On the Spectral Redundancy of Pineapple Graphs

Abstract

In this article, we explore the concept of spectral redundancy within the class of pineapple graphs, denoted as $\mathcal{P}(\alpha ,\beta )$. These graphs are constructed by attaching $\beta$ pendent edges to a single vertex of a complete graph $K_{\alpha }$. A connected graph G earns the title of being spectrally non-redundant if the spectral radii of its connected induced subgraphs are all distinct. Spectral redundancy, on the other hand, arises when there is a repetition of spectral radii among the connected induced subgraphs within G. Our study analyzes the adjacency spectrum of $\mathcal{P}(\alpha ,\beta )$, identifying distinct eigenvalues such as 0, $-1$, along with other positive and negative eigenvalues. Our investigation focuses on determining the spectral redundancy within this class of graphs, shedding light on their unique structural properties and implications for graph theory. Understanding spectral redundancy in these graphs is crucial for applications in network design, where distinct spectral radii can indicate different connectivity patterns and resilience features.

1. Introduction

Suppose that A is an n by the n real matrix. A scalar $\lambda \in R$ is a complementarity eigenvalue of A if there is a non-zero vector $x\in R^n$ satisfying $x\ge 0,Ax-\lambda x\ge 0$ and $⟨x,Ax-\lambda x⟩=0$, where $x\ge 0$ means x is non-negative, component wise. Seeger [1], introduced the complementarity spectrum of a square matrix. Later, Fernandes et al. [2] put forward the concept of complementarity spectrum of a graph G, which is the set of the complementarity eigenvalues of the adjacency matrix $A_G$. We refer the readers to Seeger [3,4], Seeger and Sossa [5,6,7], Pinheiro et al. [8] and Merajuddin et al. [9] for more recent results on the complementarity spectrum.

Let $G(V,E)$ be a simple connected graph, where $V=\{v_1,v_2,\dots ,v_n\}$ is the vertex set and $E=\{e_1,e_2,\dots ,e_m\}$ is the edge set. The graph G has a symmetric adjacency matrix $A\left(G\right)=\left(a_{ij}\right)$ of order n, where

$$a_{ij}=\left\{\begin{array}{cc}1,\hfill & \mathrm{if}{v}_i\mathrm{and}{v}_j\mathrm{are}\mathrm{adjacent}\hfill \\ 0,\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

The spectral radius of a graph G is the largest eigenvalue of the characteristic polynomial of the adjacency matrix $A\left(G\right)$. In the work by Fernandes et al. [2], the concept of the complementary spectrum of G is introduced. The notation $\Pi \left(G\right)$ is used to represent the complementary spectrum. This spectrum comprises the spectral radii of the adjacency matrices for all non-isomorphic, connected, induced subgraphs of G, which can be formally written as $\Pi \left(G\right)=\left\{\rho \right(F):F\in \mathcal{S}(G\left)\right\}$, where $\mathcal{S}\left(G\right)$ represents the set of all such subgraphs. Based on this, Fernandes et al. [2] established the following inequalities, where $b\left(G\right)$ refers to the total number of non-isomorphic induced subgraphs and $c\left(G\right)$ represents the size of the set $\Pi \left(G\right)$:

$$b\left(G\right)\ge c\left(G\right)\mathrm{and}{2}^n-1\ge b\left(G\right)\ge n.$$

While the upper bound $b\left(G\right)\le 2^n-1$ is not sharp, the authors in [2] demonstrated that it increases more rapidly than any polynomial function of n. The lower bound $b\left(G\right)\ge n$ was conclusively resolved in [4], with the equality attained just for the following so-called elementary graphs. Namely, they are cycles, stars, complete graphs, and paths. Any n-vertex graph other than these possesses over n induced subgraphs. As distinct induced subgraphs of G may share a spectral radius, it follows that $b\left(G\right)\ge c\left(G\right)$. To quantify this overlap, Seeger [3] introduced the concept of spectral redundancy for a graph G. It is given by the following

$$r\left(G\right)={\displaystyle \frac{b\left(G\right)}{c\left(G\right)}}.$$

When we have a collection $\mathcal{G}$ of connected graphs, the spectral redundancy index is naturally given by $r\left(\mathcal{G}\right)={sup}_{G\in \mathcal{G}}r\left(G\right)$.

Recent studies have explored vague graphs and their spectral properties, highlighting potential avenues for further research [10,11,12,13,14]. In this study, we focus specifically on the spectral redundancy within pineapple graphs. The pineapple graph $\mathcal{P}(\alpha ,\beta )$ is the coalescence of the star $S_{1,\beta }$ at the vertex of degree $\beta$ with the complete graph $K_{\alpha }$ at any vertex. For instance, Figure 1 illustrates the pineapple graph $\mathcal{P}(5,3)$, which is derived from combining $K_5$ and $S_{1,3}$. The star graph $S_{1,\beta +1}$ and the complete graph $K_{\alpha }$ are both special cases of the pineapple graph when $\alpha =2$ and $\beta =0$, respectively. If we include these two special cases in the family of pineapple graphs, the family of pineapple graphs is hereditary, that is, the induced subgraph of a pineapple graph is a pineapple graph as well. When $\beta >0$, all induced subgraphs with at least one pendant vertex share the same unique groupie [15].

Figure 1. Pineapple graph, $\mathcal{P}(5,3)$, with clique 5 and 3 pendant vertices.

In this article, we study the spectral aspect of pineapple graphs. It was claimed in [16] that the pineapple graphs are determined by their adjacency spectrum, but Topcu et al. [17] found some disconnected graphs with the same characteristic polynomial as the pineapple graphs. However, it is shown in [18] that the pineapple graph can determined by its adjacency spectrum if we are confined in the realm of connected graphs. The authors also determined all the disconnected graphs which are cospectral with a pineapple graph.

The structure of the rest of the paper is as follows: In Section 2, we study the spectral radius of the pineapple graphs and give some results regarding the spectral redundancy of this family.

2. Spectral Redundancy of Pineapple Graphs

First, notice that the pineapple graph $\mathcal{P}(2,\beta )$ is the star graph with $\beta +1$ pendant vertices and the pineapple graph $\mathcal{P}(\alpha ,0)$ is the complete graph isomorphic to $K_{\alpha }$. The following lemma regards the spectral radius of the pineapple graph $\mathcal{P}(\alpha ,\beta )$ with $\alpha \ge 2$ and $\beta \ge 0$.

