Generalized Randić Estrada Indices of Graphs

Abstract

Let G be a simple undirected graph on n vertices. V. Nikiforov studied hybrids of $A_G$ and $D_G$ and defined the matrix $A_{\alpha }^G$ for every real $\alpha \in [0,1]$ as $A_{\alpha }^G=\alpha D_G+(1-\alpha )A_G.$ In this paper, we define the generalized Randić matrix for graph G, and we introduce and establish bounds for the Estrada index of this new matrix. Furthermore, we find the smallest value of $\alpha$ for which the generalized Randić matrix is positive semidefinite. Finally, we present the solution to the problem proposed by V. Nikiforov. The problem consists of the following: for a given simple undirected graph G, determine the smallest value of $\alpha$ for which $A_{\alpha }^G$ is positive semidefinite.

1. Introduction

We deal with a graph G, which is a simple undirected graph with vertex set $\mathcal{V}\left(G\right)$ and edge set $\mathcal{E}\left(G\right)$. If an edge is incident to the vertices $v_i$ and $v_j$, we say that $v_i$ and $v_j$ are adjacent (i∼j), and we denote this edge by $ij$. The number of adjacent vertices to $v_i$ is denoted by $d_i$ and is called the degree of $v_i$. If $d_i=0,$ for some i, then the corresponding vertex is an isolated vertex. A cycle of order three in a graph G with vertices $i,j$ and k will be denoted by $C_3\left(ijk\right)$. The set of cycles in G will be denoted by $\Xi \left(G\right)$.

Recalling the definition of inertia of a square matrix $M,$ as a triple vector whose coordinates are negative, zero, and positive eigenvalues of M [1], respectively. The spectrum of a square matrix M (the multi-set of the eigenvalues of M) will be denoted here by $Sp\left(M\right)$. Throughout this paper, the eigenvalues of a Hermitian matrix M of order n are denoted by ${\lambda }_j\left(M\right)$ for $1\le j\le n$ and are arranged in decreasing order, that is,

$${\lambda }_1\left(M\right)\ge {\lambda }_2\left(M\right)\ge \cdots \ge {\lambda }_n\left(M\right).$$

The multiplicity $p\ge 1$ of an eigenvalue $\theta$ of M is represented by ${\theta }^{\left[p\right]}.$ If $\mathbf{u}$ is an eigenvector associated with $\theta$, the pair $(\theta ,\mathbf{u})$ is called an eigenpair. The Randić matrix of a graph G, $R=R_G$, is a matrix whose entries are indexed by the vertices of G, that is,

$$r_{ij}=\left\{\begin{array}{cc}\hfill rl{\displaystyle \frac{1}{\sqrt{d_i{d}_j}}}& \mathrm{if}i\sim j\\ \hfill 0& \mathrm{otherwise}.\end{array}\right.$$

(1)

The Randić matrix was used for the first time in 2005 by Rodríguez [2]. This matrix is closely related to the molecular structure descriptor defined as

$${\chi }^G=\sum _{i\sim j}\frac{1}{\sqrt{d_i{d}_j}},$$

(2)

which was proposed by Milan Randić in 1975 and is usually referred to as the Randić index, Refs. [2,3,4,5,6,7]. To refer to its importance, it is worth saying that this topological index is suitable for measuring the extent of the branching of the carbon-atom skeleton of saturated hydrocarbons.

Similarly, the normalized Laplacian matrix of a graph G, ${\mathcal{L}}_N^G$, is a matrix whose entries are indexed by the vertices of G, and this is described below

$$l_{ij}=\left\{\begin{array}{cc}\hfill 1& \mathrm{if}i=j\mathrm{and}{d}_j\ne 0\hfill \\ \hfill -{\displaystyle \frac{1}{\sqrt{d_i{d}_j}}}& \mathrm{if}i\sim j\hfill \\ \hfill 0& \mathrm{otherwise}.\hfill \end{array}\right.$$

Equivalently

$${\mathcal{L}}_N^G=I_n-R_G.$$

(3)

The eigenvalues of ${\mathcal{L}}_N^G$ are called normalized Laplacian eigenvalues of G, and we denote them by ${\tau }_1\ge \cdots \ge {\tau }_n$.

In 1998, Bollobás and Erdös [8] defined the generalized Randić index:

$${\mathcal{R}}_{\gamma }^G=\sum _{i\sim j}{\left(d_i{d}_j\right)}^{\gamma },$$

(4)

for every real number $\gamma \ne 0$. We note that the Randić index in (2) is a particular case of the parameter in (4) when $\gamma =-1/2.$ Throughout this paper, $A_G=\left(a_{ij}\right)$ represents the adjacency matrix of G. In this paper, the spectrum of G is the set whose elements are the eigenvalues of $A_G$. In addition, $D_G$ is the diagonal matrix of the degrees in G. If G is a graph without isolated vertices, the Randić matrix (see, e.g., [9,10]) can be written as

$$R_G=D_G^{-1/2}{A}_G{D}_G^{-1/2},$$

where $D_G^{-1/2}$ is the diagonal matrix whose i-th diagonal entry is the reciprocal of the square root of the degree $d_i$. It is also worth recalling that this matrix is symmetric and non-negative.

In [11], V. Nikiforov studies hybrids of $A_G$ and $D_G$ and defines the matrix $A_{\alpha }^G$ for every real $\alpha \in \left[0,1\right]$ as

$$A_{\alpha }^G=\alpha D_G+\left(1-\alpha \right)A_G.$$

(5)

The i-th largest eigenvalue of the matrix $A_{\alpha }^G$ will be denoted by ${\rho }_i\left(\alpha \right)\left(G\right)$, or ${\rho }_i\left(\alpha \right)$, if there is no ambiguity. Let us denote by ${\alpha }_0\left(G\right)$ the smallest value of $\alpha$ for which the matrix $A_{\alpha }^G$ is a positive semidefinite. In [11], the following problem is posed: given a simple undirected graph G, find ${\alpha }_0\left(G\right)$. This problem seems difficult in general but is worth studying because ${\alpha }_0\left(G\right)$ relates to various structural parameters of G. In [12], this problem is completely solved for bipartite graphs, and a sufficient condition for regular graphs is given. In this paper, we solve this problem for graphs in general.

