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Applied Mathematics and Computation 363 (2019) 124622
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The Aα -spectral radius of trees and unicyclic graphs with given degree sequenceR Dan Li∗, Yuanyuan Chen, Jixiang Meng College of Mathematics and System Sciences, Xinjiang University, Urumqi, Xinjiang, PR China
a r t i c l e
i n f o
a b s t r a c t For any real α ∈ [0, 1], Aα (G ) = α D(G ) + (1 − α )A(G ) is the Aα -matrix of a graph G, where A(G) is the adjacency matrix of G and D(G) is the diagonal matrix of the degrees of G. This paper presents some extremal results about the spectral radius λ1 (Aα (G)) of Aα (G) that generalize previous results about λ1 (A0 (G)) and λ1 (A 1 (G )). In this paper, we study 2 the behavior of the Aα -spectral radius under some graph transformations for α ∈ [0, 1). As applications, we show that the greedy tree has the maximum Aα -spectral radius in GD when D is a tree degree sequence firstly. Furthermore, we determine that the greedy unicyclic graph has the largest Aα -spectral radius in GD when D is a unicyclic graphic sequence, where GD = {G | G is connected with D as its degree sequence}.
Article history: Available online 2 August 2019 MSC: 05C50 05C12 Keywords: Aα -matrix Spectral radius Tree Unicyclic Degree sequence
© 2019 Elsevier Inc. All rights reserved.
1. Introduction All graphs considered in this paper are simple and undirected. Let A(G) and D(G) be the adjacency matrix and diagonal matrix of the degrees of a graph G, respectively. The matrix Q (G ) = D(G ) + A(G ) is the signless Laplacian matrix of G. For any real α ∈ [0, 1], Nikiforov [10] defined the matrix Aα (G) as
Aα ( G ) = α D ( G ) + ( 1 − α )A ( G ). Clearly, A0 (G ) = A(G ) and A 1 (G ) = 2
1 2 Q ( G ).
Let λ1 (Aα (G)) ≥ λ2 (Aα (G)) ≥ ≥ λn (Aα (G)) denote the eigenvalues of Aα (G). The
largest eigenvalue λ1 (Aα (G)) is called Aα -spectral radius of G. Let dG (v ), or d (v ) for short, be the degree of the vertex v in G. Let D = (d0 , d1 , . . . , dn−1 ) be a nonincreasing sequence of nonnegative integers, D is called graphic if there exists a simple graph G with order n having D as its vertex degree sequence. Let NG (v ) or N (v ) denote the neighbor set of vertex v of G. An internal path of G is a path P (or cycle) with vertices v1 , v2 , . . . , vk (or v1 = vk ) such that dG (v1 ) ≥ 3, dG (vk ) ≥ 3 and dG (v2 ) = · · · = dG (vk−1 ) = 2. Recently, the study of the Aα -matrix of a graph has obtained considerable attention. Nikiforov et al. [11] determined the unique graph with minimal Aα -spectral radius among all connected graphs of order n. Xue et al. [12] determined the unique graph with maximum Aα -spectral radius among all connected graphs with diameter d or given clique number. Lin et al. [7] gave some results on the eigenvalues of Aα (G). The study of question whether a graph is determined by its Aα spectrum is just in its beginning. Lin et al. [8] proved that some graphs are determined by its Aα -spectrum under some R ∗
Supported by NSFC (Nos. 11531011, 11861066), NSFXJ (No.2015KL019), XJTCDP (No.04231200746), BS (No.180201). Corresponding author. E-mail address: [email protected] (D. Li).
https://doi.org/10.1016/j.amc.2019.124622 0 096-30 03/© 2019 Elsevier Inc. All rights reserved.
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D. Li, Y. Chen and J. Meng / Applied Mathematics and Computation 363 (2019) 124622
certain conditions. Chen et al. [4] paid main attention to the bound of λ2 (Aα (G)). Chen et al. [5] considered the Aα -spectral radius of the kth power of a graph G. At the same time, Lin et al. also considered the convex combinations Dα (G) of the diagonal matrix of vertex transmissions of G and the distance matrix of G, see [9]. Zhang [13,14] proposed the following problem about λ1 (A0 (G)) and λ1 (A 1 (G )). We generalize it to 0 ≤ α < 1. 2
Problem. Let 0 ≤ α < 1 and D be a given graphic degree sequence,
GD = {G | G is connected with D as its degree sequence}. Characterize all extremal graphs such that their Aα -spectral radius reach the largest value in GD . Biyikog˘ lu et al. [2] and Zhang [13] discussed the A0 -spectral radius and A 1 -spectral radius when D is a tree degree 2
sequence, and gave the extremal trees that reach the upper bound. Belardo et al. [3] and Zhang [14] discussed the A0 spectral radius and A 1 -spectral radius when D is a unicyclic graphic sequence, and gave the extremal unicyclic graphs that 2
reach the upper bound. In this paper, the main aim is to extend their results for all α ∈ [0, 1). To generalize these results, we first consider the behavior of the Aα -spectral radius under some graph transformations for α ∈ [0, 1) in Section 2. And the following basic tool is vital. Lemma 1.1. Let G be a connected graph with α ∈ [0, 1) and uv be some edge on an internal path of G. Let Guv denote the graph obtained from G by subdivision of edge uv into edges uw and wv. Then λ1 (Aα (Guv )) < λ1 (Aα (G )). Using these basic tools, we first determine the unique tree with maximum Aα -spectral radius in GD in Section 3. Theorem 1.2. For a given nonincreasing tree degree sequence D = (d0 , d1 , . . . , dn−1 ), if α ∈ [0, 1), then TD∗ (see in Section 3) has the largest Aα -spectral radius in GD . In Section 4, we discuss the unicyclic graphs with maximum Aα -spectral radius in GD . Theorem 1.3. Let D = (d0 , d1 , . . . , dn−1 ) be a positive nonincreasing unicyclic graphic sequence. If α ∈ [0, 1), then λ1 (Aα (UD∗ )) = max{λ1 (Aα (H )) | H ∈ GD }, where UD∗ is shown in Section 4. 2. Basic tools Lemma 2.1. Let α ∈ [0, 1) and G be a connected graph with uvi ∈ E (G ) and wvi ∈ / E (G ) for i = 1, . . . , k. Let X be a unit eigenvector of Aα (G) corresponding to λ1 (Aα (G)). Let G = G − {uvi } + {wvi } for i = 1, . . . , k. If xw ≥ xu , then λ1 (Aα (G )) > λ1 (Aα (G)). Proof. Note that
λ1 (Aα (G )) − λ1 (Aα (G )) ≥ X T (Aα (G ) − Aα (G ))X = ( 2α − 1 ) dG (v )x2v + (1 − α ) ( x p + xq )2 v∈V (G )
− ( 2α − 1 )
pq∈E (G )
dG ( v )xv − ( 1 − α ) 2
v∈V (G )
( x p + xq )2
pq∈E (G )
= k(2α − 1 )(x2w − x2u ) + 2(1 − α )(xw − xu )
k
(x2w + x2u + 2xvi )
i=1
≥ 0. If λ1 (Aα (G )) = λ1 (Aα (G )), then Aα (G )X = λ1 (Aα (G ))X and
(λ1 (Aα (G )) − λ1 (Aα (G )))xu = −kα xu − (1 − α )
k
xvi = 0,
i=1
a contradiction. Thus, λ1 (Aα (G )) > λ1 (Aα (G)).
We can get the following result by Lemma 2.1 immediately. Lemma 2.2. Let α ∈ [0, 1). Let G ∈ GD and X be a unit eigenvector of Aα (G) corresponding to λ1 (Aα (G)), u, v ∈ V (G ) and dG (u ) > dG (v ). If xv ≥ xu , then there must exist G ∈ GD such that λ1 (Aα (G )) > λ1 (Aα (G)). Lemma 2.3. Let G ∈ GD and X be a unit eigenvector of Aα (G) corresponding to λ1 (Aα (G)). Let α ∈ [0, 1). Assume that uv, wy ∈ E (G ), uw, vy ∈ / E (G ) and G = G − {uv, wy} + {uw, vy}. If xu ≥ xy and xw ≥ xv , then λ1 (Aα (G )) ≥ λ1 (Aα (G)). Furthermore, if one of the two inequalities is strict, then λ1 (Aα (G )) > λ1 (Aα (G)).
D. Li, Y. Chen and J. Meng / Applied Mathematics and Computation 363 (2019) 124622
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Proof. Note that
λ1 (Aα (G )) − λ1 (Aα (G )) ≥ X T (Aα (G ) − Aα (G ))X = 2(1 − α )(xu − xy )(xw − xv ) ≥ 0. If λ1 (Aα (G )) =
λ1 (Aα (G )), then Aα (G )X = λ1 (Aα (G ))X and
(λ1 (Aα (G )) − λ1 (Aα (G )))xu = (1 − α )(xw − xv ) = 0, thus xw = xv . Similarly, we get xu = xy .
