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Abstract
Let be a graph. The Laplacian matrix of is , where is a diagonal matrix and denotes the degree of the vertex in and is the adjacency matrix of . Let and be two (unicyclic) graphs. We study the multiplicity of the Laplacian eigenvalue of where the graphs or may have perfect matching and Laplacian eigenvalue or not. We initiate the Laplacian characteristic polynomial of , and . It is also investigated that Laplacian eigenvalue of for some graphs and under the conditions.
Formatting an article for DMTCS]On the Laplacian eigenvalue of graphs

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Masoumeh Farkhondeh^{†}^{†}thanks: Corresponding email:mfarkhondeh81@gmail.com and Mohammad Habibi^{†}^{†}thanks: Corresponding email:mhabibi@tafreshu.ac.ir and Dost Ali Mojdeh^{†}^{†}thanks: Corresponding email:damojdeh@umz.ac.ir
Department of Mathematics, Tafresh University, Tafresh, Iran
Department of Mathematics, University of Mazandaran, Babolsar, Iran
1 Introduction
All graphs in this paper are finite and undirected with no loops or multiple edges. Let be a graph. The vertex set and the edge set of are denoted by and , respectively. The Laplacian matrix of is , where is a diagonal matrix and denotes the degree of the vertex in and is the adjacency matrix of . We shall use the notation to denote the th Laplacian eigenvalue of the graph and we assume that . Also, the multiplicity of the eigenvalue of is denoted by . A vertex of degree one is called a pendent vertex and a vertex is said quasi pendant (support vertex) if it is incident to a pendent vertex. Connected graphs in which the number of edges equals the number of vertices are called unicyclic graphs. Therefore, a unicyclic graph is either a cycle or a cycle with some attached trees. Let be the set of all unicyclic graphs of order with girth . A rooted tree is a tree in which one vertex has been designated the root. Furthermore, assume is a rooted tree of order attached to , where . This unicyclic graph is denoted by . The sun graph of order is a cycle with an edge terminating in a pendent vertex attached to each vertex that is the corona of . A broken sun graph is a unicyclic subgraph of a sun graph, so one can assume a sun graph is a broken sun graph too. Let be the number of vertices of degree in . A oneedge connection of two graphs and is a graph with and where and is denoted by . If and are unicyclic graphs, then is a bicyclic graph.
By Mieghem (2011) Theorem 13, due to Kelmans and Chelnokov, the Laplacian coefficient, , can be expressed in terms of subtree structures of , for . Suppose that is a spanning forest of with components of order , and . The Laplacian characteristic polynomial of is denoted by . If is a square matrix, then the determinant of is denoted by and the minor of the entry in the th row and th column is the determinant of the submatrix formed by deleting the th row and th column. This number is often denoted by . A square matrix is nonsingular if its determinant is nonzero or it has an inverse. In Section 2, the multiplicity of the Laplacian eigenvalue of oneedge connection of two graphs are investigated for some graphs while any graph has perfect matching. In section 3, we give some conditions such that under which the Laplacian eigenvalue of exists or not.
2 Multiplicity of the Laplacian eigenvalue
Let . We give a description of the some eigenvalues of and the relationship between the spectrums of , and , by the Laplacian characteristic polynomial of . The Laplacian characteristic polynomial of has been proved by Guo (2005) using Laplac Theorem. In the follow, we improve the proof of this lemma and use the Laplacian characteristic polynomial of to determining relevance among Laplacian eigenvalues of , and . For this, we need the following result.
Lemma 2.1
(Cvetkovic et al., 1995, Lemma 2.2) If is a nonsingular square matrix, then
(2.1) 
We now give a new proof for this lemma from Guo (2005).
Lemma 2.2
Let and be two graphs of order and , respectively. Then the Laplacian characteristic polynomial of is
(2.2) 
where and .
Proof: Let and be two graphs of order and , respectively. Suppose and . Let . Without lose of generality, assume that the connection edge is . Let and be the adjacency matrices of and , respectively. So and denote the degree of the vertices and in and , respectively. Let the Laplacian characteristic polynomial of be
such that
; ;
; ;
Also
Then
Furthermore, , such that
and
Consequently,
Thus
On the other hand, if
then
So thus
and the proof is complete.
Corollary 2.1
Let be a unicyclic graph on vertices. If are the Laplacian eigenvalues of , then are the Laplacian eigenvalues of .
Proof: Suppose and are the vertices of and the copy of , respectively. Let . Without lose of generality, we can assume that in . Then , by Equation (2.2) and . So and the result follows.
Now, by Mojdeh et al. (2018), Lemma 6, we characterize some bicyclic graphs that containing a perfect matching and having eigenvalue 2 with multiplicity 3.
