Eigenvectors of Z-tensors associated with least H-eigenvalue with application to hypergraphs

TL;DR

This paper investigates the eigenvectors of Z-tensors related to their least H-eigenvalue, providing explicit counts for these eigenvectors in symmetric cases and establishing relationships among Laplacian, adjacency, and signless Laplacian eigenvectors in hypergraphs.

Contribution

It characterizes the eigenvectors associated with the least H-eigenvalue of Z-tensors, especially in symmetric cases, and links these eigenvectors to hypergraph spectral properties.

Findings

Finitely many eigenvectors associated with the least H-eigenvalue.

Explicit count of eigenvectors via Smith normal form for symmetric Z-tensors.

Equivalence of the number of Laplacian, adjacency, and signless Laplacian eigenvectors in hypergraphs.

Abstract

Unlike an irreducible $Z$-matrices, a weakly irreducible $Z$-tensor $\mathcal{A}$ can have more than one eigenvector associated with the least H-eigenvalue. We show that there are finitely many eigenvectors of $\mathcal{A}$ associated with the least H-eigenvalue. If $\mathcal{A}$ is further combinatorial symmetric, the number of such eigenvectors can be obtained explicitly by the Smith normal form of the incidence matrix of $\mathcal{A}$. When applying to a connected uniform hypergraph $G$, we prove that the number of Laplacian eigenvectors of $G$ associated with the zero eigenvalue is equal to the the number of adjacency eigenvectors of $G$ associated with the spectral radius, which is also equal to the number of signless Laplacian eigenvectors of $G$ associated with the zero eigenvalue if zero is an signless Laplacian eigenvalue.

Equations

$$(\mathcal{A}{\mathbf{x}}^{m-1}{)}_i=\sum _{i_2,\cdots ,i_m\in [n]}{a}_{i{i}_2\cdots i_m}{x}_{i_2}\cdots x_{i_m},i\in [n].$$

$${\mathbb{V}}_{\lambda }={\mathbb{V}}_{\lambda }(\mathcal{A})=\{\mathbf{x}\in {\mathbb{P}}^{n-1}:\mathcal{A}{\mathbf{x}}^{m-1}=\lambda {\mathbf{x}}^{[m-1]}\}.$$

$$\mathcal{A}=s\mathcal{I}-\mathcal{B},$$

$$E(\mathcal{A})=\{(i_1,i_2,\cdots ,i_m)\in [n{]}^m:a_{i_1{i}_2\cdots i_m}\ne 0,1\le i_1\le \cdots \le i_m\le n\}.$$

$$b_{e,j}=|\{k:i_k=j,e=(i_1,i_2,\cdots ,i_m)\in E(\mathcal{A}),k\in [m]\}|$$

$$PBQ=\left(\begin{array}{ccccccc}{d}_1& 0& 0& & \cdots & & 0\\ 0& d_2& 0& & \cdots & & 0\\ 0& 0& \ddots & & & & 0\\ ⋮& & & d_r& & & ⋮\\ & & & & 0& & \\ & & & & & \ddots & \\ 0& & & \cdots & & & 0\end{array}\right),$$

$${\mathfrak{D}}^{(0)}={\mathfrak{D}}^{(0)}(\mathcal{A})=\{D:\mathcal{A}=D^{-(m-1)}\mathcal{A}D,d_{11}=1\},$$

$$D_{\mathbf{y}}=\mathrm{diag}\left(y_1/|y_1|,\dots ,y_n/|y_n|\right),$$

$$\mathbf{y}\circ \hat{\mathbf{y}}:=D_{\mathbf{y}}{D}_{\hat{\mathbf{y}}}{\mathbf{v}}_p.$$

$${\mathbb{S}}_0(\mathcal{A})=\{\mathbf{x}\in {\mathbb{Z}}_m^n:\text{mbox}{\mathrm{I}}_{\mathcal{A}}\mathbf{x}=0over{\mathbb{Z}}_m,x_1=0\},$$

$${\mathbb{S}}_0(\mathcal{A})\cong {\oplus }_{i,d_i\ne 1}{\mathbb{Z}}_{d_i}\oplus (n-1-r){\mathbb{Z}}_m.$$

$${\mathbb{V}}_{{\lambda }_{\min }(\mathcal{A})}\cong {\oplus }_{i,d_i\ne 1}{\mathbb{Z}}_{d_i}\oplus (n-1-r){\mathbb{Z}}_m.$$

$$|{\mathbb{V}}_0(\mathcal{L}(G))|=s(\mathcal{L}(G))=s(\mathcal{A}(G))=m^{n-1-r}{\Pi }_{i=1}^r{d}_i.$$

$$|{\mathbb{V}}_0(\mathcal{Q}(G))|=|{\mathbb{V}}_0(\mathcal{L}(G))|=s(\mathcal{L}(G)),$$