Lemma 2.4. Let X be a unit eigenvector of a graph G ∈ GD corresponding to λ1 (Aα (G)) and α ∈ [0, 1). Let uv ∈ E (G ), uw ∈ / E (G ) and xv < xw ≤ xu . If xu ≥ xt for any vertex t ∈ N (w ), then there exists a graph G ∈ GD such that λ1 (Aα (G )) > λ1 (Aα (G)). Proof. Let W = N (w ) \ ({v} ∪ N (v )). Claim 1. W = ∅. Otherwise, N (w ) \ {v} ⊂ N (v ) and N (w ) \ {v} ⊂ N (v ) \ {u} since u ∈ / N (w ). So dG (w ) < dG (v ). By Aα (G )X = λ1 (Aα (G ))X, we obtain that
λ1 (Aα (G ))xw = α dG (w )xw + (1 − α )
x j,
j∈N (w )
λ1 (Aα (G ))xv = α dG (v )xv + (1 − α )
x j,
j∈N (v )
then
(λ1 (Aα (G )) − α dG (w ))xw = (1 − α )
x j ≤ ( 1 − α ) xv +
j∈N (w )
< ( 1 − α ) xv +
< ( 1 − α ) xu +
xj
j∈N (w )\{v}
xj
j∈N (v )\{u}
= (1 − α )
xj
j∈N (v )\{u}
xj
j∈N (v )
= (λ1 (Aα (G )) − α dG (v ))xv . Since xv < xw , λ1 (Aα (G )) − α dG (w ) < λ1 (Aα (G )) − α dG (v ), i.e., dG (w ) > dG (v ), a contradiction. Claim 2. |N (w )| ≥ 2. Note that |N (w )| ≥ 1 by Claim 1. Suppose that N (w ) = { p} and pv ∈ / E (G ), p = v, then
λ1 (Aα (G ))xw = α xw + (1 − α )x p , λ1 (Aα (G ))xv = α dG (v )xv + (1 − α )
x j.
j∈N (v )
Since xw > xv ,
(λ1 (Aα (G )) − α )xw > (λ1 (Aα (G )) − α )xv ≥ (λ1 (Aα (G )) − α dG (v ))xv , that is,
( 1 − α )x p > ( 1 − α )xu + ( 1 − α )
x j ≥ ( 1 − α )xu ,
j∈N (v )\{u}
thus xp > xu , a contradiction. According to the connectivity of the graph G, let Puw = u · · · qw be a path between u and w. Next, we will show the Lemma from the following two cases. Case 1. uv ∈ / E (Puw ). Subcase 1.1. vq ∈ / E (G ). For the edges uv, wq ∈ E (G ), let G = G − {uv, wq} + {uw, vq}, then G ∈ GD . Combining with xu ≥ xq and xw > xv , we have λ1 (Aα (G )) > λ1 (Aα (G)) by Lemma 2.3. Subcase 1.2. vq ∈ E (G ). Then we can find a vertex p ∈ N (w )\{v} such that pv ∈ / E (G ) from claim 1. For the edges uv, wp ∈ E (G ), let G = G − {uv, wp} + {uw, v p}, then G ∈ GD . Combining with xu ≥ xp and xw > xv , we have λ1 (Aα (G )) > λ1 (Aα (G)) by Lemma 2.3. Case 2. uv ∈ E (Puw ).
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D. Li, Y. Chen and J. Meng / Applied Mathematics and Computation 363 (2019) 124622
Subcase 2.1. vq ∈ / E (G ) and vt ∈ E (G ) for any vertex t ∈ N (w )\{q}. For the edges uv, wq ∈ E (G ), let G = G − {uv, wq} + {uw, vq}, then G ∈ GD . Combining with xu ≥ xq and xw > xv , we have λ1 (Aα (G )) > λ1 (Aα (G)) by Lemma 2.3. Subcase 2.2. There exists a vertex p ∈ N (w )\{q} such that pv ∈ / E (G ). For the edges uv, wp, constructing G = G − {uv, wp} + {uw, v p}, then G ∈ GD . Combining with xu ≥ xp and xw > xv , we have λ1 (Aα (G )) > λ1 (Aα (G)) by Lemma 2.3. Lemma 2.5. Let G ∈ GD be a connected graph with V (G ) = {v0 , v1 , . . . , vn−1 } and α ∈ [0, 1). If λ1 (Aα (G )) = max{λ1 (Aα (H ))|H ∈ GD } and X is a unit eigenvector of Aα (G) corresponding to λ1 (Aα (G)), then the following assertions hold. (1) If xvi ≥ xv j , then dG (vi ) ≥ dG (v j ) for i < j; (2) If xvi = xv j , then dG (vi ) = dG (v j ). Proof. (1). Suppose that dG (vi ) < dG (v j ) for i < j, combining with xvi ≥ xv j , then we can get a connected graph G ∈ GD such that λ1 (Aα (G )) > λ1 (Aα (G)) by Lemma 2.2, a contradiction. (2). If xvi = xv j , then we have dG (vi ) ≤ dG (v j ) by the same argument as (1). So dG (vi ) = dG (v j ). Let G be a graph with root v0 and h(v ) = dG (v, v0 ) denote the height of a vertex v of G, where dG (vi , v j ) is defined to be the length of the shortest path between vi and v j . Definition 1. Let G = (V, E ) be a graph with root v0 . A well-ordering ≺ of the vertices is called a breadth-first-search ordering (BFS-ordering for short) if the following hold for all vertices u, v ∈ V : (1) u ≺ v implies h(u ) ≤ h(v ); (2) u ≺ v implies dG (u ) ≥ dG (v ); (3) Let uv ∈ E (G ), xy ∈ E(G), uy ∈ E(G), xv ∈ / E (G ) with h(u ) = h(x ) = h(v ) − 1 = h(y ) − 1. If u≺x, then v ≺ y. We call a graph that has a BFS-ordering of its vertices a BFS-graph. Theorem 2.6. Let G ∈ GD be a connected graph with V (G ) = {v0 , v1 , . . . , vn−1 } and α ∈ [0, 1). Let X be a unit eigenvector corresponding to λ1 (Aα (G)). If λ1 (Aα (G )) = max{λ1 (Aα (H ))|H ∈ GD }, then G is a BFS-graph, and u ≺ v implies xu ≥ xv . Proof. Clearly, we can number the entries of V(G) such that v0 ≺ v1 ≺ · · · ≺ vn−1 and xv0 ≥ xv1 ≥ · · · ≥ xvn−1 . Claim. h(v0 ) ≤ h(v1 ) ≤ · · · ≤ h(vn−1 ). We will prove it by induction for i. If i = 0, then h(v0 ) = 0 < h(v1 ). Assume that h(vi ) ≤ h(vi+1 ) for i ≤ k − 1. Now, we try to prove h(vk ) ≤ h(vk+1 ). Let V = V1 ∪ V2 , where V1 = {v0 , v1 , . . . , vk }, V2 = {vk+1 , . . . , vn−1 }, without loss of generality, assume that xvk > xvk+1 . There exists an edge between V1 and V2 according to the connectivity of G, and let vl be the first vertex of V1 such that vl vt ∈ E (G ) for some vt ∈ V2 . Case 1. h(vl ) ≥ h(vk ) − 1. Assume that v p ∈ V1 is the last one of Pv0 vk+1 , where Pv0 vk+1 is the shortest path between v0 and vk+1 , then h(vl ) ≤ h(v p ). And then h(vk+1 ) − 1 ≥ h(v p ) ≥ h(vl ) ≥ h(vk ) − 1. Case 2. h(vl ) < h(vk ) − 1, which implies that vl vk ∈ / E (G ). Note that xvt ≤ xvk+1 < xvk ≤ xvl . We claim that xvs ≤ xvl for any vertex vs ∈ N (vk ). If vs ∈ V2 , then the claim holds obviously. If vs ∈ V1 and xvs > xvl , then 0 ≤ s < l ≤ k and h(vs ) ≤ h(vl ). Thus, h(vk ) ≤ h(vs ) + 1 ≤ h(vl ) < h(vk ), a contradiction. Considering the vertices vl , vk and vt . Note that xvt < xvk ≤ xvl and xvs ≤ xvl for any vertex xvs ∈ N (vk ). Then by Lemma 2.4, we can find a connected graph G ∈ GD whose Aα -spectral radius is greater than λ1 (Aα (G)), a contradiction. Based on Lemma 2.5 and the Claim, we have obtained a well-ordering ≺ of V(G), which conforms to (1) and (2) of the definition of BFS-ordering. We just need to show that this ordering also satisfies (3) of the definition of BFS-ordering as follows. Let uv, xy ∈ E (G ) and uy, xv ∈ / E (G ) with h(u ) = h(x ) = h(v ) − 1 = h(y ) − 1 and u≺x, we just need to show that v ≺ y. Otherwise, might as well suppose that xv < xy . Let G = G − {uv, xy} + {uy, xv}, then G ∈ GD and λ1 (Aα (G )) > λ1 (Aα (G)) by Lemma 2.3, a contradiction. Lemma 2.7 [1]. Let M be a nonnegative irreducible symmetric matrix with spectral radius λ1 (M). If there exists a positive vector Y > 0 and a positive real β such that MY < β Y, then λ1 (M) < β . Lemma 2.8 [10,11]. Let G be a graph with maximal degree and α ∈ [0, 1). Then
λ1 (A(G )) ≤ λ1 (Aα (G )) ≤ . If the left equality holds, then G has a λ1 (A(G))-regular component. If the right equality holds, then G is regular. Proof of Lemma 1.1. Let P = u0 u1 . . . uk+1 be the internal path of G and X be a unit eigenvector of Aα (G) corresponding to λ1 (Aα (G)) in which xi is corresponding to ui . Without loss of generality, assume that x0 ≤ xk+1 . Let t be the smallest index such that xt = min0≤i≤k+1 xi , so t < k + 1. Without loss of generality, assume that u = ut and v = ut+1 . Note that ⊂ G and λ1 (A()) = 2, where the graph is shown in Fig 1. Since G is not a regular graph, λ1 (Aα (G)) > 2 by Lemma 2.8. Case 1. t > 0. A positive vector Y can be constructed as follows.
yp =
x p,
if p = w;
xu ,
if p = w.
D. Li, Y. Chen and J. Meng / Applied Mathematics and Computation 363 (2019) 124622
Fig. 1. The graph .
If xt−1 > xu , then
(Aα (Guv )Y )u = 2α xu + (1 − α )(xt−1 + xu ) ≤ 2α xu + (1 − α )(xt−1 + xv ) = λ1 (Aα (G ))xu = λ1 (Aα (G ))yu , and
(Aα (Guv )Y )w = 2α xu + (1 − α )(xu + xv ) < 2α xu + (1 − α )(xt−1 + xv ) = λ1 (Aα (G ))xu = λ1 (Aα (G ))yw . If xt−1 = xu , then
(Aα (Guv )Y )u = 2xu < λ1 (Aα (G ))xu = λ1 (Aα (G ))yu and
(Aα (Guv )Y )w = (Aα (Guv )Y )w . Therefore, Aα (Guv )Y < λ1 (Aα (G ))Y and λ1 (Aα (Guv )) < λ1 (Aα (G )) by Lemma 2.7. Case 2. t = 0, i.e., dG (u) ≥ 3. Let S = j∈NG (u )\{v} x j . Subcase 2.1. λ1 (Aα (G ))xu > 2α xu + (1 − α )xu + (1 − α )xv . A positive vector Y can be constructed as follows.
yp =
x p,
if p = w;
xu ,
if p = w.
