On the Distance Signless Laplacian Spectrum of Graphs

Abstract

Let G be a simple, connected graph, D(G) be the distance matrix of G, and Tr(G) be the diagonal matrix of vertex transmissions of G. The distance signless Laplacian matrix of G is defined as $D^Q(G)=Tr(G)+D(G)$, and the distance signless Laplacian spectral radius of G is the largest eigenvalue of $D^{Q}(G)$. In this paper, we study Nordhaus–Gaddum-type inequalities for distance signless Laplacian eigenvalues of graphs and present some new upper and lower bounds on the distance signless Laplacian spectral radius of G and of its line graph L(G), based on other graph-theoretic parameters, and characterize the extremal graphs. Further, we study the distance signless Laplacian spectrum of some graphs obtained by operations.

1 Introduction

Unless stated otherwise, we follow [12] for terminology and notations, and we consider only finite, connected, simple graphs. In particular, denote $V(G)=\{v_{1}, v_{2},\ldots ,v_{n}\}$ the vertex set of G and E(G) the edge set of G. For a graph $G=(V(G),E(G))$, two vertices are called adjacent if they are connected by an edge. $|V(G)|=n$ is always called the order of G and $|E(G)|=m$ is always called the size of G. The set of vertices adjacent to $v\in V(G)$, denoted by N(v), refers to the neighborhood of v. The degree of v,  denoted by $d_{G}(v)$ (we simply write $d_v$ if it is clear from the context), means the cardinality of N(v). A graph is called regular if each of its vertex has the same degree. The distance between two vertices $u,v\in V(G),$ denoted by d(u, v) or $d_{uv}$, is defined as the length of a shortest path between u and v in G. The diameter of G, denoted by $\mathrm{diam}(G)$, is the maximum distance between any two vertices of G. Let $A=A(G)$ be the adjacency matrix of G. The signless Laplacian matrix of G is $Q(G)=A+D^{\prime }$, where $D^{\prime }=D^{\prime }(G)$ is the diagonal matrix of vertex degrees of G. The distance matrix of G is denoted by D(G) and is defined as $D(G)=(d_{uv})_{u,v\in V(G)}$. We also denote the complement graph of G by ${\bar{G}}$.

The transmission$Tr_{G}(v)$ of a vertex v is defined to be the sum of the distances from v to all other vertices in G, that is, $Tr_{G}(v)=\sum \nolimits _{u\in V(G)}d_{uv}$. A graph G is said to be k-transmission regular if $Tr_{G}(v)=k$, for each $v\in V(G)$. The transmission regular graphs are exactly the distance-balanced graphs introduced in [11]. Clearly, any vertex-transitive graph (a graph G in which for every two vertices u and v, there exists an automorphism f on G such that $f(u)=f(v)$) is a transmission regular graph. But there are transmission regular graphs which are not degree regular and hence not vertex-transitive (see [1] and the references therein). The transmission of a graph G, denoted by $\sigma (G)$, is the sum of distances between all unordered pairs of vertices in G. Clearly, $\sigma (G)=\frac{1}{2}\sum _{v\in V(G)}Tr_{G}(v)$. For a graph G with $V(G)=\{v_{1},v_{2},\ldots ,v_{n}\}$, $Tr_G(v_i)$ has been referred as the transmission degree$Tr_{i}$ [13] and hence the transmission degree sequence is given by $\{Tr_{1},Tr_{2},\ldots ,Tr_{n}\}$. The second transmission degree of $v_{i}$, denoted by $T_{i}$ is given by $T_{i}=\sum _{j=1}^{n}d_{v_iv_j}Tr_{j}$.

Let $Tr(G)=\mathrm{diag}(Tr_1,Tr_2,\ldots ,Tr_n)$ be the diagonal matrix of vertex transmissions of G. Aouchiche and Hansen [1, 3] introduced the Laplacian and the signless Laplacian for the distance matrix of a connected graph. The matrix $D^L(G)=Tr(G)-D(G)$ is called the distance Laplacian matrix of G, while the matrix $D^{Q}(G)=Tr(G)+D(G)$ is called the distance signless Laplacian matrix ofG. If G is connected, then $D^{Q}(G)$ is symmetric, nonnegative and irreducible. Hence, all the eigenvalue of $D^{Q}(G)$ can be arranged as $\rho _{1}\ge \rho _{2}\ge \cdots \ge \rho _{n},$ where $\rho _{1}$ is called the distance signless Laplacian spectral radius of G. From now onward, we will denote $\rho _1(G)$ by $\rho (G)$. As $D^{Q}(G)$ is irreducible, by the Perron–Frobenius theorem, $\rho (G)$ is positive and simple, and there is a unique positive unit eigenvector X corresponding to $\rho (G)$, which is called the distance signless Laplacian Perron vector of G. A column vector $X=(x_{1},x_{2},\ldots ,x_{n})^{T}\in \mathbb {R}^{n}$ can be considered as a function defined on V(G) which maps vertex $v_{i}$ to $x_{i}$, that is, $X(v_{i})=x_{i}$ for $i=1,2,\ldots ,n$. Then,

$$\begin{aligned} X^{T}D^{Q}(G)X= \sum _{\{u,v\}\subseteq V(G)}d_{uv}(x_u+x_v)^{2}, \end{aligned}$$

and $\lambda$ is an eigenvalue of $D^{Q}(G)$ corresponding to the eigenvector X if and only if $X\ne \mathbf 0$ and for each $v\in V(G)$,

$$\begin{aligned} \lambda x_v=\sum _{u\in V(G)}d_{uv}(x_u+x_v). \end{aligned}$$

These equations are called the $(\lambda , x)$-eigenequations of G. For a normalized column vector $X\in \mathbb {R}^{n}$ with at least one nonnegative component, by the Rayleigh’s principle, we have

$$\begin{aligned} \rho (G)\ge X^{T}D^{Q}(G)X, \end{aligned}$$

with equality if and only if X is the distance signless Laplacian Perron vector of G.

The investigation of matrices related to various graphical structures is a very large and growing area of research. In particular, distance signless Laplacian matrix have attracted serious attention in the literature. In  [26], Xing et al. determined the tree with minimum distance signless Laplacian spectral radius among all the trees with fixed number of vertices. In  [25], Xing et al. determined the unique graphs with minimum and second minimum distance signless Laplacian spectral radii among bicyclic graphs with fixed number of vertices. In [7], bounds for distance signless Laplacian spectral radius are given using vertex transmissions and in [18], lower bound for distance signless Laplacian spectral radius is given in terms of chromatic number.

The paper is organized as follows. In Sect. 2, we give Nordhaus–Gaddum-type results for the spectral radius of the distance signless Laplacian matrix of a graph. In Sect. 3, we obtain some results on eigenvalues of the distance signless Laplacian matrix of a graph and establish some upper and lower bounds on the distance signless Laplacian spectral radius of G. In Sect. 4, we study the line graphs L(G) of simple connected graphs and determine some upper and lower bounds on the distance signless Laplacian spectral radius of L(G) based on some graph invariants, and characterize the extremal graphs. Finally, in Sect. 5, we consider operations (like Cartesian product and lexicographic product) on graphs and determine the distance signless Laplacian spectrum of some graphs obtained by these operations.