Lemma 1

([17]). If $\mathcal{P}(\alpha ,\beta )$ is the pineapple graph, then the characteristic equation of the adjacency matrix of $\mathcal{P}(\alpha ,\beta )$ is given by

$$P(\alpha ,\beta ,x)=x^{\beta -1}{(x+1)}^{\alpha -2}[x^3-(\alpha -2)x^2-(\alpha +\beta -1)x+\beta (\alpha -2)]=0.$$

As the spectral radius of a connected graph is always positive, so the spectral radius of the pineapple graph $\mathcal{P}(\alpha ,\beta )$ is the largest root of the equation

$$P(\alpha ,\beta ,x)=x^3-(\alpha -2)x^2-(\alpha +\beta -1)x+\beta (\alpha -2)=0$$

(1)

In analyzing the spectral redundancy of pineapple graphs, it is necessary to know the number of induced subgraphs. The following lemma provides a fundamental result that counts the number of induced subgraphs for the family of pineapple graphs $\mathcal{P}(\alpha ,\beta )$.

Lemma 2.

For all $\alpha \ge 2$ and $\beta \ge 0$, we have $b\left(\mathcal{P}\right(\alpha ,\beta \left)\right)=(\alpha -1)(\beta +1)+1.$

Proof. 

Clearly, an induced subgraph of $\mathcal{P}(\alpha ,\beta )$ is also a pineapple graph (considering complete graphs and star graphs as pineapple graphs) with a lesser number of vertices. There are exactly $\beta +1$ induced subgraphs with a fixed clique number of k$(2\le k\le \alpha )$. All of them are non-isomorphic. Therefore, the number of non-empty connected induced subgraphs of $\mathcal{P}(\alpha ,\beta )$ is $(\alpha -1)(\beta +1)$. There is an induced subgraph with one vertex as well. So, $b\left(\mathcal{P}\right(\alpha ,\beta \left)\right)=(\alpha -1)(\beta +1)+1$. □

Example 1.

For instance, consider the pineapple graph $\mathcal{P}(4,3)$. According to the lemma, the number of connected induced subgraphs is calculated as $(4-1)(3+1)+1=13$. Figure 2 illustrates all these induced subgraphs, excluding the trivial graph with a single vertex.

Figure 2. A list of the non-empty connected non-isomorphic induced subgraphs of $\mathcal{P}(4,3)$.

To establish a necessary condition for two pineapple graphs to have the same spectral radius, we need to rely on some preliminary results. The next lemma provides an essential result that will be used to prove this condition.

Lemma 3

([19]). Suppose that H is a proper induced subgraph of G. We have $\rho \left(H\right)<\rho \left(G\right)$.

Now, we present the necessary condition for two pineapple graphs to have the same spectral radius.

Lemma 4.

If ${\mathcal{P}}_1({\alpha }_1,{\beta }_1)$ and ${\mathcal{P}}_2({\alpha }_2,{\beta }_2)$ are two non-isomorphic pineapple graphs with the same spectral radius, then $({\alpha }_2-{\alpha }_1)({\beta }_2-{\beta }_1)<0$.

Proof. 

Let ${\mathcal{P}}_1({\alpha }_1,{\beta }_1)$ and ${\mathcal{P}}_2({\alpha }_2,{\beta }_2)$ be two non-isomorphic pineapple graphs, such that $\rho \left({\mathcal{P}}_1({\alpha }_1,{\beta }_1)\right)=\rho \left({\mathcal{P}}_2({\alpha }_2,{\beta }_2)\right)$. Without loss of generality, let us assume $({\alpha }_2-{\alpha }_1)\ge 0$, and on the contrary, let us also assume $({\beta }_2-{\beta }_1)\ge 0$. This leads to ${\alpha }_2\ge {\alpha }_1$ and ${\beta }_2\ge {\beta }_1$, thereby implying that ${\mathcal{P}}_1({\alpha }_1,{\beta }_1)$ is isomorphic to some induced subgraph of ${\mathcal{P}}_2({\alpha }_2,{\beta }_2)$, or both are isomorphic. Hence, utilizing Lemma 3, we deduce $\rho \left({\mathcal{P}}_1({\alpha }_1,{\beta }_1)\right)<\rho \left({\mathcal{P}}_2({\alpha }_2,{\beta }_2)\right)$, which is a contradiction. Therefore, two non-isomorphic pineapple graphs ${\mathcal{P}}_1({\alpha }_1,{\beta }_1)$ and ${\mathcal{P}}_2({\alpha }_2,{\beta }_2)$ have the same spectral radius then $({\alpha }_2-{\alpha }_1)({\beta }_2-{\beta }_1)<0$. □

To illustrate the application of this lemma, consider the following example.

Example 2.

The pineapple graphs $\mathcal{P}(7,110)$ and $\mathcal{P}(9,99)$ share a common spectral radius of 11. The expression $({\alpha }_2-{\alpha }_1)({\beta }_2-{\beta }_1)$ evaluates to $(9-7)(99-110)=-22$, which is less than zero. This result substantiates the claim made by the preceding lemma.

The following lemma provides a sufficient condition for two pineapple graphs to share their two largest common eigenvalues.

Lemma 5.

Let $P({\alpha }_1,{\beta }_1,x)$ and $P({\alpha }_2,{\beta }_2,x)$ be two polynomials defined by the Equation (1). The polynomials $P({\alpha }_1,{\beta }_1,x)$ and $P({\alpha }_2,{\beta }_2,x)$ have two common roots if and only if ${\beta }_i={\displaystyle \frac{k({\alpha }_j-1)({\alpha }_j-2)}{k-1}}$ for $i\ne j$ and $i,j=1,2$, and ${\alpha }_1+{\alpha }_2-2=k$, where $k={\displaystyle \frac{{\beta }_2-{\beta }_1}{{\alpha }_1-{\alpha }_2}}$. Additionally, k must be an integer and $k-1$ divides $({\alpha }_i-1)({\alpha }_i-2)$. The common roots are given by

$${\displaystyle \frac{(k-1)\pm \sqrt{{(k-1)}^2+4({\beta }_i-k({\alpha }_j-2))}}{2}},$$

where $i,j=1$ or 2 and $i\ne j$.

Proof. 