Theorem 2.1
Let and be two cycles where . If , then .
Proof: Assume , then , by Akbari et al. (2014), Theorem 12. Therefore and . On the other hand, is a bicyclic graph and so , by Mojdeh et al. (2018), Lemma 6. According to Equation (2.2), it is enough to show that and have the factor . Let be a cyclic graph and
So,
Thus we have
such that . By Equation (2.1) and consequently has a as a factor. With a similar method, one can check that has a as a factor. So and we are done.
The following example shows that the converse of Theorem 2.1 does not necessarily true in general.
Example 2.1
Let be a unicyclic graph same as Figure 1.
Then has an eigenvalue with multiplicity . But has an eigenvalue with multiplicity (Figure 2).
Theorem 2.2
Let and be unicyclic graphs containing a perfect matching with . Then .
Proof: Let and such that and ; and
So and , for , by (Akbari et al., 2014, Theorem 13). If and are cycles, then the result follows from Theorem 2.1. So, one can assume that is not a cycle. First assume that
Since , there exist at least vertices on the cycle of with degree . Also, at least a quasi pendent vertex of has degree , since has a perfect matching. Suppose and without lose of generality . Assume that
and
By above argument, at least vertices of have degree in . Without lose of generality, named them by , and . According to Equation 2.1, it is enough to show that has a as a factor. Suppose
where is a square matrix of order . So will be one of the following matrices:
Clearly, all above determinants have a as a factor. Thus , by Equation 2.1. If is a cycle, with the similar method in Theorem 2.1, we can prove that . If is a unicyclic graph like , at least vertices of have degree in . Without lose of generality, named them by , and . Suppose
where is a square matrix of order . With the similar method of , we can show that has a as a factor. Now, in general, for any unicyclic graph; by omiting and ( such that is a pendent vertex and ), we have . Thus , by Mohar (1991), Theorem 2.5. So by repeating in this way, () convert to the form, thus and the proof is complete.
3 The Laplacian eigenvalue of one edge connection
Mojdeh et al. (2018) has considered the necessary and sufficient conditions in the bicyclic graph for having the Laplacian eigenvalue , where and are unicyclic graphs and having eigenvalue . In this section, we study the Laplacian eigenvalue of , where and are unicyclic graphs and , or both of them does not have eigenvalue . We show that if with a perfect matching has a Laplacian eigenvalue and does not have an eigenvalue , then does not have an eigenvalue . Also, if with no perfect matching has a Laplacian eigenvalue , then may be have an eigenvalue or not.
It has been proved that is an eigenvalue of with the corresponding eigenvector if and only if and
(3.3) 
(see Fiedler (1975)).
Theorem 3.1
Let be a unicyclic graph containing a perfect matching which has also a Laplacian eigenvalue . Let be a broken sun graph which has no Laplacian eigenvalue . Then does not have any Laplacian eigenvalue .
Proof: Let and be unicyclic graphs of order and respectively, and such that and . By contrary, if has an eigenvalue , then we can assume that is an eigenvector of corresponding to the eigenvalue . All vertices of satisfy in Equation (3.3) and , by Mojdeh et al. (2018), Theorem 5. Therefore
On the other hand, has an eigenvalue . So we have . Also
Thus, by noting the fact that for the other vertices of , we have
So has an eigenvalue and this is a contraction and the proof is complete.
Theorem 3.2
Let and be two broken sun graphs of order and respectively, for which has no perfect matching and has a Laplacian eigenvalue , and does not have Laplacian eigenvalue . Let be a bicyclic graph and for which and . Then has an eigenvalue , if and only if .
Proof: For proving the “if part”, let be an eigenvector of corresponding to the eigenvalue with . Then by assigning to all vertices of , one can easily check that is an eigenvector of corresponding to the eigenvalue (all vertices of satisfy in Equation (3.3)). Conversely, if , then does not have an eigenvalue by Theorem 3.1.
Example 3.1
In the following Figure , and are broken graphs such that just one of them has a Laplacain eigenvalue . By Theorem 3.2 and by Equation (3.3), is an eigenvector of corresponding to the eigenvalue .
In the fallow, we show that two broken sun graphs and do not have a Laplacian eigenvalue , but may have a Laplacian eigenvalue .
Theorem 3.3
Let and , and , be the numbers of pendant vertices in and for which and . Let and , such that and . Then has an eigenvalue .
Proof: Let and . We may assign and to the vertices of and , and assign and to each pendent vertices of and , respectively. One may check that every vertices of satisfies in Equation3.3. Therefore is an eigenvector corresponding to the Laplacian eigenvalue , such that , for and we are done.
Example 3.2
Let and be broken sun graphs (see Figure 4), then has an eigenvalue by Theorem 3.3.
References
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