ORDERING OF MINIMAL ENERGIES IN UNICYCLIC SIGNED GRAPHS

. Let S = ( G,σ ) be a signed graph of order n and size m and let t 1 ,t 2 ,...,t n be the eigenvalues of S . The energy of S is defined as E ( S ) = P n j =1 | t j | . A connected signed graph is said to be unicyclic if its order and size are the same. In this paper we characterize, up to switching, the unicyclic signed graphs with first 11 minimal energies for all n ≥ 11. For 3 ≤ n ≤ 7, we provide complete orderings of unicyclic signed graphs with respect to energy. For 8 ≤ n ≤ 10, we determine unicyclic signed graphs with first 13 minimal energies.


Introduction
Let S = (G, σ) be a signed graph of order n, where G = (V, E) is its underlying graph and σ : E → {−1, 1} is its signature.Let A be the adjacency matrix of S. In a signed graph, a cycle is said to be positive if it contains an even number of negative edges, and negative otherwise.A signed graph is said to be balanced if all its cycles are positive.For undefined terms related to signed graphs, we refer the reader to [1].The characteristic polynomial P (S, x) of S is the characteristic polynomial of its adjacency matrix A and is given by Z(X), (1.1) where L r denotes the set of all linear signed subgraphs (also known as basic figures) on r vertices, k(l) denotes the number of components in l, c(l) denotes the set of cycles in l, and Z(X) = e∈X σ(e) is the sign of X.Let S be a signed graph with 120 T. SHAMSHER, M. A. BHAT, S. PIRZADA, AND Y. SHANG vertex set V .For any X ⊆ V , let S X denote the signed graph obtained from S by reversing the signs of the edges between X and V − X.Then, we say S X is switching equivalent to S.Here we note that switching is an equivalence relation and preserves the eigenvalues including their multiplicities.We use a single signed graph as representative of a switching class.Germina et al. [2] defined the energy of a signed graph S with eigenvalues t 1 , t 2 , . . ., t n as E(S) = n j=1 |t j |.Note that the definition of the energy of a signed graph is transferred from the domain of unsigned graphs.Signed graphs significantly enrich algebraic and geometric properties compared to unsigned graphs [7].
It is well known that even and odd coefficients of the characteristic polynomial of a unicyclic signed graph alternate in sign [1,Lemma 2.7].Putting c j (S) = |a j (S)|, we have the following integral representation for the energy of a unicyclic signed graph S: From the above integral formula, we see that the energy of a unicyclic signed graph is a monotonic increasing function of the coefficients c j , where j = 0, 1, . . ., n.For signed graphs S 1 and S 2 of the same order, say n, if c j (S 1 ) ≤ c j (S 2 ) for all j, then we write S 1 ⪯ S 2 .Moreover, if S 1 ⪯ S 2 and there is a strict inequality in c j (S 1 ) ≤ c j (S 2 ) for some j = 1, 2, . . ., n, then we write S 1 ≺ S 2 .Hence, if S 1 ⪯ S 2 , then E(S 1 ) ≤ E(S 2 ) and if S 1 ≺ S 2 , then E(S 1 ) < E(S 2 ).Also, if c j (S 1 ) = c j (S 2 ) for all j, then we write S 1 ∼ S 2 .Hence, if S 1 ∼ S 2 , then E(S 1 ) = E(S 2 ).Let S n,l denote the set of unicyclic signed graphs with n vertices and a cycle of length l ≤ n.Let e = uv be a pendant edge of a signed graph S ∈ S n,l with v as the pendant vertex.Then the following relation holds [1,Lemma 3.2] for c j 's of a signed graph S and its vertex-deleted signed subgraphs: (n-4-k) (i) For all n ≥ 11 and 0 Proof.