Then
(Aα (Guv )Y )u = α dG (u )xu + (1 − α )(S + xu ) ≤ 2α xu + (1 − α )(S + xv ) = λ1 (Aα (G ))xu = λ1 (Aα (G ))yu . (Aα (Guv )Y )w = 2α xu + (1 − α )(xu + xv ) < λ1 (Aα (G ))xu = λ1 (Aα (G ))yw . Therefore, Aα (Guv )Y < λ1 (Aα (G ))Y and λ1 (Aα (Guv )) < λ1 (Aα (G )) by Lemma 2.7. Subcase 2.2. λ1 (Aα (G ))xu ≤ 2α xu + (1 − α )xu + (1 − α )xv . We can construct a positive vector Y as follows.
yp =
⎧ x p, ⎪ ⎪ ⎨
if p = u, w;
xu ,
⎪ ⎪ ⎩
if p = w;
1 [ 1 −α
(λ1 (Aα (G )) − 2α )xu − (1 − α )xv ]
Note that
λ1 (Aα (G ))xu = α dG (u )xu + (1 − α )(S + xv ), thus
yu =
α
1−α
( dG ( u ) − 2 )xu + s > 0.
if p = u.
5
6
D. Li, Y. Chen and J. Meng / Applied Mathematics and Computation 363 (2019) 124622
Obviously, by λ1 (Aα (G)) > 2 we have
2xu − xv < yu ≤ xu , that is,
yu ≤ xu < xv . By yu ≤ xu , we observe that
S≤
1 + α − α dG ( u ) xu . 1−α
Then
(Aα (Guv )Y )w = 2α xu + (1 − α )(yu + xv ) = 2α xu + (λ1 (Aα (G )) − 2α )xu − (1 − α )xv + (1 − α )xv = λ1 (Aα (G ))xu = λ1 (Aα (G ))yw . For any vertex z ∈ NG (u ) \ {v},
(Aα (Guv )Y )z = α dG (z )xz + (1 − α )
≤
α dG ( z )xz + ( 1 − α )
x j + yu
j∈NG (z )\{u}
x j + xu
j∈N (z )\{u}
λ1 (Aα (G ))xz = λ1 (Aα (G ))yz . =
Next, we only need to check it at the vertex u. Note that
λ1 (Aα (G ))S ≥ α S + (1 − α )(dG (u ) − 1 )xu . Then
(Aα (Guv )Y )u = α dG (u )yu + (1 − α )(S + xu ) α2 = dG (u )(dG (u ) − 2 )xu + α (dG (u ) − 1 )S + S + (1 − α )xu . 1−α α λ1 (Aα (G ))yu = λ1 (Aα (G )) ( dG ( u ) − 2 )xu + S 1−α α = (dG (u ) − 2 )λ1 (Aα (G ))xu + λ1 (Aα (G ))S 1−α α2 ≥ dG (u )(dG (u ) − 2 )xu + α (dG (u ) − 1 )S 1−α + α (dG (u ) − 2 )xv + (1 − α )(dG (u ) − 1 )xu . And combining with xu < xv , we get
λ1 (Aα (G ))yu − (Aα (Guv )Y )u ≥ α (dG (u ) − 2 )xv + (1 − α )(dG (u ) − 2 )xu − S
1 + α − α dG ( u ) xu 1−α 1 + α − α dG ( u ) > α (dG (u ) − 2 )xu + (1 − α )(dG (u ) − 2 )xu − xu 1−α dG ( u ) − 3 + α = xu 1−α ≥ 0. ≥
α (dG (u ) − 2 )xv + (1 − α )(dG (u ) − 2 )xu −
Therefore, Aα (Guv )Y < λ1 (Aα (G ))Y and λ1 (Aα (Guv )) < λ1 (Aα (G )) by Lemma 2.7.
3. The Aα -spectral radius of trees The graph TD∗ has been introduced by Zhang [13], let’s go over it. For a given nonincreasing tree sequence D =
(d0 , d1 , . . . , dn−1 ) with n ≥ 3, the TD∗ can be construct as follows. Set T0 = v0 . For k ≥ 1, we construct Tk from Tk−1 by adding a new vertex vk and joining vk to the non-saturated vertex with maximum degree in Tk−1 , where a vertex vi in Tk−1 is
D. Li, Y. Chen and J. Meng / Applied Mathematics and Computation 363 (2019) 124622
7
Fig. 2. The graph Gp,q (u).
= Gn−1 . As we know, TD∗ is a greedy tree actually. Zhang [13] observed non-saturated if dTk−1 (vi ) < di . If k = n − 1, then ∗ that TD is also the only one BFS-graph in GD . And, if T1 and T2 both have BFS-ordering in GD , then T1 ∼ =T2 , see Proposition 2.2 in [13]. Therefore, by Theorem 2.6 and Proposition 2.2 in [13], a concise proof of Theorem 1.2 is given. TD∗
Proof of Theorem 1.2. Let T ∈ GD and λ1 (Aα (T )) = max{λ1 (Aα (T )) | T ∈ GD }. By Theorem 2.6, T must have a BFSordering. Then T ∼ = TD∗ by Proposition 2.2 in [13]. 4. The Aα -spectral radius of unicyclic graphs The special unicyclic graph UD∗ has been introduced by Zhang [14]. For a given nonincreasing unicyclic graphic sequence D = (d0 , d1 , . . . , dn−1 ) with n ≥ 3, UD∗ = Cn if d0 = 2 obviously. Next, we assume that d0 ≥ 3. Set G3 = C3 = v0 v1 v2 v0 . For k ≥ 4, we construct Gk from Gk−1 by adding a new vertex vk−1 and joining vk−1 to the non-saturated vertex with maximum degree in Gk−1 , where a vertex vi in Gk−1 is non-saturated if dGk−1 (vi ) < di . If k = n − 1, then UD∗ = Gn−1 . As we know, UD∗ is a greedy unicyclic graph with girth three. Zhang [14] showed that UD∗ is the only one BFS-graph in GD . The graph Gp,q (u) is the graph obtained by attaching the paths Pp and Pq to the vertex u of G, which is shown in Fig 2. Lin, Huang and Xue [6] gave the relationship between the Aα -spectral radius and the parameters p and q. Lemma 4.1 [6]. Let 0 ≤ α < 1. If p − q ≥ 2, then
λ1 (Aα (G p−1,q+1 (u ))) < λ1 (Aα (G p,q (u ))). Lemma 4.2. Let D = (d0 , d1 , . . . , dn−1 ) be a positive nonincreasing unicyclic graphic sequence with d1 = 2. If α ∈ [0, 1), then λ1 (Aα (UD∗ )) = max{λ1 (Aα (G )) | G ∈ GD }. Proof. Note that GD = ∅. Let U ∈ GD and λ1 (Aα (U )) = max{λ1 (Aα (G )) | G ∈ GD }. If d0 = 2, then U∼ =Cn . If d0 ≥ 3, then U ∼ = G(l; l1 , . . . , ld0 −2 ), where G(l; l1 , . . . , ld0 −2 ) is a unicyclic graph by attaching d0 − 2 paths Pl1 , . . . , Pld −2 to a vertex v0 ∈ Cl . 0
Then the following claims must hold. Claim 1. l = 3. On the contrary, suppose that l ≥ 4, set G1 = G(l − 1; l1 , . . . , ld0 −2 ), then λ1 (Aα (G)) < λ1 (Aα (G1 )) by Lemma 1.1. Next, we can construct a graph G2 = G(l − 1; l1 + 1, . . . , ld0 −2 ) by adding a pendant edge to a pendant vertex of G1 , then G2 ∈ GD and G1 ⊂ G2 . Then λ1 (Aα (G)) < λ1 (Aα (G1 )) ≤ λ1 (Aα (G2 )), a contradiction. Claim 2. |li − l j | ≤ 1 for 1 ≤ i, j ≤ d0 − 2. Otherwise, suppose that there exist p and q such that |l p − lq | ≥ 2, without loss of generality, assume that l p ≥ lq + 2, thus G = Gl p ,lq (v0 ). Then by Lemma 4.1, we have λ1 (Aα (Gl p ,lq (v0 ))) < λ1 (Aα (Gl p −1,lq +1 (v0 ))) and Gl p −1,lq +1 (v0 ) ∈ GD , a contradiction. By the above Claims, we immediately obtain that U ∼ = UD∗ . Lemma 4.3 [10]. Let G be a graph with maximal degree . If α ∈ [0, 1/2], then
λ1 (Aα (G )) ≥ α ( + 1 ). If α ∈ [1/2, 1), then
λ1 (Aα (G )) ≥ α +
( 1 − α )2 . α
Lemma 4.4. Let D = (d0 , d1 , . . . , dn−1 ) be a positive nonincreasing unicyclic graphic sequence with d1 ≥ 3. If α ∈ [0, 1), then λ1 (Aα (UD∗ )) = max{λ1 (Aα (H )) | H ∈ GD }. Proof. Note that GD = ∅. Let U ∈ GD and λ1 (Aα (U )) = max{λ1 (Aα (H )) | H ∈ GD }. Let X be a unit eigenvector of Aα (U) corresponding to λ1 (Aα (U)). Then by Theorem 2.6, U is a BFS-graph, i.e., a BFS-ordering of V(U) exists and it makes
v0 ≺ v1 ≺ · · · ≺ vn−1 , xv0 ≥ xv1 ≥ · · · ≥ xvn−1 , dU (v0 ) ≥ dU (v1 ) ≥ · · · ≥ dU (vn−1 ),
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D. Li, Y. Chen and J. Meng / Applied Mathematics and Computation 363 (2019) 124622
and
h(v0 ) ≤ h(v1 ) ≤ · · · ≤ h(vn−1 ). Let Vi = {v | v ∈ V (U ), h(v ) = i} for i = 0, · · · , h(vn−1 ). Clearly, V0 = {v0 } and V1 = {v1 , · · · , vd0 }. Let C be the unique cycle of U, then we claim that C = v0 v1 v2 v0 . In order to show the claim, the following result is necessary. Claim 1. xv0 > xvd . 0
