On the sum of the two largest Laplacian eigenvalues of trees

Abstract

For $S(T)$, the sum of the two largest Laplacian eigenvalues of a tree T, an upper bound is obtained. Moreover, among all trees with $n\ge 4$ vertices, the unique tree which attains the maximal value of $S(T)$ is determined.

MSC:05C50.

1 Introduction

Let $V(G)$ be the vertex set and $E(G)$ be the edge set of a graph G. The numbers of vertices and edges of G are denoted by $n(G)$ and $m(G)$, respectively. For a vertex $v\in V(G)$, let $N_G(v)$ be the set of vertices adjacent to v and $d_G(v)=|N_G(v)|$ be the degree of v. Particularly, denote by $\mathrm{\Delta }(G)$ the maximum degree of G. The diameter of a connected graph G, denoted by $d(G)$, is the maximum distance among all pairs of vertices in G. Let $A(G)$ be the adjacency matrix of G and $D(G)$ be the diagonal matrix of vertex degrees. The matrix $D(G)-A(G)$ is called the Laplacian matrix of G and its eigenvalues are called the Laplacian eigenvalues of G. Let ${\mu }_1\ge {\mu }_2\ge \cdots \ge {\mu }_n$ be the Laplacian eigenvalues of a graph G with n vertices. It is well known that ${\mu }_n=0$ and ${\sum }_{i=1}^{n-1}{\mu }_i=2m(G)$. In particular, ${\mu }_{n-1}$ is called the algebraic connectivity of G and it is denoted by $\alpha (G)$.

The Laplacian matrix is an important topic in the theory of graph spectra. Particularly, much literature has paid attention to ${\mu }_1$, ${\mu }_2$, ${\mu }_{n-1}$ or ${\mu }_1-{\mu }_{n-1}$ for trees (see, for example, [1–6]). Let $S_n$ be the star of order n, $S_{a,b}^k$ be the tree obtained from two stars $S_{a+1}$, $S_{b+1}$ by joining a path of length k between their central vertices (see Figure 1). As is well known, among all trees of order n, $S_n$ has the largest value of ${\mu }_1$ (see [7]) and $S_{n-3,1}^1$ has the second largest value of ${\mu }_1$ (see [6]). On the other hand, Guo [4] proved that these two trees also attain the first two smallest values of ${\mu }_2$, respectively. This implies that ${\mu }_1$, ${\mu }_2$ cannot attain simultaneously the maximal (or minimal) value and even the relation between them seems like a seesaw. Therefore, it is interesting to investigate the value of ${\mu }_1+{\mu }_2$. Moreover, Zhang [6] showed that the $S_{k-1,k-1}^1$, $S_{k-1,k-2}^2$, $S_{k-2,k-2}^3$ attain simultaneously the largest value of ${\mu }_2$ among all trees with 2k vertices. Then Shao et al.[5] showed that $S_{k-1,k-1}^2$ attains the largest value of ${\mu }_2$ among all trees with $2k+1$ vertices.

Figure 1

${\mathit{S}}_{\mathit{a}\mathbf{,}\mathit{b}}^{\mathit{k}}$ : a tree of order $\mathit{a}\mathbf{+}\mathit{b}\mathbf{+}\mathit{k}\mathbf{+}\mathbf{1}$ .

Another motivation to study the value of ${\mu }_1+{\mu }_2$ came from a result of Haemers et al.[8], who showed that ${\mu }_1+{\mu }_2\le m(G)+3$ for any graph G. This result implies that Brouwer’s conjecture [9],

$${\mu }_1+{\mu }_2+\cdots +{\mu }_k\le m(G)+\left(\begin{array}{c}k+1\\ 2\end{array}\right),$$

is true for $k=2$. Considering a tree T, we have ${\mu }_1+{\mu }_2\le n(T)+2$. Recently, Fritscher et al.[10] improved this bound by giving ${\mu }_1+{\mu }_2<n(T)+2-\frac{2}{n(T)}$. This paper determines the extremal tree that attains the bound of ${\mu }_1+{\mu }_2$. Moreover, for general connected graphs, we also give a conjecture on the extremal graphs for ${\mu }_1+{\mu }_2$.