Let $\rho$ be the common root of the equations $P({\alpha }_1,{\beta }_1,x)=0$ and $P({\alpha }_2,{\beta }_2,x)=0$, Then, we have $P({\alpha }_1,{\beta }_1,\rho )=0$ and $P({\alpha }_2,{\beta }_2,\rho )=0$. After subtracting these equations, we obtain

$$({\alpha }_2-{\alpha }_1){\rho }^2+({\alpha }_2-{\alpha }_1+{\beta }_2-{\beta }_2)\rho +({\alpha }_1{\beta }_1-{\alpha }_2{\beta }_2)+2({\beta }_2-{\beta }_1)=0.$$

(2)

Equation (2) can be rearranged as

$${\rho }^2-(k-1)\rho -({\beta }_1-k({\alpha }_2-2))=0\mathrm{or}{\rho }^2-(k-1)\rho -({\beta }_2-k({\alpha }_1-2))=0,$$

(3)

where $k={\displaystyle \frac{{\beta }_2-{\beta }_2}{{\alpha }_1-{\alpha }_2}}>0$. Using a division algorithm, we can write $P({\alpha }_i,{\beta }_i,\rho )=0$, $i=1,2$, as

$$(\rho -{\alpha }_i+k+1)({\rho }^2-(k-1)\rho -({\beta }_i-k({\alpha }_j-2)))+A\rho +B_i=0,$$

for $i,j=1,2$ and $i\ne j$, where $A=k({\alpha }_1+{\alpha }_2)-k(k+2)$ and $B=-k({\alpha }_i-2)({\alpha }_j-2)-(k-1)({\beta }_i-k({\alpha }_j-2))$. This implies that $A\rho +B_i=0$ if and only if the equations $P({\alpha }_i,{\beta }_i,\rho )=0$, $i=1,2,$ have two common roots, which are given by Equation (3).

Also, $A\rho +B_i=0$ further implies that

$$A=0⟺{\alpha }_1+{\alpha }_2=k+2$$

(4)

and

$$B=0⟺-k({\alpha }_i-2)({\alpha }_j-2)-(k-1)({\beta }_i-k({\alpha }_j-2))=0.$$

Using Equation (4), we obtain

$${\beta }_i={\displaystyle \frac{k({\alpha }_j-1)({\alpha }_j-2)}{(k-1)}}i,j=1,2(i\ne j)$$

(5)

Since ${\alpha }_i$ and ${\beta }_i$ are integers, the Equations (4) and (5) imply that k must be an integer. Additionally, $k-1$ must divide $({\alpha }_j-1)({\alpha }_j-2)$ for $j=1,2$. Now, from Equation (1), it follows that the product of the three roots must be negative and it is given that ${\alpha }_i-k-1=-{\alpha }_j+1<0$ Consequently, the quadratic Equation (3), which is a common factor of both $P({\alpha }_i,{\beta }_i,\rho )=0$ for $i=1,2$, must have two positive roots. Therefore, the common roots of both polynomials can be expressed as

$${\displaystyle \frac{(k-1)\pm \sqrt{{(k-1)}^2+4({\beta }_i-k({\alpha }_j-2))}}{2}},$$

where $i,j=1$ or 2 and $i\ne j$. □

According to Equation (4), we have ${\alpha }_1+{\alpha }_2=k+2$. Observe that if $a={\alpha }_1-{\alpha }_2$, then we obtain

$${\beta }_1={\displaystyle \frac{k(k-a)(k-a-2)}{4(k-1)}}\mathrm{and}{\beta }_2=ka-a^2+{\displaystyle \frac{k(k-a)(k-a-2)}{4(k-1)}}.$$

This implies that ${\displaystyle \frac{k(k-a)(k-a-2)}{4(k-1)}}$ must be an integer for two pineapple graphs to have two common largest roots. The following lemma provides the conditions under which this expression is an integer.

Lemma 6.

Let k and a be two integers with the same parity. Then ${\displaystyle \frac{k(k-a)(k-a-2)}{4(k-1)}}$ is an integer if and only if one of the following cases hold.

(a)

$k\equiv 0\left(mod4\right)$ and there exists an integer r such that $a^2-1=(4r+1)(k-1)$,

(b)

$k\equiv 1\left(mod4\right)$ and there exists an integer r such that $a^2-1=(4r+2)(k-1)$,

(c)

$k\equiv 2\left(mod4\right)$ and there exists an integer r such that $a^2-1=(4r-1)(k-1)$,

(d)

$k\equiv 3\left(mod4\right)$ and there exists an integer r such that $a^2-1=4r(k-1)$.

Proof. 

The quantity ${\displaystyle \frac{k(k-a)(k-a-2)}{4(k-1)}}$ is an integer if and only if $4(k-1)$ divides $k(k-a)(k-a-2)$, Now, the following cases arise.

(a).

If $k\equiv 0\left(mod4\right)$, then

$$\begin{array}{cc}\hfill k(k-a)(k-a-2)& =k\left({(k-1)}^2-2a(k-1)+a^2-1\right)\hfill \\ \hfill & \equiv a^2-k\left(mod4(k-1)\right)\hfill \\ \hfill & \equiv (a^2-1)-(k-1)\left(mod4(k-1)\right).\hfill \end{array}$$

Therefore, there exists an integer r such that

$$(a^2-1)-(k-1)=4r(k-1)\Rightarrow (a^2-1)=(4r+1)(k-1).$$

(b).

If $k\equiv 1\left(mod4\right)$, then

$$\begin{array}{cc}\hfill k(k-a)(k-a-2)& =k\left({(k-1)}^2-2a(k-1)+a^2-1\right)\hfill \\ \hfill & \equiv a^2-1-2(k-1)\left(mod4(k-1)\right).\hfill \end{array}$$

Therefore, there exists an integer r such that $a^2-1-2(k-1)=4r(k-1)$, which implies that $a^2-1=(4r+2)(k-1)$.

(c).

If $k\equiv 2\left(mod4\right)$, then

$$\begin{array}{cc}\hfill k(k-a)(k-a-2)& =k\left({(k-1)}^2-2a(k-1)+a^2-1\right)\hfill \\ \hfill & \equiv a^2-1+(k-1)\left(mod4(k-1)\right).\hfill \end{array}$$

Therefore, there exists an integer r such that $(a^2-1+(k-1))=4r(k-1)$, which implies that $a^2-1=(4r-1)(k-1)$.

(d).

If $k=4m+3$, for some integer m, then

$$\begin{array}{cc}\hfill k(k-a)(k-a-2)& =k\left({(k-1)}^2-2a(k-1)+a^2-1\right)\hfill \\ \hfill & \equiv a^2-1\left(mod4(k-1)\right).\hfill \end{array}$$

Therefore, there exists an integer r such that $a^2-1=4r(k-1)$. □

Now, we provide the necessary and sufficient conditions for two non-isomorphic pineapple graphs to have the two largest common eigenvalues.

