=n−2b+2−i n− a−b  j=n−b+2−t

sj

t −b j=1

sj ⎠ − 2

t −b j=i+1

 xn−2b+2−j

⎞ sj ⎠

j=1

and then xvt −b > xvn−b+2−t . Suppose that a + 1  i  t − b − 1 and xvj Then   ρ(T ) xvi − xvn−2b+2−i − ρ(T ) xvi+1 − xvn−2b+1−i ⎞ ⎛ n− a−b i   ⎝ =2 sj − sj ⎠

= 2⎝

j=1 sj

> xn−2b+2−i by induction on i for a + 1  i  t − b. For i = t − b, we have



t −b

− xvj

> 0,

> xn−2b+2−j for i + 1  j  t − b.

− xvn−2b+2−i > xvi+1 − xvn−2b+1−i > 0. This proves Claim 3.

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Claim 4. xvn−a−2b+1 For 1

< xvn−2a−b .

 i  a, by Claim 1, xvi > xvn−a−b+1−i , and by Lemma 3.1, xvi > xvn −a−b+1−i , and thus 





b b−a > sn−a−b+1−i . Therefore ai=1 si > ai=1 sn−a−b+1−i = in=−na−−2a −b+1 si . Let m =  2  and m1 = n−a−b n−a−2b+m b−a sj  j=n−2a−b+1−m sj . We prove that xvn−a−2b+i  xvn−2a−b+1−i by  2 . Suppose that j=1 induction on i for 1  i  m. For i = m, we have

si



ρ(T ) xvn−a−2b+m − xvn−2a−b+1−m ⎛

= (m1 + 1 − m) ⎝

n− a−b 

sj

j=n−2a−b+1−m

n−a −2b+m

⎞ sj ⎠

j=1

 0, and then xvn−a−2b+m  xvn−2a−b+1−m . Suppose that 1  xv i + 1  j  m. By Lemma 3.1, xv n−a−2b+j



 i  m − 1 and xvn−a−2b+j  xvn−2a−b+1−j for , and thus sn−a−2b+j  sn−2a−b+1−j . Then

n−2a−b+1−j



ρ(T ) xvn−a−2b+i − xvn−2a−b+1−i − ρ(T ) xvn−a−2b+i+1 − xvn−2a−b−i ⎛

=

n− a−b 

2⎝ ⎛

= 2⎝

sj

j=n−2a−b+1−i n− a−b 

sj

j=n−2a−b+1−m

n−a −2b+i

sj ⎠

j=1 n−a −2b+m j=1

⎞ sj ⎠ − 2

m   sn−2a−b+1−j

j=i+1

− sn−a−2b+j



 0, and thus xvn−a−2b+i − xvn−2a−b+1−i  xvn−a−2b+i+1 − xvn−2a−b−i  0. Then xvn−a−2b+i  xvn−2a−b+1−i . It  xvn −2a−b+1−i , and thus follows that for 1  i  m, xvn−a−2b+i  xvn−2a−b+1−i . By Lemma 3.1, xv n−a−2b+i n−2a−b m sn−a−2b+i  sn−2a−b+1−i . Therefore i=1 sn−a−2b+i  i=n−2a−b+1−m si . Then n−a −2b+m

si

i=1

 > =

a 

si

i=1

m  i=1

n− a−b  i=n−2a−b+1

sn−a−2b+i

si

+

n− a−b  i=n−2a−b+1−m

a contradiction. Therefore Let T 

+

n− 2a−b

si

i=n−2a−b+1−m

si ,

n−a−2b+m i=1

si

>

n−a−b

i=n−a−2b+m1 +1 si ,

and as above Claim 4 holds.

