A sharp upper bound on the spectral radius of a nonnegative k-uniform tensor and its applications to (directed) hypergraphs

Abstract

In this paper, we obtain a sharp upper bound on the spectral radius of a nonnegative k-uniform tensor and characterize when this bound is achieved. Furthermore, this result deduces the main result in [X. Duan and B. Zhou, Sharp bounds on the spectral radius of a nonnegative matrix, Linear Algebra Appl. 439:2961–2970, 2013] for nonnegative matrices; improves the adjacency spectral radius and signless Laplacian spectral radius of a uniform hypergraph for some known results in [D.M. Chen, Z.B. Chen and X.D. Zhang, Spectral radius of uniform hypergraphs and degree sequences, Front. Math. China 6:1279–1288, 2017]; and presents some new sharp upper bounds for the adjacency spectral radius and signless Laplacian spectral radius of a uniform directed hypergraph. Moreover, a characterization of a strongly connected k-uniform directed hypergraph is obtained.

1 Introduction

Let k, n be two positive integers. As in [17, 21], an order k dimension n tensor $\mathbb {A}=(a_{i_{1}\cdots i_{k}})$ over the real field $\mathbb{R}$ is a multidimensional array with $n^{k}$ entries $a_{i_{1}\cdots i_{k}}\in\mathbb{R}$, where $i_{j}\in[n]=\{1,2,\ldots, n\}$, $j\in[k]=\{ 1,2,\ldots, k\}$. Obviously, a vector is an order 1 tensor and a square matrix is an order 2 tensor.

Furthermore, we call a tensor $\mathbb{A}$ nonnegative (positive), denoted by $\mathbb{A}\geq0$ ($\mathbb{A}>0$), if every entry has $a_{i_{1}\cdots i_{k}}\geq0$ ($a_{i_{1}\cdots i_{k}}>0$). The tensor $\mathbb{A}=(a_{i_{1}\cdots i_{k}})$ is called symmetric if $a_{i_{1}\cdots i_{k}}=a_{\sigma(i_{1})\cdots\sigma(i_{k})}$, where σ is any permutation of the indices.

Let $\mathbb{A}$ be an order k dimension n tensor. If there is a complex number λ and a nonzero complex vector $x=(x_{1},x_{2},\ldots, x_{n})^{T}$ such that

$$\mathbb{A}x^{k-1}=\lambda x^{[k-1]}, $$

then λ is called an eigenvalue of $\mathbb{A}$ and x an eigenvector of $\mathbb{A}$ corresponding to the eigenvalue λ [17, 18, 21]. Here $\mathbb{A}x^{k-1}$ and $x^{[k-1]}$ are vectors, whose ith entries are

$$\bigl(\mathbb{A}x^{k-1} \bigr)_{i}=\sum _{i_{2},\ldots, i_{k}=1}^{n}a_{ii_{2}\cdots i_{k}}x_{i_{2}}\cdots x_{i_{k}} $$

and $(x^{[k-1]})_{i}=x_{i}^{k-1}$, respectively. Moreover, the spectral radius $\rho(\mathbb{A})$ of a tensor $\mathbb{A}$ is defined as

$$\rho(\mathbb{A})=\max \bigl\{ \vert \lambda \vert : \lambda \mbox{ is an eigenvalue of } \mathbb{A} \bigr\} . $$

Some properties of the spectral radius of a nonnegative tensor can be found in [3, 9, 14, 16–18, 21, 25–27].

Definition 1.1

([22])

Let $\mathbb{A}$ and $\mathbb{B}$ be two tensors with order $m\geq2$ and $k\geq1$ dimension n, respectively. The general product $\mathbb {AB}$ of $\mathbb{A}$ and $\mathbb{B}$ is the following tensor $\mathbb{C}$ with order $(m-1)(k-1)+1$ and dimension n:

$$c_{i\alpha_{1}\cdots\alpha_{m-1}}=\sum_{i_{2},\ldots, i_{m}=1}^{n}a_{ii_{2}\cdots i_{m}}b_{i_{2}\alpha_{1}} \cdots b_{i_{m}\alpha_{m-1}} \quad\bigl(i\in[n], \alpha_{1}, \ldots, \alpha_{m-1}\in[n]^{k-1} \bigr). $$

Definition 1.2

([22])

Let $\mathbb{A}=(a_{i_{1}i_{2}\cdots i_{k}})$ and $\mathbb {B}=(b_{i_{1}i_{2}\cdots i_{k}})$ be two order k dimension n tensors. We say that $\mathbb{A}$ and $\mathbb{B}$ are diagonal similar if there exists some invertible diagonal matrix $D=(d_{11},d_{22},\ldots ,d_{nn})$ of order n such that $\mathbb{B}=D^{-(k-1)}\mathbb{A}D$ with entries

$$b_{i_{1}i_{2}\cdots i_{k}}=d_{i_{1}i_{1}}^{-(k-1)}a_{i_{1}i_{2}\cdots i_{k}}d_{i_{2}i_{2}} \cdots d_{i_{k}i_{k}} . $$

Theorem 1.3

([22])

If the two orderkdimensionntensors$\mathbb{A}$and$\mathbb {B}$are diagonal similar, then they have the same eigenvalues including multiplicity and same spectral radius.