Furthermore, we also give a generalization of the Randić matrix for a graph G, $R_{\alpha }^G=\left(R_{ij}\right)$, whose entries are indexed by the vertices of G as:

$$R_{ij}=\left\{\begin{array}{cc}\hfill \alpha & \mathrm{if}i=j\hfill \\ \hfill {\displaystyle \frac{1-\alpha }{\sqrt{d_i{d}_j}}}& \mathrm{if}i\sim j\hfill \\ \hfill 0& \mathrm{otherwise}.\hfill \end{array}\right.$$

(6)

for $\alpha \in [0,1]$.

From Equations (1) and (6), it is easy to verify that

$$R_{\alpha }^G=\alpha I_n+(1-\alpha )R_G.$$

(7)

Using the notation in (5), we can write the generalization for the Randić matrix of graphs G without isolated vertices by

$$R_{\alpha }^G=D_G^{-\frac{1}{2}}{A}_{\alpha }^G{D}_G^{-\frac{1}{2}},$$

(8)

for $\alpha \in [0,1]$, and the eigenvalues of this matrix are called $\alpha$-Randić eigenvalues. The i-th largest $\alpha$-Randić eigenvalue of G will be denoted by ${\phi }_i\left(\alpha \right)\left(G\right)$, or ${\phi }_i\left(\alpha \right)$, if there is no ambiguity.

From (7), we can conclude the following remark.

Remark 1.

Let G be a graph of order n without isolated vertices and let $\alpha \in [0,1]$. Considering properties of the matrix determinant, a direct consequence is the following

$$det\left(R_{\alpha }^G\right)=\frac{det\left(A_{\alpha }^G\right)}{d_1\dots d_n}=\prod _{i=1}^n\frac{{\rho }_i\left(\alpha \right)}{d_i}$$

where $d_i$ is the degree of vertex $v_i.$

1.1. Generalized Randić Estrada Indices

The Estrada index of a graph G is an invariant that is calculated from the eigenvalues of its adjacency matrix, that is, the sum of the terms $e^{{\lambda }_i\left(A_G\right)}$, where $i\in \{1,\dots ,n\}$, ${\lambda }_i\left(A_G\right)$ represent the i-th eigenvalues of the graph G. This graph invariant appeared for the first time in the year 2000 in a paper by Ernesto Estrada dealing with the folding of protein molecules, see [13]. Many definitions of Estrada-like indices have appeared, for instance, the distance Estrada [14], and Laplacian Estrada indices [1,15]. Invariants of the form of the Estrada index can be conceived for any Hermitian matrix M with spectrum $S_p\left(M\right)$, see [16], as

$$EE\left(M\right)=\sum _{{\lambda }_i\left(M\right)\in S_p\left(M\right)}{e}^{{\lambda }_i\left(M\right)}.$$

Thereby, having this definition as motivation and considering the matrices previously presented, for $\alpha \in [0,1]$, we define the generalized Randić Estrada indices as

$$EE{R}_{\alpha }^G=\sum _{i=1}^n{e}^{{\phi }_i\left(\alpha \right)}.$$

We denote by $M_k=M_k\left(G\right)$ the k-th spectral moment of $R_{\alpha }^G$, that is,

$$M_k\left(G\right)=\sum _{i=1}^n{\phi }_i^k\left(\alpha \right).$$

Recalling the power-series expansion of $e^x$, we can write the Estrada index of $R_{\alpha }^G$ as:

$$EE{R}_{\alpha }^G=\sum _{k\ge 0}\frac{M_k}{k!}.$$

1.2. Outline of the Paper

The present paper consists of three sections. Section 2 presents bounds for the generalized Randić Estrada index of connected and regular non-bipartite graphs. Finally, in Section 3, we solve the problem proposed by V. Nikiforov in ([11], Problem 8), where the goal is to find the smallest value of $\alpha$ in which the matrix $A_{\alpha }^G$ is positive semidefinite.

2. Bounds for the Generalized Randić Estrada Indices

In this section, we introduce some bounds for the generalized Randić Estrada indices. Furthermore, we obtain extreme graphs where these bounds are attained. These bounds are found sometimes for non-bipartite graphs and, in other cases, for regular graphs using some numerical inequalities present in the literature.

We start this section by setting some properties of $\alpha$-Randić eigenvalues and spectral moments of the generalized Randić matrix.

For a graph G, we define

$${\mathrm{\Gamma }}_{-1}^G=\sum _{C_3\left(ijk\right)\in \Xi \left(G\right)}{\displaystyle \frac{1}{d_i{d}_j{d}_k}}.$$

The following lemma shows the first four spectral moments of $R_{\alpha }^G$, which depend only on the cardinality of G, the variable $\alpha$ and the vertex degrees in G.

Lemma 1.

Let G be a graph on n vertices and let $\alpha \in [0,1].$ Then

1. 

$M_0=n,$

2. 

$M_1=n\alpha$,

3. 

$M_2=n{\alpha }^2+2{(1-\alpha )}^2{\mathcal{R}}_{-1}^G$,

4. 

$M_3=n{\alpha }^3+6\alpha {(1-\alpha )}^2{\mathcal{R}}_{-1}^G+6{(1-\alpha )}^3{\Gamma }_{-1}^G$.

Proof. 

Let G be a graph on n vertices and let $\alpha \in [0,1]$. It is known that the trace of a square matrix is equal to the sum of its eigenvalues, then

$$M_k\left(G\right)=trace\left({\left(R_{\alpha }^G\right)}^k\right)$$

(9)

for $k\ge 0$.

1.