= C (n, a + 1, b − 1). It is easily seen that

ρ(T  ) − ρ(T )  x(T ) (D(T  ) − D(T ))x(T ) = 2xvn −a−2b+1 W , where

W

=r

a  i=1

(sn−a−b+1−i − si ) + r

b− a−1  i=1

sn−a−2b+1+i

+

 2r −1

 i=0



(r − 2i) xvn−a−2b+1−i − xva+1+i

Y. Wang, B. Zhou / Linear Algebra and its Applications 438 (2013) 3490–3503

and r

3497

= n − 2a − 2b. By Lemma 3.1 and Claim 1, for 1  i  a, 

si



− sn−a−b+1−i = xvi − xvn−a−b+1−i + xvi − xvn −a−b+1−i

  ρ(T ) xvi − xvn−a−b+1−i = 1+ ρ(T ) + 2 



< 2 xva+1 − xvn−2a−b . Let F

= 0 if b − a = 2 and F = r 

b−a−1

sn−a−2b+1+i if b − a

i=2

 3. By Claims 2, 3 and 4,

ρ(T ) xva+1 − xvn−a−2b+1 = W + rxvn −a−2b+1 

< W + r xvn−a−2b+2 + xvn −a−2b+2 a 

= 2W + r +

i=1

(si − sn−a−b+1−i ) − F

 2r −1

 i=0



(r − 2i) xva+1+i − xvn−a−2b+1−i

< 2W + 2ar (xva+1 − xvn−2a−b ) +

 2r −1

 i=0

(r − 2i)(xva+1 − xvn−a−2b+1 )

 2r −1

⎜  < 2W + ⎝ i=0

For 1

⎞ ⎟



(r − 2i) + 2ar ⎠ xva+1 − xvn−a−2b+1 .

 i  a,  u∈V (T )

dv u i



>

u∈V (T )

> >

n−a −2b+1 j=a+1 r +1 i=1

For n − a − 2b + 1  u∈V (T )

dv u i

>

i+

>

dva vj

+

n− a−b 



j=n−2a−b+1

n− a−b 

2r

j=n−2a−b+1

>

dva vj

+ dva vj

 2r −1

 i=0

(r − 2i) + 2ar .

 i  n − a − b, 

u∈V (T )

>

dvi u

dvi u

n−a −2b+1 j=a+1 r +1 i=1

dvn−a−2b+1 vj

i + 2ar

>

 2r −1

 i=0

+

a   j=1

dvn−a−2b+1 vj

(r − 2i) + 2ar .

+ dvn−a−2b+1 vj

3498

Y. Wang, B. Zhou / Linear Algebra and its Applications 438 (2013) 3490–3503

For a + 1

 i  n − a − 2b,

 u∈V (T )

dvi u

> >

n−a −2b+1 j=a+1 n−a −2b+1 j=a+1

dvi ,vj

dv

+

a   j=1

 r+2 1 vj

a+

dvi vj

+ dvi vn−a−b+1−j + dvi vj + dvi vn −a−b+1−j

+ 2ar =

 2r −1

 i=0

(r − 2i) + 2ar .

Since ρ(T ) is bounded below by the minimum row sum of D(T ) [11, p. 24], ρ(T ) 2ar. Then ⎛

 r −1

2  ⎜ 2W > ⎝ρ(T ) −

i=0

and thus ρ(T  ) − ρ(T )



>

 2r −1 i=0

(r − 2i) +

(r − 2i) − 2ar ⎠ xva+1 − xvn−a−2b+1 > 0,

 2xvn −a−2b+1 W > 0. 

A caterpillar is a tree in which removal of all pendent vertices gives a path. Lemma 3.3. Let T be a caterpillar on n vertices with p pendent vertices, where 2  p   2n  and each       p p vertex of T has at most one pendent neighbor. Then ρ(T )  ρ C n, 2 , 2 with equality if and                  p p p1 p1 ∼ < ρ C n, p22 , p22 . only if T = C n, 2 , 2 . Also, for p1 > p2 , ρ C n, 2 , 2 Proof. When p = n2 , T ∼ = C (n,  2 ,  2 ), the result is trivial. When p = 2, T ∼ = Pn ∼ = C (n,  2 ,  2 ), n the result is also trivial. Suppose that 2 < p < 2 . Obviously, the diameter of T is n − p + 1. Let T be a caterpillar satisfying the conditions of the lemma with maximum distance spectral radius. Let v1 v2 · · · vn−p+2 be a diametrical path of T and U = {v ∈ V (T ) : δ(v) = 2} \ {v2 , vn−p+1 }. We claim that T − U has exactly 2 nontrivial components. Otherwise, there are three vertices vi , vk , and vj in T such that δ(vk ) = 3 and {vi , vj } ⊆ U, where i < k < j. Let vk be the unique pendent neighbor of vk . Let T1 , T2 be two nontrivial components of T − vk such that vi ∈ V (T1 ) and vj ∈ V (T2 ). Suppose without loss of generality that s(T1 )  s(T2 ). Let T  = T − vk vk + vj vk . Obviously, p(T  ) = p. 2 nontrivial components, i.e., By Lemma 2.2, ρ(T  ) > ρ(T ), a contradiction. Thus T − U has  exactly  p

