# Some Results on the Signless Laplacian Spectra of Unicyclic Graphs

Authors: Muhuo Liu
Journal: ISRN Algebra
 (2011)

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## Abstract

We determine the second to fourth largest (resp. the second smallest) signless Laplacian spectral radii and the second to fourth largest signless Laplacian spreads together with the corresponding graphs in the class of unicyclic graphs withn vertices. Moreover, we prove that one class of unicyclic graphs are determined by their signless Laplacian spectra.


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## Body

## 1. Introduction

Throughout the paper,G=(V,E) is an undirected simple graph with n vertices and m edges. If G is connected with m=n+c-1, then G is called a c-cyclic graph. Especially, if c=0 or 1, then G is called a tree or a unicyclic graph, respectively. Let 𝕌n be the class of unicyclic graphs with n vertices. The neighbor set of a vertex v is denoted by N(v). We write d(v) for the degree of vertex v. In particular, let Δ(G) and δ(G) be the maximum degree and minimum degree of G, respectively. Let A(G) be the adjacency matrix and D(G) be the diagonal matrix whose (i,i)-entry is d(vi), of G, respectively. The signless Laplacian matrix of G is Q(G)=D(G)+A(G). Clearly, Q(G) is positive semidefinite [1] and its eigenvalues can be arranged as(1.1)μ1(G)≥μ2(G)≥⋯≥μn(G)≥0.
Let μ(G) be the signless Laplacian spectral radius of G, namely, μ(G)=μ1(G). The signless Laplacian spread of G is defined as SQ(G)=μ1(G)-μn(G) [1, 2]. Let Φ(G,x) be the signless Laplacian characteristic polynomial of G, that is, Φ(G,x)=det(xI-Q(G)).Recently, the research on the spectrum ofQ(G) receive much attention. Some properties of signless Laplacian spectra of graphs and some possibilities for developing the spectral theory of graphs based on Q(G) are discussed in [3–5]. The largest signless Laplacian spectral radius and the largest signless Laplacian spread among the class of unicyclic graphs with n vertices were determined in [6] and [1], respectively. The smallest signless Laplacian spectral radius among the class of unicyclic graphs with n vertices was determined in [7]. In this paper, we will determine the second to fourth largest and the second smallest signless Laplacian spectral radii together with the corresponding graphs in the class of unicyclic graphs with n vertices. Moreover, we also indentify the second to fourth largest signless Laplacian spreads together with the corresponding graphs in the class of unicyclic graphs with n vertices. In the end of this paper, we will prove that a class of unicyclic graphs are determined by their signless Laplacian spectra.