Trivial.

2.

This result is obtained from (7) and (9).

3.

From (7), we obtain

$${\left(R_{\alpha }^G\right)}^2={\alpha }^{2}I+2\alpha (1-\alpha )R_G+{(1-\alpha )}^2{R}_G^2.$$

(10)

Thereby, using the Frobenius norm, we obtain,

$$\left|\right|R_G{\left|\right|}_F^2=trace\left(R_G{R}_G^T\right)=trace\left(R_G^2\right)=2\sum _{ij\in \mathcal{E}\left(G\right)}\frac{1}{d_i{d}_j}.$$

From (1), (9) and (10), we obtain

$$\begin{array}{ccc}\hfill M_2& =& n{\alpha }^2+2{(1-\alpha )}^2\sum _{ij\in \mathcal{E}\left(G\right)}\frac{1}{d_i{d}_j}\hfill \\ \\ & =& n{\alpha }^2+2{(1-\alpha )}^2{\mathcal{R}}_{-1}^G.\hfill \end{array}$$

4.

Finally, note that

$${\left(R_{\alpha }^G\right)}^3={\alpha }^{3}I+3{\alpha }^2(1-\alpha )R_G+3\alpha {(1-\alpha )}^2{R}_G^2+{(1-\alpha )}^3{R}_G^3.$$

(11)

Using the matrix product and calculating the trace, we have:

$$trace\left(R_G^3\right)=6\sum _{C_3\left(ijk\right)\in \Xi \left(G\right)}{\displaystyle \frac{1}{d_i{d}_j{d}_k}}.$$

From (9) and (11), we obtain

$$\begin{array}{ccc}\hfill M_3& =& n{\alpha }^3+6\alpha {(1-\alpha )}^2\sum _{ij\in \mathcal{E}\left(G\right)}{\displaystyle \frac{1}{d_i{d}_j}}+6{(1-\alpha )}^3\sum _{C_3\left(ijk\right)\in \Xi \left(G\right)}{\displaystyle \frac{1}{d_i{d}_j{d}_k}}\hfill \\ \\ & =& n{\alpha }^3+6\alpha {(1-\alpha )}^2{\mathcal{R}}_{-1}^G+6{(1-\alpha )}^3{\Gamma }_{-1}^G.\hfill \end{array}$$

The proof is complete. □

The following lemmas present intervals that contain the eigenvalues of the normalized Laplacian and generalized Randić matrices.

Lemma 2

([17]). For all weighted connected graphs G,

$$0={\tau }_n<{\tau }_{n-1}\le \cdots \le {\tau }_1\le 2,$$

with ${\tau }_1=2$ if and only if G is non-trivial and bipartite.

The next result is an immediate consequence of (3) and Lemma 2.

Lemma 3.

Let G be a graph on n vertices. Then,

$$-1\le {\phi }_i\left(0\right)\le 1,$$

for $i=1,\dots ,n$. Moreover, ${\phi }_n\left(0\right)=-1$ if and only if G has a connected component non-trivial and bipartite.

The following result is a direct consequence of (7) and Lemma 3.

Lemma 4.

Let G be a graph on n vertices and let $\alpha \in [0,1]$. Then,

$$2\alpha -1\le {\phi }_i\left(\alpha \right)\le 1,$$

for $i=1,\dots ,n$. Moreover, ${\phi }_n\left(\alpha \right)=2\alpha -1$ if and only if G has a connected component non-trivial and bipartite or $\alpha =1$.

The following result is an immediate consequence of Lemmas 1 and 4.

Theorem 1.

Let G be a graph on n vertices and let $\alpha \in [0,1].$ Then

$$EE{R}_{\alpha }^G\ge n\left(1+\alpha +\frac{{\alpha }^2}{2}+\frac{{\alpha }^3}{6}\right)+(1+\alpha ){(1-\alpha )}^2{\mathcal{R}}_{-1}^G+{(1-\alpha )}^3{\Gamma }_{-1}^G.$$

(12)

Equality holds in (12) if and only if $G\cong {\overline{K}}_n$ and $\alpha =0$.

Proof. 

By Lemma 4,

$${\phi }_i\left(\alpha \right)>-2k-1,$$

for $k\ge 2$ and $i=1,\dots ,n$.

Hence,

$${\phi }_i^{2k}\left(\alpha \right)+{\displaystyle \frac{{\phi }_i^{2k+1}\left(\alpha \right)}{2k+1}}\ge 0,$$

for $k\ge 2$ and $i=1,\dots ,n$, where the equality is given if and only if ${\phi }_i\left(\alpha \right)=0$ for $i=1,\dots ,n$, which implies that

$$\sum _{k\ge 4}\frac{M_k}{k!}\ge 0$$

(13)

where the equality is attained if and only if $G\cong {\overline{K}}_n$ and $\alpha =0$.

By (13) and Lemma 1, we have

$$\begin{array}{ccc}\hfill EE{R}_{\alpha }^G& =& M_0+M_1+\frac{M_2}{2!}+\frac{M_3}{3!}+\sum _{k\ge 4}\frac{M_k}{k!}\hfill \\ & \ge & M_0+M_1+\frac{M_2}{2!}+\frac{M_3}{3!}\hfill \\ & =& n\left(1+\alpha +\frac{{\alpha }^2}{2}+\frac{{\alpha }^3}{6}\right)+(1+\alpha ){(1-\alpha )}^2{\mathcal{R}}_{-1}^G+{(1-\alpha )}^3{\Gamma }_{-1}^G,\hfill \end{array}$$

where the equality is attained if and only if $G\cong {\overline{K}}_n$ and $\alpha =0$.

The proof is complete. □

Lemma 5.

Let G be a connected graph on $n\ge 2$ vertices. Then,

$${\phi }_1\left(\alpha \right)=1 for \alpha \in [0,1].$$

Proof. 