T

p

p

∼ = C (n, a, b) with a + b = p. By Lemma 3.2, T ∼ = C n,       p p By Lemma 2.3, ρ C n, 21 , 21

<

, 2 .     2   p2 p2 ρ C n, 2 , 2 for p1 p

p

p

Lemma 3.4 [4]. Let T be a tree on n vertices with domination number diameter of T is at most 2n − 3γ + 1.

> p2 . 

γ , where γ >  n3 . Then the

Theorem 3.3. Let T be a tree on n vertices with domination number γ , where  3n < γ   2n . Then            3γ −n 3γ −n ρ(T )  ρ C n, 3γ2−n , 3γ2−n with equality if and only if T ∼ . = C n, 2 , 2 Proof. Let T be a tree with maximum distance spectral radius among the trees on n vertices with domination number at least γ . First we prove that each vertex of T has at most one pendent neighbor. Otherwise, there is a vertex u with at least two pendent neighbors, say v and w. Let T  = T − uv + vw. For a γ (T  )-set S of T  , it contains one of v or w, and thus S ∪ {u} \ {x} with x = v or w is a dominating set of T with cardinality at most γ (T  ), implying that γ (T  )  γ (T )  γ . By Lemma 2.3, ρ(T  ) > ρ(T ), a contradiction. Note that γ (T − e) − γ (T ) = 0 or 1 for e ∈ E (T ).

Y. Wang, B. Zhou / Linear Algebra and its Applications 438 (2013) 3490–3503

3499

Next we prove that if δ(u)  3, then γ (T − uz ) = γ (T ) for uz ∈ E (T ). Otherwise, γ (T − uz ) = γ (T ) + 1. Let Tv be the component of T − u containing v for v ∈ NT (u). Let {x, y} ⊆ NT (u) \ {z}. Suppose without loss of generality that s(Tx )  s(Ty ). Then T  = T − uz + yz is a tree with γ (T  )  γ (T  − yz) − 1 = γ (T − uz) − 1 = (γ (T ) + 1) − 1 = γ (T )  γ , and by Lemma 2.2, ρ(T  ) > ρ(T ), a contradiction. Now we claim that T is a caterpillar. Otherwise, there is a vertex u in T such that T − u has at least 3 nontrivial components. Let Tx , Ty , Tz be three nontrivial components in T − u containing x, y, z, respectively, where {x, y, z } ⊆ NT (u). Since δ(u)  3, we have γ (T − ux) = γ (T − uy) = γ (T − uz ) = γ (T ). Note that γ (T − w) − γ (T )  −1 for w ∈ V (T ). Suppose that γ (Tx − x) − γ (Tx ) = γ (Ty − y) − γ (Ty ) = γ (Tz − z ) − γ (Tz ) = −1. Then there is a γ (Ti )-set Si of Ti such that i ∈ Si and Si \ {i} is a γ (Ti − i)-set of Ti − i, where i = x, y, z. Let  T be the tree obtained from T by deleting the vertices in V (Tx ) T )-set of  and S a γ ( T. If u ∈ / S, then S = (S \V (Ty )∪V (Tz ))∪(Sy \{y})∪(Sz \{z})∪{u} is a dominating set of  T with cardinality |S  | = |S |−γ (Ty )−γ (Tz )+γ (Ty −y)+γ (Tz −z )+1 = |S |−1 = γ ( T )−1 < γ ( T ), a contradiction, and thus u ∈ S. It follows that S ∪ (Sx \ {x}) is a dominating set of T with cardinality T ) + γ (Tx − x). Thus γ (T )  γ ( T ) + γ (Tx − x). Note that γ (T − ux) = γ ( T ) + γ (Tx ). Hence γ ( γ (T − ux)−γ (T )  γ (Tx )−γ (Tx − x) = 1, which implies that γ (T − ux)−γ (T ) = 1, a contradiction. Thus there exists a vertex in {x, y, z }, say z such that γ (Tz − z ) − γ (Tz )  0. Suppose without loss of generality that s(Tx )  s(Ty ). Let w ∈ V (Ty ) be a quasi-pendent vertex of T. Let T  = T − uz + wz. Note that w has a unique pendent neighbor. Then there is a γ (T  )-set R of T  containing w. If z ∈ R, then R is also a dominating set of T  − wz, and thus γ (T  − wz )  γ (T  ). If z ∈ R, then R ∩ V (Tz − z ) is a γ (Tz − z )-set of Tz − z, and thus for a γ (Tz )-set R1 of Tz , (R \ R ∩ V (Tz − z )) ∪ R1 is a dominating set of T  − wz with cardinality |R| − γ (Tz − z ) + γ (Tz )  |R|, implying that γ (T  − wz )  γ (T  ). Thus γ (T  ) = γ (T  − wz) = γ (T − uz) = γ (T )  γ , and by Lemma 2.2, ρ(T  ) > ρ(T ), a contradiction. Thus we have proved that T is a caterpillar such that each vertex of T has at most one pendent neighbor. Let p be the number of pendent vertices and d be the diameter of T. Since T is a caterpillar, p = n − d + 1. Since γ (T ) >  n3 , by Lemma 3.4, d  2n − 3γ (T ) + 1. Thus p  3γ (T ) − n              3γ −n 3γ −n 3γ −n 3γ −n C n 3γ − n. Note that p C n, , = 3 γ − n and γ , , = γ . By 2 2 2 2 Lemma 3.3, we have  

