# Largest and Least H-Eigenvalues of Symmetric Tensors and Hypergraphs

Hongying Lina111E-mail<EMAIL_ADDRESS>Lu Zhengb222E-mail:
<EMAIL_ADDRESS>Bo Zhoub333Corresponding author. E-mail:
<EMAIL_ADDRESS>
aSchool of Mathematics, South China University of Technology,
Guangzhou 510641, P.R. China
bSchool of Mathematical Sciences, South China Normal University,
Guangzhou 510631, P.R. China

###### Abstract

In tensor eigenvalue problems, one is likely to be more interested in
H-eigenvalues of tensors. The largest H-eigenvalue of a nonnegative tensor or
of a uniform hypergraph is the spectral radius of the tensor or of the uniform
hypergraph. We find upper bounds and lower bounds (interlacing inequalities)
for the largest H-eigenvalue of a principal subtensor of a symmetric zero
diagonal tensor that is of even order or nonnegative, as well as lower bounds
for the largest H-eigenvalue of a uniform hypergraph with some vertices or
edges removed. We also investigate similar problems for the least
H-eigenvalues. We give examples to verify the sharpness of the bounds or in
some cases for uniform hypergraphs, we characterize the equality.
Particularly, for a connected linear $k$-uniform hypergraph $G$ with $v\in
V(G)$, we give a sharp lower bound for the spectral radius of $G-v$ in terms
of the spectral radius of $G$ and the degree of $v$ and characterize the
extremal hypergraphs, and show that the maximum spectral radius of the
subhypergraphs with one vertex removed is greater than or equal to the
spectral radius of the hypergraph minus one, which is attained if and only if
it is a Steiner system $S(2,k,n)$.

Mathematics Subject Classification (2010): 15A69, 05C50, 05C65

Key words: H-eigenvalue, interlacing inequalities, symmetric tensor, uniform
hypergraph, Steiner system

##  1 Introduction

Let $\mathbb{R}$ be the field of real numbers and $\mathbb{R}^{n}$ the
$n$-dimensional real space. For positive integers $k$ and $n$, a (real) tensor
(or hypermatrix) $\mathcal{T}=(t_{i_{1}\ldots i_{k}})$ of order $k$ and
dimension $n$ is a multidimensional array with entries $t_{i_{1}\dots
i_{k}}\in\mathbb{R}$ for $i_{j}\in[n]:=\\{1,\dots,n\\}$ and $j\in[k]$. An
entry $t_{i_{1}\ldots i_{k}}$ with $i_{1}=\dots=i_{k}=i\in[n]$ is a diagonal
entry of $\mathcal{T}$. A zero diagonal tensor is a tensor for which all
diagonal entries are equal to zero. The tensor $\mathcal{T}$ is symmetric if
each entry $t_{i_{1}\dots i_{k}}$ is invariant with respect to all
permutations of $i_{1},\dots,i_{k}$. A tensor is nonnegative if all its
entries are nonnegative.

For a tensor $\mathcal{T}$ of order $k$ and dimension $n$, and an
$n$-dimensional vector ${\bf x}=(x_{1},\ldots,x_{n})^{\top}$, $\mathcal{T}{\bf
x}^{k-1}$ is defined as an $n$-dimensional vector whose $i$-th entry is

$(\mathcal{T}{\bf
x}^{k-1})_{i}\equiv\sum_{i_{2},\dots,i_{k}\in[n]}t_{ii_{2}\dots
i_{k}}x_{i_{2}}\cdots x_{i_{k}}$

for $i\in[n]$, and $\mathcal{T}{\bf x}^{k}$ is defined as the $k$-th degree
homogeneous polynomial

$\mathcal{T}{\bf x}^{k}\equiv\sum_{i_{1},\dots,i_{k}\in[n]}t_{i_{1}\dots
i_{k}}x_{i_{1}}\cdots x_{i_{k}}.$

######  Definition 1.1.

[19, 25] Let $\mathcal{T}$ be a tensor of order $k$ and dimension $n$. For
some complex $\lambda$, if there is a nonzero vector ${\bf x}$ such that

$\lambda x_{i}^{k-1}=(\mathcal{T}{\bf x}^{k-1})_{i},$

i.e.,

$\displaystyle\lambda
x_{i}^{k-1}=\sum_{i_{2},\dots,i_{k}\in[n]}t_{ii_{2}\ldots
i_{k}}x_{i_{2}}\cdots x_{i_{k}}$ (1.1)

for $i\in[n]$, then $\lambda$ is called an eigenvalue of $\mathcal{T}$, and
${\bf x}$ is called an eigenvector of $\mathcal{T}$ corresponding to
$\lambda$. Moreover, if both $\lambda$ and ${\bf x}$ are real, then we call
$\lambda$ an H-eigenvalue and ${\bf x}$ an H-eigenvector of $\mathcal{T}$.

For more details on tensor eigenvalues and eigenvectors, we refer the readers
to [5, 17, 27]. Let $\mathcal{T}$ be a tensor of order $k$ and dimension $n$.
The spectral radius of $\mathcal{T}$ is the largest modulus of the eigenvalues
of $\mathcal{T}$, denoted by $\rho(\mathcal{T})$. Suppose that there exists at
least one H-eigenvalue of $\mathcal{T}$. For instance, $\mathcal{T}$ has at
least one H-eigenvalue if $\mathcal{T}$ is symmetric and $k$ is even, see
[25], or if $\mathcal{T}$ is nonnegative, see Proposition 1.1 below. In this
case, we denote $\lambda_{\max}(\mathcal{T})$ and
$\lambda_{\min}(\mathcal{T})$ the largest H-eigenvalue and the least
H-eigenvalue of $\mathcal{T}$, respectively. In this case, it is evident that
$\lambda_{\min}(\mathcal{T})\leq\lambda_{\max}(\mathcal{T})\leq\rho(\mathcal{T})$,
and if $\mathcal{T}$ is nonnegative, then
$\lambda_{\max}(\mathcal{T})=\rho(\mathcal{T})$. In most cases, one is likely
to be more interested in H-eigenvalues of tensors.

######  Definition 1.2.

[10] A tensor $\mathcal{T}=(t_{i_{i}\dots i_{k}})$ of order $k$ and dimension
$n$ is said to be weakly reducible if $t_{i_{1}\dots i_{k}}=0$ for some
$\emptyset\neq I\subset[n]$ and for any $i_{1}\in I$ and at least one
$j\in\\{2,\dots,k\\}$ with $i_{j}\not\in I$. Otherwise, it is weakly
irreducible.

The classic Perron-Frobenius theorem has been extended to nonnegative tensors
by the efforts scholars as follows, see [4, 10, 34, 35] with a unifying
treatment in [11].

######  Proposition 1.1.

[10] For a nonnegative tensor $\mathcal{T}$ of order $k$ and dimension $n$
with $n,k\geq 2$, $\rho(\mathcal{T})$ is an H-eigenvalue of $\mathcal{T}$ with
a positive H-eigenvector. If $\mathcal{T}$ is weakly irreducible, then there
is a unique positive H-eigenvector, up to a multiplicative constant, and
moreover, if $\lambda$ is an H-eigenvalue with a positive eigenvector, then
$\lambda=\rho(G)$.

Let $G$ be a $k$-uniform hypergraph with vertex set $V(G)=[n]$ and edge set
$E(G)$, where $n,k\geq 2$. For $u\in V(G)$, denote $E_{u}(G)$ be the set of
edges containing $u$ in $G$. The degree of $u$ in $G$ is defined as
$d_{G}(u)=|E_{u}(G)|$, we also write $d_{u}$ for $d_{G}(u)$ if there is no
confuse. For any two distinct vertices $i$ and $j$ of $G$, we write $i\sim j$
if there is an edge containing $i$ and $j$, and $i\nsim j$ otherwise. A linear
hypergraph is one in which every two distinct edges intersect in at most one
vertex.

######  Definition 1.3.

[6, 7] The adjacency tensor of $G$ is defined as the symmetric, nonnegative
tensor $\mathcal{A}(G)=(a_{i_{1}\dots i_{k}})$ of order $k$ and dimension $n$,
where

$a_{i_{1}\dots i_{k}}=\begin{cases}\frac{1}{(k-1)!}&\text{if
}\\{i_{1},\dots,i_{k}\\}\in E(G),\\\ 0&\text{otherwise.}\end{cases}$

Note that the adjacency tensor of a uniform hypergraph is nonnegative, so it
has at least one H-eigenvalue by Proposition 1.1.

######  Definition 1.4.

The spectral radius (or largest H-eigenvalue) of $\mathcal{A}(G)$ is called
the spectral radius (or the largest H-eigenvalue) of $G$, denoted by
$\rho(G)$, and the least H-eigenvalue of $\mathcal{A}(G)$ called the least
H-eigenvalue of $G$, denoted by $\lambda(G)$. That is,
$\rho(G)=\rho(\mathcal{A}(G))=\lambda_{\max}(\mathcal{A}(G))$ and
$\lambda(G)=\lambda_{\min}(\mathcal{A}(G))$.

If $G$ is an ordinary graph, then $\rho(G)$ and $\lambda(G)$ ere respectively
the largest and the least eigenvalues of (the adjacency matrix) of $G$ [1, 8,
21].

For a $k$-uniform uniform hypergraph $G$, by Proposition 1.1, $\rho(G)$ is an
H-eigenvalue of $\mathcal{A}(G)$ with an associated nonnegative H-eigenvector,
and moreover, if $G$ is connected, then $\mathcal{A}(G)$ is weakly irreducible
[24], implying that there is a unique unit positive H-eigenvector
corresponding to $\rho(G)$. In this article, we say a vector ${\bf
x}\in\mathbb{R}^{n}$ is unit if $\|{\bf
x}\|_{k}^{k}:=\sum_{i\in[n]}|x_{i}|^{k}=1$. The approach to study of
hypergraphs through tensors has been widely accepted, see, e.g., [3, 6, 7, 9,
13, 16, 20, 23, 24, 29]. It should be pointed that other treatment of the
spectral property of hypergraphs may be found, see, e,g., [15].

There are many results on the bounds for the eigenvalues (particularly the
largest one) of modified graphs by Rowlinson and coauthors, see, e.g. [2, 8,
28], where a modified graph is obtained from some given graph under small
changes such as by removing vertices or edges and moving certain edges. Li,
Wang and Van Mieghem [18] presented a new novel type lower bound for the
spectral radius of a graph when some vertices are removed. Van Mieghem et al.
[32] gave bounds for the spectral radius of a graph when some edges are
removed. In [33], an upper bound was established for the least eigenvalue of a
graph when some vertices are removed.

######  Definition 1.5.

[14, 25] For a tensor $\mathcal{T}=(t_{i_{1}\ldots i_{k}})$ of order $k$ and
dimension $n$, a principal subtensor $\mathcal{T}[I]$ of $\mathcal{T}$ with
nonempty index set $I\subseteq[n]$ is a tensor of order $k$ and dimension
$|I|$ consisting of $|I|^{k}$ elements defined by

$\mathcal{T}[I]=(t_{i_{1}\ldots i_{k}})\mbox{ with $i_{1},\dots,i_{k}\in I$.}$

######  Definition 1.6.

[31] A Steiner system $S(t,k,n)$, of order $n$ and block size $k$ with $n\geq
k\geq 2$, is a collection of $k$-sets of an $n$-set such that every $t$-set
belongs to exactly one block. In other words, $S(t,k,n)$ is a $k$-uniform
hypergraph on $n$ vertices, such that every $t$-element vertex subset is
contained in precisely one edge.

In this paper, we find upper bounds and lower bounds (interlacing
inequalities) for the largest H-eigenvalue of a principal subtensor of a
symmetric zero diagonal tensor that is of even order or nonnegative, from
which we derive new lower bounds for the largest H-eigenvalue (spectral
radius) of a uniform hypergraph in which some vertices or edges are removed.
On the other hand, we present upper bounds and lower bounds for the least
H-eigenvalue of a principal subtensor of a symmetric zero diagonal tensor of
even order, from which we derive new upper and lower bounds for the least
H-eigenvalue of a uniform hypergraph in which some vertices or edges are
removed. We also present some bounds of the components of the least
eigenvectors of hypergraphs. Some results from [13, 18, 23, 30, 32, 33] are
generalized or improved. Particularly, for any connected linear $k$-uniform
hypergraph $G$ on $n$ vertices with $n\geq k\geq 2$, we give a sharp upper
bound on $\rho(G-v)$ with $v\in V(G)$ in terms of $\rho(G)$ and $d_{G}(v)$ and
characterize the hypergraphs for which this bound is attained, and show that
$\max\\{\rho(G-v):v\in V(G)\\}\geq\rho(G)-1$ with equality if and only if $G$
is a Steiner system $S(2,k,n)$.