Notice that if we multiply the matrix (8) by $D_G^{1/2}\mathbf{1}$, we obtain that $(1,D_G^{1/2}\mathbf{1})$ is an eigenpair associated to $R_{\alpha }^G$ for all $\alpha \in [0,1]$, where $D_G^{1/2}$ is the diagonal matrix whose i-th diagonal entry is the square root of the degree $d_i$, and 1 is the all-ones vector. Notice that the $R_{\alpha }^G$ in a non-negative irreducible matrix for $0\le \alpha \le 1$, therefore ${\phi }_1\left(\alpha \right)=1.$ The proof is complete. □

From the Perron–Frobenius theory for non-negative and irreducible matrices, we have the following result.

Lemma 6.

Let G be a connected graph on $n\ge 2$ vertices. Then,

$${\phi }_j\left(\alpha \right)<1 for j=2,\dots ,n and \alpha \in [0,1)$$

Lemma 7.

Let G be a graph on n vertices, and let $\alpha \in [0,1].$ Then,

$$\sum _{i=1}^n{e}^{2{\phi }_i\left(\alpha \right)}\ge n(2\alpha +1)+e^2-3.$$

(14)

Equality holds in (14) if and only if $G\cong K_n$ and $\alpha ={\displaystyle \frac{1}{n}}$.

Proof. 

Consider the following function

$$h\left(x\right)=e^x-x-1x\ge 0$$

is straightforward and prove that function $h\left(x\right)$ is strictly increasing in the interval $[0,\infty ).$ Then,

$$e^x\ge x+1\mathrm{for}x\ge 0,$$

(15)

with equality if and only if $x=0.$

By (15) and Lemma 5, we obtain

$$\begin{array}{ccc}\hfill \sum _{i=1}^n{e}^{2{\phi }_i\left(\alpha \right)}& =& e^2+\sum _{i=2}^n{e}^{2{\phi }_i\left(\alpha \right)}\hfill \\ & \ge & e^2+\sum _{i=2}^n(2{\phi }_i\left(\alpha \right)+1)\hfill \\ & =& n(2\alpha +1)+e^2-3.\hfill \end{array}$$

Suppose equality holds in (14), then ${\phi }_i\left(\alpha \right)=0$ for $i=2,\dots ,n$, if $n\ge 2$ and ${\phi }_1\left(\alpha \right)=1$. Then, $G\cong K_1$ with $\alpha =1$ or $G\ncong {\overline{K}}_n$ with $n\alpha =1$ for $n\ge 2$. For $n\ge 2$, there two adjacent vertices $v_1$ and $v_2$ in G such that $d_1=\underset{1\le i\le n}{min}{d}_i$. Now, by Cauchy’s Interlace Theorem, the eigenvalues of the matrix

$$\left[\begin{array}{cc}\alpha & {\displaystyle \frac{1-\alpha }{\sqrt{d_1{d}_2}}}\\ {\displaystyle \frac{1-\alpha }{\sqrt{d_1{d}_2}}}& \alpha \end{array}\right]$$

interlace the eigenvalues of matrix $R_{\alpha }^G$. Then, $\alpha -{\displaystyle \frac{1-\alpha }{\sqrt{d_1{d}_2}}}=0$. As $\alpha ={\displaystyle \frac{1}{n}}$, we obtain $\sqrt{d_1{d}_2}=n-1$, that is, $d_1=d_2=n-1$, which implies $d_i=n-1$ for $i=1,\dots ,n$, i.e., $G\cong K_n$ and $\alpha ={\displaystyle \frac{1}{n}}$. Conversely, if $G\cong K_n$ and $\alpha ={\displaystyle \frac{1}{n}}$, equality holds in (14). The proof is complete. □

In the following lemma, we consider some of the numerical inequalities present in the literature.

Lemma 8

([18]). Let $a=(a_1,\dots ,a_n)$ and $b=(b_1,\dots ,b_n)$ be two positive vectors with $0<f_1\le a_i\le F_1$ and $0<f_2\le b_i\le F_2$, $i=1,\dots ,n$, for some constants $f_1$, $f_2$, $F_1$ and $F_2$. Then,

$$\frac{n^2}{4}{\left(F_1{F}_2-f_1{f}_2\right)}^2\ge \left(\sum _{i=1}^n{a}_i^2\right)\left(\sum _{i=1}^n{b}_i^2\right)-{\left(\sum _{i=1}^n{a}_i{b}_i\right)}^2.$$

Remark 2.

Note that, from the Cauchy–Schwartz inequality, the expression on the right-hand side is always positive for any suitable a and b.

Applying Lemma 8 to the generalized Randić matrix, we prove the following result.

Theorem 2.

Let G be a non-bipartite connected graph on n vertices, and let $\alpha \in [0,1]$. Then,

$$\begin{array}{ccc}\hfill {\left(EE{R}_{\alpha }^G\right)}^2& \ge & n\sum _{i=1}^n{e}^{2{\phi }_i\left(\alpha \right)}-n^2{\left({\displaystyle \frac{e^1-e^{2\alpha -1}}{2}}\right)}^2.\hfill \end{array}$$

(16)

Equality holds in (16) if and only if $\alpha =1$.

Proof. 

Let $a_i=e^{{\phi }_i\left(\alpha \right)}$ and $b_i=1$, for $i=1,\dots ,n,$ $f_1=e^{{\phi }_n\left(\alpha \right)}$, $F_1=e^1$, $f_2=F_2=1$. From Lemmas 5 and 8, we have

$$\frac{n^2}{4}{\left(e^1-e^{{\phi }_n\left(\alpha \right)}\right)}^2\ge \left(\sum _{i=1}^n{e}^{2{\phi }_i\left(\alpha \right)}\right)\left(\sum _{i=1}^{n}1\right)-{\left(\sum _{i=1}^n{e}^{{\phi }_i\left(\alpha \right)}\right)}^2.$$

(17)

From(17) and Lemma 4, we obtain

$$\frac{n^2}{4}{\left(e^1-e^{2\alpha -1}\right)}^2\ge n\sum _{i=1}^n{e}^{2{\phi }_i\left(\alpha \right)}-{\left(\sum _{i=1}^n{e}^{{\phi }_i\left(\alpha \right)}\right)}^2.$$

(18)

Suppose $\alpha =1$. Then, ${\phi }_i\left(\alpha \right)=1=2\alpha -1$ for $i=1,2,\dots ,n$. Thus, equality is attained in (18).