2 A sharp upper bound of ${\mu }_1+{\mu }_2$

Let $S_k(G)$ be the sum of the largest k Laplacian eigenvalues of a graph G. When $k=2$, we shall write $S(G)$ instead of $S_k(G)$ for simplicity. For graphs G and H, we denote by $G\cup H$ the graph with vertex set $V(G)\cup V(H)$ and edge set $E(G)\cup E(H)$. The following lemmas come from an important result as regards a real symmetric matrix.

Lemma 2.1 ([8])

Let$G_1,G_2,\dots ,G_r$be some edge-disjoint graphs. Then$S_k({\bigcup }_{i=1}^r{G}_i)\le {\sum }_{i=1}^r{S}_k(G_i)$for any k.

Lemma 2.2 ([8])

For any graph G, $S(G)\le m(G)+3$.

Lemma 2.3 Let G be a connected graph, $d_i=d_G(v_i)$and$m_i={\mathrm{\Sigma }}_{v_j\in N_G(v_i)}{d}_j/d_i$. Then

• (i) [11]${\mu }_1(G)\ge \mathrm{\Delta }(G)+1$, with equality if and only if$\mathrm{\Delta }(G)=n(G)-1$.

• (ii) [7]${\mu }_1(G)\le n(G)$, with equality if and only if the complement of G is disconnected.

• (iii) [12]${\mu }_1(G)\le max\{d_i+m_i|v_i\in V(G)\}$.

Lemma 2.4 ([6])

Let T be a tree of order n. If$T\ncong S_n$, then${\mu }_1(T)\le {\mu }_1(S_{n-3,1}^1)$, with equality if and only if$T\cong S_{n-3,1}^1$.

Corollary 2.5 Let T be a tree with n vertices and diameter$d\ge 3$. Then${\mu }_1(T)<n-0.5$.

Proof Note that any tree T has diameter $d\ge 3$ if $T\ncong S_n$. According to Lemma 2.4, ${\mu }_1(T)\le {\mu }_1(S_{n-3,1}^1)$. Further, by Lemma 2.3,

$${\mu }_1\left(S_{n-3,1}^1\right)\le max\{d_i+m_i\}=n-2+\frac{n-1}{n-2}=n-1+\frac{1}{n-2}<n-0.5$$

for $n\ge 5$. For $n=4$, a straightforward calculation shows that ${\mu }_1(S_{1,1}^1)=2+\sqrt{2}<3.5$. □

Lemma 2.6 ([2])

Let T be a tree of order n and diameter$d\ge 3$. Then$\alpha (T)\ge \alpha (S_{\lceil \frac{n-d+1}{2}\rceil ,\lfloor \frac{n-d+1}{2}\rfloor }^{d-2})$, with equality if and only if$T\cong S_{\lceil \frac{n-d+1}{2}\rceil ,\lfloor \frac{n-d+1}{2}\rfloor }^{d-2}$.

Lemma 2.7 ([11])

Let G be a graph with a vertex u of degree one. Then$\alpha (G)\le \alpha (G-u)$.

Lemma 2.7 implies that the algebraic connectivity of a tree is not greater than that of its subtree.

Lemma 2.8 ([4])

Let$T_n^k$ ($n\ge 2k+1$) be a tree obtained from a star$S_{n-k}$by replacing its k edges with k paths of length two, respectively. If$k\ge 2$, then${\mu }_2(T_n^k)=\frac{3+\sqrt{5}}{2}$.

The following lemma can be found in [13] and is known as the Interlacing Theorem of Laplacian eigenvalues.