Definition 1.4

([9, 26])

Let $\mathbb{A}$ be an order k dimensional n tensor (not necessarily nonnegative). If there exists a nonempty proper subset I of the set $[n]$, such that

$$ a_{i_{1}i_{2}\ldots i_{k}}=0 \quad\mbox{for all } i_{1}\in I \mbox{ and some } i_{j}\notin I \mbox{ where } j\in\{2,\ldots,k\}, $$

then $\mathbb{A}$ is called weakly reducible (or sometimes I-weakly reducible). If $\mathbb{A}$ is not weakly reducible, then $\mathbb{A}$ is called weakly irreducible.

The ith slice of a tensor $\mathbb{A}$ with order $k\geq2$ and dimension n, denoted by $\mathbb{A}_{i}$ in [23], is the subtensor of $\mathbb{A}$ with order $k-1$ and dimension n such that $(\mathbb{A}_{i})_{i_{2}\cdots i_{k}}=a_{ii_{2}\cdots i_{k}}$. Then the ith slice sum (also called “the ith row sum”) of $\mathbb{A}$ is defined as

$$r_{i}(\mathbb{A})=\sum_{i_{2}, \ldots, i_{k}=1}^{n}a_{ii_{2}\cdots i_{k}}\quad \bigl(i\in[n] \bigr). $$

Lemma 1.5

([13, 25])

Let$\mathbb{A}$be a nonnegative tensor with order$k\geq2$and dimensionn. Then we have

$$ \min_{1\leq i\leq n}r_{i}(\mathbb{A})\leq\rho( \mathbb{A})\leq\max_{1\leq i\leq n}r_{i}(\mathbb{A}). $$

(1.1)

Moreover, if$\mathbb{A}$is weakly irreducible, then one of the equalities in (1.1) holds if and only if$r_{1}(\mathbb{A})=r_{2}(\mathbb{A})=\cdots=r_{n}(\mathbb{A})$.

We denote by $\binom{n}{r}$ the number of r-combinations of an n-element set, and let $\binom{n}{r}=0$ if $r>n$ or $r<0$. Clearly, $\binom{n}{r}=\frac{n!}{r!(n-r)!}$ when $0\leq r\leq n$.

Lemma 1.6

([2])

Letn, k, andmbe positive integers. Then

• (1) $\sum_{r=0}^{k}\binom{n}{r}\binom{m}{k-r}=\binom{n+m}{k}$ ($n+m\geq k$);

• (2) $\binom{n}{k}=\frac{n}{k}\binom{n-1}{k-1}$ ($n\geq k \geq1$).

Let $S=\{s_{1}, s_{2},\ldots, s_{n}\}$ be an n-element set, noting that $s_{i}\neq s_{j}$ if $i\neq j$.

Definition 1.7

Let $n\geq2$, $k\geq2$, $\mathbb{A}$ be an order k dimension n tensor, we call $\mathbb{A}$ a k-uniform tensor if its entries are defined as follows: $a_{i_{1}i_{2}\cdots i_{k}} \in\mathbb{R}$ if $\{i_{1}, i_{2}, \ldots, i_{k}\}$ is a k-element set or $i_{1}=i_{2}=\cdots=i_{k}$, otherwise, $a_{i_{1}i_{2}\cdots i_{k}}=0$.

Obviously, a 2-uniform tensor is an ordinary matrix. Let $\mathbb {A}$ be a k-uniform tensor with order k dimension n. Then $a_{i_{1}i_{2}\cdots i_{k}}\neq0$ implies $\{i_{1}, i_{2},\ldots, i_{k}\}$ is a k-element set or $i_{1}=i_{2}=\cdots=i_{k}$.

In this paper, we obtain a sharp upper bound on the spectral radius of a nonnegative k-uniform tensor in Sect. 2. By applying the bound to a nonnegative matrix, we can obtain the main result in [7]. In Sect. 3, we apply the bound to the adjacency spectral radius and signless Laplacian spectral radius of a uniform hypergraph and improve some known results in [4]. Furthermore, we give a characterization of a strongly connected k-uniform directed hypergraph and obtain some new results by applying the bound to the adjacency spectral radius and the signless Laplacian spectral radius of a uniform directed hypergraph in Sect. 4.

2 Main results

In this section, we obtain a sharp upper bound on the spectral radius of a nonnegative k-uniform tensor and characterize when this bound is achieved. Furthermore, this bound deduces the main result in [7] for a nonnegative matrix.

Theorem 2.1

Let$n\geq2$, $k\geq2$, $\mathbb{A}=(a_{i_{1}i_{2}\cdots i_{k}})$be a nonnegativek-uniform tensor with orderkdimensionn, $r_{i}=r_{i}(\mathbb{A})=\sum_{i_{2},\ldots, i_{k}=1}^{n}a_{ii_{2}\cdots i_{k}}$for$i\in[n]$with$r_{1}\geq r_{2}\geq\cdots\geq r_{n}$. LetMbe the largest diagonal element andN (>0) be the largest non-diagonal element of tensor$\mathbb{A}$, $N_{1}=N(k-2)!\binom{n-2}{k-2}$, $\phi_{1}=r_{1}$, and

$$ \phi_{s}=\frac{1}{2} \Biggl\{ r_{s}+M-N_{1}+ \sqrt {(r_{s}-M+N_{1})^{2}+4N_{1} \sum_{t=1}^{s-1}(r_{t}-r_{s})} \Biggr\} $$