To the best of our knowledge, there is no such type of lower (upper,
respectively) bounds for the largest (least, respectively) H-eigenvalues of
symmetric tensors and uniform hypergraphs in the literature. We also give
examples to verify the sharpness of the bounds or in some cases for
hypergraphs, we characterize the equality.

##  2 Preliminaries

We now give some tools that will be use later.

######  Lemma 2.1.

[25, Theorem 5] Let $\mathcal{T}$ be a symmetric tensor of even order $k$ and
dimension $n$, where $n,k\geq 2$. Then
$\lambda_{\max}(\mathcal{T})=\max\\{\mathcal{T}{\bf x}^{k}:\|{\bf
x}\|_{k}=1,{\bf x}\in\mathbb{R}^{n}\\}$ and
$\lambda_{\min}(\mathcal{T})=\min\\{\mathcal{T}{\bf x}^{k}:\|{\bf
x}\|_{k}=1,{\bf x}\in\mathbb{R}^{n}\\}$.

It is not difficult to see that the previous lemma is not true if $k$ is odd.
Denote by $\mathbb{R}_{+}^{n}$ the set of all nonnegative vectors in
$\mathbb{R}^{n}$.

######  Lemma 2.2.

[26, Theorem 2] Let $\mathcal{T}$ be a symmetric nonnegative tensor of order
$k$ and dimension $n$, where $n,k\geq 2$. Then
$\lambda_{\max}(\mathcal{T})=\max\\{\mathcal{T}{\bf x}^{k}:\|{\bf
x}\|_{k}=1,{\bf x}\in\mathbb{R}_{+}^{n}\\}$. If
$\lambda_{\max}(\mathcal{T})=\mathcal{T}{\bf x}^{k}$ for some ${\bf
x}\in\mathbb{R}_{+}^{n}$ with $\|{\bf x}\|_{k}=1$, then ${\bf x}$ is an
$H$-eigenvector of $\mathcal{T}$ associated with
$\lambda_{\max}(\mathcal{T})$.

If $\mathcal{T}$ be a symmetric tensor of order $k$ and dimension $n$, where
$k$ is even or if $\mathcal{T}$ be a symmetric, essentially nonnegative tensor
of order $k$ and dimension $n$, where $n,k\geq 2$, then
$\lambda_{\max}(\mathcal{T})=\max\\{\mathcal{T}{\bf x}^{k}:\|{\bf
x}\|_{k}=1,{\bf x}\in\mathbb{R}^{n}\\}$.

Let $G$ be a $k$-uniform hypergraph with $V(G)=[n]$, and let ${\bf
x}\in\mathbb{R}^{n}$. For $U\subseteq V(G)$, let ${\bf x}^{U}=\Pi_{w\in
U}x_{w}$. Then

$\mathcal{A}(G){\bf x}^{k}=k\sum_{e\in E(G)}{\bf x}^{e}$

and for $u\in V(G)$,

$(\mathcal{A}(G){\bf x}^{k-1})_{u}=\sum_{e\in E_{u}(G)}{\bf
x}^{e\setminus\\{u\\}}.$

As $\mathcal{A}(G)$ is symmetric and nonnegative, we have from Lemma 2.2 that

$\lambda(G)\leq\mathcal{A}(G){\bf x}^{k}\leq\rho(G).$

For a hypergraph $G$ with $V_{1}\subset V(G)$, $G-V_{1}$ denotes the
hypergraph with vertex set $V(G)\setminus V_{1}$ and edge set
$E(G)\setminus\\{e:e\cap V_{1}=\emptyset\\}$. If $V_{1}=\\{v\\}$, then we
write $G-v$ for $G-\\{v\\}$.

For a hypergraph $G$ with $E_{1}\subseteq E(G)$, $G-E_{1}$ denotes the
hypergraph with vertex set $V(G)$ and edge set $E(G)\setminus E_{1}$. If
$E_{1}=\\{e\\}$, then we write $G-e$ for $G-\\{e\\}$.

For a symmetric tensor $\mathcal{T}$ of order $k$ and dimension $n$ and
$\emptyset\neq I\subset[n]$, let $\mathcal{T}_{I}$ be the tensor of order $k$
and dimension $n$ such that

$(\mathcal{T}_{I})_{i_{1}\ldots i_{k}}=\begin{cases}t_{i_{1}\ldots
i_{k}}&\hbox{if ~{}$\\{i_{1},\ldots,i_{k}\\}\subseteq I$,}\\\
0&\hbox{otherwise.}\end{cases}$

######  Lemma 2.3.

Let $\mathcal{T}$ be a zero diagonal symmetric tensor of order $k$ and
dimension $n$, where $n,k\geq 2$. If $k$ is even or $\mathcal{T}$ is
nonnegative, then $\lambda_{\max}(\mathcal{T})\geq 0$ and
$\lambda_{\min}(\mathcal{T})\leq 0$.

###### Proof.

Suppose first that $k$ is even. By Lemma 2.1, for any ${\bf
x}\in\mathbb{R}^{n}$ with $\|{\bf x}\|_{k}=1$, we have

$\lambda_{\max}(\mathcal{T})\geq\mathcal{T}{\bf
x}^{k}\geq\lambda_{\min}(\mathcal{T}).$

As $t_{1\dots 1}=0$, the coefficient of $x_{1}^{k}$ in $\mathcal{T}{\bf
x}^{k}$ is $0$. Setting ${\bf y}=(1,0,\dots,0)^{\top}\in\mathbb{R}^{n}$, we
have $\|{\bf y}\|_{k}=1$ and $\mathcal{T}{\bf y}^{k}=0$. So
$\lambda_{\max}(\mathcal{T})\geq 0\geq\lambda_{\min}(\mathcal{T})$.

Suppose next that $\mathcal{T}$ is nonnegative. By Lemma 2.2,
$\lambda_{\max}(\mathcal{T})\geq 0$. If $k$ is even, then by the above
argument, $\lambda_{\min}(\mathcal{T})\leq 0$. If $k$ is odd, then set ${\bf
y}$ as above and we have $\mathcal{T}{\bf y}^{k-1}=0{\bf y}^{k-1}$ since
$t_{1\dots 1}=0$, so $0$ is an H-eigenvalue of $\mathcal{T}$, implying that
$\lambda_{\min}(\mathcal{T})\leq 0$. ∎

######  Lemma 2.4.

Let $\mathcal{T}$ be a zero diagonal symmetric tensor of order $k$ and
dimension $n$, where $n,k\geq 2$. Let $\emptyset\neq I\subset[n]$. Suppose
that $k$ is even or $\mathcal{T}$ is nonnegative. Then
$\lambda_{\max}(\mathcal{T}[I])=\lambda_{\max}(\mathcal{T}_{I})$.

###### Proof.

Let ${\bf x}\in\mathbb{R}^{|I|}$ be a unit eigenvector corresponding to
$\lambda_{\max}(\mathcal{T}[I])$. Then
$\lambda_{\max}(\mathcal{T}[I])=\mathcal{T}[I]{\bf x}^{k}$. Set $\widehat{{\bf
x}}\in\mathbb{R}^{n}$ as a vector such that $\widehat{x}_{i}=x_{i}$ if $i\in
I$, and $\widehat{x}_{i}=0$ if $i\in[n]\setminus I$. Then it is easy to see
that $\mathcal{T}[I]{\bf x}^{k}=\mathcal{T}_{I}\widehat{{\bf x}}^{k}$. By
Lemma 2.1 or 2.2, we have $\mathcal{T}_{I}\widehat{{\bf
x}}^{k}\leq\lambda_{\max}(\mathcal{T}_{I})$. It follows that

$\lambda_{\max}(\mathcal{T}[I])=\mathcal{T}[I]{\bf
x}^{k}=\mathcal{T}_{I}\widehat{{\bf
x}}^{k}\leq\lambda_{\max}(\mathcal{T}_{I}).$ (2.1)

By Lemma 2.3, $\lambda_{\max}(\mathcal{T}[I])\geq 0$. If
$\lambda_{\max}(\mathcal{T}_{I})=0$, then we have from (2.1) that
$\lambda_{\max}(\mathcal{T}[I])=0=\lambda_{\max}(\mathcal{T}_{I})$. Suppose
that $\lambda_{\max}(\mathcal{T}_{I})>0$. From (2.1), we have
$\lambda_{\max}(\mathcal{T}[I])\leq\lambda_{\max}(\mathcal{T}_{I})$.

Next, we prove the converse inequality. Let ${\bf y}\in\mathbb{R}^{n}$ be a
unit eigenvector corresponding to $\lambda_{\max}(\mathcal{T}_{I})$. Note that
$\mathcal{T}$ is a zero diagonal tensor. For $i\in[n]\setminus I$ and
$i_{2},\ldots,i_{k}\in[n]$, as entries of $\mathcal{T}_{I}$, we have
$t_{i\ldots i}=0$ and $t_{ii_{2}\ldots i_{k}}=0$. So

$\lambda_{\max}(\mathcal{T}_{I})y_{i}^{k-1}=\sum_{i_{2},\ldots,i_{k}\in[n]}t_{ii_{2}\dots
i_{k}}y_{i_{2}}\cdots y_{i_{k}}=0$

for each $i\in[n]\setminus I$, implying that $y_{i}=0$ for each
$i\in[n]\setminus I$ as $\lambda_{\max}(\mathcal{T}_{I})>0$. Let
$\widehat{{\bf y}}\in\mathbb{R}^{|I|}$ such that $\widehat{y}_{i}=y_{i}$ for
$i\in I$. Note that $\widehat{{\bf y}}$ is unit. For each $i\in I$, we have

$\displaystyle(\mathcal{T}[I]\widehat{{\bf y}}^{k-1})_{i}$
$\displaystyle=\sum_{i_{2},\ldots,i_{k}\in I}t_{ii_{2}\ldots
i_{k}}\widehat{y}_{i_{2}}\cdots\widehat{y}_{i_{k}}$
$\displaystyle=\sum_{i_{2},\ldots,i_{k}\in I}t_{ii_{2}\ldots
i_{k}}y_{i_{2}}\cdots y_{i_{k}}$
$\displaystyle=\sum_{i_{2},\ldots,i_{k}\in[n]}t_{ii_{2}\ldots
i_{k}}y_{i_{2}}\cdots y_{i_{k}}$
$\displaystyle=\lambda_{\max}(\mathcal{T}_{I})y_{i}^{k-1}$
$\displaystyle=\lambda_{\max}(\mathcal{T}_{I})\widehat{y}_{i}^{k-1}.$

This means that $\lambda_{\max}(\mathcal{T}_{I})$ is an H-eigenvalue of
$\mathcal{T}[I]$, so
$\lambda_{\max}(\mathcal{T}[I])\geq\lambda_{\max}(\mathcal{T}_{I})$. Now it
follows that $\lambda_{\max}(\mathcal{T}[I])=\lambda_{\max}(\mathcal{T}_{I})$.
∎

######  Lemma 2.5.

Let $\mathcal{T}$ be a zero diagonal symmetric tensor of order even $k$ and
dimension $n$, where $n,k\geq 2$. Let $\emptyset\neq I\subset[n]$. Then
$\lambda_{\min}(\mathcal{T}[I])=\lambda_{\min}(\mathcal{T}_{I})$.

###### Proof.

Let ${\bf x}\in\mathbb{R}^{|I|}$ be a unit eigenvector corresponding to
$\lambda_{\min}(\mathcal{T}[I])$. Then
$\lambda_{\min}(\mathcal{T}[I])=\mathcal{T}[I]{\bf x}^{k}$. Set
$\widetilde{{\bf x}}\in\mathbb{R}^{n}$ as a vector such that
$\widetilde{x}_{i}=x_{i}$ if $i\in I$, and $\widetilde{x}_{i}=0$ if
$i\in[n]\setminus I$. Then it is easy to see that $\mathcal{T}[I]{\bf
x}^{k}=\mathcal{T}_{I}\widetilde{{\bf x}}^{k}$. By Lemma 2.1, we have
$\lambda_{\min}(\mathcal{T}_{I})\leq\mathcal{T}_{I}\widetilde{{\bf x}}^{k}$.
Therefore

$\lambda_{\min}(\mathcal{T}[I])=\mathcal{T}[I]{\bf
x}^{k}=\mathcal{T}_{I}\widetilde{{\bf
x}}^{k}\geq\lambda_{\min}(\mathcal{T}_{I}).$

By Lemma 2.3, $\lambda_{\min}(\mathcal{T}[I])\leq 0$. If
$\lambda_{\min}(\mathcal{T}_{I})=0$, then it follows from the above
inequalities that
$\lambda_{\min}(\mathcal{T}_{I})=0=\lambda_{\min}(\mathcal{T}[I])$. Suppose
that $\lambda_{\min}(\mathcal{T}_{I})<0$.