ρ(T )  ρ C n,

    p p 2

,

2

 

 ρ C n,



−n 2

     with equalities if and only if T ∼ = C n, 2p , 2p and p      3γ −n 3γ −n .  Thus T ∼ = C n, 2 , 2

 

,

−n



2 

= 3γ − n, i.e., T ∼ = C n,



3γ −n 2

 

,

3γ −n 2

 .

Similar to the proof of Corollary 3.1, we have Corollary 3.2. Let G be a connected graph on n vertices with domination number γ , where  3n < γ             3γ −n 3γ −n with equality if and only if G ∼ .  2n . Then ρ(G)  ρ C n, 3γ2−n , 3γ2−n = C n, 2 , 2

4. Distance spectral radius of bipartite graphs with given bipartition size In this section, we determine the unique tree of given bipartition size with minimum distance spectral radius and the unique connected bipartite graph (tree) of given bipartition size with maximum distance spectral radius. If T is a tree with bipartition size (1, q), then T ∼ = S1+q . Theorem 4.1. Let T be a tree with bipartition size (p, q), where 2  p 1, p − 1)) with equality if and only if T ∼ = D(p + q, q − 1, p − 1).

 q. Then ρ(T )  ρ(D(p + q, q −

3500

Y. Wang, B. Zhou / Linear Algebra and its Applications 438 (2013) 3490–3503

Proof. Let T be a tree with minimum distance spectral radius among the trees with bipartition size (p, q). Let d be the diameter of T and Pt = u1 u2 · · · ud+1 be a diametrical path of T. Let H be the and k be the number component of T − u3 u4 containing u4  of pendent neighbors of u2 . Suppose that d  4. Let T  = T − v∈NH (u4 ) u4 v + v∈NH (u4 ) u2 v. Obviously, T  is also a tree with bipartition size (p, q). Let xz be the component of x(T  ) corresponding to the vertex z ∈ V (T  ). It is easily seen that

ρ(T ) − ρ(T  )  x(T  ) (D(T ) − D(T  ))x(T  ) 



= 4s(H − u4 ) kxu1 + xu2 − xu4 . For v

∈ V (T  ) \ {u1 , u2 , u4 }, dvu1 (T  ) + dvu2 (T  ) − dvu4 (T  )  0. Note that 



ρ(T  ) xu1 + xu2 − xu4 = −2xu1 − xu2 + 5xu4 +





v∈V (T  )\{u1 ,u2 ,u4 }

dvu1 (T  ) + dvu2 (T  ) − dvu4 (T  ) xv

 −2xu1 − xu2 + 5xu4 .   Then (ρ(T  ) + 2) xu1 + xu2 − xu4  xu2 + 3xu4 > 0, and thus Therefore d  3, and hence T ∼ = D(p + q, q − 1, p − 1). 