2 Nordhaus–Gaddum-Type Inequalities for Distance Signless Laplacian Eigenvalues of a Graph

Nordhaus and Gaddum [21] studied the chromatic number of a graph G and of its complement together. They proved lower and upper bounds on the sum and on the product of chromatic number of G and that of ${\bar{G}}$ in terms of the order of G. Since then, any bound on the sum and/or the product of an invariant in a graph G and the same invariant in ${\bar{G}}$ is called a Nordhaus–Gaddum-type inequality. In [2], Nordhaus–Gaddum-type inequalities for graph parameters were surveyed. Many of those inequalities involve eigenvalues of adjacency, Laplacian and signless Laplacian matrices of graphs. In this section, we bring Nordhaus–Gaddum-type results for the spectral radius of the distance signless Laplacian matrix of a graph.

Let $\varGamma$ be the class of graphs $H=(V(H),E(H))$ such that H is connected graph of diameter d$(3\le d\le 4)$ with $|V(H)|\ge d+2$, having the following property: Let $P_{d+1}$ be the $(d+1)$-vertex path contained in H. Then, for any vertex $i\in V(H)\setminus V(P_{d+1})$ and for any vertex $j\in V(H)$, where $j\ne i$, it should be either $d(v_i,v_j)=1$ or $d(v_i,v_j)=2$.

We start this section by mentioning the following Lemma from [26], which will be useful to derive a main result of this section.

Lemma 1

[26] Let G be a connected graph on n vertices. Then,

$$\begin{aligned} \rho (G)\ge \frac{4\sigma (G)}{n}, \end{aligned}$$

with equality holding if and only if G is transmission regular.

Recall that the Wiener index W(G) of a connected graph G is defined to be the sum of all distances in G, that is,

$$\begin{aligned} W(G)= \sum _{\{u,v\} \subseteq V(G)}d_{G}(u,v). \end{aligned}$$

The following lower bound for $W(G)+W({\bar{G}})$ is presented in [6].

Lemma 2

[6] Let G be a connected graph on $n\ge 4$ vertices, diameter d, and with a connected complement ${\bar{G}}$. Then

$$\begin{aligned} W(G)+W({\bar{G}})\ge \frac{3}{2}n(n-1)+\frac{1}{6}(d-2)(d-1)d, \end{aligned}$$

with equality holding if and only if G is a graph of diameter two or $G\cong P_{n}$ or $G\in \varGamma$ and ${\bar{G}}$ is a graph of diameter 2.

Now we give a lower bound for $\rho (G)+\rho ({\bar{G}})$.

Theorem 1

Let G be a connected graph on $n\ge 4$ vertices, diameter d, and with a connected complement ${\bar{G}}$. Then

$$\begin{aligned} \rho (G)+\rho ({\bar{G}})\ge 6(n-1)+\frac{2}{3n}(d-2)(d-1)d, \end{aligned}$$

with equality holding if and only if both G and ${\bar{G}}$ are transmission regular graphs of diameter two.

Proof

Using Lemmas 1 and 2, we get the required result. Moreover, one can see easily that the equality holds if and only if both G and ${\bar{G}}$ are transmission regular graphs of diameter two. $\square$

Corollary 1

Let G be a connected graph on $n\ge 4$ vertices with a connected complement ${\bar{G}}$. Then

$$\begin{aligned} \rho (G)+\rho ({\bar{G}})\ge 6(n-1), \end{aligned}$$

with equality holding if and only if both G and ${\bar{G}}$ are transmission regular graphs of diameter two.

Proof

The proof follows directly from Theorem 1. $\square$

Corollary 2

Let G be a connected graph with $n\ge 2$ vertices and m edges. Then

$$\begin{aligned} \rho (G)\ge 4(n-1)-\frac{4m}{n}, \end{aligned}$$

(1)

with equality holding if and only if $G\cong K_{n}$ or G is a transmission regular graph with diameter two.

Proof

Note that

$$\begin{aligned} W(G)\ge m+2\left( \frac{n(n-1)}{2}-m\right) =n(n-1)-m, \end{aligned}$$

with equality holding if and only if the diameter of G is at most two. By Lemma 1, we have (1). Now suppose equality holds in (1), then the equality in Lemma 1 holds and $W(G) = m+2\left( \frac{n(n-1)}{2}-m\right) =n(n-1)-m$. So G is transmission regular and the diameter of G is at most two, that is, $G\cong K_{n}$ or G is transmission regular graph of diameter two.

Conversely, if $G\cong K_{n}$ or G is a transmission regular graph of diameter two, then it is easy to see that (1) is an equality. $\square$

Corollary 3

Let G be a triangle-and quadrangle-free connected graph with $n\ge 2$ vertices and m edges. Then

$$\begin{aligned} \rho (G)\ge 6(n-1)-\frac{2}{n}\sum _{i=1}^{n}d_{{v}_{i}}^{2}- \frac{4m}{n}, \end{aligned}$$

(2)

with equality holding if and only if the row sums of $D^{Q}(G)$ are all equal and the diameter of G is at most three.

Proof

Note that by [27], we have

$$\begin{aligned} W(G)\ge \frac{3n(n-1)}{2}-\frac{1}{2}\sum _{i=1}^{n}d_{{v}_{i}}^{2}-m, \end{aligned}$$

with equality holding if and only if the diameter is at most three. Now, the result follows from Lemma 1. $\square$

3 Bounds on Distance Signless Laplacian Spectral Radius

The distance spectral radius (the spectral radius of the distance matrix) and the Laplacian spectral radius (the spectral radius of the Laplacian matrix) of a connected graph have been studied extensively in the literature. The spectral radius of the distance signless Laplacian matrix of a graph has received much attention recently. Compared to the Laplacian spectral radius and the distance spectral radius, there are not many results available in literature on distance signless Laplacian spectral radius. For some recent works on the distance signless Laplacian spectral radius, see [18, 25, 26]. In this section, we establish some bounds on distance signless Laplacian spectral radius of a simple, connected graph G.

The following lemma is well known and will be helpful in the sequel.

Lemma 3

[19] If A is an $n\times n$ nonnegative matrix with the spectral radius $\lambda (A)$ and row sums $r_{1}, r_{2},\ldots , r_{n}$, then $\min \limits _{1\le i\le n}r_{i}\le \lambda (A)\le \max \limits _{1\le i\le n}r_{i}$. Moreover, if A is irreducible, then one of the equalities holds if and only if the row sums of A are all equal.

Now, we give the lower and upper bounds for $\rho (G)$ in terms of transmission degree and second transmission degree of G.