Theorem 1.

Let ${\mathcal{P}}_1({\alpha }_1,{\beta }_1)$ and ${\mathcal{P}}_2({\alpha }_2,{\beta }_2)$ be two non-isomorphic pineapple graphs. Their two largest eigenvalues are identical if and only if ${\beta }_i={\displaystyle \frac{k({\alpha }_j-1)({\alpha }_j-2)}{k-1}}$ for $i,j=1,2$ with $i\ne j$, ${\alpha }_1+{\alpha }_2-2=k$, and $k={\displaystyle \frac{{\beta }_2-{\beta }_1}{{\alpha }_1-{\alpha }_2}}$ is an integer. Furthermore, one of the four conditions specified in Lemma 6 must be satisfied for k and $a={\alpha }_1-{\alpha }_2$. The common eigenvalues are given by

$${\displaystyle \frac{(k-1)\pm \sqrt{{(k-1)}^2+4({\beta }_i-k({\alpha }_j-2))}}{2}},$$

where $i,j=1\mathrm{or}2$ and $i\ne j$.

Proof. 

Assume that ${\alpha }_1-{\alpha }_2=a$. Then, we obtain the following expressions

$${\beta }_1={\displaystyle \frac{k(k-a)(k-a-2)}{4(k-1)}}\mathrm{and}{\beta }_2=ka-a^2+{\displaystyle \frac{k(k-a)(k-a-2)}{4(k-1)}}.$$

Clearly, $a={\alpha }_1-{\alpha }_2$ and $k={\alpha }_1+{\alpha }_2-2$ have the same parity. So, Lemma 6 confirms that ${\beta }_i={\displaystyle \frac{k({\alpha }_j-1)({\alpha }_j-2)}{(k-1)}}$ are integers if one of the four conditions in Lemma 6 holds. The rest of the proof follows that of Lemma 5. □

Now, we present the following observation.

Corollary 1.

Corresponding to two integers a and k of the same parity satisfying one of the four conditions in Lemma 6, there exist two pineapple graphs ${\mathcal{P}}_1({\alpha }_1,{\beta }_1)$ and ${\mathcal{P}}_2({\alpha }_2,{\beta }_2)$ which have the two largest common eigenvalues, where ${\alpha }_1={\displaystyle \frac{k+a+2}{2}}$, ${\alpha }_2={\displaystyle \frac{k-a+2}{2}}$ and ${\beta }_i={\displaystyle \frac{k({\alpha }_j-1)({\alpha }_j-2)}{(k-1)}}$, $i\ne j\mathit{and}i,j=1,2$.

Proof. 

Let a and k be two integers with the same parity, satisfying one of the four conditions in Lemma 6. Then, using Lemma 6, ${\alpha }_1={\displaystyle \frac{k+a+2}{2}}$, ${\alpha }_2={\displaystyle \frac{k-a+2}{2}}$ and ${\beta }_i={\displaystyle \frac{k({\alpha }_j-1)({\alpha }_j-2)}{(k-1)}}$$,i\ne j\mathrm{and}i,j=1,2$ are integers.

Now, we have ${\alpha }_1-{\alpha }_2={\displaystyle \frac{k+a+2}{2}}-{\displaystyle \frac{k-a+2}{2}}=a$ and ${\alpha }_1+{\alpha }_2={\displaystyle \frac{k+a+2}{2}}+{\displaystyle \frac{k-a+2}{2}}=k+2$. Thus, according to Theorem 1, it follows that ${\mathcal{P}}_1({\alpha }_1,{\beta }_1)$ and ${\mathcal{P}}_2({\alpha }_2,{\beta }_2)$ have the two largest common eigenvalues. □

We illustrate this with the help of the following example.

Example 3.

First choose a, say $a=8$. Then we find $a^2-1$ and factorize it to choose the value of k, keeping Lemma 6 in mind. In our case, we have $a^2-1=63=3\times 21$. Now, choose $k=22$. Then, the pair $(a,k)=(8,22)$ satisfies the third condition of Lemma 6. Now, ${\alpha }_1={\displaystyle \frac{k+a+2}{2}}={\displaystyle \frac{22+8+2}{2}}=16$, ${\alpha }_2={\displaystyle \frac{k-a+2}{2}}={\displaystyle \frac{22-8+2}{2}}=8$ and ${\beta }_1={\displaystyle \frac{k({\alpha }_2-1)({\alpha }_2-2)}{(k-1)}}={\displaystyle \frac{22\times 7\times 6}{\left(21\right)}}=44$ and ${\beta }_2={\displaystyle \frac{k({\alpha }_1-1)({\alpha }_1-2)}{(k-1)}}={\displaystyle \frac{22\times 15\times 14}{\left(21\right)}}=220$. So, ${\mathcal{P}}_1(16,44)$ and ${\mathcal{P}}_2(8,220)$ have the two largest common eigenvalues and the common eigenvalues are 5.783 and 15.217.

Up to this point, we have established the conditions under which two pineapple graphs share their two largest eigenvalues. In the following, we focus on identifying the condition necessary for two pineapple graphs to have one common eigenvalue.

Lemma 7.

Let $P({\alpha }_1,{\beta }_1,x)$ and $P({\alpha }_2,{\beta }_2,x)$ denote the two polynomials as specified previously. They share precisely one common root ρ if and only if the following conditions hold

$$\begin{array}{c}\hfill {\beta }_i=\rho (\rho +1)+{\displaystyle \frac{r({\alpha }_j-1)}{s}}-{\displaystyle \frac{r(\rho +1)}{s}},({\alpha }_1-\rho -2)({\alpha }_2-\rho -2)={\displaystyle \frac{s\rho (\rho +1)}{r}}, \mathit{and}A\ne 0,\end{array}$$

where ρ is a positive integer, $k={\displaystyle \frac{{\beta }_2-{\beta }_1}{{\alpha }_1-{\alpha }_2}}={\displaystyle \frac{r}{s}}$, r and s are coprime, and $A=k({\alpha }_1+{\alpha }_2)-k(k+2)$.

Proof. 

Proceeding as in Lemma 5, if $\rho$ is a common root of the equations $P({\alpha }_i,{\beta }_i,\rho )=0$ $(i=1,2)$, then we have $A\rho +B_i=0$, $i=1,2$, where $A=k({\alpha }_1+{\alpha }_2)-k(k+2)$, $B=-k({\alpha }_i-2)({\alpha }_j-2)-(k-1)({\beta }_i-k({\alpha }_j-2))$ and $k={\displaystyle \frac{{\beta }_2-{\beta }_1}{{\alpha }_1-{\alpha }_2}}$.

