#### Abstract

The nullity of a graph is the multiplicity of the eigenvalue zero in its spectrum. A signed graph is a graph with a sign attached to each of its edges. In this paper, we apply the coefficient theorem on the characteristic polynomial of a signed graph and give two formulae on the nullity of signed graphs with cut-points. As applications of the above results, we investigate the nullity of the bicyclic signed graph , obtain the nullity set of unbalanced bicyclic signed graphs, and thus determine the nullity set of bicyclic signed graphs.

#### 1. Introduction

Let be a simple graph with vertex set and edge set . The adjacency matrix of is defined as follows: if there exists an edge joining and , and otherwise. The nullity of a simple graph is the multiplicity of the eigenvalue zero in the spectrum of , denoted by . The rank of is referred to the rank of and denoted by . Clearly, if has vertices.

A signed graph is a graph with a sign attached to each of its edges. Formally, a signed graph consists of a simple graph which is regarded as its underlying graph, and a mapping , the edge labeling. To avoid confusion, we also write instead of , instead of , and .

The adjacency matrix of is with , where is the adjacency matrix of the underlying graph . In the case of , which is an all-positive edge labeling, is exactly the classical adjacency matrix of . So a simple graph is always assumed as a signed graph with all edges positive.

The nullity of a signed graph is defined as the multiplicity of the eigenvalue zero in the spectrum of , and is denoted by . The rank of is referred to the rank of , and denoted by . Surely, if has vertices.

Let be a signed graph and let be a signed cycle of . The sign of is defined by . The cycle is said to be positive or negative if or . A signed graph is said to be balanced if all its cycles are positive, or equivalently, all cycles have even number of negative edges; otherwise it is called unbalanced.

About the nullity of simple graphs and its applications, there are many known results (see [1–8] for details). In 1953, Hurary [9] introduced the concept of a signed graph in connection with the study of the theory of social balance in social psychology. For more results of signed graphs and their applications, see [9–21].

In this paper, we obtain the coefficients theorem of the characteristic polynomial of a signed graph and give two formulae on the nullity of signed graphs with cut-points. As applications of the above results, we investigate the nullity of the bicyclic signed graph , obtain the nullity set of unbalanced bicyclic signed graphs, and thus determine the nullity set of bicyclic signed graphs.

#### 2. The Coefficients of

In this section, we obtain the coefficients theorem of the characteristic polynomial of a signed graph , , and the nullity of a signed cycle by using the coefficients theorem.

Theorem 1 (see [11]). *Let
**
be the characteristic polynomial of an arbitrary undirected weighted multigraph .**Call an “elementary figure”: (a) the graph or (b) every graph (loops being included with ).**Call a basic figure if all of its components are elementary figures; let and be the number of components and the number of cycles contained in , respectively, and let denote the set of all basic figures contained in having exactly vertices. Then for any ,
**
where is the set of edges of , is the weight of the edge , and
*

Corollary 2. *Let be a signed graph on vertices and let
**
be the characteristic polynomial of . Then for any ,
**
where is the number of negative edges in cycle of ; other notations are similar to Theorem 1.*

*Proof. *Since is a signed graph, or , then . Thus the result follows from Theorem 1.

Proposition 3. *(1) (see [11]). Let be a balanced cycle. Then if and otherwise.**(2) (see [13]). Let be an unbalanced signed cycle. Then if and otherwise.*

Let be the number of negative edges of . It is clear that is balanced if and only if ; is unbalanced if and only if . Then Proposition 3 is equivalent to Theorem 4. We will give a new proof by Corollary 2.

Theorem 4. *Let be a signed cycle on vertices, and is the number of negative edges of . Then
*

*Proof. *By Corollary 2, for any , we have
*Case **1. *.

Clearly, . Thus .*Case **2. *.

Clearly, or , and there exist two basic figures in . Then
If , then .

If , then we consider and . Since is even, it is clear that and by Corollary 2. Thus .

Similarly, by Corollary 2, we have the following.

Proposition 5. *Let be a signed path on vertices. Then
*

#### 3. The Nullity of a Signed Graph with Cut-Points

In this section, we deduce two concise formulae on the nullity of signed graphs with cut-points by similar technique applied in [3].

We first introduce some concepts and notations.

Let be a simple graph with vertex set . For a nonempty subset of , the subgraph with vertex set and edge set consisting of those pairs of vertices that are edges in is called the induced subgraph of , denoted by . Denote by , where , the graph obtained from by removing the vertices of together with all edges incident to them. Sometimes we use the notation instead of if is an induced subgraph of . For an induced subgraph (of ) and , the induced subgraph is simply written as . The vertex is called a cut-point of if the resultant graph is disconnected.

Let be the adjacency matrix of a graph on vertices. For , , denoted by is the submatrix of with rows corresponding to the vertices of and columns corresponding to the vertices of . To simplify, the submatrix is written as . For convenience, we usually write instead of the standard for the two induced subgraphs and of . In particular, denoted by is the row vector of corresponding to the vertex and by is the subvector of corresponding to the vertices of . We refer to Cvetković et al. [11] for more terminologies and notations not defined here.