Theorem 2

Let G be a simple, connected graph. Then

$$\begin{aligned} \rho (G)\le \frac{Tr_{\max }+\sqrt{Tr^{2}_{\max }+8T_{\max }}}{2}, \end{aligned}$$

where $Tr_{\max }$ and $T_{\max }$ are the maximum transmission degree and the maximum second transmission degree of G,  respectively. Moreover, the equality holds if and only if G is transmission regular.

Proof

Since $D^{Q}(G)=Tr(G)+D(G)$, by a simple calculation, we have $r_{v_{i}}(D^{Q}(G))=2Tr_{i}$ and $r_{v_{i}}(D Tr)=r_{v_{i}}(D^{2})=\sum _{j=1}^{n}d_{v_iv_j}Tr_{j}$. Then

$$\begin{aligned} r_{v_{i}}((D^{Q}(G))^2)= & {} r_{v_{i}}(Tr^{2}+Tr D+D Tr+D^{2})\\= & {} r_{v_{i}}(Tr(Tr+D))+r_{v_{i}}(DTr)+r_{v_{i}}(D^{2})\\= & {} Tr_{i}r_{v_{i}}(D^{Q}(G))+2\sum _{j=1}^{n}d_{v_iv_j}Tr_{j}\\\le & {} Tr_{\max }r_{v_{i}}(D^{Q}(G))+2T_{\max }. \end{aligned}$$

So we have

$$\begin{aligned} r_{v_{i}}((D^{Q}(G))^2-Tr_{\max }D^{Q}(G))\le 2T_{\max }. \end{aligned}$$

By Lemma 3, we have

$$\begin{aligned} (\rho (G))^2-Tr_{\max }\rho (G)-2T_{\max }\le 0, \end{aligned}$$

and then by Lemma 3, the result follows. In order to get the equality, all inequalities in the above should be equalities. That is $Tr_{i}=Tr_{\max }$ and $T_{i}=T_{\max }$ holds for any vertex $v_{i}$. So by Lemma 3, G is transmission regular.

Conversely, when G is transmission regular, it is easy to check that the equality holds.$\square$

Theorem 3

Let G be a simple, connected graph. Then

$$\begin{aligned} \rho (G)\ge \frac{Tr_{\min }+\sqrt{Tr^{2}_{\min }+8T_{\min }}}{2}, \end{aligned}$$

where $Tr_{\min }$ and $T_{\min }$ are the minimum transmission degree and the minimum second transmission degree of G,  respectively. Moreover, the equality holds if and only if G is transmission regular.

Proof

Similar to the proof of Theorem 2.

Corollary 4

Let G be a simple, connected graph. Then

$$\begin{aligned} \min _{v_{i}\in V(G)}\sqrt{2Tr^{2}_{i}+2T_{i}}\le \rho (G)\le \max _{v_{i}\in V(G)}\sqrt{2Tr^{2}_{i}+2T_{i}}. \end{aligned}$$

Proof

Since $D^{Q}(G)=Tr(G)+D(G)$, by a simple calculation, we have $r_{v_{i}}(D^{Q}(G))=2Tr(v_{i})$ and $r_{v_{i}}(D Tr)=r_{v_{i}}(D^{2})=\sum _{j=1}^{n}d_{v_iv_j}Tr(v_{j})$. Then

$$\begin{aligned} r_{v_{i}}((D^{Q})^2)= & {} r_{v_{i}}(Tr^{2}+Tr D+D Tr+D^{2})\\= & {} r_{v_{i}}(Tr(Tr+D))+r_{v_{i}}(DTr)+r_{v_{i}}(D^{2})\\= & {} Tr(v_{i})r_{v_{i}}(D^{Q})+2\sum _{j=1}^{n}d_{v_iv_j}Tr(v_{j})\\= & {} 2Tr^{2}_{i}+2T_{i}. \end{aligned}$$

Then, by using Lemma 3, we get

$$\begin{aligned} \rho (G)\le \max _{v_{i}\in V(G)}\sqrt{2Tr^{2}_i+2T_{i}}. \end{aligned}$$

The proof of the other part is similar. $\square$

Theorem 4

If the transmission degree sequence and the second transmission degree sequence of a connected graph G are $\{Tr_{1},Tr_{2},\ldots ,Tr_{n}\}$ and $\{T_{1},T_{2},\ldots ,T_{n}\}$, respectively, then

$$\begin{aligned} \rho (G)\ge \sqrt{\frac{\displaystyle \sum _{i=1}^{n}\left[ Tr_{i}(Tr^{2}_{i}+T_{i}) +\sum _{j=1}^{n}d_{v_iv_j}(Tr^{2}_{j}+T_{j})\right] ^{2}}{\displaystyle \sum _{i=1}^{n}(Tr^{2}_{i}+T_{i})^{2}}}. \end{aligned}$$

(3)

The equality holds in (3) if and only if there exists a positive constant number c such that, for all $i\in \{1,2,\ldots , n\}$,

$$\begin{aligned} c=\frac{Tr_{i}(Tr^{2}_{i}+T_{i})+\displaystyle \sum _{j=1}^{n}d_{v_iv_j} (Tr^{2}_{j}+T_{j})}{Tr^{2}_{i}+T_{i}}. \end{aligned}$$

In fact, $c=\rho (G)$.

Proof

The proof is analogous to [24, Theorem 1]. $\square$

Now, we present another lower bound for the distance signless Laplacian spectral radius of graphs, which improves some known results in [26].

Theorem 5

If the transmission degree sequence and the second transmission degree sequence of a connected graph G are $\{Tr_{1},Tr_{2},\ldots ,Tr_{n}\}$ and $\{T_{1},T_{2},\ldots ,T_{n}\}$, respectively, then

$$\begin{aligned} \rho (G)\ge \sqrt{\frac{\displaystyle \sum _{i=1}^{n}(Tr_{i}^{2}+T_{i})^{2}}{\displaystyle \sum _{i=1}^{n}Tr_{i}^{2}}}. \end{aligned}$$

(4)

The equality holds if and only if G is transmission regular.