(i) By (1.1), we have ).Therefore, to compare the energy of S k,r n,n for r = 1, 2 and B k,1 n,n , it is enough to compare the energy of S k,1 n,n and B k,1 n,n .Clearly, S k,1 n,n and B k,1 n,n are not quasi-order comparable.We use the integral formula (1.2) in this case, and we have Since n > k + 5, we get for n ≥ 11 and t > 0, and thus (ii) The characteristic polynomials of B k,1 n,n and S k+1,r n,n for r = 1, 2 are given by To prove the result, it is enough to show that for k = 0, 1, and therefore E(B k,1 n,n ) < E(S k+1,1 n,n ) for k = 0, 1 and for all n ≥ 11.To compare the energy of S n,n is given by are not quasi-order comparable.Proceeding similarly to part (ii), we can prove that n,n for all n > 2k + 7, and therefore  (i) For all n ≥ 6, we have Proof.(i) The characteristic polynomials of B 2,r n,n for r = 1, 2, F 1 n,n , and F 2 n,n are given by ) for all n ≥ 6.
(ii) We have n denote the signed graph obtained by identifying the center of the signed star S n−l+1 with a vertex of C σ l .The following theorem shows that among all unicyclic signed graphs with cycle length greater than 5, S 6− n has the minimal energy.
Case 1.Let S ∈ S n,l be unbalanced, where n ≥ l, n ≥ 7, and l ≥ 6.Then, by [1,Theorem 3.3], it suffices to show that c i (S 6− n ) ≤ c i (S l− n ) for all i = 4, 6, with strict inequality for at least one i.We use induction on n − l for n ≥ l, where n ≥ 7 and l ≥ 6.
If 6 and n ≥ 7. By (1.3), for i = 4, 6, we have Since l ≥ 6, P l−1 has P 5 as a subgraph, and hence c i (S 6− n ) ≤ c i (S l− n ) for all i = 4, 6, with strict inequality for at least one i.Case 2. This is similar to Case 1. □ Lemma 2.5.For all n ≥ 6, we have > 0 for all n ≥ 10.Also, g 3 (2) = 0 and g 3 ( √ n − 4) = 0.By Descartes's rule of signs, f 3 (t) has three positive and three negative zeros and g 3 (t) has two positive and two negative zeros.As the energy of a signed graph is twice the sum of its positive eigenvalues, we have ) for all n ≥ 10.We have verified the result directly for n = 6, 7, 8, 9.
Let S(n) be the set of all unicyclic signed graphs of order n.Let S ∈ S(n) and u be a vertex of S. Let T be a rooted signed tree and S u (T ) be the signed graph obtained by attaching T to S such that the root of T is u.When T is a signed path P m+1 with one endpoint as the root, we write S u (T ) as S u (m).When T is a star K 1,m with the center as its root, we write S u (T ) as S * u (m).When T is a star K 1,m with a pendant vertex as its root, we write S u (T ) as S * u (m − 1, 1).For example, if S = C − 3 , then S * u (n − 3) = S 0,2 n,n and S * u (n − 4, 1) = Q 3,2 n,n .With these notations, we have following lemmas.

Conclusion.
In this paper, we were able to provide unicyclic signed graphs with first eleven minimal energies.After that the problem becomes difficult.For example, it is easy to see that the unicyclic signed graphs Q 3,1 n,n and Q 4,1 n,n have 12 th and 13 th minimal energy, respectively, for n = 12.But for n = 1000, Q 4,1 n,n has 12 th minimal energy and Q 3,1  n,n has 13 th minimal energy.It will be interesting to provide further orderings with respect to minimal energy.Energy ordering in other families of signed graphs, such as bipartite, k-cyclic (k ≥ 2), complete signed graphs of fixed order, etc., remains a problem for future study.It will be useful to see the work on weighted graphs [3] and directed graphs [5].

Figure 1 .
Figure 1.Switching classes corresponding to unicyclic signed graphs S k n,n and B k n,n .
(v) This follows by [1, Theorem 2.10 (i)].□ Let F 1 n,n be a unicyclic graph as shown in Figure 2.There are two switching classes on the signings of F 1 n,n .Let F 1 n,n and F 2 n,n be the representatives for these two switching classes, where F 1 n,n contains a positive cycle of length 4 and F 2 n,n contains a negative cycle of length 4. With these notations, we have the following lemma.
For all n ≥ 12, we have E