ρ(T ) > ρ(T  ), a contradiction.

Let G be a connected bipartite graph with bipartition size (p, q), where 1  p  q. Let Kp,q be the complete bipartite graph with p and q vertices in its two partite sets, respectively. Note that ρ(G) > ρ(G + e) for an edge of the complement of G. Then ρ(G)  ρ(Kp,q ) with equality if and only if G ∼ = Kp,q . Let G be a connected bipartite graph with bipartition size (p, q), where p = q or p = q − 1. In this case, Pp+q also has bipartition size (p, q). By the result from [15] mentioned above, ρ(G)  ρ Pp+q with equality if and only if G ∼ = Pp+q .

ρ(T ) Theorem 4.2. Let T bea tree with    bipartition size (p, q), where 1  p  q − 2.  Then  q−p+1 q−p+1 q−p+1 q−p+1 ∼ , , with equality if and only if T = D p + q, . ρ D p + q, 2 2 2 2 Proof. Note that m(T )



 p and D p + q,

size (p, q). By Lemma 2.5, we have  

ρ(T )  ρ D p + q,



q+p+1





q−p+1 2

− m(T ),

 

,



q−p+1 2





q+p+1

is a bipartite graph with bipartition 



− m(T )

2 2        q+p+1 q+p+1  ρ D p + q, − p, −p 2 2       q−p+1 q−p+1 = ρ D p + q, , 2 2

     with equalities if and only if T ∼ = D p + q, q+p2+1 − m(T ), q+p2+1 − m(T ) and m(T )      T ∼ = D p + q, q−p2+1 , q−p2+1 . 

= p, i.e.,

Corollary 4.1. Let G be a connectedbipartite graph   on p + q vertices  with bipartition size (p, q), where 1

 p  q − 2. Then ρ(G)  ρ D p + q,



D p + q,



q−p+1 2

 

,

q−p+1 2



.

q−p+1 2

,

q−p+1 2

with equality if and only if G

∼ =

Y. Wang, B. Zhou / Linear Algebra and its Applications 438 (2013) 3490–3503

Proof. Note that for a spanning tree T of G, ρ(G)  ρ(T ) with equality if and only if G T is a tree with bipartition size (p, q). By Theorem 4.2, we have the result. 

3501

∼ = T. Obviously,

5. Trees with small and large distance spectral radius As mentioned above, among the trees on n vertices, the star Sn is the unique tree with minimum distance spectral radius [16] and the path Pn is the unique tree with maximum distance spectral radius [15]. In this section, we determine among the trees on n vertices, the unique tree with the second minimum (maximum, respectively) distance spectral radius for n  5 and the unique tree with the third minimum (maximum, respectively) distance spectral radius for n  6. Theorem 5.1. Let T  Pn , Sn be a tree on n  5 vertices. Then ρ(A(n, 2))  ρ(T )  ρ(Bn,3 ) with left (right, respectively) equality if and only if T ∼ = A(n, 2) (T ∼ = Bn,3 , respectively). Proof. Let  be the maximum degree of T. Obviously,   3. By Lemma 2.6, ρ(T )  ρ(Bn, )  ρ(Bn,3 ) with equalities if and only if T ∼ = Bn, and  = 3, i.e., T ∼ = Bn,3 . Note that m(T )  2. By Lemma 2.1, ρ(T )  ρ(A(n, m(T )))  ρ(A(n, 2)) with equalities if and only if T ∼ = A(n, m(T )) and m(T ) = 2, i.e., T ∼ = A(n, 2).  Lemma 5.1. Label the vertices of a diametrical path of B2d+k−1,k as consecutively ud , ud−1 , . . . , u1 , v, v1 , v2 , . . . , vd and the remaining pendent vertices adjacent to vd−1 as vd+1 , . . . , vd+k−2 . Let G be the graph consisting of B2d+k−1,k and a nontrivial connected graph H such that they have exactly one common vertex v, and G = G − ud−1 ud + vd−1 ud (see Fig. 1). Then ρ(G) > ρ(G ). Proof. Let xz be the component of x(G ) corresponding to the vertex z ∈ V (G ). By symmetry, xvd xvd+1 = · · · = xvd+k−2 = xud . d−1 d−1 Suppose that j=1 xuj  j=1 xvj + kxvd . We prove that xui  xvi by induction on i for 1  i d − 1. For i = 1,  