Let ${\bf z}\in\mathbb{R}^{n}$ be a unit eigenvector corresponding to
$\lambda_{\min}(\mathcal{T}_{I})$. As $t_{i\ldots i}=0$ and $t_{ii_{2}\ldots
i_{k}}=0$ for each $i\in[n]\setminus I$ and $i_{2},\ldots,i_{k}\in[n]$, we
have $\lambda_{\min}(\mathcal{T}_{I})z_{i}^{k-1}=0$ for each $i\in[n]\setminus
I$. Then $z_{i}=0$ for each $i\in[n]\setminus I$. Let $\widehat{{\bf z}}$ be a
vector in $\mathbb{R}^{|I|}$ such that $\widehat{z}_{i}=z_{i}$ for $i\in I$.
Obviously, $\widehat{{\bf z}}$ is unit. Then for each $i\in I$,

$\displaystyle(\mathcal{T}[I]\widehat{{\bf z}}^{k-1})_{i}$
$\displaystyle=\sum_{i_{2},\ldots,i_{k}\in I}t_{ii_{2}\ldots
i_{k}}z_{i_{2}}\cdots z_{i_{k}}$
$\displaystyle=\sum_{i_{2},\ldots,i_{k}\in[n]}t_{ii_{2}\ldots
i_{k}}z_{i_{2}}\cdots z_{i_{k}}$
$\displaystyle=\lambda_{\min}(\mathcal{T}_{I})\widehat{z}_{i}^{k-1}.$

This shows that $\lambda_{\min}(\mathcal{T}_{I})$ is an H-eigenvalue of
$\mathcal{T}[I]$. Thus
$\lambda_{\min}(\mathcal{T}[I])\leq\lambda_{\min}(\mathcal{T}_{I})$. It
follows that $\lambda_{\min}(\mathcal{T}[I])=\lambda_{\min}(\mathcal{T}_{I})$.
∎

We also need the well known combinatorial identity in the following lemma.

######  Lemma 2.6.

Let $n$ and $k$ be positive integers with $1\leq k\leq n$. Then

$\sum_{i=k}^{n}{i\choose k}={n+1\choose k+1}.$

##  3 A formula for homogeneous polynomials of the form of inclusion-exclusion

In this section we establish a formula for homogeneous polynomials of the type
of Principle of Inclusion-Exclusion that will be used in the proofs.

######  Lemma 3.1.

Let $\mathcal{T}$ be a zero diagonal nonzero symmetric tensor of order $k$ and
dimension $n$, where $n,k\geq 2$. Let ${\bf x}$ be a unit $n$-dimensional
eigenvector and $\emptyset\neq I\subset[n]$. For positive integers $s$ and $m$
with $s+m\leq k$,

$\sum_{{i_{1},\ldots,i_{s}\in[n]\setminus I\atop i_{s+1},\ldots,i_{s+m}\in
I}\atop i_{s+m+1},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots
x_{k}=\sum_{\ell=0}^{m}(-1)^{\ell}{m\choose\ell}\sum_{i_{1},\ldots,i_{s+\ell}\in[n]\setminus
I\atop i_{s+\ell+1},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots
x_{i_{k}}.$ (3.1)

###### Proof.

We prove identity (3.1) by induction on $m$. It is obvious that for any $s\geq
1$ with $s+1\leq k$,

$\displaystyle\sum_{{i_{1},\ldots,i_{s}\in[n]\setminus I\atop i_{s+1}\in
I}\atop i_{s+2},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots x_{k}$
$\displaystyle=\sum_{i_{1},\ldots,i_{s}\in[n]\setminus I\atop
i_{s+1},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots
x_{i_{k}}-\sum_{i_{1},\ldots,i_{s+1}\in[n]\setminus I\atop
i_{s+2},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots x_{i_{k}}$
$\displaystyle=\sum_{\ell=0}^{1}(-1)^{\ell}{1\choose\ell}\sum_{i_{1},\ldots,i_{s+\ell}\in[n]\setminus
I\atop i_{s+\ell+1},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots
x_{i_{k}},$

proving (3.1) when $m=1$. Suppose that $1\leq j<k-s$ and identity (3.1)
follows for $m=j$. Suppose in the following that $m=j+1$. Then

$\displaystyle\quad\sum_{{i_{1},\ldots,i_{s}\in[n]\setminus I\atop
i_{s+1},\ldots,i_{s+m}\in I}\atop i_{s+m+1},\ldots,i_{k}\in[n]}t_{i_{1}\ldots
i_{k}}x_{i_{1}}\cdots x_{k}$ (3.2)
$\displaystyle=\sum_{{i_{1},\ldots,i_{s}\in[n]\setminus I\atop
i_{s+1},\ldots,i_{s+j+1}\in I}\atop
i_{s+j+2},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots x_{k}$
$\displaystyle=\sum_{{i_{1},\ldots,i_{s}\in[n]\setminus I\atop
i_{s+1},\ldots,i_{s+j}\in I}\atop i_{s+j+1},\ldots,i_{k}\in[n]}t_{i_{1}\ldots
i_{k}}x_{i_{1}}\cdots x_{k}-\sum_{{{i_{1},\ldots,i_{s}\in[n]\setminus I\atop
i_{s+1},\ldots,i_{s+j}\in I}\atop i_{s+j+1}\in[n]\setminus I}\atop
i_{s+j+2},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots x_{k}$
$\displaystyle=\sum_{{i_{1},\ldots,i_{s}\in[n]\setminus I\atop
i_{s+1},\ldots,i_{s+j}\in I}\atop i_{s+j+1},\ldots,i_{k}\in[n]}t_{i_{1}\ldots
i_{k}}x_{i_{1}}\cdots x_{k}-\sum_{{i_{1},\ldots,i_{s+1}\in[n]\setminus I\atop
i_{s+2},\ldots,i_{s+j+1}\in I}\atop
i_{s+j+2},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots x_{k},$

where the last equality in (3.2) follows because $\mathcal{T}$ is symmetric.
Now applying inductive hypothesis to the two summations of the last equation
in (3.2), using combinatorial identity
${j\choose\ell}={j+1\choose\ell}-{j\choose\ell-1}$ for $\ell\geq 1$, and by
algebraic calculation to get

$\displaystyle\quad\sum_{{i_{1},\ldots,i_{s}\in[n]\setminus I\atop
i_{s+1},\ldots,i_{s+m}\in I}\atop i_{s+m+1},\ldots,i_{k}\in[n]}t_{i_{1}\ldots
i_{k}}x_{i_{1}}\cdots x_{k}$
$\displaystyle=\sum_{\ell=0}^{j}(-1)^{\ell}{j\choose\ell}\sum_{i_{1},\ldots,i_{s+\ell}\in[n]\setminus
I\atop i_{s+\ell+1},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots
x_{i_{k}}-\sum_{\ell=0}^{j}(-1)^{\ell}{j\choose\ell}\sum_{i_{1},\ldots,i_{s+\ell+1}\in[n]\setminus
I\atop i_{s+\ell+2},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots
x_{i_{k}}$ $\displaystyle=\sum_{i_{1},\ldots,i_{s}\in[n]\setminus I\atop
i_{s+1},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots
x_{i_{k}}+\sum_{\ell=1}^{j}(-1)^{\ell}\left({j+1\choose\ell}-{j\choose\ell-1}\right)\sum_{i_{1},\ldots,i_{s+\ell}\in[n]\setminus
I\atop i_{s+\ell+1},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots
x_{i_{k}}$ $\displaystyle-\sum_{i_{1},\ldots,i_{s+1}\in[n]\setminus I\atop
i_{s+2},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots
x_{i_{k}}-\sum_{\ell=2}^{j+1}(-1)^{\ell-1}{j\choose\ell-1}\sum_{i_{1},\ldots,i_{s+\ell}\in[n]\setminus
I\atop i_{s+\ell+1},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots
x_{i_{k}}$ $\displaystyle=\sum_{i_{1},\ldots,i_{s}\in[n]\setminus I\atop
i_{s+1},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots
x_{i_{k}}+\sum_{\ell=1}^{j}(-1)^{\ell}{j+1\choose\ell}\sum_{i_{1},\ldots,i_{s+\ell}\in[n]\setminus
I\atop i_{s+\ell+1},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots
x_{i_{k}}$
$\displaystyle+(-1)^{j+1}\sum_{i_{1},\ldots,i_{s+j+1}\in[n]\setminus I\atop
i_{s+j+2},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots x_{i_{k}}.$

Now we have

$\sum_{{i_{1},\ldots,i_{s}\in[n]\setminus I\atop i_{s+1},\ldots,i_{s+m}\in
I}\atop i_{s+m+1},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots
x_{k}=\sum_{\ell=0}^{j+1}(-1)^{\ell}{j+1\choose\ell}\sum_{i_{1},\ldots,i_{s+\ell}\in[n]\setminus
I\atop i_{s+\ell+1},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots
x_{i_{k}}.$

This proves (3.1) when $m=j+1$. By induction, (3.1) follows. ∎

##  4 Largest H-eigenvalues

We are now ready to give our results on the largest H-eigenvalues of symmetric
tensors and uniform hypergraphs. First, we give the interlacing inequalities
for the largest H-eigenvalues.

######  Theorem 4.1.

Let $\mathcal{T}$ be a zero diagonal nonzero symmetric tensor of order $k$ and
dimension $n$, where $n,k\geq 2$. Suppose that $k$ is even or $\mathcal{T}$ is
nonnegative. Let ${\bf x}$ be a unit eigenvector corresponding to
$\lambda_{\max}(\mathcal{T})$. Let $\emptyset\neq I\subset[n]$. Then

$\displaystyle\lambda_{\max}(\mathcal{T})\left(1-k\sum_{i\in[n]\setminus
I}x_{i}^{k}\right)-\sum_{j=1}^{k-1}(-1)^{j}{k\choose
j+1}\sum_{i_{1},\ldots,i_{j+1}\in[n]\setminus I\atop
i_{j+2},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots x_{i_{k}}$
$\displaystyle\leq\lambda_{\max}(\mathcal{T}[I])$
$\displaystyle\leq\lambda_{\max}(\mathcal{T}).$

###### Proof.

As $\mathcal{T}$ is symmetric, we have by Lemma 3.1 that, for $1\leq j\leq
k-1$, one has

$\displaystyle\sum_{{i_{1},\ldots,i_{j}\in I\atop i_{j+1}\in[n]\setminus
I}\atop i_{j+2},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots
x_{i_{k}}$ $\displaystyle=\sum_{{i_{1}\in[n]\setminus I\atop
i_{2},\ldots,i_{j+1}\in I}\atop i_{j+2},\ldots,i_{k}\in[n]}t_{i_{1}\ldots
i_{k}}x_{i_{1}}\cdots x_{i_{k}}$
$\displaystyle=\sum_{\ell=0}^{j}(-1)^{\ell}{j\choose\ell}\sum_{i_{1},\ldots,i_{\ell+1}\in[n]\setminus
I\atop i_{\ell+2},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots
x_{i_{k}}.$

For any $i_{1}\in[n]\setminus I$, as $\lambda_{\max}(\mathcal{T})$ is an
H-eigenvalue of $\mathcal{T}$ associated to eigenvector ${\bf x}$, we have

$\sum_{i_{2},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots
x_{i_{k}}=\lambda_{\max}(\mathcal{T})x_{i_{1}}^{k}$

and hence

$\sum_{i_{1}\in[n]\setminus I\atop i_{2},\ldots,i_{k}\in[n]}t_{i_{1}\ldots
i_{k}}x_{i_{1}}\cdots
x_{i_{k}}=\lambda_{\max}(\mathcal{T})\sum_{i_{1}\in[n]\setminus
I}x_{i_{1}}^{k}.$

