Some Criteria for a Signed Graph to Have Full Rank
aa r X i v : . [ m a t h . C O ] A ug Some Criteria for a Signed Graph to Have FullRank
S. Akbari ∗ , A. Ghafari, K. Kazemian, M. Nahvi Department of Mathematical SciencesSharif University of Technology, Tehran, Iran
Abstract
A weighted graph G ω consists of a simple graph G with a weight ω , which is a mapping, ω : E ( G ) → Z \{ } . A signed graph is a graph whose edges are labeled with − G , there is a sign σ so that G σ has full rank if and only if G has a { , } -factor. We also show that for a graph G , there is aweight ω so that G ω does not have full rank if and only if G has at least two { , } -factors. Keywords:
Weighted graph, Signed graph, Weighted adjacency matrix, Signed adjacency matrix,Rank.
Throughout this paper, by a graph we mean a simple, undirected and finite graph. Let G be agraph. We denote the edge set and the vertex set of G by E ( G ) and V ( G ), respectively. By order and size of G , we mean the number of vertices and the number of edges of G , respectively. Theadjacency matrix of a simple graph G is denoted by A ( G ) = [ a ij ], where a ij = 1 if v i and v j areadjacent, and a ij = 0 otherwise. We denote the complete graph of order n by K n . A { , } -factor of a graph G is a spanning subgraph of G which is a disjoint union of copies of K and cycles. Fora { , } -factor H of G , the number of cycles of H is denoted by c ( H ). The number of { , } -factorsof a graph G is denoted by t ( G ). The perrank of a graph G of order n is defined to be the orderof its largest subgraph which is a disjoint union of copies of K and cycles, and we say G has fullperrank if perrank ( G ) = n . For a graph G , a zero-sum flow is an assignment of non-zero realnumbers to the edges of G such that the total sum of the assignments of all edges incident withany vertex is zero. For a positive integer k , a zero-sum k -flow of G is a zero-sum flow of G usingthe numbers {± , . . . , ± ( k − } .We call a matrix integral if all of its entries are integers. A set X of n entries of an n × n matrix A is called a transversal , if X contains exactly one entry of each row and each column of A . Atransversal is called a non-zero transversal if all its entries are non-zero. The identity matrix isdenoted by I . Also, j n is an n × weighted graph G ω consists of a simple graph G with a weight ω , which is a mapping, ω : E ( G ) → Z \{ } . A signed graph G σ is a weighted graph where σ : E ( G ) → {− , } . The weighted ∗ Email addresses: s [email protected], ghafaribaghestani [email protected], kazemian [email protected],nahvi [email protected] adjacency matrix of the weighted graph G ω is denoted by A ( G ω ) = [ a ωij ], where a ωij = ω ( v i v j ) if v i and v j are adjacent vertices, and a ωij = 0, otherwise. The rank of a weighted graph is definedto be the rank of its weighted adjacency matrix. A bidirected graph G is a graph such that eachedge is composed of two directed half edges . Function f : E ( G ) → Z \{ } is a nowhere-zero Z -flow of G if for every vertex v of G we have P e ∈ E + ( v ) f ( e ) = P e ∈ E − ( v ) f ( e ), where E + ( v ) (resp. E − ( v )) is the set of all edges with tails (resp. heads) at v . For a positive integer k , a nowhere-zero k -flow of G is a nowhere-zero Z -flow of G using the numbers {± , . . . , ± ( k − } . For a graph G ,where E ( G ) = { e , . . . , e m } and V ( G ) = { v , . . . , v n } , we define M G ( x , . . . , x m ) = [ m ij ] to be an n × n matrix, where m ij = ( x k If e k = v i v j f G ( x , . . . , x m ) = det ( M G ( x , . . . , x m )).In order to establish our results, first we need the following well-known theorem, which has manyapplications in algebraic combinatorics. Theorem A. [3]
Let F be an arbitrary field and let f = f ( x , . . . , x n ) be a polynomial in F [ x , . . . , x n ] . Suppose the degree deg ( f ) of f is n P i =1 t i , where each t i is a nonnegative integer,and suppose the coefficient of Q ni =1 x t i i in f is non-zero. Then, if S , . . . , S n are subsets of F with | S i | > t i , there are s ∈ S , s ∈ S , . . . , s n ∈ S n so that f ( s , . . . , s n ) = 0 . The following remark is a necessary tool in proving our results.
Remark 1.1.
For a graph G , each non-zero transversal in A ( G ) corresponds to a { , } -factor of G , and each { , } -factor H of G corresponds to c ( H ) non-zero transversals in M . In this paper, we prove the following theorems:
Theorem.