ρ(G ) xu1 − xv1 =

⎛ 2⎝

d −1 j=1

xvj

+ kxvd −

d −1 j=1

⎞ xuj ⎠

 0,

Fig. 1. G and G in Lemma 5.1.

= 

3502

Y. Wang, B. Zhou / Linear Algebra and its Applications 438 (2013) 3490–3503

and thus xu1

 xv1 . Suppose that i  2 and xuj  xvj for 1  j  i − 1. Then 





ρ(G ) xui − xvi − ρ(G ) xui−1 − xvi−1 =2

d −1  j=i

=

2⎝

xvj

d −1 j=1

− xuj + 2kxvd + kxvd −

xvj

d −1 j=1

⎞ xuj ⎠ − 2

i−1   j=1

− xuj

xvj

 0,    and thus ρ(G ) xui − xvi  ρ(G ) xui−1 − xvi−1  0. It follows that xui  xvi for 1  i  d − 1. d−1 d−1 d−1 d−1 Thus j=1 xuj < j=1 xvj + kxvd , a contradiction. Therefore j=1 xuj < j=1 xvj + kxvd , and as above, we have xui > xvi for 1  i  d − 1, and xui − xvi > xui−1 − xvi−1 for 2  i  d − 1. It is easily seen that

ρ(G) − ρ(G )  x(G ) (D(G) − D(G ))x(G ) ⎛

= 2xud ⎝(k − 1)(2d − 2)xvd + 2 = (k − 1)(2d − 2)xvd + 2

Let W



d−1  i=1 i xvi

< = For u

w∈V (G )

k−1

k k−1

W

+

W

+

i=1

 i xvi

d −1

2

k − 1 i=1 d −1

2

 i xui



− xvi

 i xud−1

k − 1 i=1 d−1 2 i=1 i  k−1

− xui



− xvd−1

xud−1

− xvd−1 .

d −1  i=1

d −1  i=1

duui (G ) + duvi (G )

dvui (G ) + dvvi (G )

=2

d −1 i=1

i.

ρ(G ) is bounded below by the minimum row sum of D(G ) [11, p. 24], ρ(G ) > 2

d−1 i=1

k−1

duw (G ) > duv (G ) +

>

2

k

+

d −1

∈ V (G ), it is easily seen that 

Since

k−1

W

⎞  − xui ⎠ .



k

i=1

 i xvi

− xui . Then

ρ(G ) xud−1 − xvd−1 = k(2d − 2)xvd + 2 =

d −1

i

. It follows that

W

>

k−1 k



ρ(G ) −

2

d−1   i=1 i

k−1

xud−1

− xvd−1 > 0,

d−1 i=1

i



Y. Wang, B. Zhou / Linear Algebra and its Applications 438 (2013) 3490–3503

and thus ρ(G) − ρ(G )

3503

 2xud W > 0. 

Let Tn be the graph obtained by attaching a pendent vertex to v3 of the path Pn−1

= v1 v2 · · · vn−1 .