Therefore,

$\displaystyle\quad(\mathcal{T}-\mathcal{T}_{I}){\bf x}^{k}$
$\displaystyle=\sum_{i_{1},\ldots,i_{k}\in[n]}t_{i_{1}\ldots
i_{k}}x_{i_{1}}\cdots x_{i_{k}}-\sum_{i_{1},\ldots,i_{k}\in I}t_{i_{1}\ldots
i_{k}}x_{i_{1}}\cdots x_{i_{k}}$ $\displaystyle=\sum_{i_{1}\in[n]\setminus
I\atop i_{2},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots
x_{i_{k}}+\sum_{j=1}^{k-1}\sum_{{i_{1},\ldots,i_{j}\in I\atop
i_{j+1}\in[n]\setminus I}\atop i_{j+2},\ldots,i_{k}\in[n]}t_{i_{1}\ldots
i_{k}}x_{i_{1}}\cdots x_{i_{k}}$ $\displaystyle=\sum_{i_{1}\in[n]\setminus
I\atop i_{2},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots
x_{i_{k}}+$ $\displaystyle+\sum_{j=1}^{k-1}\left(\sum_{i_{1}\in[n]\setminus
I\atop i_{2},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots
x_{i_{k}}+\sum_{\ell=1}^{j}(-1)^{\ell}{j\choose\ell}\sum_{i_{1},\ldots,i_{\ell+1}\in[n]\setminus
I\atop i_{\ell+2},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots
x_{i_{k}}\right)$ $\displaystyle=k\sum_{i_{1}\in[n]\setminus I\atop
i_{2},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots
x_{i_{k}}+\sum_{j=1}^{k-1}\sum_{\ell=1}^{j}(-1)^{\ell}{j\choose\ell}\sum_{i_{1},\ldots,i_{\ell+1}\in[n]\setminus
I\atop i_{\ell+2},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots
x_{i_{k}}.$

By interchanging the order of summation in

$\sum_{j=1}^{k-1}\sum_{\ell=1}^{j}(-1)^{\ell}{j\choose\ell}\sum_{i_{1},\ldots,i_{\ell+1}\in[n]\setminus
I\atop i_{\ell+2},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots
x_{i_{k}},$

one has

$(\mathcal{T}-\mathcal{T}_{I}){\bf
x}^{k}=k\lambda_{\max}(\mathcal{T})\sum_{i\in[n]\setminus
I}x_{i}^{k}+\sum_{j=1}^{k-1}(-1)^{j}\sum_{\ell=j}^{k-1}{\ell\choose
j}\sum_{i_{1},\ldots,i_{j+1}\in[n]\setminus I\atop
i_{j+2},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots x_{i_{k}}.$

Now, by Lemma 2.6, one has

$\displaystyle(\mathcal{T}-\mathcal{T}_{I}){\bf x}^{k}$
$\displaystyle=k\lambda_{\max}(\mathcal{T})\sum_{i\in[n]\setminus
I}x_{i}^{k}+\sum_{j=1}^{k-1}(-1)^{j}{k\choose
j+1}\sum_{i_{1},\ldots,i_{j+1}\in[n]\setminus I\atop
i_{j+2},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots x_{i_{k}}.$
(4.1)

Thus, one has by Lemma 2.1 or 2.2 that

$\displaystyle\lambda_{\max}(\mathcal{T}_{I})$
$\displaystyle\geq\mathcal{T}_{I}{\bf x}^{k}=\mathcal{T}{\bf
x}^{k}-(\mathcal{T}-\mathcal{T}_{I}){\bf x}^{k}$
$\displaystyle=\lambda_{\max}(\mathcal{T})-\left(k\lambda_{\max}(\mathcal{T})\sum_{i\in[n]\setminus
I}x_{i}^{k}+\sum_{j=1}^{k-1}(-1)^{j}{k\choose
j+1}\sum_{i_{1},\ldots,i_{j+1}\in[n]\setminus I\atop
i_{j+2},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots
x_{i_{k}}\right)$
$\displaystyle=\lambda_{\max}(\mathcal{T})\left(1-k\sum_{i\in[n]\setminus
I}x_{i}^{k}\right)-\sum_{j=1}^{k-1}(-1)^{j}{k\choose
j+1}\sum_{i_{1},\ldots,i_{j+1}\in[n]\setminus I\atop
i_{j+2},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots x_{i_{k}}.$

Now, the first inequality follows from the above equation and Lemma 2.4.

Let ${\bf y}\in\mathbb{R}^{|I|}$ be a unit eigenvector corresponding to
$\lambda_{\max}(\mathcal{T}[I])$. Construct a unit vector ${\bf
z}\in\mathbb{R}^{n}$ such that $z_{i}=y_{i}$ if $i\in I$ and $z_{i}=0$ if
$i\in[n]\setminus I$. Note that $T_{I}{\bf z}^{k}=T[I]{\bf y}^{k}$. As
$\mathcal{T}$ is zero diagonal, and
$(\mathcal{T}-\mathcal{T}_{I})_{i_{1}\ldots i_{k}}=t_{i_{1}\ldots i_{k}}$ if
$\\{i_{1},\ldots,i_{k}\\}\cap([n]\setminus I)\neq\emptyset$ and $0$ otherwise,
we have $(\mathcal{T}-\mathcal{T}_{I}){\bf z}^{k}=0$. Thus by Lemma 2.1 or
2.2,

$\displaystyle\lambda_{\max}(\mathcal{T})\geq\mathcal{T}{\bf
z}^{k}=\mathcal{T}_{I}{\bf z}^{k}+(\mathcal{T}-\mathcal{T}_{I}){\bf
z}^{k}=T[I]{\bf y}^{k}+0=\lambda_{\max}(\mathcal{T}[I]).$

This proves the second inequality. ∎

######  Example 4.1.

Let $\mathcal{T}$ be a tensor of order $4$ and dimension $3$, where
$t_{1122}=t_{1212}=t_{1221}=t_{2121}=t_{2211}=t_{2112}=-1$,
$t_{1222}=t_{2122}=t_{2212}=t_{2221}=\frac{1}{2}$,
$t_{3222}=t_{2322}=t_{2232}=t_{2223}=1$, and otherwise, $t_{ijst}=0$. Let
$I=\\{2,3\\}$. By MATLAB, we have $\lambda_{\max}(\mathcal{T})=2.4043$ with
eigenvector ${\bf x}_{0}=(0.1632,1,0.7465)^{\top}$ and
$\lambda_{\max}(\mathcal{T}[I])=2.2795$. Note that ${\bf x}_{0}$ is not unit.
Let ${\bf x}=\frac{{\bf x}_{0}}{\|{\bf x}_{0}\|_{4}}$. The lower bound for
$\lambda_{\max}(\mathcal{T}[I])$ in Theorem 4.1 is equal to

$\displaystyle\quad\lambda_{\max}(\mathcal{T})\left(1-k\sum_{i\in[n]\setminus
I}x_{i}^{k}\right)-\sum_{j=1}^{k-1}(-1)^{j}{k\choose
j+1}\sum_{i_{1},\ldots,i_{j+1}\in[n]\setminus I\atop
i_{j+2},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots x_{i_{k}}$
$\displaystyle=\lambda_{\max}(\mathcal{T})\left(1-4x_{1}^{4}\right)-(-1)^{1}{4\choose
2}t_{1122}x_{1}^{2}x_{2}^{2}$
$\displaystyle=2.4043\times\left(1-4\times\frac{0.1632^{4}}{0.1632^{4}+1+0.7465^{4}}\right)+{4\choose
2}\times\frac{-1\times 0.1632^{2}\times 1^{2}}{0.1632^{4}+1+0.7465^{4}}$
$\displaystyle=2.2772.$

######  Example 4.2.

Let $\mathcal{T}$ be a tensor of order $3$ and dimension $5$, where
$t_{112}=t_{121}=t_{211}=\frac{1}{3}$, $t_{122}=t_{221}=t_{212}=\frac{1}{12}$,
$t_{113}=t_{131}=t_{311}=\frac{1}{6}$, $t_{223}=t_{232}=t_{322}=\frac{1}{12}$,
$t_{233}=t_{323}=t_{332}=\frac{1}{18}$,
$t_{123}=t_{132}=t_{213}=t_{231}=t_{321}=t_{312}=-\frac{1}{12}$
$t_{445}=t_{454}=t_{544}=\frac{1}{6}$, and otherwise, $t_{ijst}=0$. Let
$I=\\{1,2,4\\}$. By MATLAB, we have $\lambda_{\max}(\mathcal{T})=0.6894$ with
eigenvector ${\bf x}_{0}=(1,0.8241,0.4256,0,0)^{\top}$ and
$\lambda_{\max}(\mathcal{T}[I])=0.6387$. Let ${\bf x}=\frac{{\bf
x}_{0}}{\|{\bf x}_{0}\|_{3}}$ The lower bound for
$\lambda_{\max}(\mathcal{T}[I])$ in Theorem 4.1 is equal to

$\displaystyle\quad\lambda_{\max}(\mathcal{T})\left(1-3(x_{3}^{3}+x_{5}^{3})\right)-(-1)^{1}{3\choose
2}t_{332}x_{3}^{2}x_{2}$ $\displaystyle=0.6894\times\left(1-\frac{3\times
0.4256^{3}}{1+0.8241^{3}+0.4256^{3}}\right)+{3\choose
2}\times\frac{\frac{1}{18}\times 0.4256^{2}\times
0.8241}{1^{3}+0.8241^{3}+0.4256^{3}}$ $\displaystyle=0.6072.$

We remark that the second inequality has been known when $\mathcal{T}$ is
nonnegative [12, 14].

######  Theorem 4.2.

Let $\mathcal{T}$ be a symmetric, nonnegative zero diagonal tensor of order
$k$ and dimension $n$, where $k,n\geq 2$ and ${\bf x}$ be a unit nonnegative
eigenvector corresponding to $\lambda_{\max}(\mathcal{T})$. Let $\emptyset\neq
I\subset[n]$. If $\sum_{i\in I}x_{i}^{k}\neq 0$, then

$\displaystyle\frac{\lambda_{\max}(\mathcal{T})\left(1-k\sum_{i\in[n]\setminus
I}x_{i}^{k}\right)-\sum_{j=1}^{k-1}(-1)^{j}{k\choose
j+1}\sum_{i_{1},\ldots,i_{j+1}\in[n]\setminus I\atop
i_{j+2},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots
x_{i_{k}}}{\sum_{i\in I}x_{i}^{k}}$ $\displaystyle\leq\rho(\mathcal{T}[I])$
$\displaystyle\leq\rho(\mathcal{T}).$

###### Proof.

As $\mathcal{T}$ is nonnegative, we have by Proposition 1.1 that
$\lambda_{\max}(\mathcal{T})=\rho(\mathcal{T})$ and
$\lambda_{\max}(\mathcal{T}[I])=\rho(\mathcal{T}[I])$.

It is evident that the upper bound of $\rho(\mathcal{T}[I])$ follows from
Theorem 4.1, see also [14].

By the proof of Theorem 4.1, we have

$\displaystyle\mathcal{T}_{I}{\bf x}^{k}$
$\displaystyle=\rho(\mathcal{T})\left(1-k\sum_{i\in[n]\setminus
I}x_{i}^{k}\right)-\sum_{j=1}^{k-1}(-1)^{j}{k\choose
j+1}\sum_{i_{1},\ldots,i_{j+1}\in[n]\setminus I\atop
i_{j+2},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots x_{i_{k}}.$

Applying Lemma 2.2 by setting ${\bf y}\in\mathbb{R}^{|I|}$ with $y_{i}=x_{i}$
for $i\in I$, we have

$\displaystyle\rho(\mathcal{T}[I])$ $\displaystyle\geq\frac{\mathcal{T}[I]{\bf
y}^{k}}{\|{\bf y}\|_{k}^{k}}=\frac{\mathcal{T}_{I}{\bf x}^{k}}{\sum_{i\in
I}x_{i}^{k}}$
$\displaystyle=\frac{\rho(\mathcal{T})\left(1-k\sum_{i\in[n]\setminus
I}x_{i}^{k}\right)-\sum_{j=1}^{k-1}(-1)^{j}{k\choose
j+1}\sum_{i_{1},\ldots,i_{j+1}\in[n]\setminus I\atop
i_{j+2},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots
x_{i_{k}}}{\sum_{i\in I}x_{i}^{k}},$

proving the lower bound part. ∎

######  Example 4.3.

Let $\mathcal{T}$ be a tensor of order $3$ and dimension $3$, where
$t_{112}=t_{121}=t_{211}=\frac{1}{3}$, $t_{122}=t_{221}=t_{212}=\frac{1}{12}$,
$t_{113}=t_{131}=t_{311}=\frac{1}{6}$, $t_{223}=t_{232}=t_{322}=\frac{1}{12}$,
$t_{233}=t_{323}=t_{332}=\frac{1}{18}$ and otherwise, $t_{ijst}=0$. Let
$I=\\{1,2\\}$. By MATLAB, we have $\lambda_{\max}(\mathcal{T})=0.8143$ with
eigenvector ${\bf x}_{0}=(1,0.84,0.5866)^{\top}$ and
$\lambda_{\max}(\mathcal{T}[I])=0.6387$. Let ${\bf x}=\frac{{\bf
x}_{0}}{\|{\bf x}_{0}\|}$. The lower bound for
$\lambda_{\max}(\mathcal{T}[I])$ in Theorem 4.2 is equal to

$\displaystyle\quad\frac{\lambda_{\max}(\mathcal{T})\left(1-3x_{3}^{3}\right)-(-1)^{1}{3\choose
2}t_{331}x_{3}^{2}x_{1}}{x_{1}^{3}+x_{2}^{3}}$
$\displaystyle=\frac{0.8143\times\left(1-3\times\frac{0.5866^{3}}{1+0.84^{3}+0.5866^{3}}\right)+{3\choose
2}\times\frac{1}{18}\times\frac{0.5866^{2}\times
0.84}{1+0.84^{3}+0.5866^{3}}}{\frac{1+0.84^{3}}{1+0.84^{3}+0.5866^{3}}}$
$\displaystyle=0.6381.$

When the tensor $\mathcal{T}$ is weakly irreducible and nonnegative, by
Proposition 1.1, there is a unique unit positive eigenvector corresponding to
$\lambda_{\max}(\mathcal{T})=\rho(\mathcal{T})$. By the proof of Theorem 4.1,
$\rho(\mathcal{T}[I])<\rho(\mathcal{T})$. This together with Theorem 4.2
implies the following corollary.