Clearly, $A=0$ gives two common roots. Therefore, for the existence of exactly one common root, we should have $A\ne 0$. This implies that $\rho =-{\displaystyle \frac{B}{A}}$, which is a rational number. But according to the rational root theorem, a rational root of a monic polynomial is an integer. So, we have $\rho =-{\displaystyle \frac{B}{A}}$, which gives

$${\displaystyle \frac{k({\alpha }_1-2)({\alpha }_2-2)+(k-1)({\beta }_i-k({\alpha }_j-2))}{k({\alpha }_1+{\alpha }_2)-k(k+2)}}=\rho \left(\mathrm{integer}\right).$$

(6)

Let $k={\displaystyle \frac{r}{s}}$ such that $gcd(r,s)=1$. Using the fact that $\rho$ is also a root of the quadratic Equation (3), we have

$$\begin{array}{cccc}\hfill & \hfill & \hfill {\displaystyle \frac{(k-1)\pm \sqrt{{(k-1)}^2+4({\beta }_i-k({\alpha }_j-2))}}{2}}& =\rho ,\hfill \\ \hfill & or\hfill & \hfill {\displaystyle \frac{(r-s)\pm \sqrt{{(r-s)}^2+4s(s{\beta }_i-r({\alpha }_j-2))}}{2s}}& =\rho ,\hfill \\ \hfill & or\hfill & \hfill \pm \sqrt{{(r-s)}^2+4s(s{\beta }_i-r({\alpha }_j-2))}& =2s\rho -(r-s).\hfill \end{array}$$

After squaring and simplification, we obtain

$$\begin{array}{cccc}\hfill & \hfill & \hfill (s{\beta }_i-r({\alpha }_j-2))& =\rho (s\rho -r+s)\hfill \\ \hfill & or\hfill & \hfill {\beta }_i& =\rho (\rho +1)+{\displaystyle \frac{r({\alpha }_j-1)}{s}}-{\displaystyle \frac{r(\rho +1)}{s}},\hfill \end{array}$$

(7)

$i,j=1,2$ and $i\ne j$. Using Equations (6) and (7), we obtain the following expression

$$rs({\alpha }_1-2)({\alpha }_2-2)+(r-s)\left(\rho (s\rho -r+s)\right)=rs({\alpha }_1+{\alpha }_2)-r(r+2s)\rho .$$

Rearranging the terms, we obtain

$$({\alpha }_1-\rho -2)({\alpha }_2-\rho -2)={\displaystyle \frac{s\rho (\rho +1)}{r}}.$$

(8)

□

Again, ${\alpha }_i$ and ${\beta }_i$, obtained in the previous lemma, are not necessarily integers. Since ${\alpha }_i$ and ${\beta }_i$ must be integers, as these are the order of the graphs, the next theorem deals with the restrictions on the parameters necessary to make ${\alpha }_i$ and ${\beta }_i$ integers.

Theorem 2.

Let ${\mathcal{P}}_1({\alpha }_1,{\beta }_1)$ and ${\mathcal{P}}_2({\alpha }_2,{\beta }_2)$ represent two non-isomorphic pineapple graphs. They share exactly one common eigenvalue if and only if the following conditions hold

$$\begin{array}{ccccccc}\hfill {\beta }_i& =\rho (\rho +1)+{\displaystyle \frac{r({\alpha }_j-1)}{s}}-{\displaystyle \frac{r(\rho +1)}{s}},\hfill & & ({\alpha }_1-\rho -2)({\alpha }_2-\rho -2)={\displaystyle \frac{s\rho (\rho +1)}{r}},\hfill & & & \hfill \mathit{and}A\ne 0,\end{array}$$

where r and s are coprime such that $k={\displaystyle \frac{r}{s}}$, r divides $\rho (\rho +1)$, and s divides $\rho +1$, ${\alpha }_i-1$, for $i=1,2$, and $A=k({\alpha }_1+{\alpha }_2)-k(k+2)$.

Proof. 

According to Lemma 7, it is sufficient to prove that ${\alpha }_i$ and ${\beta }_i$ are integers. If $\rho$ is the common eigenvalue, then proceeding as in Lemma 7, we have

$$rs({\alpha }_1-2)({\alpha }_2-2)+(r-s)(s{\beta }_i-r({\alpha }_j-2))=rs({\alpha }_1+{\alpha }_2)-r(r+2s)\rho .$$

This implies that $-r^2({\alpha }_j-1)\equiv 0\left(mods\right)$, which further implies that $s\left|\right({\alpha }_j-1)$, $j=1,2.$ These conditions, along with the expression

$${\beta }_i=\rho (\rho +1)+{\displaystyle \frac{r({\alpha }_j-1)}{s}}-{\displaystyle \frac{r(\rho +1)}{s}}$$

imply that ${\beta }_i$ is an integer if and only if s divides $\rho +1$. Now, the expression

$$({\alpha }_1-\rho -2)({\alpha }_2-\rho -2)={\displaystyle \frac{s\rho (\rho +1)}{r}}$$

implies that ${\alpha }_i$ is an integer if and only if $r\left|\rho \right(\rho +1)$, as r and s are coprime. □

Equation (8) has as many solutions as the number of different ways in which ${\displaystyle \frac{s\rho (\rho +1)}{r}}$ can be factored in to two factors. However, not all the solutions give the common spectral radius. In the next theorem, we show that only the solutions of Equation (8) can give two graphs, ${\mathcal{P}}_1({\alpha }_1,{\beta }_1)$ and ${\mathcal{P}}_2({\alpha }_2,{\beta }_2)$, with a common spectral radius, which correspond to the negative factors of ${\displaystyle \frac{s\rho (\rho +1)}{r}}$. In other words, we show that ${\alpha }_i-\rho -2<0$ or ${\alpha }_i<\rho +2$, for $i=1,2$.

Theorem 3.