Lemma 2.9 Let G be a graph of order n and H be a graph obtained from G by deleting an edge. Then

$${\mu }_1(G)\ge {\mu }_1(H)\ge \cdots \ge {\mu }_n(G)\ge {\mu }_n(H)=0.$$

Next we give the main theorem of this section. Its proof is divided into several sequent claims.

Theorem 2.10 For any tree T with order$n\ge 4$, $S(T)\le S(S_{\lceil \frac{n-2}{2}\rceil ,\lfloor \frac{n-2}{2}\rfloor }^1)$. The equality holds if and only if$T\cong S_{\lceil \frac{n-2}{2}\rceil ,\lfloor \frac{n-2}{2}\rfloor }^1$.

Claim 2.11 For any tree T with order$n\ge 4$and diameter$d\le 3$, $S(T)<S(S_{\lceil \frac{n-2}{2}\rceil ,\lfloor \frac{n-2}{2}\rfloor }^1)$except that$T\cong S_{\lceil \frac{n-2}{2}\rceil ,\lfloor \frac{n-2}{2}\rfloor }^1$.

Proof If $d(T)=3$, then $T\cong S_{a,b}^1$ for some positive integers a, b with $a+b=n-2$. It is well known that the Laplacian characteristic polynomial of $S_{a,b}^1$ is $\mu {(\mu -1)}^{n-4}{f}_{a,b}(\mu )$, where

$$f_{a,b}(\mu )={\mu }^3-(n+2){\mu }^2+(ab+2n+1)\mu -n.$$

(1)

Note that $S_{a,b}^1$ contains $S_{1,1}^1$ as a subtree. By Lemma 2.9, ${\mu }_2(S_{a,b}^1)\ge {\mu }_2(S_{1,1}^1)=2$. Moreover, we know that for any tree T, $\alpha (T)\le 1$, with equality if and only if T is a star. These imply that ${\mu }_1(S_{a,b}^1)$, ${\mu }_2(S_{a,b}^1)$, and $\alpha (S_{a,b}^1)$ consist of the three roots of $f_{a,b}(\mu )$. As follows from (1), we have

$${\mu }_1\left(S_{a,b}^1\right)+{\mu }_2\left(S_{a,b}^1\right)+\alpha \left(S_{a,b}^1\right)=n+2.$$

(2)

By virtue of Lemma 2.6, we have $\alpha (S_{a,b}^1)>\alpha (S_{\lceil \frac{n-2}{2}\rceil ,\lfloor \frac{n-2}{2}\rfloor }^1)$ except that $(a,b)=(\lceil \frac{n-2}{2}\rceil ,\lfloor \frac{n-2}{2}\rfloor )$. Equivalently, $S(S_{a,b}^1)<S(S_{\lceil \frac{n-2}{2}\rceil ,\lfloor \frac{n-2}{2}\rfloor }^1)$ except that $(a,b)=(\lceil \frac{n-2}{2}\rceil ,\lfloor \frac{n-2}{2}\rfloor )$.

If $d(T)=2$, then $T\cong S_n$. We first give a lower bound of $S(S_{\lceil \frac{n-2}{2}\rceil ,\lfloor \frac{n-2}{2}\rfloor }^1)$ for $n\ge 6$:

$$S\left(S_{\lceil \frac{n-2}{2}\rceil ,\lfloor \frac{n-2}{2}\rfloor }^1\right)>n+1.5.$$

(3)

Indeed, by (2) it suffices to show $\alpha (S_{\lceil \frac{n-2}{2}\rceil ,\lfloor \frac{n-2}{2}\rfloor }^1)<0.5$. Note that for $n\ge 6$, $S_{\lceil \frac{n-2}{2}\rceil ,\lfloor \frac{n-2}{2}\rfloor }^1$ contains $S_{2,2}^1$ as a subtree. By Lemma 2.7, $\alpha (S_{\lceil \frac{n-2}{2}\rceil ,\lfloor \frac{n-2}{2}\rfloor }^1)\le \alpha (S_{2,2}^1)=\frac{5-\sqrt{17}}{2}<0.5$.