Theorem 5.2. Let T  Pn , Bn,3 , Sn , A(n, 2) be a tree on n  6 vertices. Then ρ(D(n, n−4, 2))  ρ(T )  ρ(Tn ) with left (right, respectively) equality if and only if T ∼ = D(n, n − 4, 2) (T ∼ = Tn , respectively). Proof. Let  be the maximum degree of T. Obviously,   3. If   4, then by Lemmas 2.6 and 2.3, ρ(T )  ρ Bn,  ρ(Bn,4 ) < ρ(Tn ). If  = 3, then by Lemma 2.7, ρ(T )  ρ(Tn ) with equality if and only if T ∼ = Tn . Obviously, m(T )  2. If m(T )  3, then by Lemmas 2.1 and 5.1, ρ(T )  ρ(A(n, m(T ))  ρ(A(n, 3)) > ρ(D(n, n − 4, 2)). If m(T ) = 2, then T ∼ = D(n, a, b), where a + b = n − 2 or n − 3. If a + b = n − 3, then by Lemma 2.3, ρ(D(n, a, b)) > ρ(D(n, a, b + 1)) for b = 1 and ρ(D(n, a, b)) > ρ(D(n, a + 1, b)) for b > 1. Thus by Lemma 2.4, ρ(T ) = ρ(D(n, a, b))  ρ(D(n, n − 4, 2)) with equality if and only if a = n − 4 and b = 2, i.e., T ∼ = D(n, n − 4, 2).  Acknowledgement This work was supported by the National Natural Science Foundation of China (No. 11071089) and the Guangdong Provincial Natural Science Foundation of China (No. S2011010005539). References [1] A.T. Balaban, D. Ciubotariu, M. Medeleanu, Topological indices and real number vertex invariants based on graph eigenvalues or eigenvectors, J. Chem. Inf. Comput. Sci. 31 (1991) 517–523. [2] B. Bollobás, E.J. Cockayne, Graph-theoretic parameters concerning domination, independence, and irredundance, J. Graph Theory 3 (1979) 241–249. [3] S.S. Bose, M. Nath, S. Paul, Distance spectral radius of graphs with r pendent vertices, Linear Algebra Appl. 435 (2011) 2828–2836. [4] P. Dankelmann, Average distance and domination number, Discrete Appl. Math. 80 (1997) 21–35. [5] M. Edelberg, M.R. Garey, R.L. Graham, On the distance matrix of a tree, Discrete Math. 14 (1976) 23–39. [6] R.L. Graham, L. Lovász, Distance matrix polynomials of trees, Adv. Math. 29 (1978) 60–88. [7] R.L. Graham, H.O. Pollack, On the addressing problem for loop switching, Bell System Tech. J. 50 (1971) 2495–2519. [8] M.A. Henning, S. Mukwembi, Domination, radius, and minimum degree, Discrete Appl. Math. 157 (2009) 2964–2968. [9] A. Ilic, ´ Distance spectral radius of trees with given matching number, Discrete Appl. Math. 158 (2010) 1799–1806. [10] R. Merris, The distance spectrum of a tree, J. Graph Theory 14 (1990) 365–369. [11] H. Minc, Nonnegative Matrices, John Wiley Sons, New York, 1988. [12] M. Nath, S. Paul, On the distance spectral radius of bipartite graphs, Linear Algebra Appl. 436 (2012) 1285–1296. [13] M. Nath, S. Paul, On the distance spectral radius of trees, Linear and Multilinear Algebra, in press, http://dx.doi.org/10.1080/03081087.2012.711324. [14] O. Ore, Theory of Graphs, Amer. Math. Soc. Colloq. Publ., vol. XXXVIII, Am. Math. Soc., Providence, RI, 1962. [15] S.N. Ruzieh, D.L. Powers, The distance spectrum of the path Pn and the first distance eigenvector of connected graphs, Linear and Multilinear Algebra 28 (1990) 75–81. [16] D. Stevanovic, ´ A. Ilic, ´ Distance spectral radius of trees with fixed maximum degree, Electron. J. Linear Algebra 20 (2010) 168–179. [17] G. Yu, H. Jia, H. Zhang, J. Shu, Some graft transformations and its applications on the distance spectral radius of a graph, Appl. Math. Lett. 25 (2012) 315–319. [18] G. Yu, Y. Wu, Y. Zhang, J. Shu, Some graft transformations and its application on a distance spectrum, Discrete Math. 311 (2011) 2117–2123. [19] B. Zhou, A. Ilic, ´ On distance spectral radius and distance energy of graphs, MATCH Commun. Math. Comput. Chem. 64 (2010) 261–280. [20] B. Zhou, N. Trinajstic, ´ On the largest eigenvalue of the distance matrix of a connected graph, Chem. Phys. Lett. 447 (2007) 384–387. [21] B. Zhou, N. Trinajstic, ´ Further results on the largest eigenvalues of the distance matrix and some distance-based matrices of connected (molecular) graphs, Internet Electron. J. Mol. Des. 6 (2007) 375–384.