Full text

∗The corresponding author. This work was supported by National Natural Science Foundation of China (Grant No. 11871073, 11771016, 11871071).

1. Introduction

A real tensor (also called hypermatrix) $\mathcal{A}=(a_{i_{1}i_{2}\ldots i_{m}})$ of order $m$ and dimension $n$ refers to a multiarray of entries $a_{i_{1}i_{2}\ldots i_{m}}\in\mathbb{R}$ for all $i_{j}\in[n]:=\{1,2,\ldots,n\}$ and $j\in[m]$, which can be viewed to be the coordinates of the classical tensor under an orthonormal basis. For a vector ${\mathbf{x}}=(x_{1},\cdots,x_{n})\in\mathbb{C}^{n}$, $\mathcal{A}{\mathbf{x}}^{m-1}\in\mathbb{C}^{n}$, which is defined as

[TABLE]

Let $\mathcal{I}=(i_{i_{1}i_{2}\cdots i_{m}})$ be the identity tensor of order $m$ and dimension $n$, that is, $i_{i_{1}i_{2}\cdots i_{m}}=1$ if $i_{1}=i_{2}=\cdots=i_{m}$ and $i_{i_{1}i_{2}\cdots i_{m}}=0$ otherwise.

Definition 1.1** ([14, 17, 2]).**

Let $\mathcal{A}$ be an $m$-th order $n$-dimensional tensor. For some $\lambda\in\mathbb{C}$, if the polynomial system $(\lambda\mathcal{I}-\mathcal{A})x^{m-1}=0$, or equivalently $\mathcal{A}x^{m-1}=\lambda x^{[m-1]}$, has a solution $x\in\mathbb{C}^{n}\backslash\{0\}$, then $\lambda$ is called an eigenvalue of $\mathcal{A}$ and $x$ is an eigenvector of $\mathcal{A}$ associated with $\lambda$, where $x^{[m-1]}:=(x_{1}^{m-1},x_{2}^{m-1},\ldots,x_{n}^{m-1})$.

For a real tensor $\mathcal{A}$, an eigenvalue $\lambda$ of $\mathcal{A}$ is called an H-eigenvalue if there is a real eigenvector associated with $\lambda$, implying that $\lambda$ is real. Denote by $\lambda_{\min}(\mathcal{A})$ the least H-eigenvalue of $\mathcal{A}$, and $\rho(\mathcal{A})$ the spectral radius of $\mathcal{A}$ (i.e. the largest modulus of the eigenvalues of $\mathcal{A}$). Let $\mathbb{P}^{n-1}$ be the complex projective spaces of dimension $n-1$. Consider the projective variety

[TABLE]

which is called the projective eigenvariety of $\mathcal{A}$ associated with $\lambda$ [7]. In this paper the number of eigenvectors of $\mathcal{A}$ is considered in $\mathbb{V}_{\lambda}(\mathcal{A})$.

By the Perron-Frobenius theorem of nonnegative tensors [1, 10, 21, 22, 23], for an irreducible (or weakly irreducible) nonnegative tensor $\mathcal{A}$, the spectral radius $\rho(\mathcal{A})$ is an eigenvalue of $\mathcal{A}$ associated with a unique nonnegative eigenvector (or positive eigenvector) up to a scalar, which is called the Perron vector of $\mathcal{A}$. In [6] the authors investigate the spectral symmetry of $\mathcal{A}$ by using the eigenvalues with modulus $\rho(\mathcal{A})$, which generalizes some spectral properties of nonnegative irreducible matrices. But, different from the matrices case, $\mathcal{A}$ can have more than one eigenvector associated with $\rho(\mathcal{A})$, including the Perron vector.