(2.1)

for$2\leq s\leq n$. Then

$$\rho(\mathbb{A})\leq\min_{1\leq s\leq n}\phi_{s}. $$

Let$\phi_{s}=\min_{1\leq l\leq n}\phi_{l}$. If$\mathbb{A}$is weakly irreducible, then

• (1) when$k=2$, $\rho(\mathbb{A})=\phi_{s}$if and only if$r_{1}=r_{2}=\cdots=r_{n}$or for somet ($2\leq t\leq s$), $\mathbb{A}$ satisfies the following conditions:

  • (i) $a_{ii}=M$for$1\leq i\leq t-1$;

  • (ii) $a_{ii_{2}}=N$for$1\leq i\leq n$, $1\leq i_{2}\leq t-1$, and$i\neq i_{2}$;

  • (iii) $r_{t}=r_{t+1}=\cdots=r_{n}$;

• (2) when$k\geq3$, $\rho(\mathbb{A})=\phi_{s}$if and only if$r_{1}=r_{2}=\cdots=r_{n}$.

Proof

Firstly, we show $\rho(\mathbb{A})\leq\phi_{s}$ for $1\leq s\leq n$.

If $s=1$, then by Lemma 1.5 we have $\rho(\mathbb{A})\leq r_{1}=\phi_{1}$. Now we only consider the cases of $2\leq s\leq n$.

Let

$$U=\operatorname{diag}(x_{1},\ldots,x_{s-1},x_{s}, \ldots,x_{n}), $$

where $x_{i}>0$ for $1\leq i\leq n$, $x_{i}^{k-1}=1+\frac{r_{i}-r_{s}}{\phi _{s}+N_{1}-M}$ for $1\leq i\leq s-1$, and $x_{s}=\cdots=x_{n}=1$.

Now we show $x_{i}\geq1$ for $1\leq i\leq s-1$. By $r_{1}\geq r_{2}\geq \cdots\geq r_{n}$, we only need to show $\phi_{s}+N_{1}-M>0$.

If $\sum_{t=1}^{s-1}(r_{t}-r_{s})>0$, then by (2.1) we have

$$\phi_{s}>\frac{1}{2} \bigl(r_{s}+M-N_{1}+ \vert r_{s}-M+N_{1} \vert \bigr)\geq \frac {1}{2} \bigl(r_{s}+M-N_{1}-(r_{s}-M+N_{1}) \bigr)=M-N_{1}, $$

and thus $\phi_{s}-M+N_{1}>0$.

If $\sum_{t=1}^{s-1}(r_{t}-r_{s})=0$, then $r_{1}=r_{2}=\cdots=r_{s}$. Thus $\phi_{s}-M+N_{1}>0$ by $r_{1}\geq M$ and $\phi_{s}=r_{s}$ from (2.1).

Combining the above arguments, we know $x_{i}\geq1$, and then U is an invertible diagonal matrix. Let $\mathbb{B}=U^{-(k-1)}\mathbb{A}U=(b_{i_{1}\cdots i_{k}})$. By Theorem 1.3, we have

$$ \rho(\mathbb{A})=\rho(\mathbb{B}). $$

(2.2)

By (2.1), it is easy to see that

$$\phi_{s}^{2}-(r_{s}+M-N_{1}) \phi_{s}+(M-N_{1})r_{s}-N_{1}\sum _{t=1}^{s-1}(r_{t}-r_{s}) =0. $$

Then

$$\begin{aligned} (\phi_{s}-M+N_{1}) (\phi_{s}-r_{s})&=N_{1} \sum_{t=1}^{s-1}(r_{t}-r_{s}) =N_{1}\sum_{t=1}^{s-1}( \phi_{s}-M+N_{1}) \bigl(x_{t}^{k-1}-1 \bigr) \\&=N_{1}(\phi_{s}-M+N_{1}) \Biggl(\sum _{t=1}^{s-1}x_{t}^{k-1}-(s-1) \Biggr).\end{aligned} $$

Therefore, $\phi_{s}=r_{s}+ N_{1}\sum_{t=1}^{s-1}x_{t}^{k-1}-N_{1}(s-1)$ and thus

$$ \sum_{t=1}^{s-1}x_{t}^{k-1}= \frac{\phi_{s}-r_{s}+(s-1)N_{1}}{N_{1}}. $$

(2.3)

In the following we show $r_{i}(\mathbb{B})\leq\phi_{s}$ for any $i\in [n]=\{1,2,\ldots,n\}$.

Let $S(\mathbb{A})=\{\{i, i_{2}, \ldots, i_{k}\}|a_{ii_{2}\cdots i_{k}}\neq0\}$. Since M is the largest diagonal element and $N>0$ is the largest non-diagonal element of tensor $\mathbb{A}$, by Definition 1.2, we have

