Abstract
Let represent the family of square power graphs of order and size , obtained from the family of graphs of order and size , with . In this paper, we discussed the least eigenvalue of graph in the family . All graphs considered here are undirected, simple, connected, and not a complete for positive integer .
1. Introduction
Let be a simple graph with vertex set and edge set . The adjacency matrix of a graph is defined to be the matrix of order , where if is adjacent to , otherwise . As the graphs are undirected, is real and symmetric and the eigenvalues are also real and can be arranged as . The eigenvalues of are referred to as the eigenvalues of , and one can easily find that is the largest absolute eigenvalue among all eigenvalues of .
A graph is called minimizing in a certain graph class if its least eigenvalue attains the minimum among all the least eigenvalues of the graphs of that class. We will denote the least eigenvalue of a graph by and for the corresponding least eigenvector. A graph is said to be a square power graph denoted by of order , if and , where denotes the distance between vertices and in ; also, the bipartite square power graph is denoted by , where and are the partitions of the vertex set .
We limit our discussion to only connected graphs because we know that the spectrum (the set of eigenvalues of a graph) of a disconnected graph is the union of spectra of its connected components and so the least eigenvalue of a disconnected graph will be the minimum eigenvalue of its component among all the minimum eigenvalues of its components.
There are many results in the literature regarding the least eigenvalue of a graph. Fan et al. [1] discussed the least eigenvalue of complement of trees. Hoffman [2] discussed the limiting point of the least eigenvalue of connected graphs. Hong and Shu [3] discussed the sharp lower bounds of the least eigenvalue of planar graphs. Javid [4, 5] discussed the minimizing graphs of the connected graphs whose complements are bicyclic (with two cycles) and explained the characterization of the minimizing graph of the connected graphs whose complement is bicyclic.
Theorem 1. (see [6]). For a graph of order ,with equality iff .
According to Powers et al. [7], if is the size of a graph , then
Theorem 2. (see [8]). If is a graph whose least eigenvalue is minimal among all the connected graphs, then(i) is either a bipartite graph, or(ii) is a join of two nested split graphs
Theorem 3. (see [9]). Let denote the family of all connected bipartite graphs of order and size . Let be the graph such that its least eigenvalue is minimal among all the connected bipartite graphs, then must be a double nested graph.
2. Main Results
We begin with some definitions. Given a graph of order , a vector is defined on if there exists an injective function from to the entries of , simply written as for each . If is an eigenvector of , then it is naturally defined on and every entry of corresponds to the vertex . One can easily prove thatand is an eigenvalue of corresponding to the eigenvector if and only if , and is called the open neighborhood of in :
Equation (4) is called the eigen equation for the graph . Now, for any arbitrary unit vector ,and it will be equal iff is a least eigenvector of . Now, let denote the complement of a graph . One can easily prove that , where and denote the identity matrix and the ones matrix of the same size as . Now, for any vector ,
A graph is said to be a tree iff there is a path between any pair of vertices and has no cycle, and a tree is said to be a star iff there exists a vertex of degree and all other vertices have degree one. The former is called the central vertex, and the later is called the pendant vertex. If a graph is a path of order less than or equal to 4 then . If a graph contains as an induced subgraph, then . If a graph is complete or its complement is a null graph or union of disjoint complete graph, then . So, we are limiting our discussion to , where denotes all those graphs whose is .
Special tree graph denoted by is obtained from two stars and of orders and , respectively, by joining any two vertices of degree one from the two stars. is a square power graph of this tree and is its complement graph, as shown in Figure 1; the order of is , and its size is .
Lemma 1. Given , for any positive integers such that and ,with equality iff and .
Proof. For , let be the least eigenvalue of and be the corresponding least eigenvector. Let denote the entries of corresponding to the vertices of . Now, by eigen equation, all the vertices have the same values, say and denote the entry of corresponding to the vertex , for the vertex , and all the vertices have the same value . represents the entry of corresponding to the vertex , and for the vertex , the entry of is . Now, . So, by an eigen equation,Transform (8) into a matrix equation , where and the matrix of order 6 is omitted. We getNote that is a least root of . In addition, as , which implies .
If , by the above discussion, for , we haveIn particular, , which impliesThis complete the proof.
Considering the family of square power graphs of order and size obtained from the family of graphs of order and size with , we have the following result.
Lemma 2. Let be a graph in and let be its least eigenvector, then contains at least two positive and two negative entries.
Before we state the other results, we need some definitions.
Let be a bipartite graph with colour classes and such that ; is said to be a double nested graph if and such that each is adjacent to every vertex in , for .
Now, as we know that, from [9], if two vertices and belong to the same colour class and , then , where is the entry in eigenvector corresponding to the vertex .
Let and be the colour classes such that and , where and for are the entries of the corresponding eigenvector of .
Lemma 3. (see [9]). Let be a graph with the above assumptions, then(i)The vertices and are adjacent(ii)The degree of and is complete(iii)If the vertex is adjacent to , then is adjacent to for , and if the vertex is adjacent to , then is adjacent to for .
Observation 1. If is a bipartite graph of order and size , then goes to its lower bound by increasing the size of .
Bell et al. [9] discussed the behavior of maximum eigenvalue by increasing the number of edges and kept fixed. We here show the behaviour of least eigenvalue of by fixing the number of edges at and increasing the number of vertices. In Figure 2, the horizontal axes show vertices and vertical axes represent least eigenvalues.
Theorem 4. Let be a graph of order from the family . Then,
Proof. As we know thatby Theorems 2 and 3, it is found that is a largest double nested graph. Let be the graph such that is the least one among all the square power graphs and let be the least eigenvector. Now, let be the graph obtained from by rotating the edge to such that for , then and similarly if and belong to the same colour class such that , then deg (s) deg (t). So, by Lemma 2.3, it will be a double nested graph. Now, we need to show that it is the largest possible double nested graph. The size of is . So, all we need to show is the size cannot exceed . On the contrary, let . Then, either or . In both cases, the connected square power graph is not possible because if we take with the size greater than , then will be a disconnected graph, e.g., for which is not allowed. So, this completes the proof [10].
Data Availability
The data used to support the findings of this study are cited at relevant places within the text as references.
Conflicts of Interest
The authors declare that they have no conflicts of interest.