$Z$-matrices and $M$-matrices are the generalization of nonnegative matrices. Recently they were generalized to $Z$-tensors and $M$-tensors respectively [24, 4].

Definition 1.2** ([24, 4]).**

A real tensor $\mathcal{A}$ is called a $Z$-tensor if all of its off-diagonal entries are non-positive, or equivalently it can be written as

[TABLE]

where $s>0$ and $\mathcal{B}$ is nonnegative. If $s\geq\rho(\mathcal{B})$, then $\mathcal{A}$ is called an $M$-tensor; if $s>\rho(\mathcal{B})$, then $\mathcal{A}$ is called a nonsingular $M$-tensor or strong $M$-tensor.

Zhang et al. [24] showed that the minimum real part of all eigenvalues an $M$-tensor is the least H-eigenvalue. Some characterization of a $Z$-tensor being an $M$-tensor is given in [24, 4]. Most of these results are generalization of $M$-matrices. Let $\mathcal{A}$ be a weakly irreducible $Z$-tensor as in (1.1). Then $\mathbb{V}_{\lambda_{\min}(\mathcal{A})}=\mathbb{V}_{\rho(\mathcal{B})}$. As $\mathcal{B}$ is nonnegative and weakly irreducible, it can have more than one eigenvector associated with $\rho(\mathcal{B})$, which implies that $\mathcal{A}$ can have more than one eigenvector associated with $\lambda_{\min}(\mathcal{A})$. This is different from the case of irreducible $Z$-matrices.

In this paper, we show that for a weakly irreducible $Z$-tensor $\mathcal{A}$, $\mathbb{V}_{\lambda_{\min}(\mathcal{A})}$ is a finite abelian group, and admits a $\mathbb{Z}_{m}$-module if $\mathcal{A}$ is further combinatorial symmetric. For the latter case, the cardinality of $\mathbb{V}_{\lambda_{\min}(\mathcal{A})}$ can obtained explicitly by solving the Smith normal form of the incidence matrix of $\mathcal{A}$. For a connected uniform hypergraph $G$, we show that $\mathbb{V}_{0}(\mathcal{L}(G))$ has the same cardinality as $\mathbb{V}_{\rho(\mathcal{A}(G))}$, which is also equal to that of $\mathbb{V}_{0}(\mathcal{Q}(G))$ if [math] is an eigenvalue of $\mathcal{Q}(G)$, where $\mathcal{A}(G)$, $\mathcal{L}(G)$ and $\mathcal{Q}(G)$ are the adjacency, Laplaican and signless Laplacian of $G$ respectively.

2. Preliminaries

Let $\mathcal{A}$ be an $m$-th order $n$-dimensional real tensor. $\mathcal{A}$ is called symmetric if its entries are invariant under any permutation of their indices. The irreducibility or weakly irreducibility of a tensor can be referred to [1, 10]. The support of $\mathcal{A}$, denoted by $\mbox{\rm supp}(\mathcal{A})=(s_{i_{1}i_{2}\ldots i_{m}})$, is defined as a tensor with same order and dimension as $\mathcal{A}$, such that $s_{i_{1}\ldots i_{m}}=1$ if $a_{i_{1}\ldots i_{m}}\neq 0$, and $s_{i_{1}\ldots i_{m}}=0$ otherwise. $\mathcal{A}$ is called combinatorial symmetric if $\mbox{\rm supp}(\mathcal{A})$ is symmetric. Let $\mathcal{A}$ be a combinatorial symmetric tensor of order $m$ and dimension $n$. Set

[TABLE]

Define

[TABLE]

and obtain an $|E(\mathcal{A})|\times n$ matrix $\mbox{\rm I}_{\mathcal{A}}=(b_{e,j})$, called the incidence matrix of $\mathcal{A}$.