Suppose that xv0 = xvd , then xv0 = xv1 = · · · = xvd . By Aα (U )X = λ1 (Aα (U ))X, we have 0
0
λ1 (Aα (U ))xv0 = d0 xv0 , then λ1 (Aα (U )) = d0 and U is a regular graph by Lemma 2.8, a contradiction. Now, suppose that C = v0 v1 v2 v0 . Case 1. v0 ∈ V (C ). Then v1 v2 ∈ / E (U ). Subcase 1.1 vi v j ∈ E (U ) for 1 ≤ i < j ≤ d0 . If i = 1 and 3 ≤ j ≤ d0 . Without loss of generality, assume that xv2 > xv j (otherwise, we can exchange v2 and v j , then C = v0 v1 v2 v0 .), then dU (v2 ) ≥ dU (v j ) ≥ 2 by Lemma 2.5. Then we can find a vertex vk ∈ V2 such that v2 vk ∈ E (U ) and v j vk ∈/ E (U ). Let U1 = U − {v1 v j , v2 vk } + {v1 v2 , v j vk } and U1 ∈ GD obviously. Since xv1 > xvk and xv2 > xv j , λ1 (Aα (U )) < λ1 (Aα (U1 )) by Lemma 2.3, a contradiction. If 1 < i < j ≤ d0 , without loss of generality, assume that xv1 > xvi (otherwise, we can exchange v1 and vi , then it is as same as the above case). Since dU (v1 ) = d1 ≥ 3, there exists a vertex vk ∈ V2 such that v1 vk ∈ E (U ) and v j vk ∈ / E (U ) since U is a unicyclic graph. Let U2 = U − {v1 vk , vi v j } + {v1 vi , vk v j } and U2 ∈ GD obviously. Since xv1 > xvi and xv j ≥ xvk , λ1 (Aα (U )) < λ1 (Aα (U2 )) by Lemma 2.3, a contradiction. Subcase 1.2 vi v j ∈ / E (U ) for 1 ≤ i < j ≤ d0 . Then there must exist vk ∈ V1 and vl ∈ V2 such that vk vl ∈ E (C ) and k ≥ 2. From dU (v1 ) = d1 ≥ 3, we can get a vertex vt ∈ V2 such that v1 vt ∈ E (U )\E (C ) and xvt ≤ xu , where u ∈ N (v1 )\V (C ). Then vk vt ∈ / E (U ) and vl vt ∈ / E (U ) since U is a unicyclic graph. The following claim about U holds. Claim. xv1 > xvl or xvk > xvt . Otherwise, let xv1 = xvl and xvk = xvt , then xv1 = xvl = xvk = xvt , xu = xv1 for u ∈ N (v1 )\{v0 } and xv1 = · · · = xvd . By 0
Aα (U )X = λ1 (Aα (U ))X, we have
λ1 (Aα (U ))xvt = α dU (vt )xvt + (1 − α )
xu
u∈N (vt )
≤
α dU (vt )xv1 + (1 − α )dU (vt )xv1
= dU (vt )xv1 that is, λ1 (Aα (U )) ≤ dU (vt ) = dU (v1 ) = d1 by Lemma 2.5. Note that
λ1 (Aα (U ))xv0 = α dU (v0 )xv0 + (1 − α )dU (v0 )xv1 that is, (λ1 (Aα (U )) − α d0 )xv0 = (1 − α )d0 xv1 . Note that
λ1 (Aα (U ))xv1 = α dU (v1 )xv1 + (1 − α )
xu
u∈N ( v 1 )
α dU (v1 )xv1 + (1 − α )(dU (v1 ) − 1 )xv1 + (1 − α )xv0 = d1 xv1 + (1 − α )(xv0 − xv1 ) =
that is, (λ1 (Aα (U )) − d1 − α + 1 )xv1 = (1 − α )xv0 . By the above equalities, we have
d (α d1 − (1 − α )) d1 − d0 (λ1 (Aα (U )) − d1 ) λ1 (Aα (U )) − 0 = λ1 (Aα (G ))(α − 1 ) . d1
Note that λ1 (Aα (U )) >
d0 (α d1 −(1−α )) d1
d1
(1 − α ) dd0 1
= α d0 − by Lemma 4.3. Combining with λ1 (Aα (U)) ≤ d1 , we have λ1 (Aα (U )) = d1 , and then d1 = d0 , which implies that U is a regular graph by Lemma 2.8, a contradiction. Thus, the claim must hold. Let U3 = U − {v1 vt , vk vl } + {v1 vk , vl vt } and U3 ∈ GD obviously. Since xv1 > xvl or xvk > xvt , we have λ1 (Aα (U )) < λ1 (Aα (U3 )) by Lemma 2.3, a contradiction. Case 2. v0 ∈ / V (C ). Then vi v j ∈ / E (U ) for 1 ≤ i < j ≤ d0 . Let vq ∈ V (C ) and h(vq ) ≤ h(u ) and dU (vq ) ≥ dU (u ) for any u ∈ V(C). Subcase 2.1. vq = v1 . Then there must exist vs ∈ V (C ) ∩ V2 such that v1 vs ∈ E (U ), then dU (v2 ) ≥ dU (vs ) ≥ 2. So we can also find a vertex vt ∈ V2 such that v2 vt ∈ E (U ) and vt ∈ / V (C ) since v0 ∈ / V (C ). By the same argument as the claim of subcase 1.2, it can be shown that xv1 > xvt or xv2 > xvs . Let U4 = U − {v1 vs , v2 vt } + {v1 v2 , vs vt } and U4 ∈ GD . Then λ1 (Aα (U )) < λ1 (Aα (U4 )) by Lemma 2.3, a contradiction. Subcase 2.2. vq = v1 . Claim. h(vq ) < h(u ) for any u ∈ V (C ) \ {vq }. Otherwise, suppose that there is a vertex vr ∈ V (C ) \ {vq } such that h(vq ) = h(vr ), then there exist two internal disjoint paths Pl1 and Pl2 between vq and vr such that C = Pl1 ∪ Pl2 . However, there also exist two paths Pl3 and Pl4 from v0 to vq and vr , respectively. And Pl3 ∪ Pl4 ∪ Pl1 contains another cycle of U, which contradicts U is a unicyclic graph.
D. Li, Y. Chen and J. Meng / Applied Mathematics and Computation 363 (2019) 124622
9
Then there is a vertex vt ∈ V (C ) such that vq vt ∈ E (U ) and h(vq ) = h(vt ) − 1, thus dU (vk ) ≥ dU (vt ) ≥ 2 for any vk ∈ Vh , h ≤ h(vq ). And we can find two vertices vi ∈ Vh(vq ) and v j ∈ Vh(vq )+1 such that vi v j ∈ E (U ) \ E (C ). Obviously, vq v j , vt v j ∈ / E (U ). Let U5 = U − {vq vt , vi v j } + {vq vi , vt v j } and U5 ∈ GD . Combining with xvq ≥ xv j and xvi ≥ xvt , we have λ1 (Aα (U )) ≤ λ1 (Aα (U5 )) by Lemma 2.3. Furthermore, the smallest height of vertices of the cycle in U5 is less than h(vq ). By repeating the use of subcase 2.2 or Cases 1 and subcase 2.1, we can get λ1 (Aα (U )) ≤ λ1 (Aα (U5 )) < λ1 (Aα (UD∗ )), a contradiction. Therefore, C = v0 v1 v2 v0 , and then U ∼ = UD∗ obviously. By Lemmas 4.2 and 4.4, Theorem 1.3 can be obtained immediately. References [1] A. Berman, J.S. Plemmons, Nonnegative Matrices in the Mathematical Sciences, Academic Press, New York, 1994. Reprinted, SIAM [2] T. Biyikog˘ lu, J. Leydold, Graphs with given degree sequence and maximal spectral rradius, Electron. J. Combin. 15 (2008) 525–573. [3] F. Belardo, E.M.L. Marzi, S.K. Simic´ , J.F. Wang, On the spectral radius of unicyclic graphs with prescribed degree sequence, Linear Algebra Appl. 432 (2010) 2323–2334. [4] Y.Y. Chen, D. Li, J.X. Meng, On the second largest Aα -eigenvalues of graphs, Linear Algebra Appl. 580 (2019) 343–358. [5] Y.Y. Chen, D. Li, J.X. Meng, aα -spectral radius of the second power of a graph, Appl. Math. Comput. 359 (2019) 418–425. [6] H.Q. Lin, X. Huang, J. Xue, A note on the Aα -spectral radius of graphs, Linear Algebra Appl. 557 (2018) 430–437. [7] H.Q. Lin, J. Xue, J.L. Shu, On the Aα -spectra of graphs, Linear Algebra Appl. 556 (2018) 210–219. [8] H.Q. Lin, X.G. Liu, J. Xue, Graphs determined by their Aα -spectra, Discret. Math. 342 (2019) 441–450. [9] H.Q. Lin, J. Xue, J.L. Shu, On the Dα -spectra of graphs, Linear Multilinear Algebra. 10.1080/03081087.2019.1618236. [10] V. Nikiforov, Merging the A-and Q-spectral theories, Appl. Anal. Discret. Math. 11 (2017) 81–107. [11] V. Nikiforov, G. Pasten, O. Rojo, R.L. Soto, On the Aα -spectra of trees, Linear Algebra Appl. 520 (2017) 286–305. [12] J. Xue, H.Q. Lin, S.T. Liu, J.L. Shu, On the Aα -spectral radius of a graph, Linear Algebra Appl. 550 (2018) 105–120. [13] X.D. Zhang, The laplacian spectral radii of trees with degree sequences, Discret. Math. 308 (2008) 3143–3150. [14] X.D. Zhang, The signless Laplacian spectral radius of graphs with given degree sequences, Discret. Appl. Math. 157 (2009) 2928–2937.
