1. Introduction
Suppose that
A is an
n by the
n real matrix. A scalar
is a complementarity eigenvalue of
A if there is a non-zero vector
satisfying
and
, where
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
. 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
be a simple connected graph, where
is the vertex set and
is the edge set. The graph
G has a symmetric adjacency matrix
of order
n, where
The spectral radius of a graph
G is the largest eigenvalue of the characteristic polynomial of the adjacency matrix
. In the work by Fernandes et al. [
2], the concept of the complementary spectrum of
G is introduced. The notation
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
, where
represents the set of all such subgraphs. Based on this, Fernandes et al. [
2] established the following inequalities, where
refers to the total number of non-isomorphic induced subgraphs and
represents the size of the set
:
While the upper bound
is not sharp, the authors in [
2] demonstrated that it increases more rapidly than any polynomial function of
n. The lower bound
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
. To quantify this overlap, Seeger [
3] introduced the concept of
spectral redundancy for a graph
G. It is given by the following
When we have a collection of connected graphs, the spectral redundancy index is naturally given by .
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
is the coalescence of the star
at the vertex of degree
with the complete graph
at any vertex. For instance,
Figure 1 illustrates the pineapple graph
, which is derived from combining
and
. The star graph
and the complete graph
are both special cases of the pineapple graph when
and
, 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
, all induced subgraphs with at least one pendant vertex share the same unique groupie [
15].
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 is the star graph with pendant vertices and the pineapple graph is the complete graph isomorphic to . The following lemma regards the spectral radius of the pineapple graph with and .
Lemma 1 ([
17])
. If is the pineapple graph, then the characteristic equation of the adjacency matrix of is given by As the spectral radius of a connected graph is always positive, so the spectral radius of the pineapple graph
is the largest root of the equation
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 .
Lemma 2. For all and , we have
Proof. Clearly, an induced subgraph of is also a pineapple graph (considering complete graphs and star graphs as pineapple graphs) with a lesser number of vertices. There are exactly induced subgraphs with a fixed clique number of k. All of them are non-isomorphic. Therefore, the number of non-empty connected induced subgraphs of is . There is an induced subgraph with one vertex as well. So, . □
Example 1. For instance, consider the pineapple graph . According to the lemma, the number of connected induced subgraphs is calculated as . Figure 2 illustrates all these induced subgraphs, excluding the trivial graph with a single vertex. 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 . Now, we present the necessary condition for two pineapple graphs to have the same spectral radius.
Lemma 4. If and are two non-isomorphic pineapple graphs with the same spectral radius, then .
Proof. Let and be two non-isomorphic pineapple graphs, such that . Without loss of generality, let us assume , and on the contrary, let us also assume . This leads to and , thereby implying that is isomorphic to some induced subgraph of , or both are isomorphic. Hence, utilizing Lemma 3, we deduce , which is a contradiction. Therefore, two non-isomorphic pineapple graphs and have the same spectral radius then . □
To illustrate the application of this lemma, consider the following example.
Example 2. The pineapple graphs and share a common spectral radius of 11. The expression evaluates to , 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 and be two polynomials defined by the Equation (1). The polynomials and have two common roots if and only if for and , and , where . Additionally, k must be an integer and divides . The common roots are given bywhere or 2 and . Proof. Let
be the common root of the equations
and
, Then, we have
and
. After subtracting these equations, we obtain
Equation (
2) can be rearranged as
where
. Using a division algorithm, we can write
,
, as
for
and
, where
and
. This implies that
if and only if the equations
,
have two common roots, which are given by Equation (
3).
Also,
further implies that
and
Using Equation (
4), we obtain
Since
and
are integers, the Equations (
4) and (
5) imply that
k must be an integer. Additionally,
must divide
for
. Now, from Equation (
1), it follows that the product of the three roots must be negative and it is given that
Consequently, the quadratic Equation (
3), which is a common factor of both
for
, must have two positive roots. Therefore, the common roots of both polynomials can be expressed as
where
or 2 and
. □
According to Equation (
4), we have
. Observe that if
, then we obtain
This implies that 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 is an integer if and only if one of the following cases hold.
- (a)
and there exists an integer r such that ,
- (b)
and there exists an integer r such that ,
- (c)
and there exists an integer r such that ,
- (d)
and there exists an integer r such that .
Proof. The quantity is an integer if and only if divides , Now, the following cases arise.
- (a).
If
, then
Therefore, there exists an integer
r such that
- (b).
If
, then
Therefore, there exists an integer r such that , which implies that .
- (c).
If
, then
Therefore, there exists an integer r such that , which implies that .
- (d).
If
, for some integer
m, then
Therefore, there exists an integer r such that . □
Now, we provide the necessary and sufficient conditions for two non-isomorphic pineapple graphs to have the two largest common eigenvalues.
Theorem 1. Let and be two non-isomorphic pineapple graphs. Their two largest eigenvalues are identical if and only if for with , , and is an integer. Furthermore, one of the four conditions specified in Lemma 6 must be satisfied for k and . The common eigenvalues are given bywhere and . Proof. Assume that
. Then, we obtain the following expressions
Clearly,
and
have the same parity. So, Lemma 6 confirms that
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 and which have the two largest common eigenvalues, where , and , .
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, , and are integers.
Now, we have and . Thus, according to Theorem 1, it follows that and have the two largest common eigenvalues. □
We illustrate this with the help of the following example.