A hypergraph $G=(V(G),E(G))$ consists of a vertex set $V(G)=\{v_{1},v_{2},\ldots,v_{n}\}$ and an edge set $E(G)=\{e_{1},e_{2},\ldots,e_{l}\}$, where $e_{j}\subseteq V(G)$ for $j\in[l]$. If $|e_{j}|=m$ for each $j\in[l]$, then $G$ is called an $m$-uniform hypergraph. The adjacency tensor of an $m$-uniform hypergraph $G$ is defined to be $\mathcal{A}(G)=(a_{i_{1}i_{2}\ldots i_{m}})$, where $a_{i_{1}i_{2}\ldots i_{m}}=\frac{1}{(m-1)!}$ if $\{v_{i_{1}},v_{i_{2}},\ldots,v_{i_{m}}\}\in E(G)$, and is [math] otherwise [3]. Let $\mathcal{D}(G)$ be an $m$-th order $n$-dimensional diagonal tensor, where $d_{i\ldots i}=d_{v_{i}}$, the degree of the vertex $v_{i}$, for each $i\in[n]$. Then $\mathcal{L}(G)=\mathcal{D}(G)-\mathcal{A}(G)$ and $\mathcal{Q}(G)=\mathcal{D}(G)+\mathcal{A}(G)$ are called the Laplacian tensor and the signless Laplacian tensor of $G$, respectively [18]. The adjacency, Laplacian or signless Laplacian tensor of $G$ is symmetric, and it is weakly irreducible if and only if $G$ is connected [16, 23]. The incidence matrix of $G$, denoted by $\mbox{\rm I}_{G}=(b_{e,v})$, coincides with that of $\mathcal{A}(G)$, namely $b_{e,v}=1$ if $v\in e$, and $b_{e,v}=0$ otherwise.

For a matrix $B\in\mathbb{Z}_{m}^{k\times n}$, there exist two invertible matrices $P\in\mathbb{Z}_{m}^{k\times k}$ and $Q\in\mathbb{Z}_{m}^{n\times n}$ such that

[TABLE]

where $r\geq 0$, $1\leq d_{i}\leq m-1$, $d_{i}|d_{i+1}$ for $i\in[r-1]$, and $d_{i}|m$ for all $i\in[r]$. The matrix in (2.1) is called the Smith normal form of $B$ over $\mathbb{Z}_{m}$.

Let $\mathcal{A}$ be a tensor of order $m$ and dimension $n$. Define

[TABLE]

where $D$ is an $n\times n$ invertible diagonal matrix and the product is defined as in [19].

Definition 2.1** ([7]).**

For a general tensor $\mathcal{A}$, the cardinality of $\mathfrak{D}^{(0)}(\mathcal{A})$, denoted by $s(\mathcal{A})$, is called the stabilizing index of $\mathcal{A}$.

By [7, Lemma 2.5(1)], $\mathfrak{D}^{(0)}(\mathcal{A})$ is an abelian group under the usual matrix multiplication, which is determined by the support of $\mathcal{A}$ by [7, Lemma 2.6]. Suppose that $\mathcal{A}$ is a nonnegative weakly irreducible tensor. By [23, Theorem 3.7], for each ${\mathbf{y}}\in\mathbb{V}_{\rho(\mathcal{A})}$, $|{\mathbf{y}}|=:{\mathbf{v}}_{p}$ is the unique positive Perron vector of $\mathcal{A}$. Therefore, we can assume each ${\mathbf{y}}\in\mathbb{V}_{\rho(\mathcal{A})}$ satisfies $y_{1}=1$. Define

[TABLE]

and a quasi-Hadamard product $\circ$ in $\mathbb{V}_{\rho(\mathcal{A})}$ as follows:

[TABLE]

Lemma 2.2** ([7], Lemma 2.5, Lemma 3.1).**

Let $\mathcal{A}$ be a nonnegative weakly irreducible tensor. Then the following results hold.

1. (1)

$\mathfrak{D}^{(0)}(\mathcal{A})=\{D_{{\mathbf{y}}}:{\mathbf{y}}\in\mathbb{V}_{\rho(\mathcal{A})},y_{1}=1\}$, and hence $s(\mathcal{A})=|\mathbb{V}_{\rho(\mathcal{A})}|$. 2. (2)

$(\mathbb{V}_{\rho(\mathcal{A})},\circ)$* is an abelian group isomorphism to $\mathfrak{D}^{(0)}(\mathcal{A})$.*

Further assume $\mathcal{A}$ is also combinatorial symmetric of order $m$. By [7, Lemma 2.5(3)], $D^{m}=\mathcal{I}$ for each $D\in\mathfrak{D}^{(0)}$. Hence for each ${\mathbf{y}}\in\mathbb{V}_{\rho(\mathcal{A})}$, ${\mathbf{y}}^{\circ m}=D_{{\mathbf{y}}}^{m}{\mathbf{v}}_{p}={\mathbf{v}}_{p}$ (the identity), which implies that $(\mathbb{V}_{\rho(\mathcal{A})},\circ)$ admits a $\mathbb{Z}_{m}$-module. Define

[TABLE]

where $\mbox{\rm I}_{\mathcal{A}}$ is the incidence matrix of $\mathcal{A}$.