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Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
The Aα -spectral radius of trees and unicyclic graphs with given degree sequenceR Dan Li∗, Yuanyuan Chen, Jixiang Meng College of Mathematics and System Sciences, Xinjiang University, Urumqi, Xinjiang, PR China
a r t i c l e
i n f o
a b s t r a c t For any real α ∈ [0, 1], Aα (G ) = α D(G ) + (1 − α )A(G ) is the Aα -matrix of a graph G, where A(G) is the adjacency matrix of G and D(G) is the diagonal matrix of the degrees of G. This paper presents some extremal results about the spectral radius λ1 (Aα (G)) of Aα (G) that generalize previous results about λ1 (A0 (G)) and λ1 (A 1 (G )). In this paper, we study 2 the behavior of the Aα -spectral radius under some graph transformations for α ∈ [0, 1). As applications, we show that the greedy tree has the maximum Aα -spectral radius in GD when D is a tree degree sequence firstly. Furthermore, we determine that the greedy unicyclic graph has the largest Aα -spectral radius in GD when D is a unicyclic graphic sequence, where GD = {G | G is connected with D as its degree sequence}.
Article history: Available online 2 August 2019 MSC: 05C50 05C12 Keywords: Aα -matrix Spectral radius Tree Unicyclic Degree sequence
© 2019 Elsevier Inc. All rights reserved.
1. Introduction All graphs considered in this paper are simple and undirected. Let A(G) and D(G) be the adjacency matrix and diagonal matrix of the degrees of a graph G, respectively. The matrix Q (G ) = D(G ) + A(G ) is the signless Laplacian matrix of G. For any real α ∈ [0, 1], Nikiforov [10] defined the matrix Aα (G) as
Aα ( G ) = α D ( G ) + ( 1 − α )A ( G ). Clearly, A0 (G ) = A(G ) and A 1 (G ) = 2
1 2 Q ( G ).
Let λ1 (Aα (G)) ≥ λ2 (Aα (G)) ≥ ≥ λn (Aα (G)) denote the eigenvalues of Aα (G). The
largest eigenvalue λ1 (Aα (G)) is called Aα -spectral radius of G. Let dG (v ), or d (v ) for short, be the degree of the vertex v in G. Let D = (d0 , d1 , . . . , dn−1 ) be a nonincreasing sequence of nonnegative integers, D is called graphic if there exists a simple graph G with order n having D as its vertex degree sequence. Let NG (v ) or N (v ) denote the neighbor set of vertex v of G. An internal path of G is a path P (or cycle) with vertices v1 , v2 , . . . , vk (or v1 = vk ) such that dG (v1 ) ≥ 3, dG (vk ) ≥ 3 and dG (v2 ) = · · · = dG (vk−1 ) = 2. Recently, the study of the Aα -matrix of a graph has obtained considerable attention. Nikiforov et al. [11] determined the unique graph with minimal Aα -spectral radius among all connected graphs of order n. Xue et al. [12] determined the unique graph with maximum Aα -spectral radius among all connected graphs with diameter d or given clique number. Lin et al. [7] gave some results on the eigenvalues of Aα (G). The study of question whether a graph is determined by its Aα spectrum is just in its beginning. Lin et al. [8] proved that some graphs are determined by its Aα -spectrum under some R ∗
Supported by NSFC (Nos. 11531011, 11861066), NSFXJ (No.2015KL019), XJTCDP (No.04231200746), BS (No.180201). Corresponding author. E-mail address: [email protected] (D. Li).
https://doi.org/10.1016/j.amc.2019.124622 0 096-30 03/© 2019 Elsevier Inc. All rights reserved.
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D. Li, Y. Chen and J. Meng / Applied Mathematics and Computation 363 (2019) 124622
certain conditions. Chen et al. [4] paid main attention to the bound of λ2 (Aα (G)). Chen et al. [5] considered the Aα -spectral radius of the kth power of a graph G. At the same time, Lin et al. also considered the convex combinations Dα (G) of the diagonal matrix of vertex transmissions of G and the distance matrix of G, see [9]. Zhang [13,14] proposed the following problem about λ1 (A0 (G)) and λ1 (A 1 (G )). We generalize it to 0 ≤ α < 1. 2
Problem. Let 0 ≤ α < 1 and D be a given graphic degree sequence,
GD = {G | G is connected with D as its degree sequence}. Characterize all extremal graphs such that their Aα -spectral radius reach the largest value in GD . Biyikog˘ lu et al. [2] and Zhang [13] discussed the A0 -spectral radius and A 1 -spectral radius when D is a tree degree 2
sequence, and gave the extremal trees that reach the upper bound. Belardo et al. [3] and Zhang [14] discussed the A0 spectral radius and A 1 -spectral radius when D is a unicyclic graphic sequence, and gave the extremal unicyclic graphs that 2
reach the upper bound. In this paper, the main aim is to extend their results for all α ∈ [0, 1). To generalize these results, we first consider the behavior of the Aα -spectral radius under some graph transformations for α ∈ [0, 1) in Section 2. And the following basic tool is vital. Lemma 1.1. Let G be a connected graph with α ∈ [0, 1) and uv be some edge on an internal path of G. Let Guv denote the graph obtained from G by subdivision of edge uv into edges uw and wv. Then λ1 (Aα (Guv )) < λ1 (Aα (G )). Using these basic tools, we first determine the unique tree with maximum Aα -spectral radius in GD in Section 3. Theorem 1.2. For a given nonincreasing tree degree sequence D = (d0 , d1 , . . . , dn−1 ), if α ∈ [0, 1), then TD∗ (see in Section 3) has the largest Aα -spectral radius in GD . In Section 4, we discuss the unicyclic graphs with maximum Aα -spectral radius in GD . Theorem 1.3. Let D = (d0 , d1 , . . . , dn−1 ) be a positive nonincreasing unicyclic graphic sequence. If α ∈ [0, 1), then λ1 (Aα (UD∗ )) = max{λ1 (Aα (H )) | H ∈ GD }, where UD∗ is shown in Section 4. 2. Basic tools Lemma 2.1. Let α ∈ [0, 1) and G be a connected graph with uvi ∈ E (G ) and wvi ∈ / E (G ) for i = 1, . . . , k. Let X be a unit eigenvector of Aα (G) corresponding to λ1 (Aα (G)). Let G = G − {uvi } + {wvi } for i = 1, . . . , k. If xw ≥ xu , then λ1 (Aα (G )) > λ1 (Aα (G)). Proof. Note that
λ1 (Aα (G )) − λ1 (Aα (G )) ≥ X T (Aα (G ) − Aα (G ))X = ( 2α − 1 ) dG (v )x2v + (1 − α ) ( x p + xq )2 v∈V (G )
− ( 2α − 1 )
pq∈E (G )
dG ( v )xv − ( 1 − α ) 2
v∈V (G )
( x p + xq )2
pq∈E (G )
= k(2α − 1 )(x2w − x2u ) + 2(1 − α )(xw − xu )
k
(x2w + x2u + 2xvi )
i=1
≥ 0. If λ1 (Aα (G )) = λ1 (Aα (G )), then Aα (G )X = λ1 (Aα (G ))X and
(λ1 (Aα (G )) − λ1 (Aα (G )))xu = −kα xu − (1 − α )
k
xvi = 0,
i=1
a contradiction. Thus, λ1 (Aα (G )) > λ1 (Aα (G)).
We can get the following result by Lemma 2.1 immediately. Lemma 2.2. Let α ∈ [0, 1). Let G ∈ GD and X be a unit eigenvector of Aα (G) corresponding to λ1 (Aα (G)), u, v ∈ V (G ) and dG (u ) > dG (v ). If xv ≥ xu , then there must exist G ∈ GD such that λ1 (Aα (G )) > λ1 (Aα (G)). Lemma 2.3. Let G ∈ GD and X be a unit eigenvector of Aα (G) corresponding to λ1 (Aα (G)). Let α ∈ [0, 1). Assume that uv, wy ∈ E (G ), uw, vy ∈ / E (G ) and G = G − {uv, wy} + {uw, vy}. If xu ≥ xy and xw ≥ xv , then λ1 (Aα (G )) ≥ λ1 (Aα (G)). Furthermore, if one of the two inequalities is strict, then λ1 (Aα (G )) > λ1 (Aα (G)).
D. Li, Y. Chen and J. Meng / Applied Mathematics and Computation 363 (2019) 124622
3
Proof. Note that
λ1 (Aα (G )) − λ1 (Aα (G )) ≥ X T (Aα (G ) − Aα (G ))X = 2(1 − α )(xu − xy )(xw − xv ) ≥ 0. If λ1 (Aα (G )) =
λ1 (Aα (G )), then Aα (G )X = λ1 (Aα (G ))X and
(λ1 (Aα (G )) − λ1 (Aα (G )))xu = (1 − α )(xw − xv ) = 0, thus xw = xv . Similarly, we get xu = xy .