Conversely, suppose that equality in (18) holds, then

$$\frac{n^2}{4}{\left(e^1-e^{2\alpha -1}\right)}^2=\frac{n^2}{4}{\left(e^1-e^{{\phi }_n\left(\alpha \right)}\right)}^2=n\sum _{i=1}^n{e}^{2{\phi }_i\left(\alpha \right)}-{\left(\sum _{i=1}^n{e}^{{\phi }_i\left(\alpha \right)}\right)}^2.$$

Hence,

$${\phi }_n\left(\alpha \right)=2\alpha -1.$$

From the hypothesis and Lemma 4,

$$\alpha =1.$$

Therefore, from (18), we can conclude

$$\begin{array}{ccc}\hfill {\left(EE{R}_{\alpha }^G\right)}^2& \ge & n\sum _{i=1}^n{e}^{2{\gamma }_i\left(\alpha \right)}-n^2{\left(\frac{e^1-e^{2\alpha -1}}{2}\right)}^2\hfill \end{array}$$

where equality holds if and only if $\alpha =1$.

The proof is complete. □

As a direct consequence of Lemma 7 and Theorem 2, we obtain the following result.

Corollary 1.

Let G be a non-bipartite connected graph on n vertices, and let $\alpha \in [0,1]$. Then,

$${\left(EE{R}_G^{\alpha }\right)}^2\ge n^2\left(2\alpha +1-{\left({\displaystyle \frac{e^1-e^{2\alpha -1}}{2}}\right)}^2\right)+n(e^2-3).$$

(19)

Equality holds in (19) if and only if $G\cong K_1$ and $\alpha =1$.

The following lower bound is obtained by using Lemma 4 and Theorem 2.

Corollary 2.

Let G be a non-bipartite connected graph on n vertices. Let ${\displaystyle \frac{2-ln3}{2}}\le \alpha \le 1$. Then,

$$EE{R}_{\alpha }^G\ge ne\sqrt{e^{4\alpha -4}-\frac{{(1-e^{2\alpha -2})}^2}{4}}.$$

(20)

Equality holds in (20) if and only if $\alpha =1$.

Proof. 

By Lemma 4 and Theorem 2:

$$\begin{array}{ccc}\hfill {\left(EE{R}_{\alpha }^G\right)}^2& \ge & n\sum _{i=1}^n{e}^{2{\phi }_i\left(\alpha \right)}-\frac{n^2}{4}{\left(e-e^{2\alpha -1}\right)}^2\hfill \\ & \ge & n\sum _{i=1}^n{e}^{2(2\alpha -1)}-\frac{n^2{e}^2}{4}{\left(1-e^{2\alpha -2}\right)}^2\hfill \\ & =& n^2{e}^2\left(e^{4\alpha -4}-\frac{{\left(1-e^{2\alpha -2}\right)}^2}{4}\right)\ge 0.\hfill \end{array}$$

For $\alpha =1$, the equality in the inequality (20) is clear.

Reciprocally, if the equality in (20) holds, then the last two inequalities return to equalities. By Theorem 2, the first of these equalities implies that $\alpha =1$. By Lemmas 4 and 5, the second of these equalities implies that $\alpha =1$. The proof is complete. □

For a graph G on n vertices and $\alpha \in [0,1]$, let

$$\mathcal{C}(n,\alpha )=n\left(1+2\alpha +2{\alpha }^2+\frac{4{\alpha }^3}{3}\right)+4(1+2\alpha ){(1-\alpha )}^2{\mathcal{R}}_{-1}^G+8{(1-\alpha )}^3{\Gamma }_{-1}^G.$$

Considering Lemma 1 and the power series of the exponential function, we have the following result.

Theorem 3.

Let G be a graph on n vertices and let $\alpha \in [0,1].$ Then,

$$EE{R}_{\alpha }^G\ge \sqrt{\mathcal{C}(n,\alpha )+n(n-1)e^{2\alpha }}.$$

(21)

Equality holds in (21) if and only if $G\cong {\overline{K}}_n$ and $\alpha =0.$

Proof. 

By definition, we obtain

$${\left(EE{R}_{\alpha }^G\right)}^2=\sum _{i=1}^n{e}^{2{\phi }_i\left(\alpha \right)}+2\sum _{i<j}{e}^{{\phi }_i\left(\alpha \right)}{e}^{{\phi }_j\left(\alpha \right)}.$$

(22)

It is known in [19], for $q\ge 1$

$${\displaystyle \frac{{\displaystyle \sum _{i=1}^q}{x}_i}{q}}\ge \sqrt[q]{\prod _{i=1}^q}{x}_i,$$

where $x_i\ge 0$ for $i=1,\dots ,q$, with equality if and only if $x_i=x_j$ for $i\ne j$.

Thus,

$$\begin{array}{ccc}\hfill 2\sum _{i<j}{e}^{{\phi }_i\left(\alpha \right)}{e}^{{\phi }_j\left(\alpha \right)}& \ge & n(n-1){\left(\prod _{i<j}{e}^{{\phi }_i\left(\alpha \right)}{e}^{{\phi }_j\left(\alpha \right)}\right)}^{\frac{2}{n(n-1)}}\hfill \\ & =& n(n-1){\left[{\left(\prod _{i<j}{e}^{{\phi }_i\left(\alpha \right)}\right)}^{n-1}\right]}^{\frac{2}{n(n-1)}}\hfill \\ & =& n(n-1){\left(e^{{\displaystyle \sum _{i<j}}{\phi }_i\left(\alpha \right)}\right)}^{\frac{2}{n}}\hfill \\ & =& n(n-1)e^{2\alpha }\hfill \end{array}$$

where the equality holds if and only if ${\phi }_1\left(\alpha \right)=\cdots ={\phi }_n\left(\alpha \right)=\alpha$, i.e., $G\cong {\overline{K}}_n$ or $\alpha =1$.

