On the eigenvalues of some signed graphs

TL;DR

This paper investigates the eigenvalues of the Seidel matrix for complete multipartite graphs, showing that under certain conditions, the Seidel spectrum uniquely determines the graph up to switching.

Contribution

It provides a characterization of the Seidel eigenvalues for complete multipartite graphs and proves their spectral uniqueness under specific conditions.

Findings

Seidel eigenvalues of complete multipartite graphs are studied.

The Seidel spectrum determines the graph up to switching if at least three parts have size n_i.

The Seidel characteristic polynomial is explicitly investigated.

Abstract

Let $G$ be a simple graph and $A(G)$ be the adjacency matrix of $G$. The matrix $S(G) = J -I -2A(G)$ is called the Seidel matrix of $G$, where $I$ is an identity matrix and $J$ is a square matrix all of whose entries are equal to 1. Clearly, if $G$ is a graph of order $n$ with no isolated vertex, then the Seidel matrix of $G$ is also the adjacency matrix of a signed complete graph $K_n$ whose negative edges induce $G$. In this paper, we study the Seidel eigenvalues of the complete multipartite graph $K_{n_1,\ldots,n_k}$ and investigate its Seidel characteristic polynomial. We show that if there are at least three parts of size $n_i$, for some $i=1,\ldots,k$, then $K_{n_1,\ldots,n_k}$ is determined, up to switching, by its Seidel spectrum.

Equations

$$S_{G}(\lambda)=(\lambda+1)^{n-k}\Big{(}1+\sum_{i=1}^{k}\frac{n_{i}}{\lambda+1-2n_{i}}\Big{)}\prod_{i=1}^{k}(\lambda+1-2n_{i}).$$

$$\Big{(}1+\sum_{i=1}^{k}\frac{n_{i}}{\lambda+1-2n_{i}}\Big{)}\prod_{i=1}^{k}(\lambda+1-2n_{i}).$$

$${\phi }_B(\lambda )=\det \left[\begin{array}{ccccc}\lambda +1-n_1& n_2& \dots & n_{k-1}& n_k\\ n_1& \lambda +1-n_2& \dots & n_{k-1}& n_k\\ ⋮& ⋮& \ddots & ⋮& ⋮\\ n_1& n_2& \dots & \ddots & n_k\\ n_1& n_2& \dots & n_{k-1}& \lambda +1-n_k\end{array}\right].$$

$${\phi }_B(\lambda )=\det \left[\begin{array}{cccccc}\lambda +1-2n_1& 0& \dots & \dots & 0& -\lambda -1+2n_k\\ 0& \lambda +1-2n_2& & & & -\lambda -1+2n_k\\ ⋮& & \ddots & & 0& ⋮\\ ⋮& 0& & \ddots & & ⋮\\ 0& & & & \lambda +1-2n_{k-1}& -\lambda -1+2n_k\\ n_1& n_2& \dots & \dots & n_{k-1}& \lambda +1-n_k\end{array}\right].$$

$${\phi }_B(\lambda )=(\lambda +1-2n_1)\det \left[\begin{array}{ccccc}\lambda +1-2n_2& & & & -\lambda -1+2n_k\\ & \ddots & & 0& ⋮\\ 0& & \ddots & & ⋮\\ & & & \lambda +1-2n_{k-1}& -\lambda -1+2n_k\\ n_2& \dots & \dots & n_{k-1}& \lambda +1-n_k\end{array}\right]$$

$$\varphi_{B}(\lambda)=\Big{(}1+\sum_{i=2}^{k}\frac{n_{i}}{\lambda+1-2n_{i}}\Big{)}\prod_{i=1}^{k}(\lambda+1-2n_{i})+n_{1}\prod_{i=2}^{k}(\lambda+1-2n_{i}),$$

$$S_{G}(\lambda)=(\lambda+1)^{n-k}\prod_{i=1}^{s}(\lambda+1-2n_{i})^{r_{i}-1}\Big{(}\prod_{i=1}^{s}(\lambda+1-2n_{i})+\sum_{i=1}^{s}r_{i}n_{i}\prod_{j\neq i}(\lambda+1-2n_{j})\Big{)}.$$

$$2n_k-1\le {\lambda }_{k-1}\le \cdots \le 2n_2-1\le {\lambda }_1\le 2n_1-1.$$

$${\lambda }_n\le \frac{n}{k}-1-\frac{2}{k}\sum _{1\le i<j\le k}\sqrt{n_i{n}_j}.$$