Lemma 2.4. Let X be a unit eigenvector of a graph G ∈ GD corresponding to λ1 (Aα (G)) and α ∈ [0, 1). Let uv ∈ E (G ), uw ∈ / E (G ) and xv < xw ≤ xu . If xu ≥ xt for any vertex t ∈ N (w ), then there exists a graph G ∈ GD such that λ1 (Aα (G )) > λ1 (Aα (G)). Proof. Let W = N (w ) \ ({v} ∪ N (v )). Claim 1. W = ∅. Otherwise, N (w ) \ {v} ⊂ N (v ) and N (w ) \ {v} ⊂ N (v ) \ {u} since u ∈ / N (w ). So dG (w ) < dG (v ). By Aα (G )X = λ1 (Aα (G ))X, we obtain that
λ1 (Aα (G ))xw = α dG (w )xw + (1 − α )
x j,
j∈N (w )
λ1 (Aα (G ))xv = α dG (v )xv + (1 − α )
x j,
j∈N (v )
then
(λ1 (Aα (G )) − α dG (w ))xw = (1 − α )
x j ≤ ( 1 − α ) xv +
j∈N (w )
< ( 1 − α ) xv +
< ( 1 − α ) xu +
xj
j∈N (w )\{v}
xj
j∈N (v )\{u}
= (1 − α )
xj
j∈N (v )\{u}
xj
j∈N (v )
= (λ1 (Aα (G )) − α dG (v ))xv . Since xv < xw , λ1 (Aα (G )) − α dG (w ) < λ1 (Aα (G )) − α dG (v ), i.e., dG (w ) > dG (v ), a contradiction. Claim 2. |N (w )| ≥ 2. Note that |N (w )| ≥ 1 by Claim 1. Suppose that N (w ) = { p} and pv ∈ / E (G ), p = v, then
λ1 (Aα (G ))xw = α xw + (1 − α )x p , λ1 (Aα (G ))xv = α dG (v )xv + (1 − α )
x j.
j∈N (v )
Since xw > xv ,
(λ1 (Aα (G )) − α )xw > (λ1 (Aα (G )) − α )xv ≥ (λ1 (Aα (G )) − α dG (v ))xv , that is,
( 1 − α )x p > ( 1 − α )xu + ( 1 − α )
x j ≥ ( 1 − α )xu ,
j∈N (v )\{u}
thus xp > xu , a contradiction. According to the connectivity of the graph G, let Puw = u · · · qw be a path between u and w. Next, we will show the Lemma from the following two cases. Case 1. uv ∈ / E (Puw ). Subcase 1.1. vq ∈ / E (G ). For the edges uv, wq ∈ E (G ), let G = G − {uv, wq} + {uw, vq}, then G ∈ GD . Combining with xu ≥ xq and xw > xv , we have λ1 (Aα (G )) > λ1 (Aα (G)) by Lemma 2.3. Subcase 1.2. vq ∈ E (G ). Then we can find a vertex p ∈ N (w )\{v} such that pv ∈ / E (G ) from claim 1. For the edges uv, wp ∈ E (G ), let G = G − {uv, wp} + {uw, v p}, then G ∈ GD . Combining with xu ≥ xp and xw > xv , we have λ1 (Aα (G )) > λ1 (Aα (G)) by Lemma 2.3. Case 2. uv ∈ E (Puw ).
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D. Li, Y. Chen and J. Meng / Applied Mathematics and Computation 363 (2019) 124622
Subcase 2.1. vq ∈ / E (G ) and vt ∈ E (G ) for any vertex t ∈ N (w )\{q}. For the edges uv, wq ∈ E (G ), let G = G − {uv, wq} + {uw, vq}, then G ∈ GD . Combining with xu ≥ xq and xw > xv , we have λ1 (Aα (G )) > λ1 (Aα (G)) by Lemma 2.3. Subcase 2.2. There exists a vertex p ∈ N (w )\{q} such that pv ∈ / E (G ). For the edges uv, wp, constructing G = G − {uv, wp} + {uw, v p}, then G ∈ GD . Combining with xu ≥ xp and xw > xv , we have λ1 (Aα (G )) > λ1 (Aα (G)) by Lemma 2.3. Lemma 2.5. Let G ∈ GD be a connected graph with V (G ) = {v0 , v1 , . . . , vn−1 } and α ∈ [0, 1). If λ1 (Aα (G )) = max{λ1 (Aα (H ))|H ∈ GD } and X is a unit eigenvector of Aα (G) corresponding to λ1 (Aα (G)), then the following assertions hold. (1) If xvi ≥ xv j , then dG (vi ) ≥ dG (v j ) for i < j; (2) If xvi = xv j , then dG (vi ) = dG (v j ). Proof. (1). Suppose that dG (vi ) < dG (v j ) for i < j, combining with xvi ≥ xv j , then we can get a connected graph G ∈ GD such that λ1 (Aα (G )) > λ1 (Aα (G)) by Lemma 2.2, a contradiction. (2). If xvi = xv j , then we have dG (vi ) ≤ dG (v j ) by the same argument as (1). So dG (vi ) = dG (v j ). Let G be a graph with root v0 and h(v ) = dG (v, v0 ) denote the height of a vertex v of G, where dG (vi , v j ) is defined to be the length of the shortest path between vi and v j . Definition 1. Let G = (V, E ) be a graph with root v0 . A well-ordering ≺ of the vertices is called a breadth-first-search ordering (BFS-ordering for short) if the following hold for all vertices u, v ∈ V : (1) u ≺ v implies h(u ) ≤ h(v ); (2) u ≺ v implies dG (u ) ≥ dG (v ); (3) Let uv ∈ E (G ), xy ∈ E(G), uy ∈ E(G), xv ∈ / E (G ) with h(u ) = h(x ) = h(v ) − 1 = h(y ) − 1. If u≺x, then v ≺ y. We call a graph that has a BFS-ordering of its vertices a BFS-graph. Theorem 2.6. Let G ∈ GD be a connected graph with V (G ) = {v0 , v1 , . . . , vn−1 } and α ∈ [0, 1). Let X be a unit eigenvector corresponding to λ1 (Aα (G)). If λ1 (Aα (G )) = max{λ1 (Aα (H ))|H ∈ GD }, then G is a BFS-graph, and u ≺ v implies xu ≥ xv . Proof. Clearly, we can number the entries of V(G) such that v0 ≺ v1 ≺ · · · ≺ vn−1 and xv0 ≥ xv1 ≥ · · · ≥ xvn−1 . Claim. h(v0 ) ≤ h(v1 ) ≤ · · · ≤ h(vn−1 ). We will prove it by induction for i. If i = 0, then h(v0 ) = 0 < h(v1 ). Assume that h(vi ) ≤ h(vi+1 ) for i ≤ k − 1. Now, we try to prove h(vk ) ≤ h(vk+1 ). Let V = V1 ∪ V2 , where V1 = {v0 , v1 , . . . , vk }, V2 = {vk+1 , . . . , vn−1 }, without loss of generality, assume that xvk > xvk+1 . There exists an edge between V1 and V2 according to the connectivity of G, and let vl be the first vertex of V1 such that vl vt ∈ E (G ) for some vt ∈ V2 . Case 1. h(vl ) ≥ h(vk ) − 1. Assume that v p ∈ V1 is the last one of Pv0 vk+1 , where Pv0 vk+1 is the shortest path between v0 and vk+1 , then h(vl ) ≤ h(v p ). And then h(vk+1 ) − 1 ≥ h(v p ) ≥ h(vl ) ≥ h(vk ) − 1. Case 2. h(vl ) < h(vk ) − 1, which implies that vl vk ∈ / E (G ). Note that xvt ≤ xvk+1 < xvk ≤ xvl . We claim that xvs ≤ xvl for any vertex vs ∈ N (vk ). If vs ∈ V2 , then the claim holds obviously. If vs ∈ V1 and xvs > xvl , then 0 ≤ s < l ≤ k and h(vs ) ≤ h(vl ). Thus, h(vk ) ≤ h(vs ) + 1 ≤ h(vl ) < h(vk ), a contradiction. Considering the vertices vl , vk and vt . Note that xvt < xvk ≤ xvl and xvs ≤ xvl for any vertex xvs ∈ N (vk ). Then by Lemma 2.4, we can find a connected graph G ∈ GD whose Aα -spectral radius is greater than λ1 (Aα (G)), a contradiction. Based on Lemma 2.5 and the Claim, we have obtained a well-ordering ≺ of V(G), which conforms to (1) and (2) of the definition of BFS-ordering. We just need to show that this ordering also satisfies (3) of the definition of BFS-ordering as follows. Let uv, xy ∈ E (G ) and uy, xv ∈ / E (G ) with h(u ) = h(x ) = h(v ) − 1 = h(y ) − 1 and u≺x, we just need to show that v ≺ y. Otherwise, might as well suppose that xv < xy . Let G = G − {uv, xy} + {uy, xv}, then G ∈ GD and λ1 (Aα (G )) > λ1 (Aα (G)) by Lemma 2.3, a contradiction. Lemma 2.7 [1]. Let M be a nonnegative irreducible symmetric matrix with spectral radius λ1 (M). If there exists a positive vector Y > 0 and a positive real β such that MY < β Y, then λ1 (M) < β . Lemma 2.8 [10,11]. Let G be a graph with maximal degree and α ∈ [0, 1). Then
λ1 (A(G )) ≤ λ1 (Aα (G )) ≤ . If the left equality holds, then G has a λ1 (A(G))-regular component. If the right equality holds, then G is regular. Proof of Lemma 1.1. Let P = u0 u1 . . . uk+1 be the internal path of G and X be a unit eigenvector of Aα (G) corresponding to λ1 (Aα (G)) in which xi is corresponding to ui . Without loss of generality, assume that x0 ≤ xk+1 . Let t be the smallest index such that xt = min0≤i≤k+1 xi , so t < k + 1. Without loss of generality, assume that u = ut and v = ut+1 . Note that ⊂ G and λ1 (A()) = 2, where the graph is shown in Fig 1. Since G is not a regular graph, λ1 (Aα (G)) > 2 by Lemma 2.8. Case 1. t > 0. A positive vector Y can be constructed as follows.
yp =
x p,
if p = w;
xu ,
if p = w.
D. Li, Y. Chen and J. Meng / Applied Mathematics and Computation 363 (2019) 124622
Fig. 1. The graph .
If xt−1 > xu , then
(Aα (Guv )Y )u = 2α xu + (1 − α )(xt−1 + xu ) ≤ 2α xu + (1 − α )(xt−1 + xv ) = λ1 (Aα (G ))xu = λ1 (Aα (G ))yu , and
(Aα (Guv )Y )w = 2α xu + (1 − α )(xu + xv ) < 2α xu + (1 − α )(xt−1 + xv ) = λ1 (Aα (G ))xu = λ1 (Aα (G ))yw . If xt−1 = xu , then
(Aα (Guv )Y )u = 2xu < λ1 (Aα (G ))xu = λ1 (Aα (G ))yu and
(Aα (Guv )Y )w = (Aα (Guv )Y )w . Therefore, Aα (Guv )Y < λ1 (Aα (G ))Y and λ1 (Aα (Guv )) < λ1 (Aα (G )) by Lemma 2.7. Case 2. t = 0, i.e., dG (u) ≥ 3. Let S = j∈NG (u )\{v} x j . Subcase 2.1. λ1 (Aα (G ))xu > 2α xu + (1 − α )xu + (1 − α )xv . A positive vector Y can be constructed as follows.
yp =
x p,
if p = w;
xu ,
if p = w.