Theorem 2.3** ([7]).**

Let $\mathcal{A}$ be a nonnegative combinatorial symmetric weakly irreducible tensor of order $m$. Then $\mathbb{V}_{\rho(\mathcal{A})}$ is $\mathbb{Z}_{m}$-module isomorphic to $\mathbb{S}_{0}(\mathcal{A})$.

Theorem 2.4** ([7]).**

Let $\mathcal{A}$ be a nonnegative combinatorial symmetric weakly irreducible tensor of order $m$ and dimension $n$. Suppose that $\mbox{\rm I}_{\mathcal{A}}$ has a Smith normal form over $\mathbb{Z}_{m}$ as in (2.1). Then $1\leq r\leq n-1$, and

[TABLE]

Theorem 2.5** ([8]).**

Let $\mathcal{A}$ be a weakly irreducible nonnegative tensor. Then $\mathbb{V}_{\rho(\mathcal{A})}$ has dimension zero, i.e. there are finite many eigenvectors of $\mathcal{A}$ corresponding to $\rho(\mathcal{A})$ up to a scalar.

3. $Z$-tensors, Laplacian tensors and signless Laplacian tensors

Lemma 3.1**.**

Let $\mathcal{A}=s\mathcal{I}-\mathcal{B}$ be a $Z$-tensor, where $s>0$ and $\mathcal{B}\geq 0$. Then

1. (1)

$\lambda_{\min}(\mathcal{A})=s-\rho(\mathcal{B})$, which is the eigenvalue of $\mathcal{A}$ with the least real part. 2. (2)

$\mathbb{V}_{\lambda_{\min}(\mathcal{A})}=\mathbb{V}_{\rho(\mathcal{B})}$. 3. (3)

$\mathfrak{D}^{(0)}(\mathcal{A})=\mathfrak{D}^{(0)}(\mathcal{B})$, $s(\mathcal{A})=s(\mathcal{B})$. 4. (4)

For any diagonal tensor $\mathcal{D}$, $s(\mathcal{A})=s(\mathcal{D}+\mathcal{A})$.

Proof.

The first result follows by a similar discussion to [24, Theorem 3.3]. Obviously, ${\mathbf{x}}$ is an eigenvector of $\mathcal{A}$ associated with $\lambda_{\min}(\mathcal{A})$ if and only if it is an eigenvector of $\mathcal{B}$ associated with $\rho(\mathcal{B})$. So the second result follow. The last two results follow from the definition. ∎

Lemma 3.2**.**

Let $\mathcal{A}$ be a weakly irreducible $Z$-tensor. Then

1. (1)

$\mathbb{V}_{\lambda_{\min}(\mathcal{A})}$* is finite, i.e. there are finitely many eigenvectors of $\mathcal{A}$ associated with $\lambda_{\min}(\mathcal{A})$.* 2. (2)

$\mathbb{V}_{\lambda_{\min}(\mathcal{A})}$* is an abelian group isomorphic to $\mathfrak{D}^{(0)}(\mathcal{A})$.* 3. (3)

$s(\mathcal{A})=|\mathbb{V}_{\lambda_{\min}(\mathcal{A})}|$.

Proof.

Suppose $\mathcal{A}=s\mathcal{I}-\mathcal{B}$, where $s>0$ and $\mathcal{B}\geq 0$. Then $\mathbb{V}_{\lambda_{\min}(\mathcal{A})}=\mathbb{V}_{\rho(\mathcal{B})}$ by Lemma 3.1. As $\mathcal{B}$ is nonnegative and weakly irreducible, the finiteness of $\mathbb{V}_{\lambda_{\min}(\mathcal{A})}$ follows from Theorem 2.5. By Lemma 2.2, $\mathbb{V}_{\rho(\mathcal{B})}$, as well as $\mathbb{V}_{\lambda_{\min}(\mathcal{A})}$, is an abelian group isomorphic to $\mathfrak{D}^{(0)}(\mathcal{B})$, which is equal to $\mathfrak{D}^{(0)}(\mathcal{A})$ by Lemma 3.1. ∎