$$\lambda^{n-k}\Big{(}\lambda^{k}-\sum_{m=2}^{k}(m-1)\sigma_{m}\lambda^{k-m}\Big{)},$$

$$S_{G}(\lambda)=(\lambda+1)^{n-k}\Big{(}\prod_{i=1}^{k}(\lambda+1-2n_{i})+\sum_{i=1}^{k}n_{i}\prod_{j\neq i}(\lambda+1-2n_{j})\Big{)},$$

$$\sum _{i=0}^m(-1{)}^i{2}^i(\genfrac{}{}{0ex}{}{k-i}{m-i}){\sigma }_i,$$

$$\sum _{i=1}^m(-1{)}^{i-1}{2}^{i-1}i(\genfrac{}{}{0ex}{}{k-i}{m-i}){\sigma }_i.$$

$$(\lambda+1)^{n-3}\Big{(}\lambda^{3}+(3-n)\lambda^{2}+(3-2n)\lambda+1-n+4\sigma_{3}\Big{)},$$

$$(\lambda+1)^{n-4}\Big{(}\lambda^{4}+(4-n)\lambda^{3}+(6-3n)\lambda^{2}+(4-3n+4\sigma_{3})\lambda+1-n+4\sigma_{3}-16\sigma_{4}\Big{)},$$

$$S_{G}(\lambda)=(\lambda+1)^{n-k}\prod_{i=1}^{s}(\lambda+1-2n_{i})^{r_{i}-1}\Big{(}\prod_{i=1}^{s}(\lambda+1-2n_{i})+\sum_{i=1}^{s}r_{i}n_{i}\prod_{j\neq i}(\lambda+1-2n_{j})\Big{)},$$

$$\sum _{i=0}^m(-1{)}^i{2}^i(\genfrac{}{}{0ex}{}{s-i}{m-i}){\sigma }_i,$$

$$\sum _{i=1}^m(-1{)}^{i-1}{2}^{i-1}(\genfrac{}{}{0ex}{}{s-i}{m-i})(\sum _{l=1}^s{r}_l{\sigma }_{li}),$$

$$n_k{\sigma }_{i-1}(n_1,\dots ,n_{k-1})+{\sigma }_i(n_1,\dots ,n_{k-1})=q_k{\sigma }_{i-1}(q_1,\dots ,q_{k-1})+{\sigma }_i(q_1,\dots ,q_{k-1}),$$

$$2n_l-1\le {\lambda }_{t_l+2}\le 2n_l-1\le {\lambda }_{t_l+1}\le 2n_l-1.$$

$${\lambda }_{t_l+2}\le 2q_{t_l+2}-1\le {\lambda }_{t_l+1}.$$

Full text

a**Department of Mathematics, Karaj Branch, Islamic Azad University, Karaj, Iran

Keywords: signed graph, eigenvalues, complete multipartite graphs,

        Seidel matrix, $S$-determined.

MSC(2010): 05C22; 05C31; 05C50.

1 Introduction

Let $G$ be a simple graph with the vertex set $V(G)$ and the edge set $E(G)$. The order of $G$ is defined $|V(G)|$. A signed graph $\Gamma$ is an ordered pair $(G,\sigma)$, where $\sigma:E(G)\longrightarrow\{-1,+1\}$ is a sign function. If all edges of a signed graph are positive (resp., negative), then we denote it by $(G,+)$ (resp., $(G,-)$). The signed graphs have been studied by many authors, for instance see [1, 2, 3].

Let $A(G)=(a_{ij})$ be the adjacency matrix of $G$, and let $S(G)=J-I-2A(G)$ be its Seidel matrix, where $I$ is an identity matrix and $J$ is a square matrix all of whose entries are equal to 1. The adjacency matrix of a signed graph $\Gamma=(G,\sigma)$ is a square matrix $A(\Gamma)=(a_{ij}^{\sigma})$, where $a_{ij}^{\sigma}=\sigma(v_{i}v_{j})a_{ij}$. As usual, $K_{n}$ and $\overline{K_{n}}$ will denote the complete graph and the null graph of order $n$, respectively. Clearly, if $G$ is a graph of order $n$ and $\Gamma=(K_{n},G^{-})$ denotes a signed complete graph $K_{n}$ in which the edges of $G$ are negative and the edges not in $G$ are positive, then $A(\Gamma)=S(G)$. The polynomial $S_{G}(\lambda)=\det(\lambda I-S(G))$ is called the Seidel characteristic polynomial of $G$ and its roots often referred to as the Seidel eigenvalues or Seidel spectrum of $G$.