Then
(Aα (Guv )Y )u = α dG (u )xu + (1 − α )(S + xu ) ≤ 2α xu + (1 − α )(S + xv ) = λ1 (Aα (G ))xu = λ1 (Aα (G ))yu . (Aα (Guv )Y )w = 2α xu + (1 − α )(xu + xv ) < λ1 (Aα (G ))xu = λ1 (Aα (G ))yw . Therefore, Aα (Guv )Y < λ1 (Aα (G ))Y and λ1 (Aα (Guv )) < λ1 (Aα (G )) by Lemma 2.7. Subcase 2.2. λ1 (Aα (G ))xu ≤ 2α xu + (1 − α )xu + (1 − α )xv . We can construct a positive vector Y as follows.
yp =
⎧ x p, ⎪ ⎪ ⎨
if p = u, w;
xu ,
⎪ ⎪ ⎩
if p = w;
1 [ 1 −α
(λ1 (Aα (G )) − 2α )xu − (1 − α )xv ]
Note that
λ1 (Aα (G ))xu = α dG (u )xu + (1 − α )(S + xv ), thus
yu =
α
1−α
( dG ( u ) − 2 )xu + s > 0.
if p = u.
5
6
D. Li, Y. Chen and J. Meng / Applied Mathematics and Computation 363 (2019) 124622
Obviously, by λ1 (Aα (G)) > 2 we have
2xu − xv < yu ≤ xu , that is,
yu ≤ xu < xv . By yu ≤ xu , we observe that
S≤
1 + α − α dG ( u ) xu . 1−α
Then
(Aα (Guv )Y )w = 2α xu + (1 − α )(yu + xv ) = 2α xu + (λ1 (Aα (G )) − 2α )xu − (1 − α )xv + (1 − α )xv = λ1 (Aα (G ))xu = λ1 (Aα (G ))yw . For any vertex z ∈ NG (u ) \ {v},
(Aα (Guv )Y )z = α dG (z )xz + (1 − α )
≤
α dG ( z )xz + ( 1 − α )
x j + yu
j∈NG (z )\{u}
x j + xu
j∈N (z )\{u}
λ1 (Aα (G ))xz = λ1 (Aα (G ))yz . =
Next, we only need to check it at the vertex u. Note that
λ1 (Aα (G ))S ≥ α S + (1 − α )(dG (u ) − 1 )xu . Then
(Aα (Guv )Y )u = α dG (u )yu + (1 − α )(S + xu ) α2 = dG (u )(dG (u ) − 2 )xu + α (dG (u ) − 1 )S + S + (1 − α )xu . 1−α α λ1 (Aα (G ))yu = λ1 (Aα (G )) ( dG ( u ) − 2 )xu + S 1−α α = (dG (u ) − 2 )λ1 (Aα (G ))xu + λ1 (Aα (G ))S 1−α α2 ≥ dG (u )(dG (u ) − 2 )xu + α (dG (u ) − 1 )S 1−α + α (dG (u ) − 2 )xv + (1 − α )(dG (u ) − 1 )xu . And combining with xu < xv , we get
λ1 (Aα (G ))yu − (Aα (Guv )Y )u ≥ α (dG (u ) − 2 )xv + (1 − α )(dG (u ) − 2 )xu − S
1 + α − α dG ( u ) xu 1−α 1 + α − α dG ( u ) > α (dG (u ) − 2 )xu + (1 − α )(dG (u ) − 2 )xu − xu 1−α dG ( u ) − 3 + α = xu 1−α ≥ 0. ≥
α (dG (u ) − 2 )xv + (1 − α )(dG (u ) − 2 )xu −
Therefore, Aα (Guv )Y < λ1 (Aα (G ))Y and λ1 (Aα (Guv )) < λ1 (Aα (G )) by Lemma 2.7.
3. The Aα -spectral radius of trees The graph TD∗ has been introduced by Zhang [13], let’s go over it. For a given nonincreasing tree sequence D =
(d0 , d1 , . . . , dn−1 ) with n ≥ 3, the TD∗ can be construct as follows. Set T0 = v0 . For k ≥ 1, we construct Tk from Tk−1 by adding a new vertex vk and joining vk to the non-saturated vertex with maximum degree in Tk−1 , where a vertex vi in Tk−1 is
D. Li, Y. Chen and J. Meng / Applied Mathematics and Computation 363 (2019) 124622
7
Fig. 2. The graph Gp,q (u).
= Gn−1 . As we know, TD∗ is a greedy tree actually. Zhang [13] observed non-saturated if dTk−1 (vi ) < di . If k = n − 1, then ∗ that TD is also the only one BFS-graph in GD . And, if T1 and T2 both have BFS-ordering in GD , then T1 ∼ =T2 , see Proposition 2.2 in [13]. Therefore, by Theorem 2.6 and Proposition 2.2 in [13], a concise proof of Theorem 1.2 is given. TD∗
Proof of Theorem 1.2. Let T ∈ GD and λ1 (Aα (T )) = max{λ1 (Aα (T )) | T ∈ GD }. By Theorem 2.6, T must have a BFSordering. Then T ∼ = TD∗ by Proposition 2.2 in [13]. 4. The Aα -spectral radius of unicyclic graphs The special unicyclic graph UD∗ has been introduced by Zhang [14]. For a given nonincreasing unicyclic graphic sequence D = (d0 , d1 , . . . , dn−1 ) with n ≥ 3, UD∗ = Cn if d0 = 2 obviously. Next, we assume that d0 ≥ 3. Set G3 = C3 = v0 v1 v2 v0 . For k ≥ 4, we construct Gk from Gk−1 by adding a new vertex vk−1 and joining vk−1 to the non-saturated vertex with maximum degree in Gk−1 , where a vertex vi in Gk−1 is non-saturated if dGk−1 (vi ) < di . If k = n − 1, then UD∗ = Gn−1 . As we know, UD∗ is a greedy unicyclic graph with girth three. Zhang [14] showed that UD∗ is the only one BFS-graph in GD . The graph Gp,q (u) is the graph obtained by attaching the paths Pp and Pq to the vertex u of G, which is shown in Fig 2. Lin, Huang and Xue [6] gave the relationship between the Aα -spectral radius and the parameters p and q. Lemma 4.1 [6]. Let 0 ≤ α < 1. If p − q ≥ 2, then
λ1 (Aα (G p−1,q+1 (u ))) < λ1 (Aα (G p,q (u ))). Lemma 4.2. Let D = (d0 , d1 , . . . , dn−1 ) be a positive nonincreasing unicyclic graphic sequence with d1 = 2. If α ∈ [0, 1), then λ1 (Aα (UD∗ )) = max{λ1 (Aα (G )) | G ∈ GD }. Proof. Note that GD = ∅. Let U ∈ GD and λ1 (Aα (U )) = max{λ1 (Aα (G )) | G ∈ GD }. If d0 = 2, then U∼ =Cn . If d0 ≥ 3, then U ∼ = G(l; l1 , . . . , ld0 −2 ), where G(l; l1 , . . . , ld0 −2 ) is a unicyclic graph by attaching d0 − 2 paths Pl1 , . . . , Pld −2 to a vertex v0 ∈ Cl . 0
Then the following claims must hold. Claim 1. l = 3. On the contrary, suppose that l ≥ 4, set G1 = G(l − 1; l1 , . . . , ld0 −2 ), then λ1 (Aα (G)) < λ1 (Aα (G1 )) by Lemma 1.1. Next, we can construct a graph G2 = G(l − 1; l1 + 1, . . . , ld0 −2 ) by adding a pendant edge to a pendant vertex of G1 , then G2 ∈ GD and G1 ⊂ G2 . Then λ1 (Aα (G)) < λ1 (Aα (G1 )) ≤ λ1 (Aα (G2 )), a contradiction. Claim 2. |li − l j | ≤ 1 for 1 ≤ i, j ≤ d0 − 2. Otherwise, suppose that there exist p and q such that |l p − lq | ≥ 2, without loss of generality, assume that l p ≥ lq + 2, thus G = Gl p ,lq (v0 ). Then by Lemma 4.1, we have λ1 (Aα (Gl p ,lq (v0 ))) < λ1 (Aα (Gl p −1,lq +1 (v0 ))) and Gl p −1,lq +1 (v0 ) ∈ GD , a contradiction. By the above Claims, we immediately obtain that U ∼ = UD∗ . Lemma 4.3 [10]. Let G be a graph with maximal degree . If α ∈ [0, 1/2], then
λ1 (Aα (G )) ≥ α ( + 1 ). If α ∈ [1/2, 1), then
λ1 (Aα (G )) ≥ α +
( 1 − α )2 . α
Lemma 4.4. Let D = (d0 , d1 , . . . , dn−1 ) be a positive nonincreasing unicyclic graphic sequence with d1 ≥ 3. If α ∈ [0, 1), then λ1 (Aα (UD∗ )) = max{λ1 (Aα (H )) | H ∈ GD }. Proof. Note that GD = ∅. Let U ∈ GD and λ1 (Aα (U )) = max{λ1 (Aα (H )) | H ∈ GD }. Let X be a unit eigenvector of Aα (U) corresponding to λ1 (Aα (U)). Then by Theorem 2.6, U is a BFS-graph, i.e., a BFS-ordering of V(U) exists and it makes
v0 ≺ v1 ≺ · · · ≺ vn−1 , xv0 ≥ xv1 ≥ · · · ≥ xvn−1 , dU (v0 ) ≥ dU (v1 ) ≥ · · · ≥ dU (vn−1 ),
8
D. Li, Y. Chen and J. Meng / Applied Mathematics and Computation 363 (2019) 124622
and
h(v0 ) ≤ h(v1 ) ≤ · · · ≤ h(vn−1 ). Let Vi = {v | v ∈ V (U ), h(v ) = i} for i = 0, · · · , h(vn−1 ). Clearly, V0 = {v0 } and V1 = {v1 , · · · , vd0 }. Let C be the unique cycle of U, then we claim that C = v0 v1 v2 v0 . In order to show the claim, the following result is necessary. Claim 1. xv0 > xvd . 0
Suppose that xv0 = xvd , then xv0 = xv1 = · · · = xvd . By Aα (U )X = λ1 (Aα (U ))X, we have 0