######  Corollary 4.1.

Let $\mathcal{T}$ be a weakly irreducible, symmetric, nonnegative zero
diagonal tensor of order $k$ and dimension $n$, where $k,n\geq 2$ and ${\bf
x}$ be a unit positive eigenvector corresponding to $\rho(\mathcal{T})$. For
$\emptyset\neq I\subset[n]$,

$\displaystyle\frac{\rho(\mathcal{T})\left(1-k\sum_{i\in[n]\setminus
I}x_{i}^{k}\right)-\sum_{j=1}^{k-1}(-1)^{j}{k\choose
j+1}\sum_{i_{1},\ldots,i_{j+1}\in[n]\setminus I\atop
i_{j+2},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots
x_{i_{k}}}{\sum_{i\in I}x_{i}^{k}}$ $\displaystyle\leq\rho(\mathcal{T}[I])$
$\displaystyle<\rho(\mathcal{T}).$

For a tensor $\mathcal{T}$ of order $k$ and dimension $n$ with $k,n\geq 2$, we
denote by $R_{i}(\mathcal{T})=\sum_{i_{2},\dots,i_{k}\in[n]}t_{ii_{2}\dots
i_{k}}$ for $i\in[n]$, which is called the $i$th row sum of $\mathcal{T}$.

######  Example 4.4.

Let $\mathcal{T}$ be a weakly irreducible, symmetric, nonnegative zero
diagonal tensor of order $k$ and dimension $n$, where $k,n\geq 2$. Suppose
that $R_{i}(\mathcal{T})=\dots=R_{n}(\mathcal{T})=r$. By Proposition 1.1,
${\bf x}=n^{-\frac{1}{k}}(1,\dots,1)^{\top}$ is the unit positive eigenvector
corresponding to $\rho(\mathcal{T})$. Let $\emptyset\neq I\subset[n]$. Then

$\sum_{i\in[n]\setminus I}x_{i}^{k}=\frac{n-|I|}{n}$

and

$\sum_{j=1}^{k-1}(-1)^{j}{k\choose
j+1}\sum_{i_{1},\ldots,i_{j+1}\in[n]\setminus I\atop
i_{j+2},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots
x_{i_{k}}=\frac{1}{n}\sum_{j=1}^{k-1}(-1)^{j}{k\choose
j+1}\sum_{i_{1},\ldots,i_{j+1}\in[n]\setminus I\atop
i_{j+2},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}.$

By Corollary 4.1,

$\displaystyle\rho(\mathcal{T}[I])\geq\frac{\rho(\mathcal{T})\left(n-k(n-|I|)\right)-\sum_{j=1}^{k-1}(-1)^{j}{k\choose
j+1}\sum_{i_{1},\ldots,i_{j+1}\in[n]\setminus I\atop
i_{j+2},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}}{|I|}.$ (4.2)

Suppose further that $\mathcal{T}$ is a tensor of order $3$ and dimension $3$,
where $t_{121}=t_{112}=t_{211}=t_{233}=t_{323}=t_{332}=\frac{1}{3}$,
$t_{131}=t_{113}=t_{311}=t_{232}=t_{223}=t_{322}=\frac{1}{6}$, and otherwise,
$t_{ij\ell}=0$. It is easily checked that $\mathcal{T}$ is weakly irreducible
with $R_{1}(\mathcal{T})=R_{2}(\mathcal{T})=R_{3}(\mathcal{T})=1$. Then
$\rho(\mathcal{T})=1$ with eigenvector
$(\frac{1}{\sqrt[3]{3}},\frac{1}{\sqrt[3]{3}},\frac{1}{\sqrt[3]{3}})^{\top}$.
Let $I=\\{1,2\\}$. Let ${\bf y}=(y_{1},y_{2})$ be a unit positive eigenvector
corresponding to $\rho(\mathcal{T}[I])$. Then
$\rho(\mathcal{T}[I])y_{1}^{2}=\frac{2}{3}y_{1}y_{2}$ and
$\rho(\mathcal{T}[I])y_{2}^{2}=\frac{1}{3}y_{1}^{2}$, from which it follows
that $\rho(\mathcal{T}[I])^{3}=\frac{4}{27}$. So
$\rho(\mathcal{T}[I])=\frac{\sqrt[3]{4}}{3}\approx 0.5291$. This is compared
to the lower bound for $\rho(\mathcal{T}[I])$ given by (4.2), which is equal
to $0.5$.

Note that the spectral radius of an uniform hypergraph is at least $0$. For
hypergraphs, we have the following result, which generalizes/improves the
result of [18, Theorem 1].

######  Theorem 4.3.

Let $G$ be a $k$-uniform hypergraph with vertex set $[n]$, and ${\bf x}$ be a
unit nonnegative eigenvector corresponding to $\rho(G)$, where $n\geq k\geq
2$. Let $\emptyset\neq I\subset[n]$. Then

$\rho(G)\left(1-k\sum_{i\in I}x_{i}^{k}\right)+k\sum_{j=2}^{k}\sum_{e:|e\cap
I|=j}(j-1){\bf x}^{e}\leq\rho(G-I)\leq\rho(G).$

Moreover, if $\sum_{i\in I}x_{i}^{k}\neq 1$ (for example, if $G$ is
connected), then

$\rho(G)\frac{\left(1-k\sum_{i\in
I}x_{i}^{k}\right)+k\sum_{j=2}^{k}\sum_{e:|e\cap I|=j}(j-1){\bf
x}^{e}}{1-\sum_{i\in I}x_{i}^{k}}\leq\rho(G-I)\leq\rho(G).$

###### Proof.

Since $\mathcal{A}(G)$ and $\mathcal{A}(G-I)$ is nonnegative, we have
$\rho(G)=\lambda_{\max}(\mathcal{A}(G))$ and
$\rho(G-I)=\lambda_{\max}(\mathcal{A}(G-I))$. By Theorem 4.1, we have

$\rho(G)\left(1-k\sum_{i\in
I}x_{i}^{k}\right)+\sum_{j=1}^{k-1}(-1)^{j}{k\choose
j+1}\sum_{i_{1},\ldots,i_{j+1}\in I\atop
i_{j+2},\ldots,i_{k}\in[n]}a_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots
x_{i_{k}}\leq\rho(G-I)\leq\rho(G).$ (4.3)

Let $e=\\{i_{1},\ldots,i_{k}\\}\in E(G)$ with $e\cap
I=\\{i_{1},\ldots,i_{\ell}\\}$, where $0\leq\ell\leq k$. The coefficient of
${\bf x}^{e}$ in the lower bound given by (4.3) is $0$ if $\ell=0,1$. If
$\ell\geq 2$, it is

$\displaystyle-\sum_{j=1}^{\ell-1}(-1)^{j}{k\choose j+1}\left({\ell\choose
j+1}(j+1)!(k-j-1)!\right)\frac{1}{(k-1)!}$
$\displaystyle=-k\sum_{j=1}^{\ell-1}(-1)^{j}{\ell\choose j+1}$
$\displaystyle=k\sum_{j=2}^{\ell}(-1)^{j}{\ell\choose j}$
$\displaystyle=k(\ell-1).$

Thus

$-\sum_{j=1}^{k-1}(-1)^{j}{k\choose j+1}\sum_{i_{1},\ldots,i_{j+1}\in I\atop
i_{j+2},\ldots,i_{k}\in[n]}a_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots
x_{i_{k}}=k\sum_{\ell=2}^{k}\sum_{e:|e\cap I|=\ell}(\ell-1){\bf x}^{e}.$ (4.4)

Now, by (4.3) and (4.4), the first part follows.

Suppose that $\sum_{i\in I}x_{i}^{k}\neq 1$. By Theorem 4.1 and the above
argument, the second part follows. ∎

######  Example 4.5.

Let $G$ be a connected $k$-uniform regular hypergraph with vertex set $[n]$.
Then ${\bf x}=n^{-\frac{1}{k}}(1,\dots,1)^{\top}$ is the unit nonnegative
eigenvector corresponding to $\rho(G)$. Let $\emptyset\neq I\subset[n]$. Then

$\rho(G-I)\geq\rho(G)\frac{\left(n-k|I|\right)+k\sum_{j=2}^{k}\sum_{e:|e\cap
I|=j}(j-1)}{n-|I|}$

######  Corollary 4.2.

Let $G$ be a connected $k$-uniform hypergraph with vertex set $[n]$, and ${\bf
x}$ be a unit positive eigenvector corresponding to $\rho(G)$, where $n\geq
k\geq 2$. Let $\emptyset\neq I\subseteq[n]$. Then

$\sum_{i\in
I}x_{i}^{k}\leq\frac{1}{k}+\frac{1}{\rho(G)}\sum_{j=2}^{k}\sum_{e:|e\cap
I|=j}(j-1){\bf x}^{e}.$

In particular, for any $v\in V(G)$,

$\displaystyle x_{v}\leq\sqrt[k]{\frac{1}{k}}.$ (4.5)

###### Proof.

Let $\mathcal{T}=\mathcal{A}(G)$. By (4.1) and (4.4), we have

$(\mathcal{T}-\mathcal{T}_{[n]\setminus I}){\bf x}^{k}=k\rho(G)\sum_{i\in
I}x_{i}^{k}-k\sum_{j=2}^{k}\sum_{e:|e\cap I|=j}(j-1){\bf x}^{e}.$

Then

$\displaystyle 0$ $\displaystyle\leq\mathcal{T}_{[n]\setminus I}{\bf x}^{k}$
$\displaystyle=(\mathcal{T}-(\mathcal{T}-\mathcal{T}_{[n]\setminus I})){\bf
x}^{k}$ $\displaystyle=\rho(G)-\left(k\rho(G)\sum_{i\in
I}x_{i}^{k}-k\sum_{j=2}^{k}\sum_{e:|e\cap I|=j}(j-1){\bf x}^{e}\right)$
$\displaystyle=\left(1-k\sum_{i\in
I}x_{i}^{k}\right)\rho(G)+k\sum_{j=2}^{k}\sum_{e:|e\cap I|=j}(j-1){\bf
x}^{e}.$

The result follows. ∎

We remak that (4.5) has been observed in [22, Proposition 7.21].

######  Theorem 4.4.

Let $G$ be a $k$-uniform hypergraph with vertex set $[n]$, and ${\bf x}$ be a
unit nonnegative eigenvector corresponding to $\rho(G)$, where $n\geq k\geq
2$. Let $v\in V(G)$ and $\mathbf{y}$ be the restriction of $\mathbf{x}$ on
$[n]\setminus\\{v\\}$. Then

$\left(1-kx_{v}^{k}\right)\rho(G)\leq\rho(G-v)\leq\rho(G).$

Moreover, if $x_{v}^{k}\neq 1$, then

$\rho(G-v)\geq\frac{1-kx_{v}^{k}}{1-x_{v}^{k}}\rho(G),$ (4.6)

if $G$ is connected, then equality holds in (4.6) if and only if $\mathbf{y}$
is an eigenvector of $G-v$ associated with $\rho(G-v)$.

###### Proof.

Taking $I=\\{v\\}\subseteq V(G)$ in Theorem 4.3, we immediately have the
inequalities.

