TheOptimal GraphWhose Least Eigenvalue isMinimal amongAll Graphs via 1-2 Adjacency Matrix

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                  </jats:inline-formula>. Here in this paper, we discussed new type of adjacency matrix known by 1-2 adjacency matrix defined as <jats:inline-formula>
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                  </jats:inline-formula>, from eigenvalues of the graph, we mean eigenvalues of the 1-2 adjacency matrix. Let <jats:inline-formula>
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                  </jats:inline-formula> be the set of the complement of trees of order <jats:inline-formula>
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                  </jats:inline-formula>. In this paper, we characterized a unique graph whose least eigenvalue is minimal among all the graphs in <jats:inline-formula>
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Introduction
Let G � (V, E) be a simple graph with vertex set V � V(G) and edge set E � E(G). Its order is |V(G)|, denoted by n, and its size is |E(G)|, denoted by m. e distance between two vertices u and v of a graph G is the length of a shortest path between them. e 1-2 adjacency matrix of G is defined to be the matrix A 1,2 (G) � [a ij ] of order n with a ij � 1 if 1 ≤ d(v i , v j ) ≤ 2 and a ij is zero for the rest cases. e solution of det(xI − A 1,2 (G)) is the eigenvalues of G. Since A 1,2 (G) is always symmetric and real, all the eigenvalues can be arranged as λ 1 (G) ≤ λ 2 (G) ≤ . . . ≤ λ n (G). We write λ min (G) for the least eigenvalue of G. One can find that λ n (G) is exactly the spectral radius of G. e least eigenvalue of any graph is nonpositive. For only disconnected graph, it is equal to zero. Otherwise, for graphs with at least one edge, it is less than or equal to −1 (by the interlacing theorem, see [1], p-19); it is equal to −1 if and only if G is a disjoint union of complete graphs or equivalently G c is a null graph or a complete multipartite graph. If G contains K 1,2 as an induced subgraph, then (again by the interlacing theorem). For 1 − 2 adjacency matrix, λ min (K 1,2 ) � −1, and if a graph G containing P 5 is an induced subgraph, then λ min (G) ≤ 1.
ere are many results in the literature concerning the largest eigenvalue (spectral radius) of simple graph; see, e.g., [2] or [1]. Javaid examined different families of graphs to pick optimal graph with respect to least eigenvalues via usual adjacency matrix in their respective graph classes in [3][4][5]. Lubna et al. examined square power graph of G for their least eigenvalue [6]. Recently, the problem of minimizing the least eigenvalue of graphs subject to one or more parameters has received more attention. Bell et al. [7,8] discussed the least eigenvalue of connected graphs with prescribed order and size. Fan et al. [9] discussed the least eigenvalue of complement of trees. Liu et al. [10] discussed the least eigenvalue of unicyclic graphs with pendant vertices. Fan et al. [9] discussed the least eigenvalue of unicyclic graphs with given girth. Petrovic et al. [11] discussed the least eigenvalue of bicyclic graphs.
In a family of graphs, a graph is called minimizing if the least eigenvalue of its adjacency matrix is minimum in the set of the least eigenvalues of all the graphs. Denote by T n , T c n the set of trees of order n and the set of the complements of trees order n, respectively. In this paper, we determine the unique minimizing graph in T c n for n ≥ 6, which is not the complement of the star graph, and we are limiting our discussion to the set T c n − K c n where K n c denotes all those graphs whose λ min is −1.
Theorem 1 (see [7]). If G is a graph whose least eigenvalue is minimal among all the connected graphs, then (i) G is either a bipartite graph (ii) G is a join of two nested split graphs Theorem 2 (see [12]). Let K(n, m) denote the family of all connected bipartite graphs and let G ∈ K(n, m) be the graph such that its least eigenvalue is minimal among all the connected bipartite graphs, then G must be a double nested graph.