By Lemma 4,

$${\phi }_i\left(\alpha \right)>-{\displaystyle \frac{2k+1}{2}},$$

for $k\ge 2$ and $i=1,\dots ,n$.

Hence,

$${\phi }_i^{2k}\left(\alpha \right)+{\displaystyle \frac{2{\phi }_i^{2k+1}\left(\alpha \right)}{2k+1}}\ge 0,$$

for $k\ge 2$ and $i=1,\dots ,n$, where the equality is given if and only if ${\phi }_i\left(\alpha \right)=0$ for $i=1,\dots ,n$, which implies that

$$\sum _{k\ge 4}\frac{2^k{M}_k}{k!}\ge 0$$

(23)

where the equality is attained if and only if $G\cong {\overline{K}}_n$ and $\alpha =0$.

From (23) and Lemma 1, we have

$$\begin{array}{ccc}\hfill \sum _{i=1}^n{e}^{2{\phi }_i\left(\alpha \right)}& =& M_0+2M_1+2M_2+{\displaystyle \frac{4M_3}{3}}+\sum _{k\ge 4}\frac{2^k{M}_k}{k!}\hfill \\ & \ge & M_0+2M_1+2M_2+{\displaystyle \frac{4M_3}{3}}\hfill \\ & =& n\left(1+2\alpha +2{\alpha }^2+\frac{4{\alpha }^3}{3}\right)+4(1+2\alpha ){(1-\alpha )}^2{\mathcal{R}}_{-1}^G+8{(1-\alpha )}^3{\Gamma }_{-1}^G,\hfill \end{array}$$

where the equality is given if and only if $G\cong {\overline{K}}_n$ and $\alpha =0$. From (22), the proof is complete. □

Lemma 9

([20]). For non-negative real numbers $x_1,x_2,\dots ,x_n$ and $k\ge 2$, we have

$$\sum _{i=1}^n{x}_i^k\le {\left(\sum _{i=1}^n{x}_i^2\right)}^{k/2}.$$

Remark 3.

Expanding the powers of the previous expression, it can be verified that the inequality in the above lemma is strict for k, even provided that there are two positive real numbers $x_i$, $x_j$ such that $x_i\ne x_j$ for some $i,j=1,\dots ,n$.

The following theorem gives an upper bound for $EE{R}_{\alpha }^G$, as a function of the variable $\alpha$ and the degree of G, when G is a regular connected graph.

Theorem 4.

Let G be a regular connected graph on $n\ge 2$ vertices and degree r. Let $\alpha \in [0,1]$ and $\alpha (r+1)\le 1$. Then,

$$EE{R}_{\alpha }^G\le n-1+e^{\sqrt{n{\alpha }^2+\frac{n{(1-\alpha )}^2}{r}-{\left(\frac{\alpha (r+1)-1}{r}\right)}^2}}.$$

Proof. 

Let $n_{+}$ be the positive $\alpha$-Randić eigenvalues number. It is known that $f\left(x\right)=e^x$ is a monotonically increasing function in the interval $(-\infty ,+\infty )$.

By Lemma 9, we obtain

$$\begin{array}{c}\hfill \begin{array}{ccc}\hfill \sum _{i=1}^n{e}^{{\phi }_i\left(\alpha \right)}& \le \hfill & n-n_{+}+\sum _{i=1}^{n_{+}}{e}^{{\phi }_i\left(\alpha \right)}\\ & =\hfill & n-n_{+}+\sum _{i=1}^{n_{+}}\sum _{k\ge 0}\frac{{\phi }_i^k\left(\alpha \right)}{k!}\\ & =\hfill & n+\sum _{k\ge 1}\frac{1}{k!}\sum _{i=1}^{n_{+}}{\phi }_i^k\left(\alpha \right)\\ & \le \hfill & n+\sum _{k\ge 1}\frac{1}{k!}{\left(\sum _{i=1}^{n_{+}}{\phi }_i^2\left(\alpha \right)\right)}^{k/2}.\end{array}\end{array}$$

(24)

By Remark 3 and taking into account the techniques for the derivation of the inequalities in (24), it is clear that equalities hold in previous inequalities if and only if ${\phi }_i\left(\alpha \right)\ge 0$ for $i=1,\dots ,n$, and ${\phi }_i\left(\alpha \right)={\phi }_j\left(\alpha \right)$ for $i\ne j$ with $i,j=1,\dots ,n_{+}$. By Lemma 6, we conclude that the equalities are given in (24) if and only if $G\cong K_1$.

By (24) and Lemma 1, we obtain

$$EE{R}_{\alpha }^G\le n+\sum _{k\ge 1}\frac{1}{k!}{\left(n{\alpha }^2+{\displaystyle \frac{n{(1-\alpha )}^2}{r}}-\sum _{i=n_{+}+1}^n{\phi }_i^2\left(\alpha \right)\right)}^{k/2}.$$

(25)

Now, it is easy to see that the eigenvalues of the matrix $\alpha I_2+{\displaystyle \frac{(1-\alpha )}{r}}{A}_{P_2}$ interlace the eigenvalues of $R_{\alpha }^G$. Since $\alpha (r+1)\le 1$, it follows from this interlacing inequalities that

$${\phi }_n\left(\alpha \right)\le {\displaystyle \frac{\alpha (r+1)-1}{r}}\le 0,$$

which implies that

$$\sum _{i=n_{+}+1}^n{\phi }_i^2\left(\alpha \right)\ge {\left({\displaystyle \frac{\alpha (r+1)-1}{r}}\right)}^2.$$

(26)

From (25) and (26), we have

$$\begin{array}{ccc}\hfill EE{R}_{\alpha }^G& \le & n+\sum _{k\ge 1}\frac{1}{k!}{\left(n{\alpha }^2+{\displaystyle \frac{n{(1-\alpha )}^2}{r}}-{\left(\frac{\alpha (r+1)-1}{r}\right)}^2\right)}^{k/2}\hfill \\ & \le & n-1+e^{\sqrt{n{\alpha }^2+\frac{n{(1-\alpha )}^2}{r}-{\left(\frac{\alpha (r+1)-1}{r}\right)}^2}}.\hfill \end{array}$$

The proof is complete. □

The following result can be obtained from the arithmetic–geometric mean inequality, which can be found in [19].