Proof

Let $X=(x_{1},x_{2},\ldots ,x_{n})^T$ be the distance signless Laplacian Perron vector of G and $C=\frac{1}{\sqrt{\displaystyle \sum _{i=1}^{n}Tr_{i}^{2}}}(Tr_{1},Tr_{2},\ldots ,Tr_{n})^T.$ Then

$$\begin{aligned} \rho (G)=\sqrt{ \rho (G)^{2}}=\sqrt{X^T D^{Q}(G)^{2} X}\ge \sqrt{C^T D^{Q}(G)^{2} C}. \end{aligned}$$

We now have

$$\begin{aligned} C^T D= & {} \frac{1}{\sqrt{\displaystyle \sum _{i=1}^{n}Tr_{i}^{2}}}(T_{1},T_{2},\ldots , T_{n}), \end{aligned}$$

and

$$\begin{aligned} C^T Tr= & {} \frac{1}{\sqrt{\displaystyle \sum _{i=1}^{n}Tr_{i}^{2}}}(Tr^{2}_{1},Tr^{2}_{2},\ldots , Tr^{2}_{n}). \end{aligned}$$

Thus $C^T D^{2} C=\frac{\displaystyle \sum _{i=1}^nT_i^{2}}{\displaystyle \sum _{i=1}^nTr_i^{2}},$$C^T Tr^{2} C=\frac{\displaystyle \sum _{i=1}^nTr_i^{4}}{\displaystyle \sum _{i=1}^nTr_i^{2}}$ and $C^T D \cdot Tr C=\frac{\displaystyle \sum _{i=1}^nT_iTr_i^{2}}{\displaystyle \sum _{i=1}^nTr_i^{2}}.$ Therefore,

$$\begin{aligned} \rho (G)\ge \sqrt{\frac{\displaystyle \sum _{i=1}^{n}(T_{i}^{2}+Tr_{i}^{4}+2T_{i}Tr_{i}^{2})}{\displaystyle \sum _{i=1}^{n}Tr_{i}^{2}}}=\sqrt{\frac{\displaystyle \sum _{i=1}^{n}(Tr_{i}^{2}+T_{i})^{2}}{\displaystyle \sum _{i=1}^{n}Tr_{i}^{2}}}. \end{aligned}$$

Now assume that G is k-transmission regular. Then

$$\begin{aligned} \rho (G)=2k= \sqrt{\frac{nk^{4}+nk^{4}+2nk^{4}}{nk^{2}}}=\sqrt{4k^{2}}. \end{aligned}$$

Thus the equality in (4) holds.

Conversely, if equality in (4) holds, then we get C is the eigenvector corresponding to $\rho (G)$. From $\rho (G)C=D^{Q}(G)C,$ it follows that $\rho (G) Tr_{i}=Tr_{i}^{2}+T_{i},$ for $i=1,2,\ldots ,n.$ Thus $Tr_{i}+\frac{T_{i}}{Tr_{i}}=Tr_{j}+\frac{T_{j}}{Tr_{j}},$ for all $i\ne j.$ Let $Tr_{\max }$ and $Tr_{\min }$ denote the maximum and minimum vertex transmission, respectively. Without loss of generality, assume that $Tr_{i}=Tr_{\max }$ and $Tr_{j}=Tr_{\min }.$ Therefore, $Tr_{\max }+\frac{T_{i}}{Tr_{\max }}=Tr_{\min }+\frac{T_{j}}{Tr_{\min }}.$ Since $T_{i}\ge Tr_{\max }Tr_{\min }$ and $T_{j}\le Tr_{\max }Tr_{\min }$,

$$\begin{aligned} Tr_{\max }+Tr_{\min }\le Tr_{\max }+\frac{T_{i}}{Tr_{\max }}=Tr_{\min }+\frac{T_{j}}{Tr_{\min }}\le Tr_{\max }+Tr_{\min }. \end{aligned}$$

Thus we must have $T_{i}=Tr_{\max }Tr_{\min }=T_{j}$ and hence $Tr_{\max }^{2}+Tr_{\max }Tr_{\min }=Tr_{\min }^2+Tr_{\max }Tr_{\min }.$ From which it implies that $Tr_{\max }=Tr_{\min }$. Hence G is a transmission regular graph. $\square$

Remark 1

The lower bound for $\rho (G)$ given by Theorem 5 is always better than the lower bound given by Lemma 1. First observe that $\sum _{i=1}^{n}T_{i}=\sum _{i=1}^{n}Tr_{i}^{2}$. Also, by Cauchy–Schwartz inequality, we have $\left( \displaystyle \sum _{i=1}^{n}T_{i}\right) ^{2}\le n\displaystyle \sum _{i=1}^{n}T_{i}^{2}$ and $\left( \displaystyle \sum _{i=1}^{n}Tr_{i}\right) ^{2}\le n\displaystyle \sum _{i=1}^{n}Tr_{i}^{2}$. Now by Theorem 5, we get

$$\begin{aligned} \rho (G)\ge \sqrt{\frac{\displaystyle \sum _{i=1}^{n}(Tr_{i}^{2}+T_{i})^{2}}{\displaystyle \sum _{i=1}^{n}Tr_{i}^{2}}}\ge & {} \sqrt{\frac{\left( \displaystyle \sum _{i=1}^{n}(Tr_{i}^{2}+T_{i})\right) ^{2}}{n\displaystyle \sum _{i=1}^{n}Tr_{i}^{2}}}\\= & {} \sqrt{\frac{\left( \displaystyle \sum _{i=1}^{n}2Tr_{i}^{2}\right) ^{2}}{n\displaystyle \sum _{i=1}^{n}Tr_{i}^{2}}}\\= & {} 2\sqrt{\frac{\displaystyle \sum _{i=1}^{n}Tr_{i}^{2}}{n}}\ge 2\sqrt{\frac{(\displaystyle \sum _{i=1}^{n}Tr_{i})^{2}}{n^{2}}}= \frac{4\sigma (G)}{n}. \end{aligned}$$

In the next theorem, we show the equivalence between the signless Laplacian spectrum and the distance signless Laplacian spectrum of a connected regular graph of diameter two.

Theorem 6

Let G be a connected r-regular graph on n vertices with diameter $d\le 2$. If $\{2r, \mu _{2},\ldots ,\mu _{n}\}$ are the eigenvalues of the signless Laplacian matrix Q(G) of G, then the distance signless Laplacian eigenvalues of G are $4n-2r-4$ and $2n-4-\mu _{i},$$i=2,3,\ldots ,n.$

Proof

Concerning the zero eigenvalue and in the case of $d=1$ (corresponding to the complete graph), the result is trivial.

Now, let G be a graph of diameter 2. Then, we know that the transmission of each vertex $v\in V(G)$ is

$$\begin{aligned} Tr(v)=d(v)+2(n-1-d(v))=2n-d(v)-2=2n-r-2, \end{aligned}$$

where d(v) denotes the degree of v in G. Also, the distance matrix is $D(G)=2J-2I-A(G)$, where J is the all ones matrix. Thus the distance signless Laplacian matrix can be written as

$$\begin{aligned} D^{Q}(G)= & {} Tr(G)+D(G)=(2n-2)I-rI+2J-2I-A(G)\\= & {} (2n-4)I+2J-Q(G). \end{aligned}$$

Since G is an r-regular graph, $\mathbf{1} = [1, 1, \ldots , 1]^T$ is an eigenvector of the signless Laplacian matrix $Q = Q(G)$ corresponding to the eigenvalue 2r.