The following lemma is obvious.

Lemma 6. *Let and be a signed graph, where are the connected components of . Then are the connected components of , , and .*

Theorem 7. *Let be a connected signed graph on vertices, let be a cut-point of , and let be all the components of . If there exists a component, say , among such that , then .*

*Proof. *Let be the adjacency matrix of . For each , denote by the adjacency matrix of the subgraph and by the subvector of corresponding to the vertices of ; then the matrix can be partitioned as follows:
where , the transpose of , for each .

Note that , then , thus the row vector is not linear combination of the row vectors of , therefore the row vector is not linear combination of any other row vectors of . Since is a symmetric matrix, the column vector is not linear combination of any other column vectors of , which implies that . Then by Lemma 6,

Theorem 8. *Let be a connected signed graph on vertices, let be a cut-point of , and let be a component of . If , then .*

*Proof. *Let be the adjacency matrix of . Then

Because , and thus the row vector is linear combination of the row vectors of . Similarly, the column vector is linear combination of the column vectors of . Thus can be obtained from by row and column linear transformations, where
It is easy to see that is the adjacency matrix of the union of and . Then we have , which implies that .

#### 4. The Nullity of the Bicyclic Signed Graph

Firstly, we introduce some definitions and notations which will be used in the following.

A bicyclic graph is a simple connected graph in which the number of edges equals the number of vertices plus one. There are two basic bicyclic graphs: -graph and -graph. An -graph, denoted by , is obtained from two vertex-disjoint cycles and by connecting one vertex of and one of with a path of length (in the case of , identifying the above two vertices); and a -graph, denoted by , is a union of three internally disjoint paths of length , respectively, with common end vertices, where and at most one of them is 1. Observe that any bicyclic graph is obtained from an -graph or a -graph (possibly) by attaching trees to some of its vertices.

Lemma 9 (see [13]). *Let be a signed graph containing a pendant vertex, and let be the induced subgraph of obtained by deleting this pendant vertex together with the vertex adjacent to it. Then .*

Theorem 10. *Let be integers with , and .*(1)*If and are odd, then
*(2)*If and have different parities, without loss of generality, let be even; then
*(3)*If and are even, then
*

*Proof. *The proof is as follows.*Case **1* (both and are odd). By Corollary 2, we know
*Subcase **1.1* ( is even). Consider
So .*Subcase **1.2* ( is odd).

Clearly, if , ; , . It is obvious that when ; thus when .*Case **2* ( is even). Let be the vertex of joining and , then is a cut-point of . Note that by Proposition 5 and or by Theorem 4.*Subcase **2.1* (). It is clear that . Then by Theorem 7 and Lemma 9,
*Subcase **2.2* (). It is clear that . Then by Theorem 8 and Lemma 9,

By Theorem 4,
Then by Proposition 5 and previous arguments, (2) and (3) hold.

#### 5. The Nullity Set of Bicyclic Signed Graphs

Denoted by is the set of all bicyclic graphs on vertices. Obviously, consists of three types of graphs: first type denoted by is the set of those graphs each of which is an -graph, , with trees attached when ; second type denoted by is the set of those graphs each of which is an -graph, , with trees attached when ; third type denoted by is the set of those graphs each of which is a -graph, , with trees attached. Then .

Let be the set of all bicyclic signed graphs on vertices. Clearly, .

Let be a signed graph on vertices. Suppose is a sign function. Switching by means forming a new signed graph whose underlying graph is the same as , but whose sign function is defined on an edge by . Note that switching does not change the signs or balance of the cycles of . If we define a diagonal signature matrix with for each , then . Two graphs, and are called switching equivalent, denoted by , if there exists a switching function such that , or equivalently, .

Theorem 11 (see [16]). *Let be a signed graph. Then is balanced if and only if .*

Note that switching equivalence is a relation of equivalence, and two switching equivalent graphs have the same nullity. Therefore, when we discuss the nullity of signed graphs, we can choose an arbitrary representative of each switching equivalent class. If a signed graph is balanced, by Theorem 11, it is switching equivalent to one with all edges positive, that is, the underlying graph. Thus we only need to consider the case of unbalanced.

Hu et al. [6], Li et al. [7] characterize the maximal nullity of bicyclic graphs and determine the the nullity set of . Recently, Fan et al. [14] characterize the maximal, the second maximal nullity of bicyclic signed graphs.

Theorem 12 (see [7]). *Let be a positive integer, . Then*(1)*let , the nullity set of is ;*(2)*let , the nullity set of is ;*(3)*let , the nullity set of is .*

In Section 5.1–Section 5.3, we firstly obtain an upper bound of the nullity of bicyclic signed graphs in and , and then we obtain the nullity set of unbalanced bicyclic signed graphs in , , , respectively, and determine the nullity set of (unbalanced) bicyclic signed graphs.

##### 5.1. The Nullity Set of Unbalanced Bicyclic Signed Graphs in

Theorem 13. *Let , . Then .*

*Proof. *Let be a bicyclic graph with trees attached on an -graph, , where .*Case **1* (). Consider the following subcases.*Subcase **1.1* (). Note that
Then by (3) of Theorem 10, .