Let G be a graph. Then there exists a sign σ for G so that G σ has full rank if and onlyif G has full perrank. Theorem.
Let G be a graph. Then there exists a weight ω for G so that G ω does not havefull rank if and only if t ( G ) ≥ . Our first result is the following theorem.
Theorem 2.1.
Let G be a graph. Then there exists a sign σ for G so that G σ has full rank if andonly if G has full perrank.Proof. First, assume that G has full perrank. Let m = | E ( G ) | , n = | V ( G ) | . Suppose that f ( x , . . . , x m ) is the polynomial obtained by replacing x i by 1 in f G ( x , . . . , x m ), for all i , 1 ≤ i ≤ m . Clearly, deg x i f ≤ i , 1 ≤ i ≤ m .Let U be the set of all { , } -factors of G . We have U = ∅ . For each H ∈ U , we define a ( H ) asthe number of K components of H . We choose F in U so that a ( F ) = max H ∈ U ( a ( H )). Note that F does not contain any even cycles. So F is a disjoint union of a ( F ) copies of K , and c ( F )odd cycles. Assume that all edges in the cycles of F are e , . . . , e k . There are 2 c ( F ) non-zerotransversals in M corresponding to F , and all terms in f ( x , . . . , x m ) created by these non-zerotransversals are ( − a ( F ) x . . . x k .Let X be a non-zero transversal in M which makes the term ax . . . x k in f ( x , . . . , x m ) for some a ∈ R and does not correspond to F . Let F be the { , } -factor of G associated with X . Theedges in the cycles of F are e , . . . , e k . Therefore a ( F ) = a ( F ), which is maximum. So F hasno even cycles and a = ( − a ( F ) . So x . . . x k has a non-zero coefficient in f ( x , . . . , x m ). Thus f c Q mi =1 x t i i with the maximum degree m P i =1 t i , where t i ∈ { , } for each i ,1 ≤ i ≤ m . Let S i = {− , } for each i , 1 ≤ i ≤ m , so | S i | ≥
2. By Theorem A there exists( s , . . . , s m ) ∈ S × · · · × S m so that f ( s , . . . , s m ) = 0. By defining σ as a sign assigning the samesign as s i to e i for each i , 1 ≤ i ≤ m , one can see that det ( A ( G σ )) = 0.Now, assume that there exists such a sign for G . It is clear that A ( G σ ) has a non-zero transver-sal, and the associated edges with this transversal, regardless of their signs, form a { , } -factor of G . Corollary 2.2.
Let G be a graph. Then max σ ( rank ( G σ )) = perrank ( G ) . In the sequel we propose two following problems.
Problem 1.
Let G be a graph. Determine min σ ( rank ( G σ )) . Problem 2.
Find an efficient algorithm that can lead us to the desirable sign in Theorem . . In order to establish our next result, first we need the following lemma and theorems.
Lemma A. [2]
Let G be a -edge connected bipartite graph. Then G has a zero-sum -flow. Theorem B. [2]
Suppose G is not a bipartite graph. Then G has a zero-sum flow if and only iffor any edge e of G , G \{ e } has no bipartite component. Theorem C. [4]
Every bidirected graph with a nowhere-zero Z -flow has a nowhere-zero -flow. According to [1], if we orient all edges of a simple graph in a way that all edges adjacent toeach vertex v belong to E + ( v ), then a nowhere-zero bidirected flow corresponds to a zero-sum flow.Therefore, the following corollary is a result of Theorem C. Corollary A.
Every graph with a zero-sum flow has a zero-sum -flow. Now, we can prove the following theorem.
Theorem 3.1.