When dealing with Seidel matrices, the concept of Seidel switching has an important role. Let $U\subset V(G)$. A Seidel switching with respect to $U$ transforms $G$ to a graph $H$ by deleting the edges between $U$ and $V(G)\setminus U$ and adding all edges between them that were not present in $G$. It is easy to see that two Seidel matrices $S(G)$ and $S(H)$ are similar and so have the same spectrum. Seidel switching is an equivalence relation and two graphs $G$ and $H$ are called switching equivalent. We say that a graph $G$ is Seidel determined, up to switching (or $S$-determined) if the only graphs with same Seidel spectrum are switching equivalent to a graph isomorphic to $G$.

The complete multipartite graph $K_{n_{1},\ldots,n_{k}}$ is a graph of order $n=\sum_{i=1}^{k}n_{i}$. The set of vertices is partitioned into parts $V_{1},\ldots,V_{k}$ such that $|V_{i}|=n_{i}$, for $i=1,\ldots,k$ and an edge joins two vertices if and only if they belong to different parts. In [4], the authors proved that any graph which has the same Seidel spectrum as a complete $k$-partite graph is switching equivalent to a complete $k$-partite graph, and if the different partition sets sizes are $n_{1},\ldots,n_{s}$, and there are at least three parts of each size $n_{i},$ $i=1,\ldots,s$, then it is determined, up to switching, by its Seidel spectrum. In this paper, we investigate the Seidel characteristic polynomial of the complete multipartite graph $K_{n_{1},\ldots,n_{k}}$. We show that if two complete $k$-partite graphs $K_{n_{1},\ldots,n_{k}}$ and $K_{q_{1},\ldots,q_{k}}$ are Seidel cospectral and $n_{i}=q_{j}$ for some $i,j$, $1\leq i,j\leq k$, then there is a permutation $\pi$ such that $n_{i}=q_{\pi(i)}$ for $i=1,\ldots,k$. As a corollary, we prove that if there are at least three parts of size $n_{i}$, for some $i=1,\ldots,k$, then $K_{n_{1},\ldots,n_{k}}$ is $S$-determined. Among other results, we find an upper bound for the least Seidel eigenvalue of $K_{n_{1},\ldots,n_{k}}$.

2 The Seidel characteristic polynomial

Let $G$ be the complete multipartite graph $K_{n_{1},\ldots,n_{k}}$ of order $n$. The Seidel characteristic polynomial of $G$ was found in [8]. We include here a different proof by considering the Seidel matrix of $G$ as the adjacency matrix of a signed complete graph $K_{n}$ whose negative edges induce $G$. First we need the following definition.

Definition 1

([5, p.24])* *Suppose $A$ is a symmetric real matrix whose rows and columns are indexed by $X=\{1,\ldots,n\}$. Let $\{X_{1},\ldots,X_{k}\}$ be a partition of $X$ and $A$ be partitioned according to $\{X_{1},\ldots,X_{k}\}$, that is,

$\begin{bmatrix}A_{1,1}&\ldots&A_{1,k}\\ \vdots&&\vdots\\ A_{k,1}&\ldots&A_{k,k}\\ \end{bmatrix},$

where $A_{i,j}$ denote the submatrix of $A$ formed by rows in $X_{i}$ and the columns in $X_{j}$. Let $b_{i,j}$ denote the average row sum of $A_{i,j}$. Then the matrix $B=(b_{i,j})$ is called the quotient matrix of $A$ with respect to the given partition. If the row sum of each $A_{i,j}$ is constant, then the partition is called equitable. **

So we have the next theorem.

Theorem 1

Let $G$ be the complete multipartite graph $K_{n_{1},\ldots,n_{k}}$ of order $n$. Then

[TABLE]

Proof.

Let $V_{1},\ldots,V_{k}$ be the parts of $V(K_{n_{1},\ldots,n_{k}})$ and $|V_{i}|=n_{i}$ for $i=1,\ldots,k$. The partition $\sqcap=\{V_{1},\ldots,V_{k}\}$ is an equitable partition. By induction on $k$, we prove that the characteristic polynomial of the quotient matrix $B$ of $S(G)$ with respect to $\sqcap$ is as follows:

[TABLE]

One can easily see that the assertion is true for $k=2$. Let $\varphi_{B}(\lambda)=\det(\lambda I-B)$. So we have

[TABLE]

If we subtract the last row from the other rows, then

[TABLE]

Thus

[TABLE]