Clearly, is an induced subgraph of ; then . Therefore .*Subcase **1.2* ( or ). In this case, there must exist a graph on vertices as an induced subgraph of , where or , shown in Figure 1. By Lemma 9 repeatedly we obtain , then . Thus and .*Case **2* ( or ). Without loss of generality, we assume that . There must exist a graph on vertices shown in Figure 1 as an induced subgraph of . By Lemma 9 and Proposition 5, it is easy to check that
Hence
Since , . Then . Thus .

Theorem 14. *Let . Then the nullity set of unbalanced bicyclic signed graphs in is .*

*Proof. *It suffices to show that for each , there exists an unbalanced bicyclic signed graph such that .*Case **1 *. It is clear that there exists an unbalanced bicyclic signed graph satisfying by Theorem 10, where .*Case **2 *(). Let shown in Figure 2, where contains a balanced quadrangle and an unbalanced triangle. Thus by Theorem 4.

If , then and by (2) of Theorem 10.

If , then by Lemmas 9 and 6, we have .*Case **3 *. Let shown in Figure 2, where contains two unbalanced triangles. By using Lemma 9 repeatedly, after steps, we have

Hence by Lemma 6,

##### 5.2. The Nullity Set of Unbalanced Bicyclic Signed Graphs in

Theorem 15. *Let , . Then .*

*Proof. *Let be a bicyclic graph with trees attached on an -graph, , where .*Case **1* (). In this case, there must exist a graph on vertices as an induced subgraph of , where with , or with , or with shown in Figure 3. By Lemma 9 repeatedly and Theorem 4, we obtain , , , and let be the quadrangle containing no pendent edge of ; we have

Hence for each , we have , so . Thus .*Case **2* ( or ). Without loss of generality, we assume that . There must exist a graph on vertices shown in Figure 1 as an induced subgraph of . Similar to the proof of Case 2 in Theorem 13, we have . Then and thus .

Lemma 16 (see [14]). *Let be a graph on vertices as shown in Figure 4, and the two triangles of have the same balance. Then .*

Theorem 17. *Let ; then the nullity set of unbalanced bicyclic signed graphs in is .*

*Proof. *It suffices to show that for each , there exists an unbalanced bicyclic signed graph such that .*Case **1* (). It is clear that there exists an unbalanced bicyclic signed graph satisfying by Theorem 10, where .*Case **2* (). Let shown in Figure 4, where the two triangles of are unbalanced. Then by Lemmas 9 and 6, we have .*Case **3* (). Let shown in Figure 5, where the two triangles of are unbalanced.*Subcase **3.1* ( is odd). By using Lemma 9 repeatedly, after steps, we obtain the graph , where the two triangles of are unbalanced. Hence by Lemma 16.*Subcase **3.2* ( is even). By using Lemma 9 repeatedly, after steps, we obtain the graph . Hence .

##### 5.3. The Nullity Set of Unbalanced Bicyclic Signed Graphs in

Lemma 18 (see [14]). *Let be an unbalanced bicyclic signed graph on vertices. Then , with equality if and only if and the two triangles of are both unbalanced.*

By Lemma 18, we obtain the following result immediately.

Proposition 19. *Let and let be an unbalanced bicyclic signed graph in . Then .*

Theorem 20. *Let . Then the nullity set of unbalanced bicyclic signed graphs in is .*

*Proof. *It suffices to show that for each , there exists an unbalanced bicyclic signed graph such that .*Case **1* (). Let shown in Figure 6, where contains at least an unbalanced triangle. By Lemma 9 and Theorem 4 (when is odd) or Proposition 5 (when is even), we have .*Case **2* (). Let shown in Figure 6, where contains at least an unbalanced triangle. By Lemma 9 and Proposition 5, we have .*Case **3* (). Consider the following subcases. *Subcase **3.1* ( is odd). Let shown in Figure 6, where the two triangles of are unbalanced. By using Lemma 9 repeatedly, after steps, we obtain the graph . Hence by Lemmas 9 and 18.*Subcase **3.2* ( is even). Let shown in Figure 6, where the triangle of is unbalanced and the quadrangle is balanced. By using Lemma 9 repeatedly, after steps, we obtain the graph . Hence by Lemma 9 and Theorem 4.

##### 5.4. In Conclusion

From the above discussion, by Theorems 12, 14, 17 and 20 we can obtain the following results immediately.

Theorem 21. *Let . Then the nullity set of unbalanced bicyclic signed graphs is .*

Theorem 22. *Let . Then the nullity set of bicyclic signed graphs is .*

When , the nullity set of bicyclic signed graphs is easy to obtain by known results and direct calculation, so we omit it.

#### Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgments

The authors would like to thank the referees for their valuable comments, corrections, and suggestions, which lead to an improvement of the original paper. Research was supported by National Natural Science Foundation of China (no. 10901061), the Zhujiang Technology New Star Foundation of Guangzhou (no. 2011J2200090), and Program on International Cooperation and Innovation, Department of Education, Guangdong Province (no. 2012gjhz0007).