Let G be a graph. Then there exists a weight ω for G so that G ω does not havefull rank if and only if t ( G ) ≥ .Proof. Define X = { H | t ( H ) ≥ ω, rank ( A ( H ω )) = | V ( H ) |} . By contradiction assumethat X = ∅ . Let n = min H ∈ X | V ( H ) | and m = min H ∈ X, | V ( H ) | = n | E ( H ) | and G ∈ X be a graph of order n and size m . Let E ( G ) = { e , . . . , e m } . Obviously, G is connected. For all a , . . . , a m ∈ Z \{ } ,we have f G ( a , . . . , a m ) = 0. For each i , 1 ≤ i ≤ m , we can write f G = x i g G i + x i h G i + l G i , where g G i , h G i and l G i are polynomials in variables x , . . . , x i − , x i +1 , . . . , x m . One can see that g G i isthe zero polynomial if and only if e i belongs to no K component of any { , } -factor of G , h G i isthe zero polynomial if and only if e i belongs to no cycle of any { , } -factor of G , and l G i is thezero polynomial if and only if e i belongs to all { , } -factors of G .If there exists an edge e i of G belonging to no { , } -factor of G, then t ( G \{ e i } ) ≥ ω we have det ( A (( G \{ e i } ) ω )) = 0, which is a contradiction. So each edge of G is containedin at least one { , } -factor. Moreover, if there exists an edge e i = uv which belongs to no cycle ofany { , } -factor of G but belongs to all { , } -factors of G , then we have t ( G \{ u, v } ) ≥ ω , det ( A (( G \{ u, v } ) ω )) = 0, a contradiction. Furthermore we show that if e i appearsin a cycle of a { , } -factor, then neither of the polynomials g G i and l G i is the zero polynomial. Bycontradiction assume that g G i l G i ≡
0. If g G i ≡ l G i
0, then according to Theorem A thereare a , . . . , a i − , a i +1 , . . . , a m ∈ Z \{ } so that h G i l G i ( a , . . . , a i − , a i +1 , . . . , a m ) = 0. It can beseen that f G ( a , . . . , a m ) = 0, where a i = − l G i h G i ( a , . . . , a i − , a i +1 , . . . , a m ). Note that for each j ,1 ≤ j ≤ m , a j is a non-zero rational number. Since f G is a homogeneous polynimial, it has a root in( Z \{ } ) m , a contradiction. If l G i ≡ g G i
0, then the same argument leads to a contradiction.If g G i = l G i ≡ e i , e j , . . . , e j p are all edges of a cycle C of a { , } -factor of G , then for each k , 1 ≤ k ≤ p , we have h G i = x j k p k + q k , where p k and q k are polynomials in variables { x j } j ∈ I ,where I = { , . . . , n }\{ i, j k } . Obviously, p k q k
0, then using Theorem A one can see that h G i has a root in ( Z \{ } ) m − , a contradiction. Therefore, one can see that h G i = x j · · · x j p h ,where h is a polynomial in variables { x j } j ∈ J , where J = { , . . . , m }\{ i, j , . . . , j p } . So C is asubgraph of every { , } -factor of G . Now, by considering the graph G \ V ( C ) and noting that t ( G \ V ( C )) ≥
2, we obtain a contradiction. So we have g G i , l G i i , 1 ≤ i ≤ m . So,for each edge e i of G , there exists a { , } -factor of G not containing e i , and also there exists a { , } -factor of G containing e i in a K component.If G has a zero-sum flow, then M G ( a , . . . , a m ) j n = 0, and as a result f G ( a , . . . , a m ) = 0, wherefor each i , 1 ≤ i ≤ m , a i is the non-zero integer assigned to the edge e i in the flow, a contradiction.Thus, assume that G has no zero-sum flow. Now, we have two cases:1. The graph G is not bipartite. According to Theorem B, G \{ e i } has a bipartite componentfor some i , 1 ≤ i ≤ m . We have two cases:(a) The edge e i is not a cut edge. The graph G \{ e i } is a bipartite graph, say G \{ e i } =( X, Y ), where the vertices adjacent to e i belong to X . There is a { , } -factor of G having e i in a K component, so | X | − | Y | . Also, there exists a { , } -factor of G not having e i , therefore we have | X | = | Y | , a contradiction.(b) Now, assume that e i is a cut edge. The graph G \{ e i } has two components H and F ,where F = ( X, Y ) is bipartite.2. Now, suppose that G is bipartite. According to Lemma A, G has a cut edge e i . We denotethe bipartite connected components of the graph G \{ e i } by H and F = ( X, Y ).In both Cases 1b and 2, there exists a { , } -factor of G having e i in a K component. Therefore, F \ u has a { , } -factor, where u ∈ X is the vertex in F adjacent to e i , so we have | X | = | Y | + 1.On the other hand, there exists a { , } -factor of G which does not contain e i . Hence, F has a { , } -factor. So we have | X | = | Y | , a contradiction.Now, let G w be a weighted graph which has full rank. Let E ( G ) = { e , . . . , e m } . By contra-diction assume that t ( G ) <
2. Then f G has one monomial and therefore for some i , 1 ≤ i ≤ m , w ( e i ) = 0, a contradiction. Remark 3.2.
Let G be a graph with t ( G ) ≥ . According to Lemma A, if G is bipartite, thenthere exists a weight ω : E ( G ) → {± , . . . , ± } such that G ω does not have full rank. If G is notbipartite, then according to Corollary A, there exists a weight ω : E ( G ) → {± , . . . , ± } such that G ω does not have full rank. Acknowledgement.
The authors are deeply grateful to Mohammad Javad Moghadamzadeh forhis fruitful comments in the preparation of this paper.