## 2. The Signless Laplacian Spectral Radii of Unicyclic Graphs

As usually, letK1,n-1, Pn, and Cn be the star, path, and cycle with n vertices, respectively. In the following, let Sn3(n≥4) be the unicyclic graph obtained by adding one edge to two pendant vertices of K1,n-1, and let Fn, Hn, Sn4 be the unicyclic graphs with n vertices as shown in Figure 1.Figure 1
The unicyclic graphsFn,Hn, and Sn4.In [6], the largest signless Laplacian spectral radius in the class of unicyclic graphs was determined, and it was proved as follows.Theorem 2.1 (see [6]).
IfU∈𝕌n, and n≥4, then μ(U)≤μ(Sn3), where the equality holds if and only if U≅Sn3.Theorem 2.2.
SupposeU∈𝕌n, and n≥8. (1) If U≇Sn3, then μ(U)≤μ(Fn), where the equality holds if and only if U≅Fn, and μ(Fn) equals the maximum root of the equation x5-(n+5)x4+(6n+3)x3-(9n-1)x2+(3n+8)x-4=0. (2) If U≇Sn3 and U≇Fn, then μ(U)≤μ(Hn), where the equality holds if and only if U≅Hn, and μ(Hn) equals the maximum root of the equation x5-(n+5)x4+(6n+4)x3-(10n-2)x2+(3n+12)x-4=0. (3) If U≇Sn3, U≇Fn and U≇Hn, then μ(U)≤μ(Sn4), where the equality holds if and only if U≅Sn4, and μ(Sn4) equals the maximum root of the equation x3-(n+3)x2+(4n-2)x-2n=0.In order to prove Theorem2.2, the following lemmas are needed.Lemma 2.3 (see [8]).
μ(G)≤max{d(v)+m(v),v∈V(G)}, where m(v)=∑u~vd(u)/d(v).Proposition 2.4.
Supposec≥1 and G is a c-cyclic graph on n vertices with Δ≤n-3. If n≥2c+5, then μ(G)≤n-1.Proof.
We only need to prove thatmax{d(v)+m(v):v∈V}≤n-1 by Lemma 2.3. Suppose d(u)+m(u)=max{d(v)+m(v):v∈V}. We consider the next three cases.
Case 1 (d(u)=1).
Supposev∈N(u). Then, d(u)+m(u)=1+d(v)≤1+Δ≤n-2<n-1.
Case 2 (d(u)=2).
Suppose thatv,w∈N(u). Then,
(2.1)d(u)+m(u)=2+d(v)+d(w)2≤2+2Δ2=Δ+2≤n-1.
Case 3 (3≤d(u)≤n-3).
Note thatG has n+c-1 edges and 3≤d(u)≤n-3. Then,
(2.2)d(u)+m(u)≤d(u)+2(n+c-1)-d(u)-2d(u)=d(u)-1+2n+2c-4d(u).
Next we will prove that d(u)-1+((2n+2c-4)/d(u))≤n-1, equivalently, d(u)(n-d(u))≥2n+2c-4. Let f(x)=(n-x)x, where 3≤x≤n-3. Since f′(x)=n-2x and 3≤x≤n-3, we have f(x)≥min{f(3),f(n-3)}=3(n-3)≥2n+2c-4.
By combining the above arguments, the result follows.Corollary 2.5.
SupposeU∈𝕌n. If n≥7 and Δ≤n-3, then μ(U)≤n-1.Lemma 2.6 (see [6]).
IfG is a connected graph of order n≥4, then μ(G)≥Δ+1, where the equality holds if and only if G≅K1,n-1.SupposeB is a square matrix, let aii(B) be the entry appearing in the ith row and the ith column of B. The next result gives a new method to calculate the signless Lapalacian characteristic polynomial of an n-vertex graph via the aid of computer.Lemma 2.7 (see [9]).