Theorem 5.

Let G be a connected graph on $n\ge 2$ vertices, and let $\alpha \in [0,1]$. Then,

$$EE{R}_{\alpha }^G\ge e+(n-1)e^{\frac{n\alpha -1}{n-1}}.$$

(27)

Equality holds in (27) if and only if $G\cong K_n$ or $\alpha =1$.

Proof. 

The relation between the arithmetic and geometric means allows us to obtain

$$EE{R}_{\alpha }^G-e=\sum _{i=2}^n{e}^{{\phi }_i\left(\alpha \right)}\ge (n-1){\left(\prod _{i=2}^n{e}^{{\phi }_i\left(\alpha \right)}\right)}^{1/n-1}=(n-1)e^{\frac{n\alpha -1}{n-1}}.$$

(28)

Suppose the equality is given in (28). Since the sum of the $\alpha$-Randić eigenvalues is equal to $n\alpha -1$, then ${\phi }_2\left(\alpha \right)={\phi }_3\left(\alpha \right)=\cdots ={\phi }_n\left(\alpha \right)={\displaystyle \frac{n\alpha -1}{n-1}}$. Suppose $d_1=\underset{1\le i\le n}{min}{d}_i$ and $v_i$ and $v_2$ are adjacent. By the interlacing of the $\alpha$-Randić eigenvalues and the eigenvalues of the matrix $\alpha I_2+{\displaystyle \frac{1-\alpha }{\sqrt{d_1{d}_2}}}{A}_{P_2}$, we can conclude that $\alpha -{\displaystyle \frac{1-\alpha }{\sqrt{d_1{d}_2}}}={\displaystyle \frac{n\alpha -1}{n-1}}$, i.e., $\sqrt{d_1{d}_2}=n-1$ or $\alpha =1$, i.e., $G\cong K_n$ or $\alpha =1$. Conversely, if $G\cong K_n$ or $\alpha =1$, the equality is given in (28). The proof is complete. □

3. On the Positive Semidefiniteness of ${\mathit{R}}_{\mathit{\alpha }}^{\mathit{G}}$

In this section, we present the solution to the problem proposed in ([11], Problem 8). The problem consists of the following: for a given undirected graph, G determines the smallest $\alpha$ for which $A_{\alpha }^G$ is positive semidefinite. In [12], this problem is completely solved for bipartite graphs, and a sufficient condition for regular graphs is given.

From now on, ${\alpha }_1\left(G\right)$ is the smallest value of $\alpha$ for which $R_{\alpha }^G$ is positive semidefinite.

Lemma 10.

Let G be a connected graph on $n\ge 2$ vertices and let $\alpha ,\beta \in [0,1]$. If $\alpha \le \beta$ then

$${\phi }_j\left(\alpha \right)\le {\phi }_j\left(\beta \right)$$

(29)

for $j=2,\dots ,n$. Equality holds in (29) if and only if $\alpha =\beta$.

Proof. 

Let $0\le \alpha \le \beta \le 1$. From (7) and Lemma 6, we obtain

$${\phi }_j\left(\beta \right)-{\phi }_j\left(\alpha \right)=(\beta -\alpha )(1-{\phi }_j\left(0\right))$$

(30)

for $j=2,\dots ,n$. The proof is complete. □

Remark 4.

The adjacency matrix is positive semidefinite only in the case G is empty.

Let us continue by observing a general property of graphs with respect to ${\alpha }_1\left(G\right)$, we present the following result

Lemma 11.

If G is a graph on n vertices, then

$${\alpha }_1\left(G\right)=max\{{\alpha }_1\left(H\right):H\mathit{is}a\mathit{connected}\mathit{component}\mathit{of}G\}.$$

Lemma 12

([21]). (Horn and Johnson 1985, Theorem 4.5.8) Let A and B be Hermitian matrices of order n. There is a nonsingular matrix S of order n such that $A=SB{S}^{*}$ if and only if A and B have the same inertia, that is, the same number of positive, negative, and zero eigenvalues, counting multiplicities.

Theorem 6.

Let G be a graph on n vertices and let $\alpha \in [0,1]$. Then,

$${\alpha }_1\left(G\right)={\displaystyle \frac{-{\phi }_n\left(0\right)}{1-{\phi }_n\left(0\right)}} \mathit{where} 0\le {\alpha }_1\left(G\right)\le {\displaystyle \frac{1}{2}}.$$

The first inequality is given as equality if and only if $G\cong {\overline{K}}_n$, and the second inequality is given as equality if and only if G has a non-trivial bipartite connected component.

Proof. 

Let G be a graph on n vertices, and let $\alpha \in [0,1]$. First consider $G\cong {\overline{K}}_n$, then ${\alpha }_1\left(G\right)=0$ and ${\phi }_n\left(0\right)=0$, and the result is verified. Now, let G be a non-empty graph. Let $G_1,\dots ,G_l$ be the connected components in G of order $n_j\ge 2$ for $j=1,\dots ,l$. From (30), the function $f_j:[0,1]⟼\mathbb{R}$, where $f_j\left(\alpha \right)={\phi }_{n_j}\left(\alpha \right)$, is continuous for $j=1,\dots ,l$. By Lemma 10, we conclude $f_j$ is increasing strictly for $j=1,\dots ,l$. Then,

$$f_j\left(a_j\right)={\phi }_n\left(a_j\right)=0$$

(31)

for a single value $a_j\in [0,1]$ where $j=1,\dots ,l$.