Main Results
We begin with some definitions. Given a graph G of order n, a vector X ∈ R n is called defined on G if there exists an injective function f from V(G) to the entries of X; simply written , then it is naturally defined on V(G) and every entry of X corresponds to the vertex v ∈ V(G). One can easily prove that and λ is an eigenvalue of G corresponding to the eigenvector X iff X ≠ 0 and where N G (u) denotes the neighbourhood of u in the graph G. Equation (2) is called the eigen equation for the graph G. Now, for any arbitrary unit vector X ∈ R n , and it will be equal if and only if X is a least eigenvector of G. Now, let G c denote the complement of a graph G. One can easily proof that where I and J denote the identity matrix and ones matrix of the same size as A 1,2 (G). Now, for any vector X, A graph G is said to be a tree if and only if there is a path between any pair of vertices and has no cycle. A tree is said to be a star if and only if there exists a vertex of degree n − 1 and all other vertices have degree 1. We introduce a special tree, denoted by T(p, q), which is obtained from two disjoint stars K 1,p (p ≥ 1) and K 1,q (q ≥ 1) by joining one pendant vertex of K 1,p and one pendant vertex of K 1,q . e complement of T(p, q) is G c � (T(p, q)) c , as shown in Figure 1.
2 Journal of Mathematics Lemma 2. Let T c be a graph in T n c − K n c and let X be the least eigenvector of T, then X contains at least two positive and two negative entries.
Proof. Assume on contrary that X contains exactly one positive entry corresponding to the vertex w, i.e., X w > 0. Since T is not a star, there exists a vertex u ∈ N T (w) such that N T (u)∖w ≠ 0. Considering the eigen equation (2) on the vertex u for the graph T c , we have is implies that X u � 0 and X v � 0 for each v ∈ N c T (u), now as Now, let v 0 ∈ N T (u), w. en, by (2) and (11), is implies X v 0 � X w . Since λ 1 (T c ) < 1 as T is not a star, a contradiction, so our supposition is wrong and hence X contains at least two positive entries. Similarly, if we consider −X, we also get X which contains at least two negative entries. e result follows. □ Before we state other results, we need some definitions. Let G be a bipartite graph with colour classes U and V such that is adjacent to each and every vertex in V, and similarly, each u i ∈ U 2 is adjacent to each and every vertex in }. Now, as we know from [12], if the two vertices u and w belong to the same colour class and x u ≥ x w , then deg(u) ≥ deg(w).
Let U � u 1 , u 2 , . . . , u k and V � v 1 , v 2 , . . . , v k be the colour classes such that } are the entries of the corresponding eigenvector X of G.
Lemma 3 (see [12]). Let G be a graph with the above assumptions. en, (1) e vertices u 1 and v 1 are adjacent (2) e degree of u 1 and v 1 is complete (3) If the vertex u is adjacent to v i , then u is adjacent to v j for j < k, and if the vertex v is adjacent to u i , then v is adjacent to u j for j < k Observation 1. If G is a bipartite of order n and size m, then λ min G goes to its lower bound by increasing the size of G.
Bell [12] discussed the behavior of maximum eigenvalue by increasing the number of edges and kept V(G) fixed. We here show the behavior of least eigenvalue of a (T(p, q)) c graph by fixing the number of edges at m � pq + p + q − 2 and let the number of vertices be increased. In Figure 2, the horizontal axes show vertices and vertical axes represent least eigenvalues. Theorem 3. Let T be a tree of order n ≥ 4. en, Proof. As we know, By eorems 1 and 2, it is enough to show that (T(p, q) c ) is a largest double nested graph at 1 − 2 distance. Let (T(p, q) c ) be the graph such that λ min (T(p, q) c ) is the least one among all the graphs and let X be the least eigenvector. Now, let G be the graph obtained from (T(p, q) c ) by rotating the edge r i s to r i t such that s≁t for i � 1, 2, . . . , k { }, then λ min G ≤ λ min (T(p, q) c ), and similarly, if s and t belong to the same colour class s.t x s ≥ x t , then deg(s) ≥ deg(t), and so by Lemma 3, it will be a double nested graph. Next, we will show that it is the largest possible double nested graph at 1 − 2 distance. As the size (T(p, q) c ) is pq − 3, so all we need to show is that the size cannot exceed Journal of Mathematics pq − 3. Suppose on contrary that m > (pq − 3), then either m � pq − 2 or m � pq − 1. In both cases, the connected T(p, q) at 1 − 2 distance is not possible, which completes the proof.

Conflicts of Interest
e authors declare that they have no conflicts of interest.