$$\begin{aligned} r_{i}(\mathbb{B}) ={}&r_{i} \bigl(U^{-(k-1)}\mathbb{A}U \bigr) \\ ={}&\sum_{i_{2},\ldots, i_{k}=1}^{n} \bigl(U^{-(k-1)} \bigr)_{ii}a_{ii_{2}\cdots i_{k}}U_{i_{2}i_{2}}\cdots U_{i_{k}i_{k}} \\ ={}&\frac{1}{x_{i}^{k-1}}\sum_{i_{2},\ldots,i_{k}=1}^{n}a_{ii_{2}\cdots i_{k}}x_{i_{2}} \cdots x_{i_{k}} \\ ={}&\frac{1}{x_{i}^{k-1}} \Biggl\{ r_{i}+\sum_{i_{2},\ldots ,i_{k}=1}^{n}a_{ii_{2}\cdots i_{k}}(x_{i_{2}} \cdots x_{i_{k}}-1) \Biggr\} \\ =&\frac{1}{x_{i}^{k-1}} \Biggl\{ r_{i}+a_{i\cdots i} \bigl(x_{i}^{k-1}-1 \bigr) \\ &+\sum_{i_{2},\ldots,i_{k}=1}^{n}a_{ii_{2}\cdots i_{k}}(x_{i_{2}} \cdots x_{i_{k}}-1)-a_{i\cdots i} \bigl(x_{i}^{k-1}-1 \bigr) \Biggr\} \\ \leq{}&\frac{1}{x_{i}^{k-1}} \Biggl\{ r_{i}+M \bigl(x_{i}^{k-1}-1 \bigr) \\ &+\sum_{i_{2},\ldots,i_{k}=1}^{n}a_{ii_{2}\cdots i_{k}}(x_{i_{2}} \cdots x_{i_{k}}-1)-a_{i\cdots i} \bigl(x_{i}^{k-1}-1 \bigr) \Biggr\} \\ \leq{}&\frac{1}{x_{i}^{k-1}} \biggl\{ r_{i}+M \bigl(x_{i}^{k-1}-1 \bigr) +N(k-1)!\sum_{\{i,i_{2},\ldots, i_{k}\}\in S(\mathbb {A})}(x_{i_{2}}\cdots x_{i_{k}}-1) \biggr\} \\ \leq{}&\frac{1}{x_{i}^{k-1}} \biggl\{ r_{i}+M \bigl(x_{i}^{k-1}-1 \bigr) \\ &+N(k-1)!\sum_{\{i,i_{2},\ldots, i_{k}\}\in S(\mathbb{A})} \biggl(\frac {x_{i_{2}}^{k-1}+\cdots+x_{i_{k}}^{k-1}}{k-1}-1 \biggr) \biggr\} \\ \leq{}& \frac{1}{x_{i}^{k-1}} \Biggl\{ r_{i}+M \bigl(x_{i}^{k-1}-1 \bigr) \\ &+N(k-1)!\sum_{r=0}^{k-1}\sum _{ \{i_{2},\ldots, i_{k}\}\in N_{r}^{s}} \biggl(\frac{x_{i_{2}}^{k-1}+\cdots +x_{i_{k}}^{k-1}}{k-1}-1 \biggr) \Biggr\} \\ ={}&\frac{1}{x_{i}^{k-1}} \Biggl\{ r_{i}+M \bigl(x_{i}^{k-1}-1 \bigr) \\ &+N(k-1)!\sum_{r=0}^{k-2}\sum _{ \{i_{2},\ldots, i_{k}\}\in N_{r}^{s}} \biggl(\frac{x_{i_{2}}^{k-1}+\cdots +x_{i_{k}}^{k-1}}{k-1}-1 \biggr) \Biggr\} , \end{aligned}$$

where $N_{r}^{s}=\{\{i_{2},\ldots,i_{k}\}\mid i_{2},\ldots,i_{k}\in\{ 1,2,\ldots, n\}\setminus\{i\}$, and there are exactly r elements in $\{i_{2},\ldots,i_{k}\}$ such that they are not less than s} for $0\leq r\leq k-1$. Obviously, the family of all $(k-1)$-element subsets of $\{1,2,\ldots, n\}\setminus\{i\}$ is just equal to $\bigcup_{r=0}^{k-1}N_{r}^{s}$. Thus we have

$$ r_{i}(\mathbb{B})\leq M+\frac{1}{x_{i}^{k-1}} \Biggl\{ r_{i}-M+N(k-1)!\sum_{r=0}^{k-2} \sum_{ \{i_{2},\ldots, i_{k}\}\in N_{r}^{s}} \biggl(\frac{x_{i_{2}}^{k-1}+\cdots +x_{i_{k}}^{k-1}}{k-1}-1 \biggr) \Biggr\} , $$

(2.4)

and the equality holds in (2.4) if and only if (a), (b), (c), and (d) hold:

• (a) $x_{i}^{k-1}=1$ or $a_{i\cdots i}=M$ for $x_{i}>1$;

• (b) for any $\{i,i_{2},\ldots,i_{k}\}\in S(\mathbb{A})$, $x_{i_{2}}\cdots x_{i_{k}}=1$ or $a_{ii_{2}\cdots i_{k}}=N$ for $x_{i_{2}}\cdots x_{i_{k}}>1$;

• (c) $x_{i_{2}}=\cdots=x_{i_{k}}$ for any $\{i,i_{2},\ldots,i_{k}\}\in S(\mathbb{A})$;

• (d) $\sum_{\{i,i_{2},\ldots, i_{k}\}\in S(\mathbb{A})} (\frac {x_{i_{2}}^{k-1}+\cdots+x_{i_{k}}^{k-1}}{k-1}-1 )=\sum_{r=0}^{k-1}\sum_{ \{i_{2},\ldots, i_{k}\}\in N_{r}^{s}} (\frac {x_{i_{2}}^{k-1}+\cdots+x_{i_{k}}^{k-1}}{k-1}-1 )$.

Case 1:$s\leq i\leq n$.