Let ${\mathcal{P}}_1({\alpha }_1,{\beta }_1)$ and ${\mathcal{P}}_2({\alpha }_2,{\beta }_2)$ be two non-isomorphic pineapple graphs. They have exactly one largest common eigenvalue ρ if and only if the following conditions hold

$$({\alpha }_1-\rho -2)({\alpha }_2-\rho -2)={\displaystyle \frac{s\rho (\rho +1)}{r}}{\beta }_i=\rho (\rho +1)+{\displaystyle \frac{r({\alpha }_j-1)}{s}}-{\displaystyle \frac{r(\rho +1)}{s}}\mathit{and}A\ne 0,$$

where r and s are coprime such that $k={\displaystyle \frac{r}{s}}$ and r divides $\rho (\rho +1)$ and s divides $\rho +1$, ${\alpha }_i-1$ and ${\alpha }_i<\rho +2$$(i=1,2)$, $A=k({\alpha }_1+{\alpha }_2)-k(k+2)$.

Proof. 

Let ${\mathcal{P}}_1({\alpha }_1,{\beta }_1)$ and ${\mathcal{P}}_2({\alpha }_2,{\beta }_2)$ be two non-isomorphic pineapple graphs. Using Theorem 2, it is sufficient to prove that one common eigenvalue of both the graphs is largest if and only if ${\alpha }_i<\rho +2$$(i=1,2)$. Let $\rho$ be the common eigenvalue of the graphs ${\mathcal{P}}_1({\alpha }_1,{\beta }_1)$ and ${\mathcal{P}}_2({\alpha }_2,{\beta }_2)$. Now, transfer the polynomial in the negative direction by distance $\rho$. Substitute $(y+\rho )$ in place of x in $P({\alpha }_i,{\beta }_i,x)=0$, we obtain

$$y(y^2+(3\rho -{\alpha }_i+2)y+(3{\rho }^2-2\rho ({\alpha }_i-2)-{\alpha }_i-{\beta }_i+1))=0.$$

(9)

So, $\rho$ is the largest eigenvalue of the graphs if and only if the product of the two non-zero roots of Equation (9) is positive. Thus, we have

$$3{\rho }^2-2\rho ({\alpha }_i-2)-{\alpha }_i-{\beta }_i+1>0.$$

(10)

Now, let

$$({\alpha }_1-\rho -2)({\alpha }_2-\rho -2)={\displaystyle \frac{s\rho (\rho +1)}{r}}=m.n.$$

Then, without loss of generality, we have

$${\alpha }_1=\rho +2+m\mathrm{and}{\alpha }_2=\rho +2+n.$$

From Equation (10) for $i=1$, the graphs have a common spectral radius if and only if

$$\begin{array}{cc}\hfill 3{\rho }^2-2\rho ({\alpha }_1-2)-{\alpha }_1-{\beta }_1+1& =3{\rho }^2-2\rho (\rho +2+m)-\rho -2-m-\rho (\rho +1)\hfill \\ \hfill & +{\displaystyle \frac{r}{s}}[\rho +1-\rho +2-m+1]+1,\hfill \\ \hfill or3{\rho }^2-2\rho ({\alpha }_1-2)-{\alpha }_1-{\beta }_1+1& =-(m-4)(2\rho +{\displaystyle \frac{r}{s}})-(\rho -2+m)+1>0\hfill \end{array}$$

if and only if $m<0$, which implies that $n<0$, Hence, we have $({\alpha }_i-\rho -2)<0$, which implies that ${\alpha }_i<\rho +2(i=1,2)$. □

Now, we have the following observation.

Corollary 2.

If $\rho >2$ is any positive integer, then ρ is the spectral radius of at least one pineapple graph.

Proof. 

Let $\rho >2$ be a positive integer and $s=1$, $r={\displaystyle \frac{\rho (\rho +1)}{2}}$. Then, the equation

$$({\alpha }_1-\rho -2)({\alpha }_2-\rho -2)={\displaystyle \frac{s\rho (\rho +1)}{r}}=2$$

has a solution ${\alpha }_1-\rho -2=-1$, that is, ${\alpha }_1=\rho +1$ and ${\alpha }_1-\rho -2=-2$, that is, ${\alpha }_2=\rho$ and the corresponding ${\beta }_i^{\prime }s$ are

$${\beta }_1=\rho (\rho +1)+{\displaystyle \frac{r({\alpha }_2-1)}{s}}-{\displaystyle \frac{r(\rho +1)}{s}}=0,{\beta }_2=\rho (\rho +1)+{\displaystyle \frac{r({\alpha }_1-1)}{s}}-{\displaystyle \frac{r(\rho +1)}{s}}={\displaystyle \frac{\rho (\rho +1)}{2}}.$$

Therefore, $\mathcal{P}(\rho +1,0)$ and $\mathcal{P}(\rho ,{\displaystyle \frac{\rho (\rho +1)}{2}})$ have the same spectral radius $\rho$. Clearly, $\mathcal{P}(\rho +1,0)$ is isomorphic to the complete graph $K_{\rho +1}$ and $\mathcal{P}(\rho ,{\displaystyle \frac{\rho (\rho +1)}{2}})$ is the pineapple graph. □

Example 4.

To create an example satisfying all the conditions of Theorem 3, choose a positive integer $\rho =11$ (say). Now, choose s and r such that $s\left|\right(11+1)$ and $r\left|11\right(11+1)$. So, we choose $s=2$ and $r=11$. Now, we need to find ${\alpha }_i$ satisfying the condition $({\alpha }_1-13)({\alpha }_2-13)={\displaystyle \frac{2\times 11\times 12}{11}}=24.$ We find factors of 24, which are multiples of s. Such a pair of factors are $(2,12)$, $(-2,-12)$, $(4,6)$ and $(-4,-6)$. First, we choose $({\alpha }_1-13)({\alpha }_2-13)=4\times 6$, and we obtain ${\alpha }_1=13+4=17$, ${\alpha }_2=13+6=19$, ${\beta }_1=165$ and ${\beta }_2=154$. We can check that $\mathcal{P}(17,165)$ and $\mathcal{P}(19,154)$ have a common eigenvalue of 11. However, it is not the spectral radius of any graph. Now, we choose the pair of factors $(-4,-6)$. Then, we obtain ${\alpha }_1=7$ ${\alpha }_2=9$, ${\beta }_1=110$ and ${\beta }_2=99$. In this case $11>{\displaystyle \frac{({\alpha }_1-2)+\sqrt{{({\alpha }_1-2)}^2+3({\alpha }_1+{\beta }_1-1)}}{3}}\approx 8.74$ and $11>{\displaystyle \frac{({\alpha }_2-2)+\sqrt{{({\alpha }_2-2)}^2+3({\alpha }_2+{\beta }_2-1)}}{3}}\approx 8.10$. Hence, we obtain the common spectral radius of 11 for $\mathcal{P}(7,110)$ and $\mathcal{P}(9,99)$.