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +(-1)^{k+1}n_{1}\det\begin{bmatrix}0&\ldots&\ldots&0&-\lambda-1+2n_{k}\\ \lambda+1-2n_{2}&&&&-\lambda-1+2n_{k}\\ &\ddots&&\mathbb{0}&\vdots\\ \mathbb{0}&&\ddots&&\vdots\\ &&&\lambda+1-2n_{k-1}&-\lambda-1+2n_{k}\\ \end{bmatrix}.$

By induction hypothesis, we have

[TABLE]

and hence the result holds. On the other hand if $u$ and $v$ are in the same part, the vector $e_{u}-e_{v}$ whose coordinates on $u,v$ and elsewhere are respectively 1, $-1$ and 0 is an eigenvector for the eigenvalue $-1$. This explains the factor $(\lambda+1)^{n-k}$. Clearly, two graphs $K_{n_{1},n_{2}}$ and $\overline{K_{n}}$ are switching equivalent, where $n=n_{1}+n_{2}$. Thus two matrices $S(K_{n_{1},n_{2}})$ and $S(\overline{K_{n}})=A((K_{n},+))$ are similar. Therefore, $K_{n_{1},n_{2}}$ has $-1$ as a Seidel eigenvalue with the multiplicity $n-1$. If $k>2$, then $-1$ is not an eigenvalue of $B$. Hence by [5, Lemma 2.3.1], the proof is complete.

$\square$

Suppose that the number of distinct integers of $n_{1},\ldots,n_{k}$ is $s$. Without loss of generality, assume that $n_{1},\ldots,n_{s}$ are distinct. Suppose that $r_{i}$ is the multiplicity of $n_{i}$, for $i=1,\ldots,s$. The complete multipartite graph $K_{n_{1},\ldots,n_{k}}=K_{n_{1},\ldots,n_{1},\ldots,n_{s},\ldots,n_{s}}$ is also denoted by $K_{r_{1}.n_{1},\ldots,r_{s}.n_{s}}$, where $k=\sum_{i=1}^{s}r_{i}$ and $n=\sum_{i=1}^{s}r_{i}n_{i}$. From the previous theorem, we have the next immediate result.

Corollary 1

Let $G$ be the complete multipartite graph $K_{n_{1},\ldots,n_{k}}=K_{r_{1}.n_{1},\ldots,r_{s}.n_{s}}$ of order $n$. Then

[TABLE]

Remark 1

By [4, Lemma 3.2(1)], the complete $k$-partite graph $K_{n_{1},\ldots,n_{k}}$ has exactly $k-1$ (not necessarily distinct) positive Seidel eigenvalues. Moreover, if $n_{1}\geq n_{2}\geq\cdots\geq n_{k}$ and $\lambda_{1}\geq\cdots\geq\lambda_{k-1}$ are the positive Seidel eigenvalues of $K_{n_{1},\ldots,n_{k}}$, then by [4, Lemma 3.2(2)],

[TABLE]

Also if $2<k<n=\sum_{i=1}^{k}n_{i}$, then $K_{n_{1},\ldots,n_{k}}$ has exactly two distinct negative Seidel eigenvalues $\lambda_{n},-1$ ($\lambda_{n}<-1$) with multiplicity $1$ and $n-k$, respectively. (By the proof of Theorem 1, if $2<k<n=\sum_{i=1}^{k}n_{i}$, then the multiplicity of $-1$ is $n-k$.) Note that $S(K_{n})=A((K_{n},-))$ and $S(K_{n_{1},n_{2}})$ is similar to $S(\overline{K_{n}})=A((K_{n},+))$, where $n=n_{1}+n_{2}$. **

By the Rayleigh Principle [5, p.31], we have the following upper bound for $\lambda_{n}$.

Theorem 2

Let $\lambda_{n}$ be the least Seidel eigenvalue of the complete multipartite graph $K_{n_{1},\ldots,n_{k}}$ of order $n$. Then

[TABLE]

Proof. Let $V_{1},\ldots,V_{k}$ be the parts of $V(K_{n_{1},\ldots,n_{k}})$ and $|V_{i}|=n_{i}$ for $i=1,\ldots,k$. Suppose that $B$ is the quotient matrix of $S(K_{n_{1},\ldots,n_{k}})$ with respect to the equitable partition $\sqcap=\{V_{1},\ldots,V_{k}\}$. The matrix $B$ is similar to the symmetric matrix $\tilde{B}=D^{-1}BD$, where $D={\rm diag}(1/\sqrt{n_{1}},\ldots,1/\sqrt{n_{k}})$. Now by the Rayleigh quotient, the vector $[1,\ldots,1]$ gives the upper bound $\frac{n}{k}-1-\frac{2}{k}\sum_{1\leq i<j\leq k}\sqrt{n_{i}n_{j}}$.