Set $\mathbf{z} = \frac{1}{\sqrt{n}}{} \mathbf{1}$ and let P be an orthogonal matrix with its first column equal to $\mathbf{z}$ such that $P^TQP = \text{ diag }(2r, \mu _2, \ldots , \mu _n)$. Therefore,

$$\begin{aligned} \begin{array}{ccl} P^T(D^{Q}(G))P &{} = &{} P^T[(2n-4)I + 2J - Q]P \\ &{} = &{} (2n-4)I - 2P^TJP - P^TQP \\ &{} = &{} \text{ diag }(4n - 4 - 2r, 2n-4 -\mu _2, \ldots , 2n-4-\mu _n). \end{array} \end{aligned}$$

Hence the result follows. $\square$

Corollary 5

Let G be an r-regular graph of diameter 2, and let its adjacency spectrum be $\{r,\lambda _{2},\ldots ,\lambda _{n}\}$. Then the distance signless Laplacian spectrum of G is $\{2(2n-r-2), 2n-\lambda _{2}-r-4, \ldots , 2n-\lambda _{n}-r-4\}$.

Proof

Since $Q(G)=rI(G)+A(G)$, from Theorem 6 we get

$$\begin{aligned} D^{Q}(G)= & {} (2n-4)I+2J-(rI(G)+A(G))\\= & {} (2n-4-r)I+2J-A(G), \end{aligned}$$

and the result follows. $\square$

4 Results on the Spectral Radius of Line Graphs

Recall that, the line graph L(G) of a graph G is a graph such that the vertices of L(G) are the edges of G and two vertices of L(G) are adjacent if and only if their corresponding edges in G share a common vertex [12]. This concept has various applications in physical chemistry (see, e.g., [9, 10]).

Fig. 1

The forbidden induced subgraphs

Let $F_{1}$ be the 5-vertex path, $F_{2}$ the graph obtained by identifying a vertex of a triangle with an end vertex of the 3-vertex path, and $F_{3}$ the graph obtained by identifying a vertex of a triangle with a vertex of another triangle (see Fig. 1).

Theorem 7

[22, 23] For a connected graph G, $\mathrm{diam}(L(G)) \le 2$ if and only if none of the three graphs $F_1$, $F_2$ and $F_3$ of Fig. 1 is an induced subgraph of G.

Theorem 8

Let G be a connected graph with n vertices, m edges and $d_{i}=\mathrm{deg}(v_{i})$. If $\mathrm{diam}(G)\le 2$ and G does not contain $F_{i}$, $i=1,2,3$ as an induced subgraph, then

$$\begin{aligned} \rho (L(G))\ge \frac{4m^{2}-2\sum _{i=1}^{n}d_{i}^{2}}{m}. \end{aligned}$$

Proof

Let G be a connected graph of diameter 2, which does not contain $F_{i}$ for $i=1,2,3$ as an induced subgraph, and let its vertices be labeled as $v_{1}, v_{2}, \ldots , v_{n}.$ Let $d_{i}$ denote the degree of $v_{i}.$ Then, as G is of diameter 2, it is easy to observe that the ith row of $D^{Q}(G)$ consists of $d_{i}$ one’s, $n-d_{i}-1$ two’s and diagonal entry $2n-d_{i}-2$. Let $X=(1,1,\ldots ,1)^T$ be the all one vector. Then by the Raleigh’s principle,

$$\begin{aligned} \rho (G)\ge \frac{X^{T} D^{Q}(G)X}{X^{T} X}=\frac{1}{n}\sum _{i=1}^{n}(4n-2d_{i}-4)=\frac{4n^{2}-4m-4n}{n}. \end{aligned}$$

The number of vertices of L(G) is $n_{1}=m$ and the number of edges of L(G) is $m_{1}=-m+\frac{1}{2}\sum _{i=1}^{n}d^{2}_{i}$. Now since G has no $F_{i}$ for $i=1,2,3$ as its induced subgraph, by Theorem 7, we get $\mathrm{diam}(L(G))\le 2$. Therefore,

$$\begin{aligned} \rho (L(G))\ge & {} \frac{4n_{1}(n_{1}-1)-4m_{1}}{n_{1}}\\= & {} \frac{4m(m-1)-4\Big (-m+\frac{1}{2}\sum _{i=1}^{n}d^{2}_{i}\Big )}{m}\\= & {} \frac{4m^{2}-2\sum _{i=1}^{n}d_{i}^{2}}{m}. \end{aligned}$$

$\square$

Corollary 6

If G is a connected r-regular graph on n vertices and none of $F_{i}, i=1,2,3$ is an induced subgraph of G, then

$$\begin{aligned} \rho (L(G))\ge 2r(n-2). \end{aligned}$$

Proof

Since G is an r-regular graph on n vertices, the number of edges of G is $m=\frac{nr}{2}$ and $d_{i}=\mathrm{deg}(v_{i})=r$. Then from Theorem 8, we get

$$\begin{aligned} \rho (L(G))\ge & {} \frac{4m^{2}-2\sum _{i=1}^{n}d_{i}^{2}}{m}\\= & {} \frac{4(\frac{nr}{2})^{2}-2\sum _{i=1}^{n}r^{2}}{\frac{nr}{2}}\\= & {} 2r(n-2). \end{aligned}$$

$\square$

Let $e=uv$ be an edge of a graph G where u and v are the end vertices of e. The degree of a vertex $e\in V(L(G))$ is $\mathrm{deg}_{L(G)}(e)=\mathrm{deg}_{G}(u)+\mathrm{deg}_{G}(v)-2$.

Theorem 9

Let G be a connected graph with vertex set $V(G)=\{v_{1},v_{2},\ldots , v_{n}\}$ and edge set $E(G)=\{e_{1}, e_{2},\ldots , e_{m}\}$. Let $d_{i}=\mathrm{deg}(v_{i})$. Then

$$\begin{aligned} \rho (L(G))\ge \frac{2}{m}\sum _{i=1}^{n}d_{i}(d_{i}-1)+4(m-1) -\frac{4}{m}\sum _{i=1}^{m}\mathrm{deg}(e_{i}). \end{aligned}$$

Proof

If $d_{i}=\mathrm{deg}(v_{i})$ then for each vertex $v_{i}$, there are $d_{i}$ edges incident to $v_{i}$. These $d_{i}$ edges form a complete graph on $d_{i}$ vertices in L(G). Consider an edge $e=uv$ which is adjacent to $\mathrm{deg}(u)+\mathrm{deg}(v)-2=\mathrm{deg}(e)$ edges at u and v taken together. Hence the edge e is not adjacent to remaining $m-1-\mathrm{deg}(e)$ edges of G. In L(G) the distance between e and the remaining these $m-1-\mathrm{deg}(e)$ vertices is more than 1. Hence each edge $e=uv$ contributes the distance at least $2(m-1-\mathrm{deg}(e))$ in L(G). This is done analogously to the way in proof of Theorem 8, and hence the proof is complete. $\square$

5 Distance Signless Laplacian Spectrum of Some Graphs Obtained by Operations

In this section, we describe the distance signless Laplacian spectrum of some graphs obtained by operations. We consider join of the graph, the double graph, the Cartesian product, and lexicographic product of G and H (see [4]).