Note that $S(S_n)=n+1$ for $n\ge 3$. According to (3), we have $S(S_n)<S(S_{\lceil \frac{n-2}{2}\rceil ,\lfloor \frac{n-2}{2}\rfloor }^1)$ for $n\ge 6$. As for $n\in \{4,5\}$, a straightforward calculation shows that

$$S\left(S_{1,1}^1\right)\approx 5.4142,S\left(S_{2,1}^1\right)\approx 6.4811.$$

(4)

Also we have $S(S_n)<S(S_{\lceil \frac{n-2}{2}\rceil ,\lfloor \frac{n-2}{2}\rfloor }^1)$. □

Claim 2.12 For any tree T with order n and diameter$d\ge 5$, $S(T)<S(S_{\lceil \frac{n-2}{2}\rceil ,\lfloor \frac{n-2}{2}\rfloor }^1)$.

Proof Since $d(T)\ge 5$, then $n\ge 6$ and there is a path of length 5 in T. By inequality (3), it suffices to show $S(T)\le n+1.5$. First suppose that there is a path $v_0{v}_1\cdots v_5$ in T such that either $max\{d_T(v_0),d_T(v_5)\}\ge 2$ or $max\{d_T(v_2),d_T(v_3)\}\ge 3$. Let $T_1$, $T_2$ be the two components of $T-v_2{v}_3$. Clearly, both $T_1$ and $T_2$ have at least two edges.

If ${\mu }_1$, ${\mu }_2$ of $T_1\cup T_2$ attain at the same component, say $T_1$, then by Lemma 2.2,

$$S(T_1\cup T_2)=S(T_1)\le m(T_1)+3\le m(T_1\cup T_2)+1.$$

(5)

Note that $S(v_2{v}_3)=S(S_2)=2$. By Lemma 2.1,

$$S(T)\le S(T_1\cup T_2)+S(v_2{v}_3)\le m(T_1\cup T_2)+3=m(T)+2=n+1.$$

(6)

Otherwise, $S(T_1\cup T_2)={\mu }_1(T_1)+{\mu }_1(T_2)$. Whether $max\{d_T(v_0),d_T(v_5)\}\ge 2$ or $max\{d_T(v_2),d_T(v_3)\}\ge 3$, we can observe that $max\{d(T_1),d(T_2)\}\ge 3$. Say $d(T_2)\ge 3$, then by Corollary 2.5, ${\mu }_1(T_2)<n(T_2)-0.5$. By Lemma 2.3(ii), ${\mu }_1(T_1)\le n(T_1)$. Hence,

$$S(T)\le S(T_1\cup T_2)+S(v_2{v}_3)=n(T_1)+n(T_2)-0.5+2=n+1.5.$$

Next, we may assume that each path $v_0{v}_1\cdots v_5$ of length 5 in T has $d_T(v_0)=d_T(v_5)=1$ and $d_T(v_2)=d_T(v_3)=2$. This implies that $d(T)=5$ and $T\cong S_{a,b}^3$ for some integers a, b with $a+b=n-4$. If $a=b=1$, then T is isomorphic to a path of order 6 and a straightforward calculation shows that $S(T)=5+\sqrt{3}<n+1.5$, as claimed. Otherwise, assume without loss of generality that $a\ge 2$. Then $d_T(v_1)\ge 3$. Let $T_3$, $T_4$ be the two components of $T-v_1{v}_2$ with $v_0{v}_1\in E(T_3)$. Then both $T_3$ and $T_4$ have at least two edges. If ${\mu }_1$, ${\mu }_2$ of $T_3\cup T_4$ attain at the same component, say $T_3$, then by Lemmas 2.1 and 2.2,