Clearly, $\{i_{2},\ldots, i_{k}\}\in N_{r}^{s}$ implies that we should choose r elements from the set $\{s,\ldots, n\}\setminus\{i\}$ and choose $k-1-r$ elements from the set $\{1,2,\ldots, s-1\}$, then we have

$$ \sum_{r=0}^{k-2}\sum _{\{i_{2},\ldots, i_{k}\}\in N_{r}^{s}} 1=\sum_{r=0}^{k-2} \binom{s-1}{k-1-r}\binom{n-s}{r}. $$

(2.5)

Similarly, we have

$$ \begin{aligned}[b] &\sum_{r=0}^{k-2}\sum _{\{i_{2},\ldots, i_{k}\}\in N_{r}^{s}} \bigl(x_{i_{2}}^{k-1}+ \cdots+x_{i_{k}}^{k-1} \bigr) \\ &\quad=\sum_{r=0}^{k-2} \binom{s-2}{k-2-r}\binom{n-s}{r} \Biggl(\sum_{t=1}^{s-1}x_{t}^{k-1} \Biggr)\\&\qquad+\sum_{r=0}^{k-2} \binom{s-1}{k-1-r}\binom{n-s-1}{r-1} \Biggl(\sum_{t=s}^{n}x_{t}^{k-1}-x_{i}^{k-1} \Biggr).\end{aligned} $$

(2.6)

We note $x_{s}=\cdots=x_{n}=1$ and $r_{1}\geq\cdots\geq r_{s}\geq\cdots \geq r_{i}\geq\cdots\geq r_{n}$, then by (2.3), (2.4), (2.5), and (2.6), we have

$$\begin{aligned} r_{i}(\mathbb{B}) \leq {}&r_{i}+N(k-1)!\sum _{r=0}^{k-2}\sum_{ \{i_{2},\ldots, i_{k}\}\in N_{r}^{s}} \biggl(\frac{x_{i_{2}}^{k-1}+\cdots+x_{i_{k}}^{k-1}}{k-1}-1 \biggr) \\ \leq{}&r_{s}+N(k-2)!\sum_{r=0}^{k-2} \binom{s-2}{k-2-r}\binom{n-s}{r} \Biggl(\sum_{t=1}^{s-1}x_{t}^{k-1} \Biggr) \\ &+N(k-2)!\sum_{r=0}^{k-2}\binom{s-1}{k-1-r} \binom{n-s-1}{r-1} \Biggl(\sum_{t=s}^{n}x_{t}^{k-1}-x_{i}^{k-1} \Biggr) \\ &-N(k-1)!\sum_{r=0}^{k-2}\binom{s-1}{k-1-r} \binom{n-s}{r} \\ ={}& r_{s}+N(k-2)!\binom{n-2}{k-2}\sum_{t=1}^{s-1}x_{t}^{k-1} +N(k-2)!\sum_{r=0}^{k-2}\binom{s-1}{k-1-r} \binom{n-s-1}{r-1}(n-s) \\ & -N(k-1)!\sum_{r=0}^{k-2}\binom{s-1}{k-1-r} \binom{n-s}{r} \\ ={}& r_{s}+N_{1}\sum_{t=1}^{s-1}x_{t}^{k-1} \\ &+N(k-2)!\sum_{r=0}^{k-2}\binom{s-1}{k-1-r} \biggl[\binom{n-s-1}{r-1}(n-s) -(k-1)\binom{n-s}{r} \biggr] \\ ={}& r_{s}+N_{1}\sum_{t=1}^{s-1}x_{t}^{k-1}-N(k-2)! \sum_{r=0}^{k-2}\binom {s-1}{k-1-r} \binom{n-s}{r}(k-1-r) \\ ={}& r_{s}+N_{1}\sum_{t=1}^{s-1}x_{t}^{k-1}-N(k-2)! \sum_{r=0}^{k-2}(s-1)\binom{s-2}{k-2-r} \binom{n-s}{r} \\ ={}& r_{s}+N_{1}\sum _{t=1}^{s-1}x_{t}^{k-1}-N(k-2)!(s-1) \binom{n-2}{k-2} \\ = {}&r_{s}+N_{1}\sum _{t=1}^{s-1}x_{t}^{k-1}-(s-1)N_{1} \\ ={}&\phi_{s}, \end{aligned}$$

where equality holds if and only if the following condition (e) holds: (e) $r_{i}=r_{s}$.

Case 2:$1\leq i\leq s-1$.

Subcase 2.1:$s\geq3$.

Clearly, $\{i_{2},\ldots, i_{k}\}\in N_{r}^{s}$ implies that we should choose r elements from the set $\{s,\ldots, n\}$ and choose $k-1-r$ elements from the set $\{1,2,\ldots, s-1\}\setminus \{i\}$, then $\sum_{r=0}^{k-2}\sum_{\{i_{2},\ldots, i_{k}\}\in N_{r}^{s}} 1=\sum_{r=0}^{k-2}\binom{s-2}{k-1-r}\binom{n-s+1}{r}$. Similarly, we have

$$\begin{aligned} &\sum_{r=0}^{k-2}\sum _{ \{i_{2},\ldots, i_{k}\}\in N_{r}^{s}} \bigl(x_{i_{2}}^{k-1}+ \cdots+x_{i_{k}}^{k-1} \bigr) \\ &\quad=\sum_{r=0}^{k-2}\binom{s-3}{k-r-2} \binom{n-s+1}{r} \Biggl(\sum_{t=1}^{s-1}x_{t}^{k-1}-x_{i}^{k-1} \Biggr) \\ &\qquad+\sum_{r=0}^{k-2}\binom{s-2}{k-1-r} \binom{n-s}{r-1} \Biggl(\sum_{t=s}^{n}x_{t}^{k-1} \Biggr) \\ &\quad=\binom{n-2}{k-2} \Biggl(\sum_{t=1}^{s-1}x_{t}^{k-1}-x_{i}^{k-1} \Biggr) +\sum_{r=0}^{k-2}\binom{s-2}{k-1-r} \binom{n-s}{r-1}(n-s+1). \end{aligned}$$