LetG be a graph on  n-k (1≤k≤n-2) vertices with V(G)={vn,vn-1,…,vk+1}. If G′ is obtained from G by attaching k new pendant vertices, say v1,…,vk, to vk+1, then
(2.3)Φ(Q(G′),x)=(x-1)k⋅det⁡(xIn-k-Q(G)-Bn-k),
where a11(Q(G)) is corresponding to the vertex vk+1, and Bn-k=diag{k+(k/(x-1)),0,…,0}.Example 2.8.
LetFn be the unicyclic graph as shown in Figure 1. By Lemma 2.7, we have
(2.4)Φ(Fn,x)=(x-1)n-4det⁡(B),whereB=(x-(n-2)-n-4x-1-1-10-1x-2-10-1-1x-3-100-1x-1).
By using “Matlab”, it easily follows that
(2.5)Φ(Fn,x)=(x-1)n-5(x5-(n+5)x4+(6n+3)x3-(9n-1)x2+(3n+8)x-4).
With the similar method, by Lemma 2.7 we have
(2.6)Φ(Hn,x)=(x-1)n-5(x5-(n+5)x4+(6n+4)x3-(10n-2)x2+(3n+12)x-4),(2.7)Φ(Sn4,x)=x(x-1)n-5(x-2)(x3-(n+3)x2+(4n-2)x-2n).Proof of Theorem2.2.
Note thatSn3 is the unique unicyclic graph with Δ=n-1, and Fn, Hn, Sn4 are all the unicyclic graphs with Δ=n-2. Now suppose U∈𝕌n∖{Sn3,Fn,Hn,Sn4}. By Lemma 2.6 and Corollary 2.5, we have μ(Sn4)>n-1≥μ(U) because Δ(U)≤n-3.
To finish the proof of Theorem2.2, we only need to show that μ(Sn4)<μ(Hn)<μ(Fn) by Theorem 2.1. By Lemma 2.6, it follows that μ(Fn)>n-1,μ(Hn)>n-1, and μ(Sn4)>n-1. When x≥n-1 and n≥8, by (2.5), (2.6) and (2.7), it follows that
(2.8)Φ(Hn,x)-Φ(Fn,x)=(x-1)n-5x(x2-(n-1)x+4)>0,Φ(Sn4,x)-Φ(Hn,x)=(x-1)n-5(2x2+(n-12)x+4)≥(x-1)n-5(x(3n-14)+4)>0.
Therefore, we have μ(Fn)>μ(Hn)>μ(Sn4). Thus, Theorem 2.2 follows.In [7], the smallest signless Laplacian spectral radius among all unicyclic graphs with n vertices was determined, and that is as follows.Theorem 2.9 (see [7]).
IfU∈𝕌n, then μ(U)≥4, where the equality holds if and only if U≅Cn.Thelollipop graph, denoted by Wn,p, is obtained by appending a cycle Cp to a pendant vertex of a path Pn-p. The next result extends the order of Theorem 2.9.Theorem 2.10.
For anyn, if U∈𝕌n∖{Cn,Wn,n-1}, then μ(U)>μ(Wn,n-1)>μ(Cn).To prove Theorem2.10, we will introduce more useful lemmas and notations.LetG be a connected graph, and uv∈E(G). The graph Gu,v is obtained from G by subdividing the edge uv, that is, adding a new vertex w and edges wu,wv in G-uv. An internal path, say v1v2⋯vs+1(s≥1), is a path joining v1 and vs+1 (which need not be distinct) such that v1 and vs+1 have degree greater than 2, while all other vertices v2,…,vs are of degree 2.Lemma 2.11 (see [3, 10]).
Letuv be an edge of a connected graph G. If uv belongs to an internal path of G, then μ(G)>μ(Gu,v).ByG⊂G′, we mean that G is a subgraph of G′ and G≇G′.Lemma 2.12 (see [10]).