Now, we have the following result.

Theorem 4.

If $\beta \ge 8$, then $c\left(\mathcal{P}\right(3,\beta \left)\right)=2\beta +1$.

Proof. 

Any induced subgraph of $\mathcal{P}(3,\beta )$ is a pineapple graph with clique number 3 or a star graph $S_{1,n}(n\le \beta +1)$ or $K_3$. The star graph $S_{1,n+1}$ is isomorphic to $\mathcal{P}(2,n)$ and $K_3$ is isomorphic to $\mathcal{P}(3,0)$. In other words, we can say that the set of the non-empty induced subgraphs of $\mathcal{P}(3,\beta )$ is given by $\mathcal{S}\left(\mathcal{P}\right(3,\beta \left)\right)=\left\{\mathcal{P}\right(i,j\left)\right|(2\le i\le 3,0\le j\le \beta )\}.$ Let $\mathcal{P}({\alpha }_1,{\beta }_1)$, $\mathcal{P}({\alpha }_2,{\beta }_2)$$\in \mathcal{S}\left(\mathcal{P}\right(3,\beta \left)\right)$ have the same spectral radius. Then, without loss of generality, we have ${\alpha }_2-{\alpha }_1=1$. We leave the possibility ${\alpha }_2-{\alpha }_1=0$, since in that case one of the graphs becomes the induced subgraph of the other (see Lemma 3).

Case 1. Suppose that both the graphs have two common eigenvalues. Since ${\alpha }_2=3$ and ${\alpha }_1=2$, we have $k=2+3-2=3$. According to Corollary 1, we find only one pair of two graphs with two common eigenvalues, which are $\mathcal{P}(2,3)$ and $\mathcal{P}(3,0)$.

Case 2. Suppose that both the graphs have only one eigenvalue in common. Then, ${\alpha }_1$ and ${\alpha }_2$ must satisfy

$$({\alpha }_1-\rho -2)({\alpha }_2-\rho -2)={\displaystyle \frac{s\rho (\rho +1)}{r}},$$

which implies that

$$(2-\rho -2)(3-\rho -2)={\displaystyle \frac{s\rho (\rho +1)}{r}},$$

which gives $\rho =0,{\displaystyle \frac{r+s}{r-s}}$. So, now we investigate the existence of pineapple graphs with a radius ${\displaystyle \frac{r+s}{r-s}}$.

Given that $s|{\alpha }_i\Rightarrow s=1$, it follows that $\rho =(r+1)/(r-1)$. As $\rho$ is an integer, $(r-1)\left|\right(r+1)$, which is possible only when $r=2$ or 3.

If $r=2$, then $\rho =3$. Accordingly, Theorem 3 provides one pair of pineapple graphs with a radius $\rho =3$, namely $\mathcal{P}(2,8)$ and $\mathcal{P}(3,6)$.

If $r=3$, then $\rho =2$, which implies $A=0$. Consequently, no such pair exists in this case. This, combined with Lemma 2, completes the proof. □

Theorem 5.

(a) For fixed $\alpha \ge 2$, we have $\underset{\beta \to \infty }{lim}r\left(\mathcal{P}(\alpha ,\beta )\right)=1.$; (b) For fixed $\beta \ge 0$, we have $\underset{\alpha \to \infty }{lim}r\left(\mathcal{P}(\alpha ,\beta )\right)=1$.

Proof. 

(a) For fixed $\alpha \ge 2$, let $\mathcal{S}\left(\mathcal{P}\right(\alpha ,\beta \left)\right)$ be the set of the connected induced subgraphs of $\mathcal{P}(\alpha ,\beta )$. If any two graphs in $\mathcal{S}\left(\mathcal{P}\right(\alpha ,\beta \left)\right)$ have the common spectral radius, then either they have two common eigenvalues or they have only one common eigenvalue.

Case 1. When the graphs $\mathcal{P}({\alpha }_1,{\beta }_1)$ and $\mathcal{P}({\alpha }_2,{\beta }_2)$ have two common eigenvalues, then ${\beta }_i={\displaystyle \frac{k({\alpha }_j-1)({\alpha }_j-2)}{(k-1)}}$ is fixed for fixed ${\alpha }_1$ and ${\alpha }_2$. This implies that there are finite number of possible pairs of graphs, $\mathcal{P}({\alpha }_1,{\beta }_1)$ and $\mathcal{P}({\alpha }_2,{\beta }_2)$, with two common eigenvalues.

Case 2. The graphs $\mathcal{P}({\alpha }_1,{\beta }_1)$ and $\mathcal{P}({\alpha }_2,{\beta }_2)$ have exactly one common eigenvalue, that is, the spectral radius. In this case, they have the common spectral radius $\rho$, satisfying the equation

$$({\alpha }_1-\rho -2)({\alpha }_2-\rho -2)={\displaystyle \frac{s\rho (\rho +1)}{r}}.$$

(11)

If ${\displaystyle \frac{s}{r}}\ge 1$, then one of the negative factors of ${\displaystyle \frac{s\rho (\rho +1)}{r}}$ will be lesser than $-(\rho +1)$. Without loss of generality, we have ${\alpha }_1-\rho -2\le -\rho -1$. This gives ${\alpha }_1\le 1$, which is a contradiction.

If ${\displaystyle \frac{s}{r}}<1$, then Equation (11) can be written as

$$\left(1-{\displaystyle \frac{s}{r}}\right){\rho }^2-\rho \left({\alpha }_1+{\alpha }_2-4-{\displaystyle \frac{s}{r}}\right)+{\alpha }_1{\alpha }_2-2({\alpha }_1+{\alpha }_2)+4=0.$$

If ${\rho }_1$ and ${\rho }_2$ are the two roots of this quadratic equation, then

$$\begin{array}{cc}\hfill {\rho }_1+{\rho }_2& ={\displaystyle \frac{{\alpha }_1+{\alpha }_2+{\displaystyle \frac{s}{r}}-4}{1-{\displaystyle \frac{s}{r}}}}=({\alpha }_1+{\alpha }_2-4)+{\displaystyle \frac{s({\alpha }_1+{\alpha }_2-3)}{(r-s)}}\hfill \\ \hfill & \le ({\alpha }_1+{\alpha }_2-3)(s+1)-1\le \alpha (2\alpha -3)-1.\hfill \end{array}$$