Lemma 4

[8] Let

$$\begin{aligned} A=\begin{bmatrix} A_{0}&A_{1}\\ A_{1}&A_{0}\\ \end{bmatrix} \end{aligned}$$

be a $2\times 2$ block symmetric matrix. Then the eigenvalues of A are those of $A_{0}+A_{1}$ together with those of $A_{0}-A_{1}$.

Given a graph G, the graph $G \nabla G$ is obtained by joining every vertex of G to every vertex of another copy of G.

Theorem 10

Let G be a connected, r-regular graph on n vertices. If $r, \lambda _{2}, \ldots , \lambda _{n}$ are the eigenvalues of the adjacency matrix of G, then the eigenvalues of the distance signless Laplacian matrix of $G \nabla G$ are

$$\begin{aligned} \begin{array}{ll} 6n-2r-4, &{} \\ 4n-2r-4, &{} \text {and}\\ 3n-\lambda _{i}-r-4, &{} 2 \; \text {times}, \quad i=2, 3, \ldots , n. \end{array} \end{aligned}$$

Proof

As G is an r-regular graph, the distance signless Laplacian matrix of $G\nabla G$ can be written as

$$\begin{aligned} \begin{bmatrix} A+2{\overline{A}}+\left( 3n-r-2\right) I&J \\ J&A+2{\overline{A}}+\left( 3n-r-2\right) I \end{bmatrix}, \end{aligned}$$

where A is the adjacency matrix of G, ${\overline{A}}$ is the adjacency matrix of ${\bar{G}}$, J is a matrix whose all entries are equal to 1 and I is an identity matrix. Since ${\overline{A}} = J - I - A$, then by applying Lemma 4, we get the result. $\square$

Definition 1

[16] Let G be a graph with vertex set $V(G)=\{v_{1}, v_{2},\ldots , v_{n}\}$. Take another copy of G with the vertices labeled by $\{u_{1}, u_{2},\ldots , u_{n}\}$, where $u_{i}$ corresponds to $v_{i}$ for each i. Make $u_{i}$ adjacent to all the vertices in $N(v_{i})$ in G, for each i. The resulting graph, denoted by $D_{2}(G)$, is called the double graph of G.

Theorem 11

Let G be a connected r-regular graph on n vertices with diameter 2 and let $r, \lambda _{2}, \lambda _{3}, \ldots , \lambda _{n}$ be the eigenvalues of the adjacency matrix of G. Then the eigenvalues of the distance signless Laplacian matrix of $D_2(G)$ are

$$\begin{aligned} \begin{array}{ll} 8n-4r-4, &{} \\ 4n-2r-4, &{} n \; \text {times}, \;\; \text {and}\\ 4n-2\lambda _{i}-2r-4, &{} i=2, 3, \ldots , n. \end{array} \end{aligned}$$

Proof

By definition of $D_{2}(G)$, the distance signless Laplacian matrix of $D_{2}(G)$ is of the form

$$\begin{aligned} \begin{bmatrix} A+ 2{\overline{A}}+ 2(2n-r-1)I&A +2{\overline{A}} +2 I \\ A + 2{\overline{A}} + 2 I&A+2{\overline{A}}+ 2(2n-r-1)I \end{bmatrix}, \end{aligned}$$

where A is the adjacency matrix of G, ${\overline{A}}$ is the adjacency matrix of ${\bar{G}}$ and I is an identity matrix. Since ${\overline{A}} = J - I - A$, then by applying Lemma 4, the result follows.$\square$

Next, we derive the distance signless Laplacian spectrum of the Cartesian product of two distance regular graphs.

Definition 2

[4] The Cartesian product of two graphs G and H, denoted by $G+H$, is the graph with vertex set $V(G+H)= V(G) \times V(H)$ and two vertices $(u_1,u_2)$ and $(v_1,v_2)$ in $G+H$ are adjacent whenever $u_1=v_1$ and $u_2$ is adjacent to $v_2$ in H or $u_2=v_2$ and $u_1$ is adjacent to $v_1$ in G.

The following lemma of [17] will be useful to derive our next result.

Lemma 5

[17] Let G and H be two connected graphs, and let $u=(u_{1}, u_{2}), v=(v_{1}, v_{2})\in V(G)\times V(H)$. If $G+H$ denotes their Cartesian product, then

$$\begin{aligned} d_{G+H}(u,v)=d_{G}(u_{1}, v_{1})+d_{H}(u_{2}, v_{2}). \end{aligned}$$

Theorem 12

Let G and H be two distance regular graphs on p and n vertices with transmission regularity k and t, respectively. Let the distance spectrums of G and H be $Spec_{D}(G)=\{k, \mu _{2},\mu _{3},\ldots ,\mu _{p}\}$ and also $Spec_{D}(H)=\{t,\eta _{2},\eta _{3},\ldots ,\eta _{n}\}$, respectively. Then the distance signless Laplacian spectrum of $G+H$ is

$$\begin{aligned} Spec_{D^{Q}}(G+H)=\{2nk+2pt, n\mu _{i}+nk+pt,p\eta _{j}+nk+pt,nk+pt\}, \end{aligned}$$

where $i=2,\ldots ,p$ and $j=2,\ldots ,n$ and $nk+pt$ is with multiplicity $(p-1)(n-1)$.

Proof

Let $D_{G}$ and $D_{H}$ be the distance matrices of G and H, respectively. Let $V(G)=\{u_{1},u_{2},\ldots , u_{p}\}$ and $V(H)=\{v_{1},v_{2},\ldots , v_{n}\}$. Then $D_{G}=[d_{ij}]$ and $D_{H}=[e_{ij}]$, where $d_{ij}=d_{G}(u_{i},u_{j})$ and $e_{ij}=d_{H}(v_{i},v_{j})$. Since G and H are distance transmission regular graphs with transmission regularities k and t, respectively, we have

$$\begin{aligned} \sum _{j=1}^{p}d_{rj}=k \qquad \text {and} \qquad \sum _{j=1}^{n}e_{qj}=t. \end{aligned}$$

Also, since G is transmission regular, the all-one column vector of order $p\times 1$ is the eigenvector corresponding to the greatest eigenvalue k of $D_{G}$. As $D_{G}$ is real and symmetric, it is diagonalizable and hence admits an orthogonal basis $B_{G}$ consisting of eigenvectors corresponding to its eigenvalues. Thus if $\mu _{i}$ is an eigenvalue of $D_{G}$ which is different from k with an eigenvector $X_{i}=[x^{1}_{i}, x^{2}_{i}, \ldots , x^{p}_{i}]^{T}\in B_{G}$, then $\sum _{j=1}^{p}x^{j}_{i}=0$.