Then

$$\begin{aligned} &N(k-1)!\sum_{r=0}^{k-2}\sum _{ \{i_{2},\ldots, i_{k}\}\in N_{r}^{s}} \biggl(\frac{x_{i_{2}}^{k-1}+\cdots +x_{i_{k}}^{k-1}}{k-1}-1 \biggr) \\ &\quad=N_{1} \Biggl(\sum_{t=1}^{s-1}x_{t}^{k-1}-x_{i}^{k-1} \Biggr) +N(k-2)!\sum_{r=0}^{k-2} \binom{s-2}{k-1-r}\binom{n-s}{r-1}(n-s+1) \\ &\qquad-N(k-1)!\sum_{r=0}^{k-2}\binom{s-2}{k-1-r} \binom{n-s+1}{r} \\ &\quad=N_{1} \Biggl(\sum_{t=1}^{s-1}x_{t}^{k-1}-x_{i}^{k-1} \Biggr) -N(k-2)!\sum_{r=0}^{k-2}(k-1-r) \binom{s-2}{k-1-r}\binom{n-s+1}{r} \\ &\quad=N_{1} \Biggl(\sum_{t=1}^{s-1}x_{t}^{k-1}-x_{i}^{k-1} \Biggr) -N(k-2)!\sum_{r=0}^{k-2}(s-2) \binom{s-3}{k-r-2}\binom{n-s+1}{r} \\ &\quad=N_{1} \Biggl(\sum_{t=1}^{s-1}x_{t}^{k-1}-x_{i}^{k-1} \Biggr) -N(k-2)!(s-2)\binom{n-2}{k-2} \\ &\quad=N_{1} \Biggl(\sum_{t=1}^{s-1}x_{t}^{k-1}-x_{i}^{k-1} \Biggr)-(s-2)N_{1}. \end{aligned}$$

Thus, by (2.3), (2.4), and the definition of $x_{i}^{k-1}$ for $1\leq i\leq s-1$, we have

$$\begin{aligned} r_{i}(\mathbb{B}) &\leq M+\frac{1}{x_{i}^{k-1}} \Biggl\{ r_{i}-M+N_{1} \Biggl(\sum_{t=1}^{s-1}x_{t}^{k-1}-x_{i}^{k-1} \Biggr)-(s-2)N_{1} \Biggr\} \\ &=M-N_{1}+\frac{1}{x_{i}^{k-1}} \Biggl\{ r_{i}-M+N_{1} \sum_{t=1}^{s-1}x_{t}^{k-1}-(s-2)N_{1} \Biggr\} \\ &=\phi_{s}. \end{aligned}$$

Subcase 2.2:$s=2$.

In this case, we need to show $r_{1}(\mathbb{B})\leq\phi_{2}$. Noting that $x_{2}=\cdots=x_{n}=1$, by (2.4) and the definition of $N_{r}^{2}$, we have

$$\begin{aligned} r_{1}(\mathbb{B}) \leq{}& M+\frac{1}{x_{1}^{k-1}} \Biggl\{ r_{1}-M+N(k-1)!\sum _{r=0}^{k-2}\sum_{\{i_{2},\ldots, i_{k}\}\in N_{r}^{2}} \biggl(\frac{x_{i_{2}}^{k-1}+\cdots +x_{i_{k}}^{k-1}}{k-1}-1 \biggr) \Biggr\} \\ ={}&M+\frac{1}{x_{1}^{k-1}}(r_{1}-M). \end{aligned}$$

By (2.3), we have $x_{1}^{k-1}=\frac{\phi_{2}-r_{2}+N_{1}}{N_{1}}$. Then, by (2.1) and the definition of $\phi_{2}$, we have

$$\begin{aligned} &\frac{1}{x_{1}^{k-1}}(r_{1}-M) \\ &\quad=\frac{N_{1}(r_{1}-M)}{\phi_{2}-r_{2}+N_{1}} \\ &\quad=\frac{2N_{1}(r_{1}-M)}{N_{1}+M-r_{2}+\sqrt{(N_{1}-M+r_{2})^{2}+4N_{1}(r_{1}-r_{2})}} \\ &\quad=\frac{2N_{1}(r_{1}-M)(N_{1}+M-r_{2}-\sqrt {(N_{1}-M+r_{2})^{2}+4N_{1}(r_{1}-r_{2})})}{(N_{1}+M-r_{2})^{2}-((N_{1}-M+r_{2})^{2}+4N_{1}(r_{1}-r_{2}))} \\ &\quad=-\frac{N_{1}+M-r_{2}-\sqrt{(N_{1}-M+r_{2})^{2}+4N_{1}(r_{1}-r_{2})}}{2}. \end{aligned}$$

Thus

$$\begin{aligned} r_{1}(\mathbb{B})\leq M+\frac{1}{x_{1}^{k-1}}(r_{1}-M)= \phi_{2}. \end{aligned}$$

Combining Subcases 2.1 and 2.2, we have $r_{i}(\mathbb{B})\leq\phi_{s}$ for $1\leq i\leq s-1$, and combining Cases 1 and 2, we have $r_{i}(\mathbb{B})\leq\phi_{s}$ for $1\leq i\leq n$. Then $\rho(\mathbb{A})=\rho(\mathbb{B})\leq\max_{1\leq i\leq n}r_{i}(\mathbb{B})\leq\phi_{s}$ for $2\leq s\leq n$ by (2.2) and Lemma 1.5.