$\square$

Now, we compute the coefficients of the Seidel characteristic polynomial of the complete multipartite graph $K_{n_{1},\ldots,n_{k}}$.

Theorem 3

[6, Theorem 1]* The characteristic polynomial of the complete multipartite graph $K_{n_{1},\ldots,n_{k}}$ is*

[TABLE]

where $n=\sum_{i=1}^{k}n_{i}$, and $\sigma_{m}=\sum_{1\leq j_{1}<\cdots<j_{m}\leq k}n_{j_{1}}\ldots n_{j_{m}}$ for $m=2,\ldots,k$.

Theorem 4

Let $G$ be the complete multipartite graph $K_{n_{1},\ldots,n_{k}}$. Then

S_{G}(\lambda)=(\lambda+1)^{n-k}\Big{(}\lambda^{k}+(k-n)\lambda^{k-1}+(\binom{k}{2}-(k-1)n)\lambda^{k-2}**

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\sum_{m=3}^{k}\sum_{i=0}^{m}(-1)^{i-1}2^{i-1}\binom{k-i}{m-i}(i-2)\sigma_{i}\lambda^{k-m}\Big{)},**

where $\sigma_{0}=1$, $n=\sigma_{1}=\sum_{i=1}^{k}n_{i}$, and $\sigma_{i}=\sum_{1\leq j_{1}<\cdots<j_{i}\leq k}n_{j_{1}}\ldots n_{j_{i}}$ for $i=2,\ldots,k$.

Proof. By Theorem 1, we have

[TABLE]

where $n=\sum_{i=1}^{k}n_{i}$. Clearly, $\prod_{i=1}^{m}(1-2x_{i})=\sum_{i=0}^{m}(-1)^{i}2^{i}\tilde{\sigma}_{i}$, where $\tilde{\sigma}_{0}=1$ and $\tilde{\sigma_{i}}=\sum_{1\leq j_{1}<\cdots<j_{i}\leq m}x_{j_{1}}\ldots x_{j_{i}}$ for $i=1,\ldots,m$. Now, $\binom{k}{m}\binom{m}{i}=\binom{k}{i}\binom{k-i}{m-i}$ implies that the coefficient of $\lambda^{k-m}$ in $\prod_{i=1}^{k}(\lambda+1-2n_{i})$ is

[TABLE]

where $\sigma_{0}=1$ and $\sigma_{i}=\sum_{1\leq j_{1}<\cdots<j_{i}\leq k}n_{j_{1}}\ldots n_{j_{i}}$ for $i=1,\ldots,k$.

Also, $x_{1}\prod_{i=2}^{m}(1-2x_{i})=\sum_{i=1}^{m}(-1)^{i-1}2^{i-1}x_{1}\tilde{\sigma}_{i-1}$, where $\tilde{\sigma}_{0}=1$ and $\tilde{\sigma_{i}}=\sum_{2\leq j_{1}<\cdots<j_{i}\leq m}x_{j_{1}}\ldots x_{j_{i}}$ for $i=1,\ldots,m-1$. Hence $k\binom{k-1}{m-1}\binom{m-1}{i-1}=i\binom{k}{i}\binom{k-i}{m-i}$ yields that the coefficient of $\lambda^{k-m}$ in $\sum_{i=1}^{k}n_{i}\prod_{j\neq i}(\lambda+1-2n_{j})$ is

[TABLE]

The proof is complete.

$\square$

In [2] and [7], the Seidel characteristic polynomials of two graphs $K_{n_{1},n_{2},n_{3},n_{4}}$ and $K_{n_{1},n_{2},n_{3}}$ have been studied. Here, we have the next example.