Theorem 3.3**.**

Let $\mathcal{A}$ be a combinatorial symmetric weakly irreducible $Z$-tensor of order $m$ and dimension $n$. Suppose that $\mbox{\rm I}_{\mathcal{A}}$ has a Smith normal form over $\mathbb{Z}_{m}$ as in (2.1). Then

1. (1)

$1\leq r\leq n-1$. 2. (2)

$\mathbb{V}_{\lambda_{\min}(\mathcal{A})}$* is a $\mathbb{Z}_{m}$-module with the decomposition*

[TABLE] 3. (3)

$s(\mathcal{A})=|\mathbb{V}_{\lambda_{\min}(\mathcal{A})}|=m^{n-1-r}\Pi_{i=1}^{r}d_{i}$.

Proof.

Let $\mathcal{A}=s\mathcal{I}-\mathcal{B}$, where $s>0$ and $\mathcal{B}\geq 0$. Note that $\mathcal{B}$ is combinatorial symmetric and weakly irreducible, $\mathcal{A}$ and $\mathcal{B}$ have the same incidence matrices, i.e. $\mbox{\rm I}_{\mathcal{A}}=\mbox{\rm I}_{\mathcal{B}}$ over $\mathbb{Z}_{m}$. So, by Theorem 2.4, $1\leq r\leq n-1$. By Theorem 2.3, $\mathbb{V}_{\rho(\mathcal{B})}$ is $\mathbb{Z}_{m}$-module isomorphic to $\mathbb{S}_{0}(\mathcal{B})$, which has a decomposition as in (2.6). As $\mathbb{V}_{\lambda_{\min}(\mathcal{A})}=\mathbb{V}_{\rho(\mathcal{B})}$ by Lemma 3.1, the second result follows. The last result follows from (2) and Lemma 3.2. ∎

Lemma 3.4**.**

The Laplacian tensor $\mathcal{L}(G)$ of a uniform hypergraph $G$ is a singular $M$-tensor.

Proof.

As $\mathcal{L}(G)$ is diagonal dominant, $\mathcal{L}(G)$ is an $M$-tensor by [24, Theorem 3.15]. Note that [math] is the least $H$-eigenvalue of $\mathcal{L}(G)$ associated with an all-ones eigenvector. So $\mathcal{L}(G)$ is singular. ∎

Theorem 3.5**.**

Let $G$ be a connected $m$-uniform hypergraph. Suppose that $\mbox{\rm I}_{G}$ has a Smith normal form over $\mathbb{Z}_{m}$ as in (2.1). Then

[TABLE]

Proof.

As $G$ is connected, $\mathcal{L}(G)$ is a combinatorial symmetric weakly irreducible $Z$-tensor. The incidence matrix of $\mathcal{L}(G)$ is same as that of $\mathcal{A}(G)$, i.e. $\mbox{\rm I}_{G}$ over $\mathbb{Z}_{m}$. So, by Theorem 3.3(3), $s(\mathcal{L}(G))=|\mathbb{V}_{0}(\mathcal{L}(G))|=m^{n-1-r}\Pi_{i=1}^{r}d_{i}$. The result now follows as $s(\mathcal{L}(G))=s(\mathcal{A}(G))$ by definition. ∎

Let $G$ be an $m$-uniform hypergraph on $n$ vertices, where $m$ is even. $G$ is called odd-colorable [15] if there exists a map $f:V(G)\to[m]$ such that for each $e\in E(G)$, $\sum_{v\in e}f(v)\equiv\frac{m}{2}\mod m$.

Lemma 3.6** ([9]).**

Let $G$ be an $m$-uniform connected hypergraph on $n$ vertices. Then the following are equivalent.

1. (1)

[math]* is an eigenvalue of $\mathcal{Q}(G)$.* 2. (2)

$m$* is even and $G$ is odd-colorable.* 3. (3)

$\mathcal{Q}(G)=D^{-(m-1)}\mathcal{L}(G)D$* for some diagonal matrix $D$ with $|D|=\mathcal{I}$.* 4. (4)

$\rho(\mathcal{L}(G))=\rho(\mathcal{Q}(G))$.