Let $u=(u_{1},u_{2}), v=(v_{1}, v_{2})\in V(G)\times V(H)$. Then by Lemma 4,

$$\begin{aligned} d_{G+H}(u,v)=d_{G}(u_{1},v_{1})+d_{H}(u_{2},v_{2}). \end{aligned}$$

By a suitable ordering of vertices in $G+H$ and by virtue of Lemma 4, its $D^{Q}$-matrix, C can be written in the form

$$\begin{aligned} C= & {} \begin{bmatrix} d(u_{1},u_{1}) J_{n}+D_{H}&d(u_{1},u_{2}) J_{n}+D_{H}&\cdots&d(u_{1},u_{p}) J_{n}+D_{H}\\ \vdots&\vdots&\\ d(u_{p},u_{1}) J_{n}+D_{H}&d(u_{p},u_{2}) J_{n}+D_{H}&\cdots&d(u_{p},u_{p}) J_{n}+D_{H}\\ \end{bmatrix}\\&+ \begin{bmatrix} (nk+pt)I_{n}&O&\cdots&O \\ \vdots&\vdots&\\ O&O&\cdots&(nk+pt)I_{n}\\ \end{bmatrix}\\ \quad \\&= D_{G}\otimes J_{n}+J_{p}\otimes D_{H}+I_{p}\otimes (nk+pt) I_{n}, \end{aligned}$$

where O denotes the zero matrix and $\otimes$ denotes the tensor product of matrices. Now we find the eigenvalues of C by considering eigenvectors associated with them. The following relation for matrices is well known by [20]. For the matrices A, B, C and D

$$\begin{aligned} (A\otimes B)(C \otimes D)=(AC) \otimes (BD), \end{aligned}$$

whenever the products AC and BD exist.

Let ${{\mathbf {1}}_{G}}$ denote the all one eigenvector corresponding to the eigenvalue k of G and ${{\mathbf {1}}_{H}}$ the all one eigenvector corresponding to the eigenvalue t of H. Then $D_{G} {{\mathbf {1}}_{G}}=k{{\mathbf {1}}_{G}}$ and $D_{H} {{\mathbf {1}}_{H}}=t{{\mathbf {1}}_{H}}$. Therefore,

$$\begin{aligned} C ({{\mathbf {1}}_{G}}\otimes {{\mathbf {1}}_{H}})= & {} (D_{G}\otimes J_{n}+J_{p}\otimes D_{H}+I_{p}\otimes (nk+pt)I_{n}) ({{\mathbf {1}}_{G}}\otimes {{\mathbf {1}}_{H}})\\= & {} (D_{G} \mathbf {1_{G}})\otimes (J_{n}{{\mathbf {1}}_{H}})+(J_{p}{{\mathbf {1}}_{G}})\otimes (D_{H} {{\mathbf {1}}_{H}})\\&+(I_{p}{{\mathbf {1}}_{G}})\otimes ((nk+pt)I_{n}{{\mathbf {1}}_{H}})\\= & {} k{{\mathbf {1}}_{G}}\otimes (n{{\mathbf {1}}_{H}})+p{{\mathbf {1}}_{G}}\otimes (t{{\mathbf {1}}_{H}}) +(nk+pt)({{\mathbf {1}}_{G}}\otimes {{\mathbf {1}}_{H}})\\= & {} nk({{\mathbf {1}}_{G}}\otimes {{\mathbf {1}}_{H}})+pt({{\mathbf {1}}_{G}}\otimes {{\mathbf {1}}_{H}})+(nk+pt)({{\mathbf {1}}_{G}}\otimes {{\mathbf {1}}_{H}})\\= & {} 2(nk+pt)({{\mathbf {1}}_{G}}\otimes {{\mathbf {1}}_{H}}), \end{aligned}$$

showing that ${{\mathbf {1}}_{G}}\otimes {{\mathbf {1}}_{H}}$ is the eigenvector corresponding to the eigenvalue $2(nk+pt)$ of C.

Let $X_{i}$ be the eigenvector corresponding to the eigenvalue $\mu _{i}$ of $D_{G}$. Then $X_{i} \otimes {{\mathbf {1}}_{H}}$ is the eigenvector corresponding to the eigenvalue $n\mu _{i}+nk+pt$ of C, since

$$\begin{aligned} C (X_{i} \otimes {{\mathbf {1}}_{H}})= & {} [D_{G}\otimes J_{n}+J_{p}\otimes D_{H}+I_{p}\otimes (nk+pt) I_{n}] (X_{i} \otimes {{\mathbf {1}}_{H}})\\= & {} (D_{G} X_{i})\otimes (J_{n}{{\mathbf {1}}_{H}})+(J_{p}X_{i})\otimes (D_{H} {{\mathbf {1}}_{H}})\\&+(I_{p}X_{i})\otimes ((nk+pt)I_{n}{{\mathbf {1}}_{H}})\\= & {} \mu _{i}X_{i}\otimes n{{\mathbf {1}}_{H}}+0\otimes t{{\mathbf {1}}_{H}}+(nk+pt)(X_{i}\otimes {{\mathbf {1}}_{H}})\\= & {} (n\mu _{i}+nk+pt)(X_{i} \otimes {{\mathbf {1}}_{H}}) . \end{aligned}$$

Similarly, if $Z_{j}$ is an eigenvector corresponding to the eigenvalue $\eta _{j}$ of $D_{H}$, then ${{\mathbf {1}}_{G}}\otimes Z_{j}$ is an eigenvector corresponding to the eigenvalue $p\eta _{j}+nk+pt$ of C.

In addition to these eigenvalues, we can see that $nk+pt$ appears to be an eigenvalue with multiplicity $(p-1)(n-1)$. For, let $R^{i}_{p}, i=2,3, \ldots ,p$ be the $(p-1)$ linearly independent eigenvectors corresponding to the eigenvalue 0 of $J_{p}$ and $T^{j}_{n}, j=2,3,\ldots , n-1$ be the $(n-1)$ linearly independent eigenvectors corresponding to the eigenvalue 0 of $J_{n}$. Then the $(p-1)(n-1)$ vectors $R^{i}_{p}\otimes T^{j}_{n}$ are linearly independent and are the eigenvectors corresponding to $nk+pt$ of C. For,

$$\begin{aligned} C (R^{i}_{p} \otimes T^{j}_{n})= & {} [D_{G}\otimes J_{n}+J_{p}\otimes D_{H}+I_{p}\otimes (nk+pt)I_{n}) ] (R^{i}_{p} \otimes T^{j}_{n})\\= & {} (D_{G} R^{i}_{p})\otimes (J_{n} T^{j}_{n})+(J_{p}R^{i}_{p})\otimes (D_{H} T^{j}_{n})\\&+(I_{p}R^{i}_{p})\otimes ((nk+pt)I_{n}T^{j}_{n})\\= & {} (D_{G} R^{i}_{p})\otimes 0+0 \otimes (D_{H} T^{j}_{n})+(nk+pt)\otimes (R^{i}_{p} \otimes T^{j}_{n}) \\= & {} (nk+pt)(R^{i}_{p} \otimes T^{j}_{n}). \end{aligned}$$

Now, the pn vectors ${{\mathbf {1}}_{G}}\otimes {{\mathbf {1}}_{H}}$, $X_{i}\otimes {{\mathbf {1}}_{H}}$, ${{\mathbf {1}}_{G}}\otimes Z_{j}$ and $R^{i}_{p}\otimes T^{j}_{n}$ are linearly independent, and as C has a basis consisting of linearly independent eigenvectors, the result follows.$\square$

Next, we obtain the distance signless Laplacian spectrum of the lexicographic product G[H] of two graphs G and H.