Suppose that $G$ is connected. By Proposition 1.1, $\mathbf{x}$ is positive,
and then $1-x_{v}^{k}>0$. Note that

$\displaystyle\rho(G)=k\sum_{e\in E(G)\setminus
E_{v}(G)}\mathbf{x}^{e}+k\sum_{e\in
E_{v}(G)}\mathbf{x}^{e}=\mathcal{A}(G-v)\mathbf{y}^{k}+k\rho(G)x_{v}^{k}.$

From this, it is easy to see that

$(1-kx_{v}^{k})\rho(G)=\mathcal{A}(G-v)\mathbf{y}^{k}\Leftrightarrow\rho(G-v)=\frac{\mathcal{A}(G-v)\mathbf{y}^{k}}{1-x_{v}^{k}}.$

So, by Lemma 2.2, equality holds in (4.6) if and only if $\mathbf{y}$ is an
eigenvector of $G-v$ associated with $\rho(G-v)$. ∎

If $G$ is a connected graph with at least two vertices, then for any $v\in
V(G)$, $\rho(G-v)\geq\frac{1-2x_{v}^{2}}{1-x_{v}^{2}}\rho(G)$, which is known
[30]. If $v$ is the vertex such that $x_{v}$ is minimum, then the inequality
$\rho(G-v)\geq\frac{1-kx_{v}^{k}}{1-x_{v}^{k}}$ was known [13] (for $k=2$
[23]).

######  Theorem 4.5.

Let $G$ be a connected $k$-uniform linear hypergraph with vertex set $[n]$,
where $n\geq k\geq 2$. For any $v\in[n]$,

$\rho(G-v)\geq\rho(G)-\sqrt[k-1]{\frac{d_{G}(v)}{\rho(G)}}$

with equality if and only if $v$ is adjacent to any other vertex, and $G-v$ is
regular.

###### Proof.

Let $\mathbf{x}$ be the unit positive eigenvector corresponding to $\rho(G)$
and $\mathbf{y}$ be the restriction of $\mathbf{x}$ on $V(G)\setminus\\{v\\}$.
By Theorem 4.4,

$\rho(G-v)\geq\frac{1-kx_{v}^{k}}{1-x_{v}^{k}}\rho(G)$

with equality if and only if $\mathbf{y}$ is an eigenvector of $G-v$
associated with $\rho(G-v)$. Let $\rho=\rho(G)$ and $d=d_{G}(v)$. As $G$ is
linear, we have by [20, Theorem 3.1] and its proof that

$x_{v}\leq\frac{1}{\sqrt[k]{1+(k-1)\left(\frac{\rho^{k}}{d}\right)^{\frac{1}{k-1}}}}$

with equality only if the entries of $\mathbf{y}$ are all equal. So

$\rho(G-v)\geq\frac{\rho^{\frac{k}{k-1}}-d^{\frac{1}{k-1}}}{\rho^{\frac{1}{k-1}}},$

from which we have

$\rho(G)-\rho(G-v)\leq\rho-\frac{\rho^{\frac{k}{k-1}}-d^{\frac{1}{k-1}}}{\rho^{\frac{1}{k-1}}}.$

Thus we have the desired upper bound for $\rho(G)-\rho(G-v)$.

Suppose that the upper bound for $\rho(G)-\rho(G-v)$ is achieved. By the above
argument, $\mathbf{y}$ is an eigenvector of $G-v$ associated with $\rho(G-v)$,
and the entries of $\mathbf{y}$ are all equal. Thus $G-v$ is regular. Note
that $\left(\mathcal{A}(G-v)\mathbf{x}^{k-1}\right)_{w}=\rho(G-v)x_{w}^{k-1}$
for $w\in[n]\setminus\\{v\\}$. Then for any $w\in[n]\setminus\\{v\\}$, we have

$\displaystyle\rho(G)x^{k-1}_{w}$
$\displaystyle=\left(\mathcal{A}(G)\mathbf{x}^{k-1}\right)_{w}$
$\displaystyle=\left(\mathcal{A}(G-v)\mathbf{x}^{k-1}\right)_{w}+\sum_{e\in
E_{w}(G)\cap E_{v}(G)}\mathbf{x}^{e\setminus\\{w\\}}$
$\displaystyle=\rho(G-v)x_{w}^{k-1}+\sum_{e\in E_{w}(G)\cap
E_{v}(G)}\mathbf{x}^{e\setminus\\{w\\}}.$

Thus $\sum_{e\in E_{w}(G)\cap
E_{v}(G)}\mathbf{x}^{e\setminus\\{w\\}}=\left(\rho(G)-\rho(G-v)\right)x^{k-1}_{w}>0$.
This implies that $v$ is adjacent to any vertex of $G$.

Conversely, suppose that $v$ is adjacent to any other vertex, and $G-v$ is
regular. As $G$ is linear, we have $d_{G}(v)=\frac{n-1}{k-1}$. Assume the
degree of any vertex of $G-v$ is $r$.

Let $t$ be the largest positive number satisfying $t(t-r)^{k-1}=d_{G}(v)$.
Evidently, $t>r$. Let

$a=\sqrt[k]{(t-r)^{k}+n-1}$

We construct a vector $\mathbf{\hat{x}}$ on $V(G)$ with each entry
corresponding to any vertex different from $v$ to be $y$ such that
$\hat{x}_{v}=\frac{t-r}{a}$ and $y=\frac{1}{a}$, where Then
$\hat{x}_{v}^{k}+(n-1)y^{k}=1$, $t\hat{x}_{v}^{k-1}=d_{G}(v)y^{k-1}$ and
$(t-r)y=\hat{x}_{v}$. Thus, $\mathbf{\hat{x}}$ is a positive eigenvector of
$\mathcal{A}(G)$ associated with an H-eigenvalue $t$. Note that
$\mathcal{A}(G)$ is weakly irreducible as $G$ is connected. By Proposition
1.1, $t=\rho(G)$. Now by Theorem 4.4,

$\rho(G)-\rho(G-v)=\frac{\rho(G)(k-1)\hat{x}_{v}^{k}}{1-\hat{x}_{v}^{k}}=\frac{\hat{x}_{v}}{y}=\sqrt[k-1]{\frac{d_{G}(v)}{\rho(G)}}.$

That is, the upper bound for $\rho(G)-\rho(G-v)$ is attained. ∎

For a Steiner system $S(t,k,n)$, the degree (known as replication number) of
any vertex is $\frac{n-1}{k-1}$ and it possesses $\frac{n(n-1)}{k(k-1)}$
edges.

######  Theorem 4.6.

Let $G$ be a connected $k$-uniform linear hypergraph with vertex set $[n]$,
where $n\geq k\geq 2$. Let $\gamma(G)=\max\\{\rho(G-v):v\in[n]\\}$. Then

$\gamma(G)\geq\rho(G)-\sqrt[k-1]{\frac{\delta(G)}{\rho(G)}}\geq\rho(G)-1$

and either lower bound is attained if and only if $G$ is a Steiner system
$S(2,k,n)$.

###### Proof.

Let $v^{*}$ be the vertex of $G$ with the minimum degree, $\delta(G)$. Then
$\gamma(G)\geq\rho(G-v^{*})$. By Lemma 4.5, we have

$\rho(G)-\gamma(G)\leq\rho(G)-\rho(G-v^{*})\leq\sqrt[k-1]{\frac{d_{G}(v^{*})}{\rho(G)}}=\sqrt[k-1]{\frac{\delta(G)}{\rho(G)}}\leq
1.$ (4.7)

Suppose that $\rho(G)-\gamma(G)=\sqrt[k-1]{\frac{\delta(G)}{\rho(G)}}$. Then
the first and the second inequalities in (4.7) are equalities. By Theorem 4.5,
$v^{*}$ is adjacent to any other vertex and $G-v^{*}$ is regular. Since $G$ is
a linear hypergraph, the degree of $v^{*}$ in $G$ is $\frac{n-1}{k-1}$, so $G$
is $\frac{n-1}{k-1}$-regular. That is, every $2$-element vertex subset is
contained in precisely one edge. Hence, $G$ is a Steiner system $S(2,k,n)$.

Suppose that $\rho(G)-\gamma(G)=1$. Then all inequalities in (4.7) are
equalities. So $G$ is $\frac{n-1}{k-1}$-regular. As $G$ is a linear hypergraph
again, every $2$-element vertex subset is contained in precisely one edge.
Hence, $G$ is a Steiner system $S(2,k,n)$.

Conversely, suppose that $G$ is a Steiner system $S(2,k,n)$. Then the degree
of any vertex of $G$ is $\frac{n-1}{k-1}$. For any vertex $v$, the degree of
any vertex of $G-v$ is $\frac{n-1}{k-1}-1=\frac{n-k}{k-1}$. So
$\rho(G)-\gamma(G)=\sqrt[k-1]{\frac{\delta(G)}{\rho(G)}}=1$. ∎

The following theorem generalizes the result in [32] from graphs to
hypergraphs.

######  Theorem 4.7.

Let $G$ be a $k$-uniform hypergraph with vertex set $[n]$. Let $E\subseteq
E(G)$. Set ${\bf x}$ and ${\bf y}$ the unit eigenvectors of $\rho(G)$ and
$\rho(G-E)$, respectively. Then

$\rho(G)-k\sum_{e\in E}{\bf x}^{e}\leq\rho(G-E)\leq\rho(G)-k\sum_{e\in E}{\bf
y}^{e}.$

###### Proof.

We need only to show

$k\sum_{e\in E}{\bf y}^{e}\leq\rho(G)-\rho(G-E)\leq k\sum_{e\in E}{\bf
x}^{e}.$

Since $\rho(G)=\mathcal{A}(G){\bf x}^{k}$, we have

$\displaystyle\rho(G-E)\geq$ $\displaystyle\mathcal{A}(G-E){\bf x}^{k}$
$\displaystyle=k\sum_{e\in E(G)-E}{\bf x}^{e}$ $\displaystyle=k\sum_{e\in
E(G)}{\bf x}^{e}-k\sum_{e\in E}{\bf x}^{e}$ $\displaystyle=\rho(G)-k\sum_{e\in
E}{\bf x}^{e},$

from which the upper bound for $\rho(G)-\rho(G-E)$ follows.

Since $\rho(G-E)=\mathcal{A}(G-E){\bf y}^{k}=k\sum_{e\in E(G)-E}{\bf y}^{e}$,
we have

$\displaystyle\rho(G)$ $\displaystyle\geq\mathcal{A}(G){\bf y}^{k}$
$\displaystyle=k\sum_{e\in E(G)}{\bf y}^{e}$ $\displaystyle=k\sum_{e\in
E(G)-E}{\bf y}^{e}+k\sum_{e\in E}{\bf y}^{e}$
$\displaystyle=\rho(G-E)+k\sum_{e\in E}{\bf y}^{e},$

from which the lower bound for $\rho(G)-\rho(G-E)$ follows. ∎

######  Example 4.6.

Let $G$ be a connected $k$-uniform regular hypergraph with vertex set $[n]$.
Let $e\in E(G)$. Note that the unite positive vector corresponding to
$\rho(G)$ is $(\frac{1}{\sqrt[k]{n}},\ldots,\frac{1}{\sqrt[k]{n}})^{\top}$. By
Theorem 4.7, we have

$\rho(G-e)\geq\rho(G)-\frac{k}{n}.$

######  Example 4.7.

Let $n$ be a positive integer at least $3$. Let $C_{2n}^{3}$ be the
$3$-uniform hypercycle with vertex set $\\{v_{i}:1\leq i\leq 2n\\}$ and edge
set $\\{\\{v_{2i-1},v_{2i},v_{2i+1}\\}:1\leq i\leq n\\}$, where
$v_{2n+1}=v_{1}$. It is easily checked that $\rho(C_{2n}^{3})=2^{\frac{2}{3}}$
with unit eigenvector ${\bf x}$ such that $x_{v}=\sqrt[3]{\frac{2}{3n}}$ if
$v=v_{2i-1}$ for each $1\leq i\leq n$ and $x_{v}=\sqrt[3]{\frac{1}{3n}}$ if
$v=v_{2i}$ for each $1\leq i\leq n$. Let $P_{2n-1}^{3}$ be the $3$-uniform
hyperpath with vertex set $\\{v_{i}:1\leq i\leq 2n-1\\}$ and edge set
$\\{\\{v_{2i-1},v_{2i},v_{2i+1}\\}:1\leq i\leq n-1\\}$. Then
$\rho(P_{2n-1}^{3})=2^{\frac{2}{3}}\cos^{\frac{2}{3}}\frac{\pi}{n+1}$.
Obviously, $P_{2n-1}^{3}\cong C_{2n}^{3}-v_{2n}$. In Theorem 4.1 or Example
4.6, the lower bound on $\rho(P_{2n-1}^{3})$ is $\frac{3n-3}{3n-1}\cdot
2^{\frac{2}{3}}$. Obviously, the ratio of the value of $\rho(P_{2n-1}^{3})$
and the lower bound given above tends to $1$ when $n\to\infty$.