Theorem 3.7**.**

Let $G$ be an $m$-uniform connected hypergraph which is odd-colorable. Suppose that $\mbox{\rm I}_{G}$ has a Smith normal form over $\mathbb{Z}_{m}$ as in (2.1). Then

1. (1)

$s(\mathcal{Q}(G))=s(\mathcal{L}(G))=s(\mathcal{A}(G))=m^{n-1-r}\Pi_{i=1}^{r}d_{i}$. 2. (2)

$s(\mathcal{Q}(G))=|\mathbb{V}_{\rho(\mathcal{Q}(G))}|=|\mathbb{V}_{0}(\mathcal{Q}(G))|=|\mathbb{V}_{0}(\mathcal{L}(G))|$.

Proof.

The first result follows by definition and Theorem 2.4. By Lemma 3.6, as $G$ is odd-colorable, [math] is an eigenvalue of $\mathcal{Q}(G)$. Also by Lemma 3.6, there exists an invertible diagonal matrix $D$ such that ${\mathbf{y}}$ is an eigenvector of $\mathcal{Q}(G)$ associated with [math] if and only if $D{\mathbf{y}}$ is an eigenvector of $\mathcal{L}(G)$ associated with [math]. So

[TABLE]

where the last equality follows from Lemma 3.2. By Lemma 2.2, $s(\mathcal{Q}(G))=|\mathbb{V}_{\rho(\mathcal{Q}(G))}|$ as $\mathcal{Q}(G)$ is nonnegative and weakly irreducible. The result follows. ∎

Example 3.8*.*

Let $K_{n}^{[m]}$ be a complete $m$-uniform hypergraph on $n$ vertices, where $n\geq m+1$. By [7, Example 4.4], $s(\mathcal{A}(K_{n}^{[m]})=1$, so $s(\mathcal{L}(K_{n}^{[m]}))=1$ by Theorem 3.5, which implies that $\mathcal{L}(K_{n}^{[m]})$ has only one eigenvector (the all-ones vector) associated with the zero eigenvalue.

An odd bipartition $\{V_{1},V_{2}\}$ of $G$ is a bipartition of $V(G)$ such that each edge of $G$ intersects with $V_{1}$ or $V_{2}$ in an odd number of vertices. $G$ is called odd-bipartite if $G$ has an odd bipartition [11]. Shao et al. [20] proved that [math] is an $H$-eigenvalue of $\mathcal{Q}(G)$ if and only if $m$ is even and $G$ is odd-bipartite. An odd-bipartite hypergraph is odd-colorable, but the converse is not true [15]. A cored hypergraph [12] is one such that each edge contains a vertex of degree one. Obviously a cored hypergraph of even uniformity is odd-bipartite.

Example 3.9*.*

Let $G$ be a connected $m$-uniform cored hypergraph on $n$ vertices with $t$ edges. Then $s(\mathcal{A}(G))=m^{n-1-t}$ by [7, Theorem 4.1]. So $\mathcal{L}(G)$ has $m^{n-1-t}$ eigenvectors associated with the zero eigenvalue. If $m$ is even, then $G$ is odd-bipartite. So $\mathcal{Q}(G)$ has $m^{n-1-t}$ eigenvectors associated with the zero eigenvalue.

Example 3.10*.*

Let $G^{m,m/2}$ be a generalized power hypergraph [13], which is obtained from a simple graph $G$ by blowing each vertex into an $m/2$-set and preserving the adjacency relation, where $m$ is even. It is known that $G^{m,m/2}$ is non-odd-bipartite if and only if $G$ is non-bipartite [13]. Suppose that $G$ is non-bipartite. Then $\rho(\mathcal{L}(G^{m,m/2}))=\rho(\mathcal{Q}(G^{m,m/2}))$ if and only if $4$ divides $m$ [5]. Particularly, let $G$ be a triangle or $C_{3}$ and $m=4$. Then $C_{3}^{4,2}$ is non-odd-bipartite but odd-colorable by Lemma 3.6. The incidence matrix of $C_{3}^{4,2}$ has invariant divisors $1,1,2$ over $\mathbb{Z}_{4}$. So $\mathcal{Q}(C_{3}^{4,2})$ has $32$ eigenvectors associated with the zero eigenvalue.