Therefore, we know $\rho(\mathbb{A})\leq\phi_{s}$ for $1\leq s\leq n$ and thus $\rho(\mathbb{A})\leq\min_{1\leq s\leq n}\phi_{s}$.

Now suppose that $\mathbb{A}$ is weakly irreducible. Then $\mathbb{B}$ is also weakly irreducible by $\mathbb{B}=U^{-(k-1)}\mathbb{A}U$. Let $\phi_{s}=\min_{1\leq l\leq n}\phi_{l}$.

Case 1:$s=1$.

By Lemma 1.5 and the fact $r_{1}=\max_{1\leq i\leq n}r_{i}$, we have $\rho(\mathbb{A})= \phi_{1}$ if and only if $r_{1}=r_{2}=\cdots=r_{n}$.

Case 2:$2\leq s\leq n$.

Then $\rho(\mathbb{B})= \max_{1\leq i\leq n}r_{i}(\mathbb{B})$ and thus $r_{1}(\mathbb{B})=r_{2}(\mathbb{B})=\cdots=r_{n}(\mathbb{B})=\phi_{s}$ by $\phi_{s}=\rho(\mathbb{A})=\rho(\mathbb{B})\leq\max_{1\leq i\leq n}r_{i}(\mathbb{B})\leq\phi_{s}$ and Lemma 1.5. Therefore, (a), (b), (c), and (d) hold for any $i\in[n]$, (e) holds for any $i\in\{s,\ldots, n\}$.

Subcase 2.1:$r_{1}=r_{s}$.

By $r_{1}\geq r_{2}\geq\cdots\geq r_{n}$ and (e) $r_{i}=r_{s}$ for $s\leq i\leq n$, then we have $r_{1}= r_{2}=\cdots=r_{n}$.

Subcase 2.2:$r_{1}>r_{s}$.

Let t be the smallest integer such that $r_{t}=r_{s}$ for $1< t\leq s$. Since $r_{s}=r_{s+1}=\cdots=r_{n}$, we have $r_{t}=r_{t+1}=\cdots =r_{n}$ and $x_{i}>1$ for $i=1,2,\ldots,t-1$.

When $k\geq3$, (c) and (d) cannot hold at the same time. Because there are r elements in $\{i_{2},\ldots, i_{k}\}$ chosen from $\{s,\ldots, n\}$ and $k-1-r$ elements in $\{i_{2},\ldots, i_{k}\}$ chosen from $\{1,\ldots, s-1\}$, and then $x_{i_{2}}=\cdots=x_{i_{k}}$ cannot hold when $1\leq r\leq k-2$. Thus we only consider the case of $k=2$.

In the case of $k=2$, (d) implies

$$\sum_{\{i,i_{2}\}\in S(\mathbb{A})} (x_{i_{2}}-1 )=\sum _{r=0}^{1}\sum_{ \{i_{2}\}\in N_{r}^{s}} (x_{i_{2}}-1 )=\sum_{\substack{{i_{2}=1}\\ i_{2}\neq i}}^{t-1} (x_{i_{2}}-1 ). $$

Then (i)–(iii) follow from (a), (b), (c), (d) for $1\leq i\leq n$, and (e) for $s\leq i\leq n$, and thus (1) and (2) hold.

Conversely, if $r_{1}=r_{2}=\cdots=r_{n}$, then by Lemma 1.5, $\rho(\mathbb{A})=\phi_{1}=r_{1}$. If $k=2$ and (i)–(iii) hold, then (a), (b), (c), and (d) hold for $1\leq i\leq n$, (e) holds for $s\leq i\leq n$. Then we have $r_{i}(\mathbb{B})=\phi_{s}$ for $1\leq i\leq n$. Therefore, by Lemma 1.5, we have $\rho(\mathbb{A})=\rho(\mathbb {B})=\max_{1\leq i\leq n}r_{i}(\mathbb{B})=\phi_{s}$ for $s=2,\ldots,n$. □

Let $k=2$. Then $\mathbb{A}$ is a matrix, weak irreducibility for tensors corresponds to irreducibility for matrices, and slice sum for tensors corresponds to row sum for matrices. The following result follows immediately.

Corollary 2.2

([7], Theorem 2.1)

LetAbe an$n\times n$nonnegative matrix with row sums$r_{1},r_{2},\ldots,r_{n}$, where$r_{1}\geq r_{2}\geq \cdots\geq r_{n}$. LetMbe the largest diagonal element andNbe the largest non-diagonal element ofA. Suppose that$N>0$. Let$\phi _{1}=r_{1}$and, for$2\leq s\leq n$,

$$ \phi_{s}=\frac{1}{2} \Biggl(r_{s}+M-N+ \sqrt{(r_{s}-M+N)^{2}+4N \sum _{t=1}^{s-1}(r_{t}-r_{s})} \Biggr). $$

(2.7)

Then$\rho(A)\leq\min_{1\leq s\leq n}\phi_{s}$.