$$S(T)\le S(T_3\cup T_4)+S(v_1{v}_2)=S(T_3)+2\le m(T_3)+5\le m(T)+2=n+1.$$

Otherwise, $S(T_3\cup T_4)={\mu }_1(T_3)+{\mu }_1(T_4)$. Note that ${\mu }_1(T_3)\le n(T_3)$. Since $d(T_4)=3$, by Corollary 2.5, ${\mu }_1(T_4)<n(T_4)-0.5$. So

$$S(T)\le S(T_3\cup T_4)+S(v_1{v}_2)\le n(T_3)+n(T_4)-0.5+2=n+1.5.$$

 □

Claim 2.13 For any tree T with order n and diameter 4, $S(T)<S(S_{\lceil \frac{n-2}{2}\rceil ,\lfloor \frac{n-2}{2}\rfloor }^1)$.

Proof First suppose that T contains a path $v_0{v}_1\cdots v_4$ such that $max\{d_T(v_1),d_T(v_3)\}\ge 3$. Now $n\ge 6$ and it suffices to show $S(T)\le n+1.5$. Without loss of generality assume that $d_T(v_1)\ge 3$. Let $T_1$, $T_2$ be the two components of $T-v_1{v}_2$ with $v_0{v}_1\in E(T_1)$. Then both $T_1$ and $T_2$ have at least two edges.

If ${\mu }_1$, ${\mu }_2$ of $T_1\cup T_2$ attain at the same component, say $T_1$, then similarly to inequalities (5) and (6), we can observe that $S(T)\le n+1$.

Now let $S(T_1\cup T_2)={\mu }_1(T_1)+{\mu }_1(T_2)$. If $d_T(v_2)\ge 3$, then $d(T_2)\ge 3$ and hence ${\mu }_1(T_2)<n(T_2)-0.5$. So

$$S(T)\le S(T_1\cup T_2)+S(v_1{v}_2)<n(T_1)+n(T_2)-0.5+2=n+1.5.$$

If $d_T(v_2)=2$, then $T\cong S_{a,b}^2$ for some positive integers a, b with $3\le a+b=n-3$. Moreover, since $d_T(v_1)\ge 3$, then $a\ge 2$. If $(a,b)\in \{(2,1),(3,1)\}$, a straightforward calculations show that $S(S_{a,b}^2)<n+1.5$. Otherwise, $S_{a,b}^2$ contains either $S_{4,1}^2$ or $S_{2,2}^2$ as a subtree. Since

$${\mu }_3\left(S_{4,1}^2\right)\approx 1.5068,{\mu }_3\left(S_{2,2}^2\right)\approx 1.5858,$$

it follows from Lemma 2.9 that ${\mu }_3(S_{a,b}^2)>1.5$. Since $S_{a,b}^2$ is not a star, ${\mu }_{n-1}(S_{a,b}^2)<1$. On the other hand, note that the matrix $1\cdot I_n-[D(S_{a,b}^2)-A(S_{a,b}^2)]$ has a identical rows and b different identical rows, so the multiplicity of eigenvalue 1 is at least $a+b-2$ and else five eigenvalues are ${\mu }_1$, ${\mu }_2$, ${\mu }_3$, ${\mu }_{n-1}$ and ${\mu }_n=0$. Since ${\sum }_{i=1}^n{\mu }_i(S_{a,b}^2)=2(n-1)$, we have

$$\sum _{i=1}^3{\mu }_i\left(S_{a,b}^2\right)+{\mu }_{n-1}\left(S_{a,b}^2\right)+{\mu }_n\left(S_{a,b}^2\right)=2(n-1)-(a+b-2)=n+3.$$

This implies that $S(S_{a,b}^2)<n+3-{\mu }_3(S_{a,b}^2)<n+1.5$.