Example 1

The Seidel characteristic polynomials of the complete multipartite graphs $K_{n_{1},\ldots,n_{k}}$ for $k=3,4,5$ are respectively as follows:

[TABLE]

[TABLE]

(\lambda+1)^{n-5}\Big{(}\lambda^{5}+(5-n)\lambda^{4}+(10-4n)\lambda^{3}+(10-6n+4\sigma_{3})\lambda^{2}

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +(5-4n+8\sigma_{3}-16\sigma_{4})\lambda+1-n+4\sigma_{3}-16\sigma_{4}+48\sigma_{5}\Big{)},

where $n=\sum_{i=1}^{k}n_{i}$, and $\sigma_{i}=\sum_{1\leq j_{1}<\cdots<j_{i}\leq k}n_{j_{1}}\ldots n_{j_{i}}$ for $i=3,4,5$. **

Theorem 5

*Let $G$ be the complete multipartite graph $K_{n_{1},\ldots,n_{k}}=K_{r_{1}.n_{1},\ldots,r_{s}.n_{s}}$ of order $n$. Then

S_{G}(\lambda)=(\lambda+1)^{n-k}\prod_{i=1}^{s}(\lambda+1-2n_{i})^{r_{i}-1}\Big{(}\lambda^{s}+

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\sum_{m=1}^{s}\sum_{i=0}^{m}(-1)^{i-1}2^{i-1}\binom{s-i}{m-i}(\sum_{l=1}^{s}r_{l}\sigma_{li}-2\sigma_{i})\lambda^{s-m}\Big{)},**

*where $\sigma_{0}=1$, $\sigma_{l0}=0$, $\sigma_{l1}=n_{l}$, $\sigma_{i}=\sum_{1\leq j_{1}<\cdots<j_{i}\leq s}n_{j_{1}}\ldots n_{j_{i}}$, and *

$\sigma_{li}=\displaystyle\sum_{\begin{subarray}{c}1\leq j_{1}<\cdots<j_{i-1}\leq s\\ j_{1},\ldots,j_{i-1}\neq l\end{subarray}}n_{l}n_{j_{1}}\ldots n_{j_{i-1}}$* for $l,i=1,\ldots,s$.*

Proof. By Corollary 2, we have

[TABLE]

where $n=\sum_{i=1}^{s}r_{i}n_{i}$. As we did in the proof of Theorem 4, one can see that the coefficient of $\lambda^{s-m}$ in $\prod_{i=1}^{s}(\lambda+1-2n_{i})$ is

[TABLE]

where $\sigma_{0}=1$ and $\sigma_{i}=\sum_{1\leq j_{1}<\cdots<j_{i}\leq s}n_{j_{1}}\ldots n_{j_{i}}$ for $i=1,\ldots,s$.

Also, $r_{1}x_{1}\prod_{i=2}^{m}(1-2x_{i})=\sum_{i=1}^{m}(-1)^{i-1}2^{i-1}r_{1}x_{1}\tilde{\sigma}_{i-1}$, where $\tilde{\sigma}_{0}=1$ and $\tilde{\sigma_{i}}=\sum_{2\leq j_{1}<\cdots<j_{i}\leq m}x_{j_{1}}\ldots x_{j_{i}}$ for $i=1,\ldots,m-1$. Hence $\binom{s-1}{m-1}\binom{m-1}{i-1}=\binom{s-1}{i-1}\binom{s-i}{m-i}$ implies that the coefficient of $\lambda^{s-m}$ in $\sum_{i=1}^{s}r_{i}n_{i}\prod_{j\neq i}(\lambda+1-2n_{j})$ is

[TABLE]

where $\sigma_{l1}=n_{l}$, and $\sigma_{li}=\displaystyle\sum_{\begin{subarray}{c}1\leq j_{1}<\cdots<j_{i-1}\leq s\\ j_{1},\ldots,j_{i-1}\neq l\end{subarray}}n_{l}n_{j_{1}}\ldots n_{j_{i-1}}$ for $l,i=1,\ldots,s$. By setting $\sigma_{l0}=0$, for $l=1,\ldots,s$, the result holds.

$\square$

3 $S$-determined complete multipartite graphs

In [4], the authors have considered the complete $k$-partite graphs that are determined, up to switching, by their Seidel spectrum. They also proved that if the different partition sets sizes are $n_{1},\ldots,n_{s}$, and there are at least three parts of each size $n_{i},$ $i=1,\ldots,s$, then the complete multipartite graph is $S$-determined (see [4, Theorem 3.7]). Here, we extend this result. First we need the following theorems.