Example 3. First choose a, say . Then we find and factorize it to choose the value of k, keeping Lemma 6 in mind. In our case, we have . Now, choose . Then, the pair satisfies the third condition of Lemma 6. Now, , and and . So, and 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 and denote the two polynomials as specified previously. They share precisely one common root ρ if and only if the following conditions holdwhere ρ is a positive integer, , r and s are coprime, and . Proof. Proceeding as in Lemma 5, if is a common root of the equations , then we have , , where , and .
Clearly,
gives two common roots. Therefore, for the existence of exactly one common root, we should have
. This implies that
, 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
, which gives
Let
such that
. Using the fact that
is also a root of the quadratic Equation (
3), we have
After squaring and simplification, we obtain
and
. Using Equations (
6) and (
7), we obtain the following expression
Rearranging the terms, we obtain
□
Again, and , obtained in the previous lemma, are not necessarily integers. Since and must be integers, as these are the order of the graphs, the next theorem deals with the restrictions on the parameters necessary to make and integers.
Theorem 2. Let and represent two non-isomorphic pineapple graphs. They share exactly one common eigenvalue if and only if the following conditions holdwhere r and s are coprime such that , r divides , and s divides , , for , and . Proof. According to Lemma 7, it is sufficient to prove that
and
are integers. If
is the common eigenvalue, then proceeding as in Lemma 7, we have
This implies that
, which further implies that
,
These conditions, along with the expression
imply that
is an integer if and only if
s divides
. Now, the expression
implies that
is an integer if and only if
, as
r and
s are coprime. □
Equation (
8) has as many solutions as the number of different ways in which
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,
and
, with a common spectral radius, which correspond to the negative factors of
. In other words, we show that
or
, for
.
Theorem 3. Let and be two non-isomorphic pineapple graphs. They have exactly one largest common eigenvalue ρ if and only if the following conditions holdwhere r and s are coprime such that and r divides and s divides , and , . Proof. Let
and
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
. Let
be the common eigenvalue of the graphs
and
. Now, transfer the polynomial in the negative direction by distance
. Substitute
in place of
x in
, we obtain
So,
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
Then, without loss of generality, we have
From Equation (
10) for
, the graphs have a common spectral radius if and only if
if and only if
, which implies that
, Hence, we have
, which implies that
. □
Now, we have the following observation.
Corollary 2. If is any positive integer, then ρ is the spectral radius of at least one pineapple graph.
Proof. Let
be a positive integer and
,
. Then, the equation
has a solution
, that is,
and
, that is,
and the corresponding
are
Therefore, and have the same spectral radius . Clearly, is isomorphic to the complete graph and is the pineapple graph. □
Example 4. To create an example satisfying all the conditions of Theorem 3, choose a positive integer (say). Now, choose s and r such that and . So, we choose and . Now, we need to find satisfying the condition We find factors of 24, which are multiples of s. Such a pair of factors are , , and . First, we choose , and we obtain , , and . We can check that and have a common eigenvalue of 11. However, it is not the spectral radius of any graph. Now, we choose the pair of factors . Then, we obtain , and . In this case and . Hence, we obtain the common spectral radius of 11 for and .
Now, we have the following result.
Theorem 4. If , then .
Proof. Any induced subgraph of is a pineapple graph with clique number 3 or a star graph or . The star graph is isomorphic to and is isomorphic to . In other words, we can say that the set of the non-empty induced subgraphs of is given by Let , have the same spectral radius. Then, without loss of generality, we have . We leave the possibility , 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 and , we have . According to Corollary 1, we find only one pair of two graphs with two common eigenvalues, which are and .
Case 2. Suppose that both the graphs have only one eigenvalue in common. Then,
and
must satisfy
which implies that
which gives
. So, now we investigate the existence of pineapple graphs with a radius
.
Given that , it follows that . As is an integer, , which is possible only when or 3.
If , then . Accordingly, Theorem 3 provides one pair of pineapple graphs with a radius , namely and .
If , then , which implies . Consequently, no such pair exists in this case. This, combined with Lemma 2, completes the proof. □
Theorem 5. (a) For fixed , we have ; (b) For fixed , we have .
Proof. (a) For fixed , let be the set of the connected induced subgraphs of . If any two graphs in have the common spectral radius, then either they have two common eigenvalues or they have only one common eigenvalue.
Case 1. When the graphs and have two common eigenvalues, then is fixed for fixed and . This implies that there are finite number of possible pairs of graphs, and , with two common eigenvalues.
Case 2. The graphs
and
have exactly one common eigenvalue, that is, the spectral radius. In this case, they have the common spectral radius
, satisfying the equation
If , then one of the negative factors of will be lesser than . Without loss of generality, we have . This gives , which is a contradiction.
If
, then Equation (
11) can be written as
If
and
are the two roots of this quadratic equation, then
This means that, for fixed
, the spectral radius
is bounded above, so are
,
as
The last inequality is due to the fact that
. This implies that there are finite number of possible pairs of graphs
and
with the common spectral radius. Thus, there exists
, depending on
, such that the quantity
is constant for all
. This gives
as
increases unboundedly when
.
(b) Similarly, for fixed , we have the following cases.
Case 1. When the graphs and have two common eigenvalues. Then, we have . This gives , which implies that there are finite number of pairs of graphs and with two common eigenvalues.
Case 2. When the graphs
and
have exactly one common eigenvalue, that is, the spectral radius. In this case, we have
,
This implies that
, which further gives
This implies that the spectral radius is bounded above; so is
, as
. Therefore, there are finite pairs of graphs with a common spectral radius. So, there exists
, depending on
, such that the quantity
is constant for all
. This gives
as
increases unboundedly when
. □