Definition 3

[4] Let G and H be two graphs on vertex sets $V(G)=\{u_{1},u_{2},\ldots , u_{p}\}$ and $V(H)=\{v_{1},v_{2},\ldots ,v_{n}\}$, respectively. Then their lexicographic productG[H] is a graph with vertex set $V(G[H])=V(G)\times V(H)$, in which $u=(u_{1},v_{1})$ is adjacent to $v=(u_{2},v_{2})$ if and only if either

• (a) $u_{1}$ is adjacent to $u_{2}$ in G, or

• (b) $u_{1}=u_{2}$ and $v_{1}$ is adjacent to $v_{2}$ in H.

We conclude the paper by giving a distance signless Laplacian spectrum for lexicographic product G[H] of a k-transmission regular graph G with that of r-regular graph H.

Theorem 13

Let G be a k-transmission regular graph of order p. Let H be an r-regular graph on n vertices with its adjacency eigenvalues $r, \lambda _2, \ldots , \lambda _n$. Let $\mu _{1},\mu _{2},\ldots ,\mu _{p}$ be the eigenvalues of the distance matrix $D_{G}$ of G. Then the eigenvalues of the distance signless Laplacian matrix of G[H] are

$$\begin{aligned} \begin{array}{ll} n \mu _i +kn+ 4n-2r-4, &{} i = 1, 2, \ldots , p \;\; \text {and}\\ kn+2n-\lambda _{j}-r-4, &{} p \;\; \text {times}, \; j=2, 3, \ldots , n. \end{array} \end{aligned}$$

Proof

By a suitable ordering of vertices of G[H], its $D^{Q}$-matrix F, can be written in the form

$$\begin{aligned} F=D_{G}\otimes J_{n}+I_{p}\otimes (A+2{\bar{A}}+(kn+2n-2)I-D^{\prime }), \end{aligned}$$

where A denotes the adjacency matrix of H, ${\bar{A}}$ denotes the adjacency matrix of $\bar{H}$ and $D^{\prime }$ is the diagonal matrix of vertex degrees in H.

Since H is r-regular, the all one column vector ${\mathbf {1}}$ of order $n\times 1$ is an eigenvector of A with an eigenvalue r. Then the all one vector ${\mathbf {1}}$ is an eigenvector of $A+2{\bar{A}}+(kn+2n-2)I-D^{\prime }$ with an eigenvalue $kn+4n-2r-4$. Similarly, if $\lambda _{j}$ is any other eigenvalue of A with eigenvector $Y_{j}$, then $Y_{j}$ is an eigenvector of $A+2{\bar{A}}+(kn+2n-2)I-D^{\prime }$ with eigenvalue $kn+2n-\lambda _{j}-r-4$ and that $Y_{j}$ is orthogonal to ${\mathbf {1}}$.

Let $X_{i}=[x^{i}_{1}\quad x^{i}_{2}\quad \cdots \quad x^{i}_{p}]^{T}$ be an eigenvector corresponding to the eigenvalue $\mu _{i}$ of $D_{G}$. Therefore,

$$\begin{aligned} D_{G}X_{i}=\mu _{i} X_{i}. \end{aligned}$$

Now

$$\begin{aligned} F (X_{i} \otimes {{\mathbf {1}}_{n}})= & {} (D_{G}\otimes J_{n}+I_{p}\otimes (A+2{\bar{A}}+(kn+2n-2)I-D^{\prime })) (X_{i}\otimes {{\mathbf {1}}_{n}})\\= & {} (D_{G} X_{i})\otimes (J_{n} {{\mathbf {1}}_{n}})+(I_{p} X_{i})\otimes (A+2{\bar{A}}+(kn+2n-2)I-D^{\prime }) {{\mathbf {1}}_{n}} \\= & {} \mu _{i}X_{i}\otimes n{{\mathbf {1}}_{n}}+X_{i}\otimes (kn+4n-2r-4) {{\mathbf {1}}_{n}}\\= & {} n\mu _{i}(X_{i}\otimes {{\mathbf {1}}_{n}})+(kn+4n-2r-4)(X_{i}\otimes {{\mathbf {1}}_{n}})\\= & {} (n\mu _{i}+kn+4n-2r-4) (X_{i}\otimes {{\mathbf {1}}_{n}}) \end{aligned}$$

Therefore, $n\mu _{i}+kn+4n-2r-4$ is an eigenvalue of F with eigenvector $X_{i}\otimes {{\mathbf {1}}_{n}}$. As $Y_{j}$ is orthogonal to ${\mathbf {1}},$ we have $J_{n} Y_{j}=0$ for each $j=2,3,\ldots ,n.$ Let $\{Z_{k}\}$, $k=1,2,\ldots ,p$ be the family of p linearly independent eigenvectors associated with the eigenvalue 1 of $I_{p}.$ Then for each $j=2,3,\ldots ,n,$ the p vectors $Z_{k}\otimes Y_{j}$ are eigenvectors of F with eigenvalue $kn+2n-\lambda _{j}-r-4.$ For

$$\begin{aligned} F (Z_{k}\otimes Y_{j})= & {} (D_{G}\otimes J_{n}+I_{p}\otimes (A+2{\bar{A}}+(kn+2n-2)I-D^{\prime }))(Z_{k}\otimes Y_{j})\\= & {} (D_{G} Z_{k})\otimes (J_{n} Y_{j})+(I_{p} Z_{k})\otimes (A+2{\bar{A}}+(kn+2n-2)I-D^{\prime }) Y_{j}\\= & {} 0+Z_{k}\otimes (kn+2n-\lambda _{j}-r-4) Y_{j}\\= & {} (kn+2n-\lambda _{j}-r-4)(Z_{k}\otimes Y_{j}). \end{aligned}$$

Also, the pn vectors $X_{i}\otimes {{\mathbf {1}}_{n}}$ and $Z_{k}\otimes Y_{j}$ are linearly independent. (Since taking any linear combination of these vectors, it is easy to see that the linear combination is zero only if the scalars are zero by the linear independencies of the constituent vectors of the tensor product.) As the eigenvectors belonging to different eigenvalues are linearly independent and as F has a basis consisting entirely of eigenvectors, the result follows. $\square$