Theorem 6

[4, Theorem 3.3]*

Let the Seidel spectrum of $G$ be equal to that of the complete $k$-partite graph $K_{n_{1},\ldots,n_{k}}$. Then $G$ is a complete $k$-partite graph, up to switching.*

Theorem 7

[4, Theorem 3.4]*

Let $n_{1}\geq n_{2}\geq\cdots\geq n_{k}\geq 1$ and $q_{1}\geq q_{2}\geq\cdots\geq q_{k}\geq 1$ be two different $k$-tuples, $k\geq 3$, such that $\sum_{i=1}^{k}n_{i}=\sum_{i=1}^{k}q_{i}$. Then $K_{n_{1},\ldots,n_{k}}$ and $K_{q_{1},\ldots,q_{k}}$ are not switching equivalent.*

By Theorem 6, since $K_{n_{1},n_{2}}$ and $\overline{K_{n}}$ (where $n=n_{1}+n_{2}$) are switching equivalent (so Theorem 7 does not hold for $k=2$), any complete bipartite graph is $S$-determined. Throughout the rest of this paper, let $\sigma_{i}(x_{1},\ldots,x_{k})=\sum_{1\leq j_{1}<\cdots<j_{i}\leq k}x_{j_{1}}\ldots x_{j_{i}}$, for $i=1,\ldots,k$.

Theorem 8

Let two complete $k$-partite graphs $K_{n_{1},\ldots,n_{k}}$ and $K_{q_{1},\ldots,q_{k}}$ be Seidel cospectral. If $n_{i}=q_{j}$ for some $i,j$, $1\leq i,j\leq k$, then there is a permutation $\pi$ such that $n_{i}=q_{\pi(i)}$ for $i=1,\ldots,k$.

Proof. Without loss of generality, assume that $n_{k}=q_{k}$. Since $K_{n_{1},\ldots,n_{k}}$ and $K_{q_{1},\ldots,q_{k}}$ are Seidel cospectral, by Theorem 4, we have $n_{1}+\cdots+n_{k}=q_{1}+\cdots+q_{k}$ and $\sigma_{i}(n_{1},\ldots,n_{k})=\sigma_{i}(q_{1},\ldots,q_{k})$, for $i=3,\ldots,k$. Clearly, $\sigma_{i}(n_{1},\ldots,n_{k-1})=\sigma_{i}(q_{1},\ldots,q_{k-1})$ for $i=1,k-1$. Assume that $\sigma_{i}(n_{1},\ldots,n_{k-1})=\sigma_{i}(q_{1},\ldots,q_{k-1})$, where $3\leq i\leq k-1$. Now, $\sigma_{i}(n_{1},\ldots,n_{k})=\sigma_{i}(q_{1},\ldots,q_{k})$ implies that

[TABLE]

and hence $\sigma_{i-1}(n_{1},\ldots,n_{k-1})=\sigma_{i-1}(q_{1},\ldots,q_{k-1})$. Thus we conclude that $\sigma_{i}(n_{1},\ldots,n_{k-1})=\sigma_{i}(q_{1},\ldots,q_{k-1})$, for $i=1,\ldots,k-1$. Therefore two polynomials $\prod_{i=1}^{k-1}(x-n_{i})$ and $\prod_{i=1}^{k-1}(x-q_{i})$ have the same coefficients and so there is a permutation $\pi$ such that $n_{i}=q_{\pi(i)}$ for $i=1,\ldots,k-1$. $\square$

The next theorem is an extension of [4, Theorem 3.7].

Theorem 9

The complete $k$-partite graph $K_{n_{1},\ldots,n_{k}}=K_{r_{1}.n_{1},\ldots,r_{s}.n_{s}}$, where $r_{i}\geq 3$ for some $i=1,\ldots,s$, is $S$-determined.

Proof. Suppose that $n_{1}\geq n_{2}\geq\cdots\geq n_{s}$ and $r_{l}\geq 3$, where $l=1,\ldots,s$. Let $r_{0}=0$ and $t_{l}=\sum_{j=0}^{l-1}r_{j}$. If $\lambda_{1}\geq\cdots\geq\lambda_{k-1}$ are the positive Seidel eigenvalues of the complete $k$-partite graph $K_{n_{1},\ldots,n_{k}}=K_{r_{1}.n_{1},\ldots,r_{s}.n_{s}}$, then by [4, Lemma 3.2] we get that

[TABLE]

Thus $\lambda_{t_{l}+1}=\lambda_{t_{l}+2}=2n_{l}-1$. By Theorem 6, if a graph $G$ has the same Seidel spectrum, then $G$ is switching equivalent to a complete $k$-partite graph $K_{q_{1},\ldots,q_{k}}$. Assume that $q_{1}\geq q_{2}\geq\cdots\geq q_{k}$. Again by [4, Lemma 3.2],

[TABLE]

Hence $2n_{l}-1=2q_{t_{l}+2}-1$, and so $n_{l}=q_{t_{l}+2}$. Now, by Theorem 8, the result holds.