IfG⊂G′ and G′ is a connected graph, then μ(G)<μ(G′).Ifu∈V(G) and d(u)≥3, then we called u a branching point of G. Let U be a connected unicyclic graph and Tv be a tree such that Tv is attached to a vertex v of the unique cycle of U. The vertex v is called the root of Tv, and Tv is called a root tree of U. Throughout this paper, we assume that Tv does not include the root v. Clearly, U is obtained by attaching root trees to some vertices of the unique cycle of U.Proof of Theorem2.10.
In the proof of this result, we assume that the unique cycle ofU is Cp. Now choose U∈𝕌n∖{Cn} such that μ(U) is as small as possible. Since U≇Cn, U has at least one branching point. We consider the next two cases.
Case 1.
There are at least two branching points inCp.
Now supposeu and v are two branching points in Cp such that there does not exsit other barnching point between the shortest path in Cp connected u and v. Let u=v1v2⋯vs=v(s≥2) be the shortest path in Cp connected u and v. Since u is a branching point of U, there is at least one pendant vertex, say w, in Tu. Suppose z is the unique neighbor vertex of w in U (may be z=u). Let U1=U-wz and U2=U1-w. Then, U2⊂U, and hence μ(U2)<μ(U) by Lemma 2.12. Let U3=U1-v1v2+v1w+v2w. By the hypothesis, v1v2⋯vs is an internal path of U2. By Lemma 2.11, we can conclude that μ(U3)<μ(U2)<μ(U). But U3 is also a unicyclic graph with n vertices and U3≇Cn because v is a branching point of U3, it is a contradiction to the choice of U. Thus, Case 1 is impossible.
Case 2.
There is unique branching point inCp.
Subcase 1.
There is at least a branching point outsideCp.
It can be proved analogously with Case1.
Subcase 2.
There does not exist any branching point outsideCp.Suppose u is the unique brancing poing in Cp. By the hypothesis, u is also the unique branching point of U. Then, U is obtained by attaching d(u)-2 paths to the vertex u of Cp.
Ifd(u)≥4, then there are at least two paths being attaching to u. It can be proved analogously with Case 1.
Ifd(u)=3, then U is a lollipop graph, that is, U≅Wn,p. Let V(Wn,p)={v1,v2,…,vn} and E(Wn,p)={vpv1,vivi+1,1≤i≤n-1}. If n-p≥2, since Wn,p-vn-1vn⊂Wn,p, μ(Wn,p)>μ(Wn,p-vn-1vn)=μ(Wn,p-vn) by Lemma 2.12. Moreover, since Wn,p-vn-1vn-v1vp+vpvn+vnv1 is the graph obtained from Wn,p-vn by subdividing the edge v1vp. Thus, by Lemma 2.11 it follows that
(2.9)μ(Wn,p-vn-1vn)=μ(Wn,p-vn)>μ(Wn,p-vn-1vn-v1vp+vpvn+vnv1)=μ(Wn,p+1).
Therefore,μ(Wn,p)>μ(Wn,p+1). Repeating the above process, we can conclude that μ(Wn,p)>μ(Wn,p+1)>⋯>μ(Wn,n-1) holds for n-p≥2.
By combining the above arguments,U≅Wn,n-1.