##  5 Least H-eigenvalues

In this section we study least H-eigenvalues of symmetric tensors and uniform
hypergraphs. For a symmetric tensor $\mathcal{T}$ with at least H-eigenvalue,
we call a unit eigenvector of $\mathcal{T}$ associated to
$\lambda_{\min}(\mathcal{T})$ a least eigenvector of $\mathcal{T}$. In
particular, a least eigenvector of a $k$-uniform hypergraph $G$ is a least
eigenvector of $\mathcal{A}(G)$. The results in [33] are generalized to
tensors and uniform hypergraphs. We note that $k$ is always even in this
section.

######  Theorem 5.1.

Let $\mathcal{T}$ be a zero diagonal symmetric tensor of order $k$ and
dimension $n$, where $k$ is even and $n,k\geq 2$. Let ${\bf x}$ be a least
eigenvector of $\mathcal{T}$. Let $\emptyset\neq I\subset[n]$. Then

$\displaystyle\lambda_{\min}(\mathcal{T})$
$\displaystyle\leq\lambda_{\min}(\mathcal{T}[I])$
$\displaystyle\leq\lambda_{\min}(\mathcal{T})\left(1-k\sum_{i\in[n]\setminus
I}x_{i}^{k}\right)-\sum_{j=1}^{k-1}(-1)^{j}{k\choose
j+1}\sum_{i_{1},\ldots,i_{j+1}\in[n]\setminus I\atop
i_{j+2},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots x_{i_{k}}.$

Moreover, if $\sum_{i\in I}x_{i}^{k}\neq 0$, then

$\lambda_{\min}(\mathcal{T}[I])\leq\frac{\lambda_{\min}(\mathcal{T})\left(1-k\sum_{i\in[n]\setminus
I}x_{i}^{k}\right)-\sum_{j=1}^{k-1}(-1)^{j}{k\choose
j+1}\sum_{i_{1},\ldots,i_{j+1}\in[n]\setminus I\atop
i_{j+2},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots
x_{i_{k}}}{\sum_{i\in I}x_{i}^{k}}.$

###### Proof.

Since $\mathcal{T}$ is symmetric, we have by the same argument as in Theorem
4.1 that

$\displaystyle(\mathcal{T}-\mathcal{T}_{I}){\bf x}^{k}$
$\displaystyle=k\lambda_{\min}(\mathcal{T})\sum_{i\in[n]\setminus
I}x_{i}^{k}+\sum_{j=1}^{k-1}(-1)^{j}{k\choose
j+1}\sum_{i_{1},\ldots,i_{j+1}\in[n]\setminus I\atop
i_{j+2},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots x_{i_{k}}.$

Note that $\lambda_{\min}(\mathcal{T})=\mathcal{T}{\bf x}^{k}$. Thus by Lemmas
2.5 and 2.1,

$\displaystyle\lambda_{\min}(\mathcal{T}[I])$
$\displaystyle=\lambda_{\min}(\mathcal{T}_{I})$
$\displaystyle\leq\mathcal{T}_{I}{\bf x}^{k}$ $\displaystyle=\mathcal{T}{\bf
x}^{k}-(\mathcal{T}-\mathcal{T}_{I}){\bf x}^{k}$
$\displaystyle=\lambda_{\min}(\mathcal{T})-\left(k\lambda_{\min}(\mathcal{T})\sum_{i\in[n]\setminus
I}x_{i}^{k}+\sum_{j=1}^{k-1}(-1)^{j}{k\choose
j+1}\sum_{i_{1},\ldots,i_{j+1}\in[n]\setminus I\atop
i_{j+2},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots
x_{i_{k}}\right)$
$\displaystyle=\lambda_{\min}(\mathcal{T})\left(1-k\sum_{i\in[n]\setminus
I}x_{i}^{k}\right)-\sum_{j=1}^{k-1}(-1)^{j}{k\choose
j+1}\sum_{i_{1},\ldots,i_{j+1}\in[n]\setminus I\atop
i_{j+2},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots x_{i_{k}}.$

Let ${\bf y}\in\mathbb{R}^{|I|}$ be a unit eigenvector corresponding to
$\lambda_{\min}(\mathcal{T}[I])$. Construct a new unit vector ${\bf
z}\in\mathbb{R}^{n}$ such that $z_{i}=y_{i}$ if $i\in I$ and $z_{i}=0$
otherwise. Then by Lemma 2.1

$\displaystyle\lambda_{\min}(\mathcal{T})\leq\mathcal{T}{\bf
z}^{k}=\mathcal{T}_{I}{\bf z}^{k}+(\mathcal{T}-\mathcal{T}_{I}){\bf
z}^{k}=\mathcal{T}[I]{\bf y}^{k}+0=\lambda_{\min}(\mathcal{T}[I])$

as desired. This proves the first part.

Suppose that $\sum_{i\in I}x_{i}^{k}\neq 0$. By the above argument, we have

$\mathcal{T}_{I}{\bf
x}^{k}=\lambda_{\min}(\mathcal{T})\left(1-k\sum_{i\in[n]\setminus
I}x_{i}^{k}\right)-\sum_{j=1}^{k-1}(-1)^{j}{k\choose
j+1}\sum_{i_{1},\ldots,i_{j+1}\in[n]\setminus I\atop
i_{j+2},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots x_{i_{k}}.$

Let ${\bf w}$ be a vector in $\mathbb{R}^{|I|}$ such that $w_{i}=x_{i}$ if
$i\in I$. Thus by Lemma 2.1,

$\displaystyle\lambda_{\min}(\mathcal{T}[I])$
$\displaystyle\leq\frac{\mathcal{T}[I]{\bf w}^{k}}{\|{\bf
w}\|_{k}^{k}}=\frac{\mathcal{T}_{I}{\bf x}^{k}}{\sum_{i\in I}x_{i}^{k}}$
$\displaystyle=\frac{\lambda_{\min}(\mathcal{T})\left(1-k\sum_{i\in[n]\setminus
I}x_{i}^{k}\right)-\sum_{j=1}^{k-1}(-1)^{j}{k\choose
j+1}\sum_{i_{1},\ldots,i_{j+1}\in[n]\setminus I\atop
i_{j+2},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots
x_{i_{k}}}{\sum_{i\in I}x_{i}^{k}},$

proving the second part. ∎

######  Example 5.1.

Let $\mathcal{T}$ be a tensor of order $4$ and dimension $3$, where
$t_{1122}=t_{1212}=t_{1221}=t_{2121}=t_{2211}=t_{2112}=1$,
$t_{1222}=t_{2122}=t_{2212}=t_{2221}=3$,
$t_{3222}=t_{2322}=t_{2232}=t_{2223}=3$, and otherwise, $t_{ijst}=0$. Let
$I=\\{2,3\\}$. By MATLAB, we have $\lambda_{\min}(\mathcal{T})=-9.9307$ with
eigenvector ${\bf x}_{0}=(-0.5239,1,-0.671)^{\top}$ and
$\lambda_{\min}(\mathcal{T}[I])=-6.8385$. Let ${\bf x}=\frac{{\bf
x}_{0}}{\|{\bf x}_{0}\|_{4}}$. The first upper bound for
$\rho(\mathcal{T}[I])$ in Theorem 5.1 is equal to

$\displaystyle\quad\lambda_{\min}(\mathcal{T})\left(1-k\sum_{i\in[n]\setminus
I}x_{i}^{k}\right)-\sum_{j=1}^{k-1}(-1)^{j}{k\choose
j+1}\sum_{i_{1},\ldots,i_{j+1}\in[n]\setminus I\atop
i_{j+2},\ldots,i_{k}\in[n]}t_{i_{1}\ldots i_{k}}x_{i_{1}}\cdots x_{i_{k}}$
$\displaystyle=\lambda_{\min}(\mathcal{T})\left(1-4x_{1}^{4}\right)-(-1)^{1}{4\choose
2}t_{1122}x_{1}^{2}x_{2}^{2}$
$\displaystyle=-9.9307\times\left(1-\frac{4\times(-0.5239)^{4}}{(-0.5239)^{4}+1+(-0.671)^{4}}\right)-(-1)^{1}{4\choose
2}\times\frac{1\times(-0.5239)^{2}\times 1^{2}}{(-0.5239)^{4}+1+(-0.671)^{4}}$
$\displaystyle=-6.3007.$

and the second upper bound in Theorem 5.1 is equal to

$\displaystyle\frac{\lambda_{\min}(\mathcal{T})\left(1-4x_{1}^{4}\right)-(-1)^{1}{4\choose
2}t_{1122}x_{1}^{2}x_{2}^{2}}{x_{2}^{k}+x_{3}^{k}}=\frac{-6.3007}{\frac{1+(-0.671)^{4}}{(-0.5239)^{4}+1+(-0.671)^{4}}}=-6.6954.$

By similar argument as in Corollary 4.3, we have

######  Corollary 5.1.

Let $G$ be a $k$-uniform hypergraph with vertex set $[n]$, where $k$ is even,
and $n,k\geq 2$ Let ${\bf x}$ be a least eigenvector of $\mathcal{A}(G)$. If
$\emptyset\neq I\subset[n]$, then

$\lambda(G)\leq\lambda(G-I)\leq\lambda(G)\left(1-k\sum_{i\in
I}x_{i}^{k}\right)+k\sum_{j=2}^{k}\sum_{e:|e\cap I|=j}(j-1){\bf x}^{e}.$

Moreover, if $\sum_{i\in I}x_{i}^{k}\neq 0$, then

$\lambda(G-I)\leq\frac{\lambda(G)\left(1-k\sum_{i\in
I}x_{i}^{k}\right)+k\sum_{j=2}^{k}\sum_{e:|e\cap I|=j}(j-1){\bf
x}^{e}}{\sum_{i\in I}x_{i}^{k}}.$

######  Example 5.2.

Let $G$ be a $4$-uniform hypergraph with vertex set $[6]$, where
$E(G)$$=\\{\\{1,2,3,4\\}$, $\\{3,4,5,6\\}$,$\\{1,3,4,5\\}\\}$. Let
$I=\\{5,6\\}$. Obviously, $\lambda(G-I)=-1$. By MATLAB, $\lambda(G)=-2.1908$
with eigenvector ${\bf x}_{0}=(-0.9112,0.7465,1,1,0.9112,-0.7465)^{\top}$. Let
${\bf x}=\frac{{\bf x}_{0}}{\|{\bf x}_{0}\|_{4}}$. The first upper bound for
$\lambda(G-I)$ in above corollary is

$\displaystyle\quad\lambda(G)\left(1-4\sum_{i\in
I}x_{i}^{4}\right)+4\sum_{j=2}^{4}\sum_{e:|e\cap I|=j}(j-1){\bf x}^{e}$
$\displaystyle=\lambda(G)\left(1-4(x_{5}^{4}+x_{6}^{4})\right)+4(2-1)x_{3}x_{4}x_{5}x_{6}$
$\displaystyle=-2.1908\times\left(1-\frac{4\times(0.9112^{4}+(-0.7465)^{4})}{(-0.9112)^{4}+0.7465^{4}+1+1+0.9112^{4}+(-0.7465)^{4}}\right)$
$\displaystyle\quad+4\times\frac{1\times 1\times
0.9112\times(-0.7465)}{(-0.9112)^{4}+0.7465^{4}+1+1+0.9112^{4}+(-0.7465)^{4}}$
$\displaystyle=-0.6803,$

and the second one is equal to

$\frac{\lambda(G)\left(1-4(x_{5}^{4}+x_{6}^{4})\right)+4(2-1)x_{3}x_{4}x_{5}x_{6}}{x_{5}^{4}+x_{6}^{4}}=-0.9071.$

######  Theorem 5.2.

Let $G$ be a $k$-uniform hypergraph with vertex set $[n]$, where $k$ is even
and $n,k\geq 2$. Let $E\subseteq E(G)$. Let ${\bf x}$ and ${\bf y}$ be the
least eigenvectors of $G$ and $G-E$, respectively. Then

$\lambda(G)-k\sum_{e\in E}{\bf
y}^{e}\leq\lambda(G-E)\leq\lambda(G)-k\sum_{e\in E}{\bf x}^{e}.$

###### Proof.

It suffices to show that

$k\sum_{e\in E}{\bf x}^{e}\leq\lambda(G)-\lambda(G-E)\leq k\sum_{e\in E}{\bf
y}^{e}.$

Since $\lambda(G)=\mathcal{A}(G){\bf x}^{k}=k\sum_{e\in E(G)}{\bf x}^{e}$, we
have