0
λ1 (Aα (U ))xv0 = d0 xv0 , then λ1 (Aα (U )) = d0 and U is a regular graph by Lemma 2.8, a contradiction. Now, suppose that C = v0 v1 v2 v0 . Case 1. v0 ∈ V (C ). Then v1 v2 ∈ / E (U ). Subcase 1.1 vi v j ∈ E (U ) for 1 ≤ i < j ≤ d0 . If i = 1 and 3 ≤ j ≤ d0 . Without loss of generality, assume that xv2 > xv j (otherwise, we can exchange v2 and v j , then C = v0 v1 v2 v0 .), then dU (v2 ) ≥ dU (v j ) ≥ 2 by Lemma 2.5. Then we can find a vertex vk ∈ V2 such that v2 vk ∈ E (U ) and v j vk ∈/ E (U ). Let U1 = U − {v1 v j , v2 vk } + {v1 v2 , v j vk } and U1 ∈ GD obviously. Since xv1 > xvk and xv2 > xv j , λ1 (Aα (U )) < λ1 (Aα (U1 )) by Lemma 2.3, a contradiction. If 1 < i < j ≤ d0 , without loss of generality, assume that xv1 > xvi (otherwise, we can exchange v1 and vi , then it is as same as the above case). Since dU (v1 ) = d1 ≥ 3, there exists a vertex vk ∈ V2 such that v1 vk ∈ E (U ) and v j vk ∈ / E (U ) since U is a unicyclic graph. Let U2 = U − {v1 vk , vi v j } + {v1 vi , vk v j } and U2 ∈ GD obviously. Since xv1 > xvi and xv j ≥ xvk , λ1 (Aα (U )) < λ1 (Aα (U2 )) by Lemma 2.3, a contradiction. Subcase 1.2 vi v j ∈ / E (U ) for 1 ≤ i < j ≤ d0 . Then there must exist vk ∈ V1 and vl ∈ V2 such that vk vl ∈ E (C ) and k ≥ 2. From dU (v1 ) = d1 ≥ 3, we can get a vertex vt ∈ V2 such that v1 vt ∈ E (U )\E (C ) and xvt ≤ xu , where u ∈ N (v1 )\V (C ). Then vk vt ∈ / E (U ) and vl vt ∈ / E (U ) since U is a unicyclic graph. The following claim about U holds. Claim. xv1 > xvl or xvk > xvt . Otherwise, let xv1 = xvl and xvk = xvt , then xv1 = xvl = xvk = xvt , xu = xv1 for u ∈ N (v1 )\{v0 } and xv1 = · · · = xvd . By 0
Aα (U )X = λ1 (Aα (U ))X, we have
λ1 (Aα (U ))xvt = α dU (vt )xvt + (1 − α )
xu
u∈N (vt )
≤
α dU (vt )xv1 + (1 − α )dU (vt )xv1
= dU (vt )xv1 that is, λ1 (Aα (U )) ≤ dU (vt ) = dU (v1 ) = d1 by Lemma 2.5. Note that
λ1 (Aα (U ))xv0 = α dU (v0 )xv0 + (1 − α )dU (v0 )xv1 that is, (λ1 (Aα (U )) − α d0 )xv0 = (1 − α )d0 xv1 . Note that
λ1 (Aα (U ))xv1 = α dU (v1 )xv1 + (1 − α )
xu
u∈N ( v 1 )
α dU (v1 )xv1 + (1 − α )(dU (v1 ) − 1 )xv1 + (1 − α )xv0 = d1 xv1 + (1 − α )(xv0 − xv1 ) =
that is, (λ1 (Aα (U )) − d1 − α + 1 )xv1 = (1 − α )xv0 . By the above equalities, we have
d (α d1 − (1 − α )) d1 − d0 (λ1 (Aα (U )) − d1 ) λ1 (Aα (U )) − 0 = λ1 (Aα (G ))(α − 1 ) . d1
Note that λ1 (Aα (U )) >
d0 (α d1 −(1−α )) d1
d1
(1 − α ) dd0 1
= α d0 − by Lemma 4.3. Combining with λ1 (Aα (U)) ≤ d1 , we have λ1 (Aα (U )) = d1 , and then d1 = d0 , which implies that U is a regular graph by Lemma 2.8, a contradiction. Thus, the claim must hold. Let U3 = U − {v1 vt , vk vl } + {v1 vk , vl vt } and U3 ∈ GD obviously. Since xv1 > xvl or xvk > xvt , we have λ1 (Aα (U )) < λ1 (Aα (U3 )) by Lemma 2.3, a contradiction. Case 2. v0 ∈ / V (C ). Then vi v j ∈ / E (U ) for 1 ≤ i < j ≤ d0 . Let vq ∈ V (C ) and h(vq ) ≤ h(u ) and dU (vq ) ≥ dU (u ) for any u ∈ V(C). Subcase 2.1. vq = v1 . Then there must exist vs ∈ V (C ) ∩ V2 such that v1 vs ∈ E (U ), then dU (v2 ) ≥ dU (vs ) ≥ 2. So we can also find a vertex vt ∈ V2 such that v2 vt ∈ E (U ) and vt ∈ / V (C ) since v0 ∈ / V (C ). By the same argument as the claim of subcase 1.2, it can be shown that xv1 > xvt or xv2 > xvs . Let U4 = U − {v1 vs , v2 vt } + {v1 v2 , vs vt } and U4 ∈ GD . Then λ1 (Aα (U )) < λ1 (Aα (U4 )) by Lemma 2.3, a contradiction. Subcase 2.2. vq = v1 . Claim. h(vq ) < h(u ) for any u ∈ V (C ) \ {vq }. Otherwise, suppose that there is a vertex vr ∈ V (C ) \ {vq } such that h(vq ) = h(vr ), then there exist two internal disjoint paths Pl1 and Pl2 between vq and vr such that C = Pl1 ∪ Pl2 . However, there also exist two paths Pl3 and Pl4 from v0 to vq and vr , respectively. And Pl3 ∪ Pl4 ∪ Pl1 contains another cycle of U, which contradicts U is a unicyclic graph.
D. Li, Y. Chen and J. Meng / Applied Mathematics and Computation 363 (2019) 124622
9
Then there is a vertex vt ∈ V (C ) such that vq vt ∈ E (U ) and h(vq ) = h(vt ) − 1, thus dU (vk ) ≥ dU (vt ) ≥ 2 for any vk ∈ Vh , h ≤ h(vq ). And we can find two vertices vi ∈ Vh(vq ) and v j ∈ Vh(vq )+1 such that vi v j ∈ E (U ) \ E (C ). Obviously, vq v j , vt v j ∈ / E (U ). Let U5 = U − {vq vt , vi v j } + {vq vi , vt v j } and U5 ∈ GD . Combining with xvq ≥ xv j and xvi ≥ xvt , we have λ1 (Aα (U )) ≤ λ1 (Aα (U5 )) by Lemma 2.3. Furthermore, the smallest height of vertices of the cycle in U5 is less than h(vq ). By repeating the use of subcase 2.2 or Cases 1 and subcase 2.1, we can get λ1 (Aα (U )) ≤ λ1 (Aα (U5 )) < λ1 (Aα (UD∗ )), a contradiction. Therefore, C = v0 v1 v2 v0 , and then U ∼ = UD∗ obviously. By Lemmas 4.2 and 4.4, Theorem 1.3 can be obtained immediately. References [1] A. Berman, J.S. Plemmons, Nonnegative Matrices in the Mathematical Sciences, Academic Press, New York, 1994. Reprinted, SIAM [2] T. Biyikog˘ lu, J. Leydold, Graphs with given degree sequence and maximal spectral rradius, Electron. J. Combin. 15 (2008) 525–573. [3] F. Belardo, E.M.L. Marzi, S.K. Simic´ , J.F. Wang, On the spectral radius of unicyclic graphs with prescribed degree sequence, Linear Algebra Appl. 432 (2010) 2323–2334. [4] Y.Y. Chen, D. Li, J.X. Meng, On the second largest Aα -eigenvalues of graphs, Linear Algebra Appl. 580 (2019) 343–358. [5] Y.Y. Chen, D. Li, J.X. Meng, aα -spectral radius of the second power of a graph, Appl. Math. Comput. 359 (2019) 418–425. [6] H.Q. Lin, X. Huang, J. Xue, A note on the Aα -spectral radius of graphs, Linear Algebra Appl. 557 (2018) 430–437. [7] H.Q. Lin, J. Xue, J.L. Shu, On the Aα -spectra of graphs, Linear Algebra Appl. 556 (2018) 210–219. [8] H.Q. Lin, X.G. Liu, J. Xue, Graphs determined by their Aα -spectra, Discret. Math. 342 (2019) 441–450. [9] H.Q. Lin, J. Xue, J.L. Shu, On the Dα -spectra of graphs, Linear Multilinear Algebra. 10.1080/03081087.2019.1618236. [10] V. Nikiforov, Merging the A-and Q-spectral theories, Appl. Anal. Discret. Math. 11 (2017) 81–107. [11] V. Nikiforov, G. Pasten, O. Rojo, R.L. Soto, On the Aα -spectra of trees, Linear Algebra Appl. 520 (2017) 286–305. [12] J. Xue, H.Q. Lin, S.T. Liu, J.L. Shu, On the Aα -spectral radius of a graph, Linear Algebra Appl. 550 (2018) 105–120. [13] X.D. Zhang, The laplacian spectral radii of trees with degree sequences, Discret. Math. 308 (2008) 3143–3150. [14] X.D. Zhang, The signless Laplacian spectral radius of graphs with given degree sequences, Discret. Appl. Math. 157 (2009) 2928–2937.