## 3. The Signless Laplacian Spreads of Unicyclic Graphs

In [1], the largest signless Laplacian spread among all unicyclic graphs with n vrtices was determined, as follows.Theorem 3.1 (see [1]).
Ifn≥8 and U∈𝕌n∖{Sn3}, then SQ(Sn3)>SQ(U).The next result extends the order of Theorem3.1 to the first four largest values.Theorem 3.2.
Ifn≥16 and U∈𝕌n∖{Sn3,Sn4,Fn,Hn}, then
(3.1)SQ(Sn3)>SQ(Sn4)>max⁡{SQ(Fn),SQ(Hn)}≥min⁡{SQ(Fn),SQ(Hn)}>SQ(U).Remark 3.3.
With the aid of computer, we always haveSQ(Fn)<SQ(Hn). But it seems rather difficult to be proved.To prove Theorem3.2, we need to introduce more lemmas as follows.Proposition 3.4.
SupposeU is a unicyclic graph on n vertices with Δ≤n-3. If n≥9, then SQ(U)≤n-1.1.Proof.
Note thatμn(U)≥0 and SQ(U)=μ1(U)-μn(U)≤μ1(U). We only need to prove max{d(v)+m(v):v∈V}≤n-1.1 by Lemma 2.3. Suppose d(u)+m(u)=max{d(v)+m(v):v∈V}. We consider the next three cases.
Case 1 (d(u)=1).
Supposev∈N(u). Then, d(u)+m(u)=1+d(v)≤1+Δ≤n-2<n-1.1.
Case 2 (d(u)=2).
SupposeN(u)={w,v}. Note that U is a unicyclic graph. Then, |N(v)∩N(w)|≤2 and |N(v)∪N(w)|≤n. Therefore,
(3.2)d(u)+m(u)=2+d(v)+d(w)2≤2+n+22<n-1.1.
Case 3 (3≤d(u)≤n-3).
Note thatU has n edges and 3≤d(u)≤n-3. By inequality (2.2), we have
(3.3)d(u)+m(u)≤d(u)-1+2n-2d(u).
Next we will prove that d(u)-1+((2n-2)/d(u))≤n-1.1, equivalently, d(u)(n-d(u)-0.1)≥2n-2. Let g(x)=(n-x-0.1)x, where 3≤x≤n-3. Since g′(x)=n-0.1-2x and 3≤x≤n-3, we have g(x)≥min{g(3),g(n-3)}>2n-2.
By combining the above arguments, the result follows.Lemma 3.5.
Ifn≥16, then n-1.1<SQ(Fn)<n-1.Proof.
Letf1(x)=x5-(n+5)x4+(6n+3)x3-(9n-1)x2+(3n+8)x-4. Clearly,
(3.4)f1(12n)=-132n5(80n5-56n4-32n3-10n2+10n-1).
Next we will prove that f1(1/2n)<0 when n≥16. Let ψ(n)=80n5-56n4-32n3-10n2+10n-1. When n≥16, since ψ′′′(n)=4800n2-1344n-192>0, we have ψ′′(n)=1600n3-672n2-192n-20≥ψ′′(16)=6378476>0, then ψ′(n)=400n4-224n3-96n2-20n+10≥ψ′(16)=25272010>0. Thus, ψ(n)≥ψ(16)=80082591>0, and hence f1(1/2n)<0.
With the similar method, we have(3.5)f1(0.1)=0.2159n-3.18749>0,f1(0.5)=132(11-2n)<0,f1(3)=9n-52>0,f1(n-1)=-20+21n-5n2<0,f1(n-1+12n)=132n5(16n8-320n7+1232n6-1728n5+1192n4-504n3+140n2-20n+1)>0.
By (2.5), we can conclude that 1/2n<μn(Fn)<0.1 and n-1<μ1(Fn)<n-1+1/2n. Thus, n-1.1<SQ(Fn)=μ1(Fn)-μn(Fn)<n-1.Lemma 3.6.
Ifn≥16, then  n-1.1<SQ(Hn)<n-1.Proof.
Letf2(x)=x5-(n+5)x4+(6n+4)x3-(10n-2)x2+(3n+12)x-4. It is easily checked that
(3.6)f2(12n)=-132n5(80n5-112n4-40n3-14n2+10n-1)<0,f2(0.1)=0.2059n-2.77649>0,f2(0.5)=132(87-10n)<0,f2(2.8)>0.2464n-2.14>0,f2(n-1)=-24+25n-5n2<0,f2(n-1+12n)=132n5(16n8-320n7+1376n6-1888n5+1288n4-520n3+144n2-20n+1)>0.
By (2.6), we can conclude that 1/2n<μn(Hn)<0.1 and n-1<μ1(Hn)<n-1+(1/2n). Thus, n-1.1<SQ(Hn)=μ1(Hn)-μn(Hn)<n-1.Proof of Theorem3.2.
By Lemma2.6 and (2.7), SQ(Sn4)=μ1(Sn4)-μn(Sn4)=μ1(Sn4)>n-1. Note that Sn3 is the unique unicyclic graph with Δ=n-1, and Fn, Hn, Sn4 are all the unicyclic graphs with Δ=n-2. Now suppose U∈𝕌n∖{Sn3,Fn,Hn,Sn4}. Then, Δ(U)≤n-3. By Lemmas 3.5 and 3.6, Theorem 3.1, and Proposition 3.4, we can conclude that
(3.7)SQ(Sn3)>SQ(Sn4)>n-1>max⁡{SQ(Fn),SQ(Hn)}≥min⁡{SQ(Fn),SQ(Hn)}>n-1.1≥SQ(U).
This completes the proof of Theorem 3.2.