Clearly,

$$a_j={\alpha }_1\left(G_j\right)$$

for $j=1,\dots ,l$.

By (7) and (31), we obtain

$${\alpha }_1\left(G_j\right)={\displaystyle \frac{-{\phi }_{n_j}\left(0\right)}{1-{\phi }_{n_j}\left(0\right)}}$$

for $j=1,\dots ,l.$

Since $g\left(x\right)={\displaystyle \frac{x}{x-1}}$ is a strictly decreasing function in the interval $[-1,1)$, applying Lemmas 3 and 11, we conclude

$${\alpha }_1\left(G\right)={\displaystyle \frac{-{\phi }_n\left(0\right)}{1-{\phi }_n\left(0\right)}} \mathrm{where} 0\le {\alpha }_1\left(G\right)\le {\displaystyle \frac{1}{2}}$$

where ${\alpha }_1\left(G\right)={\displaystyle \frac{1}{2}}$ if and only if G has a non-trivial bipartite connected component.

Clearly, if $G\cong {\overline{K}}_n$ then ${\alpha }_1\left(G\right)=0$. Conversely, let ${\alpha }_1\left(G\right)=0$. We claim, $G\cong {\overline{K}}_n$. Suppose G has a connected component of order greater or equal to 2, denoted by $G_1$. From Lemma 11, ${\alpha }_1\left(G_1\right)=0$. Then, $R_{G_1}$ is positive semidefinite. From (8) and Lemma 12, we obtain that $A_{G_1}$ is positive semidefinite, which is a contradiction taking into account Remark 4. The proof is complete. □

Nikiforov, in [12], showed that if ${\alpha }_0\left(G\right)$ is the smallest value of $\alpha$ for which $A_{\alpha }^G$ is positive semidefinite, then ${\alpha }_0\left(G\right)\le {\displaystyle \frac{1}{2}}$. Notice that the matrix $R_{\alpha }^G$ and $A_{\alpha }^G$ have the same inertia, under this context for $A_{\alpha }^G$, as an immediate consequence of Theorem 6 and Lemma 12, we solve the following problem that was proposed in [11]:

Problem 1.

Given a graph G on n vertices, then

$${\alpha }_0\left(G\right)={\displaystyle \frac{-{\phi }_n\left(0\right)}{1-{\phi }_n\left(0\right)}} \mathit{where} 0\le {\alpha }_0\left(G\right)\le {\displaystyle \frac{1}{2}}.$$

The first inequality is given as equality if and only if $G\cong {\overline{K}}_n$, and the second inequality is given as equality if and only if G has a non-trivial bipartite connected component.

Lemma 13

([22]). (Jensen’s Inequality) Let $f:(a,b)\to \mathbb{R}$ be a convex function. Given numbers $x_1,\dots ,x_n\in (a,b)$ and non-negative numbers $c_1,\dots ,c_n$ with

$$c_1+\cdots +c_n=1,$$

we obtain

$$f(c_1{x}_1+\cdots +c_n{x}_n)\le c_{1}f\left(x_1\right)+\cdots +c_{n}f\left(x_n\right).$$

In case f is strictly convex, the inequality is strict, except in the trivial cases in which there exists k such that $c_k=1$ or $x_1=\cdots =x_n$.

Theorem 7.

Let G be a graph on n vertices and let $\alpha \in [{\alpha }_1\left(G\right),1].$ Then

$$EE{R}_{\alpha }^G\le n{e}^{\alpha }+2{(1-\alpha )}^2{\mathcal{R}}_{-1}^G.$$

(32)

Equality holds in (32) if and only if $G\cong {\overline{K}}_n$.

Proof. 

Let $\alpha \in [{\alpha }_1\left(G\right),1].$ By Lemma 10,

$${\phi }_i\left(\alpha \right)\ge {\phi }_i\left({\alpha }_1\left(G\right)\right)\ge 0$$

for $i=1,\dots ,n$. One can prove that $f\left(x\right)=x^2+x-e^x$ for x in the interval $[0,1]$ is strictly convex.

Applying Lemma 13, taking $x_i={\phi }_i\left(\alpha \right)$ and $c_i={\displaystyle \frac{1}{n}}$ for $i=1,\dots n$, we obtain

$$\begin{array}{ccc}\hfill {\left({\displaystyle \frac{{\displaystyle \sum _{i=1}^n}{\phi }_i\left(\alpha \right)}{n}}\right)}^2+\frac{{\displaystyle \sum _{i=1}^n}{\phi }_i\left(\alpha \right)}{n}-e^{\frac{{\displaystyle \sum _{i=1}^n}{\phi }_i\left(\alpha \right)}{n}}& \le & {\displaystyle \frac{1}{n}}\left[\sum _{i=1}^n{\phi }_i^2\left(\alpha \right)+\sum _{i=1}^n{\phi }_i\left(\alpha \right)-\sum _{i=1}^n{e}^{{\phi }_i\left(\alpha \right)}\right]\hfill \end{array}$$

with equality only in the cases ${\phi }_i\left(\alpha \right)=\alpha$ for $i=1,\dots ,n$, i.e., $G\cong {\overline{K}}_n$.

By Lemma 1, we have

$${\alpha }^2+\alpha -e^{\alpha }\le {\displaystyle \frac{1}{n}}\left[n{\alpha }^2+2{(1-\alpha )}^2{\mathcal{R}}_{-1}^G+n\alpha -\sum _{i=1}^n{e}^{{\phi }_i\left(\alpha \right)}\right].$$

The proof is complete. □