Next, it suffices to consider the case that each path $v_0{v}_1\cdots v_4$ of T has $d_T(v_1)=d_T(v_3)=2$. This implies that $T\cong T_n^k$ for some $k\ge 2$ and $n\ge 2k+1$, since $d(T)=4$. According to Lemma 2.8, ${\mu }_2(T_n^k)=\frac{3+\sqrt{5}}{2}$. Moreover, by Lemma 2.3,

$${\mu }_1\left(T_n^k\right)\le max\{d_i+m_i\}=n-k-1+\frac{n-1}{n-k-1}\le n-2+\frac{2}{n-3}.$$

Thus for $n\ge 6$,

$$S\left(T_n^k\right)\le n-2+\frac{2}{n-3}+\frac{3+\sqrt{5}}{2}<n+1.5<S\left(S_{\lceil \frac{n-2}{2}\rceil ,\lfloor \frac{n-2}{2}\rfloor }^1\right).$$

When $n=5$, $T_n^k$ is a path. Comparing with (4), $S(T_n^k)=4+\sqrt{5}<S(S_{2,1}^1)$. This completes the proof. □

Following from Claims 2.11-2.13, Theorem 2.10 holds and the unique tree with maximal $S(T)$ is $S_{\lceil \frac{n-2}{2}\rceil ,\lfloor \frac{n-2}{2}\rfloor }^1$. According to (2),

$$\mu \left(S_{\lceil \frac{n-2}{2}\rceil ,\lfloor \frac{n-2}{2}\rfloor }^1\right)<n+2=m\left(S_{\lceil \frac{n-2}{2}\rceil ,\lfloor \frac{n-2}{2}\rfloor }^1\right)+3.$$

Theorem 2.14 Let m, n be two positive integers with$n\le m\le 2n-3$and$G_{m,n}$be a graph of order n and size m obtained from a given edge uv by joining$m-n+1$independent vertices with u and v, respectively, and another$2n-m-3$independent vertices with u. Then$S(G_{m,n})=m+3$.

Proof Let $H_{s,t}$ be a graph obtained by joining a vertex to s vertices of a given complete graph of order $s+t$ and $H_{s,t}^c$ be its complement graph. Then $H_{s,t}^c$ is isomorphic to the union of $S_{t+1}$ and s isolated vertices. Clearly, the Laplacian eigenvalues of $H_{s,t}^c$ consist of $t+1$, 1 with multiplicity $t-1$ and 0 with multiplicity $s+1$. Recall that for any graph G with n vertices, ${\mu }_i(G)=n-{\mu }_{n-i}(G^c)$ for $1\le i\le n-1$ and ${\mu }_n(G)=0$. So the Laplacian eigenvalues of $H_{s,t}$ consist of $s+t+1$ with multiplicity s, $s+t$ with multiplicity $t-1$, s and 0.

Now $G_{m,n}^c$ is isomorphic to the union of $H_{2n-m-3,m-n+1}$ and an isolated vertex. So the Laplacian eigenvalues of $G_{m,n}^c$ consist of $n-1$ with multiplicity $2n-m-3$, $n-2$ with multiplicity $m-n$, $2n-m-3$, and 0 with multiplicity 2. Therefore, the Laplacian eigenvalues of $G_{m,n}$ consist of n, $m-n+3$, 2 with multiplicity $m-n$, 1 with multiplicity $2n-m-3$ and 0. So $S(G_{m,n})=n+(m-n+3)=m+3$. □

Recall that ${\mu }_1(G)\le n(G)$ for any graph G. When $m(G)>2n(G)-3$, Haemers’ bound is clearly not attainable. Theorem 2.14 implies that if $m(G)\le 2n(G)-3$, Haemers’ bound is always sharp for connected graphs other than trees. Ending the paper, we present a conjecture on the uniqueness of the extremal graph.

Conjecture 2.15 Among all connected graphs with n vertices and$n\le m\le 2n-3$edges, $G_{m,n}$is the unique graph with maximal value of${\mu }_1+{\mu }_2$.