## 4. A Class of Unicyclic Graphs Determined by Their Signless Laplacian Spectra

A graphG is said to be determined by its signless Laplacian spectrum if there does not exist other nonisomorphic graph H such that H and G share the same signless Laplacian spectra (see [11]). Let S3(n,k) be the unicyclic graph on n vertices obtained by attaching k, and n-k-3 pendant vetrices to two vertices of C3, respectively. By the definition, S3(n,n-3)=Sn3. The next theorem is the main result of this section.Theorem 4.1.
For anyk≥⌈n/2⌉-1, if 8k(n-3-k)≠9(n-3), then S3(n,k) is determined by its signless Laplacian spectrum.To prove Theorem4.1, we need some more lemmas as follows.Lemma 4.2 (see [12]).
IfG is a graph on n vertices with vertex degrees d1≥d2≥⋯≥dn and signless Laplacian eigenvalues μ1≥μ2≥⋯≥μn, then μ2≥d2-1. Moreover, if μ2=d2-1, then d1=d2, and the maximum and the second maximum degree vertices are adjacent.Lemma 4.3 (see [12]).
IfG is a connected graph with n vertices, then μn(G)<δ(G).Lemma 4.4 (see [7]).
In any graph, the multiplicity of the eigenvalue 0 of the signless Laplacian matrix ofG is equal to the number of bipartite components of G.Let𝕌(n,Δ) be the class of unicyclic graphs on n vertices with maximum degree Δ.Lemma 4.5 (see [13]).
For anyk≥⌈n/2⌉-1, if U∈𝕌(n,k+2), then μ(U)≤μ(S3(n,k)), where the equality holds if and only if U≅S3(n,k).Lemma 4.6 (see [13]).
LetG be the graph with the largest signless Laplacian spectral radius in 𝕌(n,Δ). If Δ≤n-2, then there must exist some graph G1∈𝕌(n,Δ+1) such that μ(G)<μ(G1).Proof of Theorem4.1.
By an elementary computation, we have(4.1)Φ(S3(n,k),x)=(x-1)n-5f3(x),
where f3(x)=x5-(n+5)x4+(kn+5n-k2-3k+7)x3-(2kn+7n-2k2-6k+7)x2+(3n+8)x-4. Now suppose that there exists another graph G such that G and S3(n,k) share the same signless Laplacian spectra. Next we will prove that G≅S3(n,k). We only need to prove the following facts.
Fact 1.
G is a connected unicyclic graph.Proof of Fact 1.
Assume thatG has exactly t connected components, say G1,…,Gt, where t≥1. By Lemma 4.4, Gi is not a bipartite graph for 1≤i≤t because S3(n,k) is not a biparite graph. Thus, Gi is a connected unicyclic graph for 1≤i≤t because G has n edges (since S3(n,k) has n edges). Moreover, since 8k(n-3-k)≠9(n-3), we have f3(4)≠0. Thus, Gi is not a cycle for 1≤i≤t because 4 is the eigenvalue of the signless Laplacian matrix of a cycle. By the a bove arguments, we can conclude that Gi is not a bipartite graph and has at least one pendant vertex. Thus, Gi has at least two signless Laplacian eigenvalues being larger than 1 by Lemma 4.2, and the smallest signless Laplacian eigenvalue of Gi is less than 1 by Lemma 4.3. Therefore, by (4.1) we can conclude that t≤1, and hence G is connected. Clearly, G is a connected unicyclic graph because G has n edges.Fact 2.
Δ(G)≤k+2.Proof of Fact 2.
Note thatk≥⌈n/2⌉-1. Then, n<2k+3. By Lemmas 2.3 and 2.6,
(4.2)Δ(G)+1≤μ(G)=μ(S3(n,k))≤max⁡{k+2+n+1k+2,n-1-k+n+1n-k-1,2+n+12}<k+4.
Thus, Δ(G)≤k+2 holds.Fact 3.
G≅S3(n,k).Proof of Fact 3.
By Fact2, we have Δ(G)≤k+2. If Δ(G)≤k+1<Δ(S3(n,k)), then μ(G)<μ(S3(n,k)) by Lemmas 4.5 and 4.6, a contradiction. Thus, Δ(G)=k+2, and hence the result follows from Lemma 4.5 because μ(G)=μ(S3(n,k)).
This completes the proof of Theorem4.1.



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