$\displaystyle\lambda(G-E)$ $\displaystyle\leq\mathcal{A}(G-E){\bf
x}^{k}=k\sum_{e\in E(G)-E}{\bf x}^{e}=k\sum_{e\in E(G)}{\bf x}^{e}-k\sum_{e\in
E}{\bf x}^{e}$ $\displaystyle=\lambda(G)-k\sum_{e\in E}{\bf x}^{e},$

and so $\lambda(G)-\lambda(G-E)\geq k\sum_{e\in E}{\bf x}^{e}$. On the other
hand, since $\lambda(G-E)=\mathcal{A}(G-E){\bf y}^{k}=k\sum_{e\in E(G)-E}{\bf
y}^{e}$, we have

$\displaystyle\lambda(G)\leq\mathcal{A}(G){\bf y}^{k}=k\sum_{e\in E(G)}{\bf
y}^{e}=k\sum_{e\in E(G)-E}{\bf y}^{e}+k\sum_{e\in E}{\bf y}^{e}$
$\displaystyle=\lambda(G-E)+k\sum_{e\in E}{\bf y}^{e},$

and so $\lambda(G)-\lambda(G-E)\leq k\sum_{e\in E}{\bf y}^{e}$. ∎

######  Example 5.3.

Let $G$ be a $4$-uniform hypergraph with vertex set $[6]$, where
$E(G)$$=\\{\\{1,2,3,4\\}$, $\\{3,4,5,6\\}$,$\\{1,3,4,5\\},\\{1,2,4,5\\}\\}$.
Let $E=\\{\\{1,2,3,4\\}\\}$. By MATLAB, $\lambda(G)=-2.8786$ with eigenvector
${\bf x}_{0}=(-0.9457,0.848,0.928,1,0.928,-0.6688)^{\top}$ and
$\lambda(G-E)=-2.1908$ with eigenvector ${\bf
y}_{0}=(-0.9112,0.7465,0.9112,1,1,-0.7465)^{\top}$. Let ${\bf x}=\frac{{\bf
x}_{0}}{\|{\bf x}_{0}\|_{4}}$ and ${\bf y}=\frac{{\bf y}_{0}}{\|{\bf
y}_{0}\|_{4}}$. The lower bound for $\lambda(G-E)$ in Theorem 5.2 is

$\displaystyle\quad\lambda(G)-k\sum_{e\in E}{\bf y}^{e}$
$\displaystyle=\lambda(G)-4y_{1}y_{2}y_{3}y_{4}$
$\displaystyle=-2.8786-4\times\frac{-0.9112\times 0.7465\times 0.9112\times
1}{(-0.9112)^{4}+0.7465^{4}+0.9112^{4}+1+1+(-0.7465)^{4}}$
$\displaystyle=-2.2587,$

and the upper bound for $\lambda(G-E)$ in Theorem 5.2 is

$\displaystyle\quad\lambda(G)-k\sum_{e\in E}{\bf x}^{e}$
$\displaystyle=\lambda(G)-4x_{1}x_{2}x_{3}x_{4}$
$\displaystyle=-2.8786-4\times\frac{(-0.9457)\times 0.848\times 0.928\times
1}{(-0.9457)^{4}+0.848^{4}+0.928^{4}+1+0.928^{4}+(-0.6688)^{4}}$
$\displaystyle=-1.4467.$

######  Theorem 5.3.

Let $G$ be a linear $k$-uniform hypergraph with at least one edge, where $k$
is even and $n,k\geq 2$. Let ${\bf x}$ be a least eigenvector of $G$. Then for
$i\in V(G)$,

$\displaystyle
x_{i}^{k}\leq\frac{d_{i}}{d_{i}+(k-1)\lambda(G)^{\frac{k}{k-1}}}$ (5.1)

with equality if and only if for $j\in V(G)\setminus\\{i\\}$,
$x_{j}^{k}=\frac{\lambda(G)^{\frac{k}{k-1}}x_{i}^{k}}{d_{i}^{2}}$ if $j\sim i$
and $x_{j}=0$ otherwise, and the sign of ${\bf x}^{e\setminus\\{i\\}}$ for
each $e\in E_{i}(G)$ is the same.

###### Proof.

Let $\lambda=\lambda(G)$. From the eigenequation of $G$ at $i$,

$\displaystyle\lambda^{\frac{k}{k-1}}x_{i}^{k}$
$\displaystyle=\left(\sum_{e\in E_{i}(G)}{\bf
x}^{e\setminus\\{i\\}}\right)^{\frac{k}{k-1}}$
$\displaystyle\leq\left(\sum_{e\in E_{i}(G)}\left|{\bf
x}^{e\setminus\\{i\\}}\right|\right)^{\frac{k}{k-1}}$
$\displaystyle\leq\left(d_{i}\left(\frac{\sum_{\\{i,i_{2},\ldots,i_{k}\\}\in
E_{i}(G)}x_{i_{2}}^{k}\cdots
x_{i_{k}}^{k}}{d_{i}}\right)^{\frac{1}{k}}\right)^{\frac{k}{k-1}}$
$\displaystyle=d_{i}\left(\sum_{\\{i,i_{2},\ldots,i_{k}\\}\in
E_{i}(G)}x_{i_{2}}^{k}\cdots x_{i_{k}}^{k}\right)^{\frac{1}{k-1}}$ (5.2)
$\displaystyle\leq d_{i}\left(\sum_{\\{i,i_{2},\ldots,i_{k}\\}\in
E_{i}(G)}\left(\frac{x_{i_{2}}^{k}+\cdots+x_{i_{k}}^{k}}{k-1}\right)^{k-1}\right)^{\frac{1}{k-1}}$
$\displaystyle\leq\frac{d_{i}}{k-1}\sum_{\\{i,i_{2},\ldots,i_{k}\\}\in
E_{i}(G)}\left(x_{i_{2}}^{k}+\cdots+x_{i_{k}}^{k}\right)$
$\displaystyle=\frac{d_{i}}{k-1}\sum_{j:j\sim i}x_{j}^{k}$
$\displaystyle\leq\frac{d_{i}}{k-1}\left(1-x_{i}^{k}\right),$

where the first and the last two inequalities follows trivially, the second
and the third inequalities follow respectively from the power mean inequality
and the arithmetic-geometric mean inequality. So (5.1) follows.

Suppose that equality holds in (5.1). Then all inequalities in (5.2) are
equalities. From the first inequality, we see that the sign of ${\bf
x}^{e\setminus\\{i\\}}$ for any $e\in E_{i}(G)$ is the same. From the third
inequality, we know that for each $\\{i,i_{2},\ldots,i_{k}\\}\in E_{i}(G)$,
$x_{i_{2}}^{k}=\cdots=x_{i_{k}}^{k}$. From the last inequality, we find that
either $j\sim i$ for all $j\in V(G)\setminus\\{i\\}$ or $x_{j}^{k}=0$ for each
$j\nsim i$. So $x_{j}^{k}=\frac{\lambda^{\frac{k}{k-1}}x_{i}^{k}}{d_{i}^{2}}$
if $j\sim i$ and $0$ otherwise.

Conversely, suppose that $j\in V(G)\setminus\\{i\\}$,
$x_{j}^{k}=\frac{\lambda(G)^{\frac{k}{k-1}}x_{i}^{k}}{d_{i}^{2}}$ if $j\sim i$
and $x_{j}=0$ otherwise, and the sign of ${\bf x}^{e\setminus\\{i\\}}$ for
each $e\in E_{i}(G)$ is the same. Then all inequalities in (5.2) are
equalities, so (5.1) is an equality. ∎

Suppose that $G$ is a $k$-uniform hypergraph with at least one edge, where $k$
is even and $k\geq 2$. Let $c_{\max}$ be the largest component among all least
eigenvectors of $\mathcal{A}(G)$.

######  Theorem 5.4.

Let $G$ be a linear $k$-uniform hypergraph with vertex set $[n]$ and at least
one edge, where $k$ is even and $n,k\geq 2$. Then

$c_{\max}\leq\sqrt[k]{\frac{n-1}{n-1+(k-1)^{2}}}$

with equality if and only if
$x_{j}^{k}=\frac{x_{i}^{k}}{\left(\frac{n-1}{k-1}\right)^{\frac{k}{k-1}}}$ if
$j\sim i$ and $x_{i}=0$ otherwise, $\lambda(G)=-1$, maximum degree is
$\frac{n-1}{k-1}$, and the sign of ${\bf x}^{e\setminus{i}}$ for each $e\in
E_{i}(G)$ is the same.

###### Proof.

Let ${\bf x}$ be a least eigenvector of $G$ containing $c_{\max}$. Suppose
without loss of generality that $c_{\max}=x_{1}$. As $G$ is linear, we have
$d_{i}\leq\frac{n-1}{k-1}$ for $i\in V(G)$. As $k$ is even and $|E(G)|\geq 1$,
we have $\lambda(G)\leq-1$ by Corollary 5.1 and the fact that the least
H-eigenvalue of the $k$-uniform hypergraph consisting of exactly one edge is
$-1$. So, by Theorem 5.3,

$x_{i}^{k}\leq\frac{d_{i}}{d_{i}+(k-1)\lambda(G)^{\frac{k}{k-1}}}\leq\frac{\frac{n-1}{k-1}}{\frac{n-1}{k-1}+(k-1)(-1)^{\frac{k}{k-1}}}=\frac{n-1}{n-1+(k-1)^{2}}$

with equalities if and only if
$x_{j}^{k}=\frac{x_{i}^{k}}{\left(\frac{n-1}{k-1}\right)^{\frac{k}{k-1}}}$ if
$j\sim i$ and $x_{i}=0$ otherwise, $\lambda(G)=-1$, $d_{1}=\frac{n-1}{k-1}$,
and the sign of ${\bf x}^{e\setminus{i}}$ for each $e\in E_{i}(G)$ is the
same. Thus the result follows. ∎

For even integer $k$, we say a $k$-uniform hypergraph $G$ is odd-bipartite if
$V(G)$ can be partitioned into two disjoint vertex set $V_{1}$ and $V_{2}$
such that each edge intersects each of $\\{V_{1},V_{2}\\}$ with odd number of
vertices.

######  Theorem 5.5.

Let $G$ be an odd-bipartite, connected $k$-uniform hypergraph with $m$ edges,
where $k$ is even, and $m\geq 1$. Then
$c_{\max}\geq\sqrt[k]{-\frac{\lambda(G)}{km}}$ with equality if and only if
$G$ is regular.

###### Proof.

Let ${\bf x}$ be the least eigenvector of $G$ containing $c_{\max}$. Since $G$
is odd-bipartite, $\lambda(G)=-\rho(G)$ [29]. Let ${\bf\widetilde{x}}$ be the
vector such that $\widetilde{x}_{i}=x_{i}$ for each $i\in V(G)$. Obviously,
${\bf\widetilde{x}}$ is unit. For any $u\in V(G)$,

$\rho(G)\widetilde{x}_{u}^{k-1}=\rho(G)|x_{u}|^{k-1}=\left|\sum_{e\in
E_{u}(G)}{\bf x}^{e\setminus\\{u\\}}\right|\leq\sum_{e\in E_{u}(G)}\left|{\bf
x}^{e\setminus\\{u\\}}\right|=\sum_{e\in
E_{u}(G)}{\bf\widetilde{x}}^{e\setminus\\{u\\}},$

i.e.,

$\rho(G)\widetilde{x}_{u}^{k-1}\leq\sum_{e\in
E_{u}(G)}{\bf\widetilde{x}}^{e\setminus\\{u\\}}.$

As ${\bf\widetilde{x}}$ is nonnegative,

$\rho(G)\widetilde{x}_{u}^{k}\leq\sum_{e\in E_{u}(G)}{\bf\widetilde{x}}^{e}.$

By Lemma 2.2, we have

$\rho(G)=\sum_{u\in V(G)}\rho(G)\widetilde{x}_{u}^{k}\leq\sum_{u\in
V(G)}\sum_{e\in E_{u}(G)}{\bf\widetilde{x}}^{e}=\sum_{e\in
E(G)}k{\bf\widetilde{x}}^{e}\leq\rho(G),$

so $\rho(G)=\sum_{e\in E(G)}k{\bf\widetilde{x}}^{e}$ and
$\rho(G)\widetilde{x}_{u}^{k-1}=\sum_{e\in
E_{u}(G)}{\bf\widetilde{x}}^{e\setminus\\{u\\}}$. That is,
${\bf\widetilde{x}}$ is a unit nonnegative eigenvector corresponding to
$\rho(G)$. As $G$ is connected, ${\bf\widetilde{x}}$ is positive. It follows
that

$-\lambda(G)=\rho(G)=k\sum_{e\in E(G)}{\bf\widetilde{x}}^{e}\leq
kmc_{\max}^{k}$

with equality if and only if $|x_{i}|=c_{\max}$, i.e., $G$ is regular. ∎

Acknowledgements. We thank Prof. Tan Zhang for kind discussions. This work was
supported by the National Natural Science Foundation of China (Nos. 12071158
and 11801410).

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