$\square$

Example 2

Let $n_{1}\geq n_{2}\geq\cdots\geq n_{k-2}\geq 1$. The graph $K_{n_{1},\ldots,n_{k-2},1,1}$ is $S$-determined. To see this assume that $\lambda_{1}\geq\cdots\geq\lambda_{k-1}$ are the positive Seidel eigenvalues of $K_{n_{1},\ldots,n_{k-2},1,1}$. By [4, Lemma 3.2], $\lambda_{k-1}=1$. If for $q_{1}\geq q_{2}\geq\cdots\geq q_{k}$ the graph $K_{q_{1},\ldots,q_{k}}$ has the same Seidel spectrum, then $2q_{k}-1\leq 1$. Hence by Theorem 8, we are done. Similarly, one can deduce that $K_{n_{1},\ldots,n_{k-3},2,2,1}$ is $S$-determined. **

In [4], it is shown that if $\max\{n_{1},n_{2},n_{3}\}$ is prime, then the complete tripartite graph $K_{n_{1},n_{2},n_{3}}$ is $S$-determined. Let us close this paper by the following result for the complete $4$-partite graphs.

Corollary 2

If $\max\{n_{1},n_{2},n_{3},n_{4}\}$ is prime, then the complete $4$-partite graph $K_{n_{1},n_{2},n_{3},n_{4}}$ is $S$-determined.

Proof. Let $n_{1}=p\geq n_{2}\geq n_{3}\geq n_{4}$, where $p$ is a prime number. By Theorem 6, if a graph $G$ is Seidel cospectral with $K_{n_{1},n_{2},n_{3},n_{4}}$, then $G$ is switching equivalent to a complete $4$-partite graph $K_{q_{1},q_{2},q_{3},q_{4}}$. Thus by Theorem 4, $\sigma_{i}(n_{1},\ldots,n_{4})=\sigma_{i}(q_{1},\ldots,q_{4})$, for $i=1,3,4$. Clearly, $\sigma_{4}(n_{1},\ldots,n_{4})=\sigma_{4}(q_{1},\ldots,q_{4})$ implies that $p|q_{1}q_{2}q_{3}q_{4}$. Without loss of generality, assume that $q_{1}=lp$ for some positive integer $l$. Hence we have $n_{2}n_{3}n_{4}=lq_{2}q_{3}q_{4}$. Since $\sigma_{1}(n_{1},\ldots,n_{4})=\sigma_{1}(q_{1},\ldots,q_{4})$, one can conclude that $4p\geq lp+3$ and so $l\leq 3$. If $l=1$, then $q_{1}=n_{1}$ and by Theorem 8, the result holds. Otherwise, $\sigma_{3}(n_{1},\ldots,n_{4})=\sigma_{3}(q_{1},\ldots,q_{4})$ yields that $p|(l-1)q_{2}q_{3}q_{4}$ and hence $p|n_{2}n_{3}n_{4}$. (Note that if $l=3$, then $p\geq 3$.) If $n_{3}=p$ or $n_{4}=p$, then by Theorem 9, we are done. Therefore suppose that $n_{2}=p$. Since $pn_{3}n_{4}=lq_{2}q_{3}q_{4}$, with no loss of generality, we have two cases: $q_{1}=l^{\prime}p^{2}$ or $q_{1}=lp,q_{2}=l^{\prime}p$, where $l,l^{\prime}$ are positive integers. If $q_{1}=l^{\prime}p^{2}$, then $n_{3}n_{4}=l^{\prime}q_{2}q_{3}q_{4}$. Now, $\sigma_{3}(n_{1},\ldots,n_{4})=\sigma_{3}(q_{1},\ldots,q_{4})$ implies that $p|q_{2}q_{3}q_{4}$. So $p|n_{3}n_{4}$, and by Theorem 9, the result holds. Finally, assume that $q_{1}=lp$ and $q_{2}=l^{\prime}p$. Thus $n_{3}n_{4}=ll^{\prime}q_{3}q_{4}$. Since $2p+n_{3}+n_{4}=(l+l^{\prime})p+q_{3}+q_{4}$, one can deduce that $l+l^{\prime}\leq 3$ which implies that $l=1$ or $l^{\prime}=1$. So $q_{1}=p$ or $q_{2}=p$, and by Theorem 8, the proof is complete.

$\square$

Acknowledgment. The authors wish to thank Professor Saieed Akbari for his fruitful comments.