This means that, for fixed $\alpha$, the spectral radius $\rho$ is bounded above, so are ${\beta }_i$, $i=1,2$ as

$${\beta }_i=\rho (\rho +1)+{\displaystyle \frac{r({\alpha }_j-1)}{s}}-{\displaystyle \frac{r(\rho +1)}{s}}=\rho (\rho +1)+{\displaystyle \frac{r({\alpha }_j-\rho -2)}{s}}\le \rho (\rho +1).$$

The last inequality is due to the fact that ${\alpha }_j-\rho -2<0$. This implies that there are finite number of possible pairs of graphs $\mathcal{P}({\alpha }_1,{\beta }_1)$ and $\mathcal{P}({\alpha }_2,{\beta }_2)$ with the common spectral radius. Thus, there exists ${\beta }^{*}\left(\alpha \right)$, depending on $\alpha$, such that the quantity $b\left(\mathcal{P}\right(\alpha ,\beta \left)\right)-c\left(\mathcal{P}\right(\alpha ,\beta \left)\right)$ is constant for all $\beta \ge {\beta }^{*}\left(\alpha \right)$. This gives

$$\underset{\beta \to \infty }{lim}r\left(\mathcal{P}(\alpha ,\beta )\right)=\underset{\beta \to \infty }{lim}\left(1+{\displaystyle \frac{b\left(\mathcal{P}\right(\alpha ,\beta \left)\right)-c\left(\mathcal{P}\right(\alpha ,\beta \left)\right)}{c\left(\mathcal{P}\right(\alpha ,\beta \left)\right)}}\right)=1$$

as $c\left(\mathcal{P}\right(\alpha ,\beta \left)\right)$ increases unboundedly when $\beta \to \infty$.

(b) Similarly, for fixed $\beta \ge 0$, we have the following cases.

Case 1. When the graphs $\mathcal{P}({\alpha }_1,{\beta }_1)$ and $\mathcal{P}({\alpha }_2,{\beta }_2)$ have two common eigenvalues. Then, we have ${\alpha }_1+{\alpha }_2-2=k={\displaystyle \frac{{\beta }_2-{\beta }_1}{{\alpha }_1-{\alpha }_2}}\le \beta$. This gives ${\alpha }_i\le \beta +2$, which implies that there are finite number of pairs of graphs $\mathcal{P}({\alpha }_1,{\beta }_1)$ and $\mathcal{P}({\alpha }_2,{\beta }_2)$ with two common eigenvalues.

Case 2. When the graphs $\mathcal{P}({\alpha }_1,{\beta }_1)$ and $\mathcal{P}({\alpha }_2,{\beta }_2)$ have exactly one common eigenvalue, that is, the spectral radius. In this case, we have $r\le {\displaystyle \frac{{\beta }_2-{\beta }_1}{{\alpha }_1-{\alpha }_2}}s\le \beta$,

$$\begin{array}{cc}\hfill {\beta }_i& =\rho (\rho +1)+{\displaystyle \frac{r({\alpha }_j-1)}{s}}-{\displaystyle \frac{r(\rho +1)}{s}}\ge \rho (\rho +1)-{\displaystyle \frac{r(\rho +1)}{s}}\hfill \\ \hfill & \ge \rho (\rho +1)-\beta (\rho +1).\hfill \end{array}$$

This implies that $\beta \ge \rho (\rho +1)-\beta (\rho +1)$, which further gives $\beta \ge {\displaystyle \frac{\rho (\rho +1)}{(\rho +2)}}\ge \rho -1.$ This implies that the spectral radius is bounded above; so is ${\alpha }_i$, as ${\alpha }_i<\rho -2$. Therefore, there are finite pairs of graphs with a common spectral radius. So, there exists ${\alpha }^{*}\left(\beta \right)$, depending on $\beta$, such that the quantity $b\left(\mathcal{P}\right(\alpha ,\beta \left)\right)-c\left(\mathcal{P}\right(\alpha ,\beta \left)\right)$ is constant for all $\alpha \ge {\alpha }^{*}\left(\beta \right)$. This gives

$$\underset{\alpha \to \infty }{lim}r\left(\mathcal{P}(\alpha ,\beta )\right)=\underset{\alpha \to \infty }{lim}\left(1+{\displaystyle \frac{b\left(\mathcal{P}\right(\alpha ,\beta \left)\right)-c\left(\mathcal{P}\right(\alpha ,\beta \left)\right)}{c\left(\mathcal{P}\right(\alpha ,\beta \left)\right)}}\right)=1$$

as $c\left(\mathcal{P}\right(\alpha ,\beta \left)\right)$ increases unboundedly when $\beta \to \infty$. □

3. Conclusions and Discussion

The study of graph spectra, particularly the spectral radius, holds significant importance in both theoretical and applied graph theory. The eigenvalues of a graph, derived from its adjacency matrix, offer valuable insights into its structural features, including connectivity, stability, and expansion. Spectra also serve as a crucial tool for distinguishing non-isomorphic graphs, aiding in the classification and analysis of complex networks. This makes spectral analysis a key aspect of understanding the behavior and properties of graphs across various disciplines.

This study investigates the spectral properties of pineapple graphs, specifically focusing on their spectral redundancy. By examining the spectral characteristics of these graphs, we established the criteria under which two non-isomorphic pineapple graphs can share common eigenvalues. The primary results indicate that, given certain conditions, the largest eigenvalues of pineapple graphs $\mathcal{P}({\alpha }_1,{\beta }_1)$ and $\mathcal{P}({\alpha }_2,{\beta }_2)$ can indeed be identical, despite the graphs themselves being non-isomorphic.

Merajuddin et al. [9] studied a collection of graphs that were non-redundant spectrally, meaning no two induced subgraphs of a graph in the family share the same spectral radius. Consequently, all graphs in the family have distinct spectral radii. Such a property makes it possible to distinguish each graph solely based on its spectral radius, with a redundancy index $r\left(\mathcal{G}\right)=1$. In contrast, our study has shown that the pineapple family of graphs exhibits spectral redundancy, meaning some graphs within the family share common eigenvalues, including the spectral radius. Despite this redundancy, the pineapple family still maintains a redundancy index of $r\left(\mathcal{G}\right)=1$ under specific conditions.

Building on these findings, a potential future direction could involve studying families of graphs that are spectrally redundant but exhibit a finite spectral redundancy index. Such investigations could have valuable applications in fields such as network theory, chemistry, and physics, where distinguishing graph structures based on their spectral properties plays a critical role in modeling and analysis.