Let$\phi_{s}=\min_{1\leq l\leq n}\phi_{l}$. IfAis irreducible, then$\rho(A)=\phi_{s}$if and only if$r_{1}=r_{2}=\cdots=r_{n}$or for somet ($2\leq t\leq s$), Asatisfies the following conditions:

• (i) $a_{ii}=M$for$1\leq i\leq t-1$;

• (ii) $a_{ii_{2}}=N$for$1\leq i\leq s-1$and$1\leq i_{2}\leq t-1$with$i\neq i_{2}$;

• (iii) $r_{t}=\cdots=r_{n}$;

• (iv) $a_{ii_{2}}=N$for$s\leq i\leq n$and$1\leq i_{2}\leq t-1$.

3 Applications to a k-uniform hypergraph

A hypergraph is a natural generalization of an ordinary graph [1].

A hypergraph $\mathcal{H}=(V(\mathcal{H}), E(\mathcal{H}))$ on n vertices is a set of vertices, say, $V(\mathcal{H})=\{1,2,\ldots,n\}$ and a set of edges, say, $E(\mathcal{H})=\{e_{1},e_{2},\ldots,e_{m}\}$, where $e_{i}=\{i_{1},i_{2},\ldots,i_{l}\}$, $i_{j}\in[n]$, $j=1,2,\ldots,l$. Let $k\geq2$, if $\mid e_{i}\mid=k$ for any $i=1,2,\ldots,m$, then $\mathcal{H}$ is called a k-uniform hypergraph. When $k=2$, then $\mathcal{H}$ is an ordinary graph. The degree $d_{i}$ of vertex i is defined as $d_{i}=|\{e_{j}:i\in e_{j}\in E(\mathcal{H})\}|$. If $d_{i}=d$ for any vertex i of a hypergraph $\mathcal{H}$, then $\mathcal{H}$ is called d-regular. A walkW of length ℓ in $\mathcal{H}$ is a sequence of alternate vertices and edges: $v_{0},e_{1},v_{1},e_{2},\ldots,e_{\ell},v_{\ell}$, where $\{v_{i},v_{i+1}\}\subseteq e_{i+1}$ for $i=0,1,\ldots,\ell-1$. The hypergraph $\mathcal{H}$ is said to be connected if every two vertices are connected by a walk.

Definition 3.1

([6, 18])

Let $\mathcal{H}=(V(\mathcal{H}),E(\mathcal{H}))$ be a k-uniform hypergraph on n vertices. The adjacency tensor of $\mathcal{H}$ is defined as the order k dimension n tensor $\mathbb{A}(\mathcal{H})$, whose $(i_{1}i_{2}\cdots i_{k})$-entry is

$$\bigl(\mathbb{A}(\mathcal{H}) \bigr)_{i_{1}i_{2}\cdots i_{k}}=\textstyle\begin{cases} \frac{1}{(k-1)!}, & \mbox{if } \{i_{1},i_{2},\ldots,i_{k}\}\in E(\mathcal{H}), \\ 0, & \mbox{otherwise} . \end{cases} $$

Let $\mathbb{D}(\mathcal{H})$ be an order k dimension n diagonal tensor with its diagonal entry $\mathbb{D}_{ii\cdots i}$ being $d_{i}$, the degree of vertex i for all $i\in V(\mathcal{H})=[n]$. Then $\mathbb{Q}(\mathcal{H})=\mathbb{D(\mathcal{H})}+ \mathbb {A(\mathcal{H})}$ is called the signless Laplacian tensor of the hypergraph $\mathcal{H}$. Clearly, the adjacency tensor and the signless Laplacian tensor of a k-uniform hypergraph $\mathcal{H}$ are nonnegative symmetric k-uniform tensors and, for any $1\leq i\leq n$,

$$r_{i} \bigl(\mathbb{A(\mathcal{H})} \bigr)=\sum _{i_{2},\ldots, i_{k}=1}^{n} \bigl(\mathbb {A(\mathcal{H})} \bigr)_{ii_{2}\cdots i_{k}} =d_{i}, r_{i} \bigl(\mathbb{Q( \mathcal{H})} \bigr)=\sum_{i_{2},\ldots, i_{k}=1}^{n} \bigl( \mathbb {Q(\mathcal{H})} \bigr)_{ii_{2}\cdots i_{k}} =2d_{i}. $$

It was proved in [9, 20] that a k-uniform hypergraph $\mathcal{H}$ is connected if and only if its adjacency tensor $\mathbb{A}(\mathcal{H})$ (and thus the signless Laplacian tensor $\mathbb{Q}(\mathcal{H})$) is weakly irreducible.

Recently, several papers studied the spectral radii of the adjacency tensor $\mathbb{A}(\mathcal{H})$ and the signless Laplacian tensor $\mathbb{Q}(\mathcal{H})$ of a k-uniform hypergraph $\mathcal{H}$ (see [4, 6, 18, 19, 27, 28] and so on). In this section, we apply Theorem 2.1 to the adjacency tensor $\mathbb{A}(\mathcal{H})$ and the signless Laplacian tensor $\mathbb{Q}(\mathcal{H})$ of a k-uniform hypergraph $\mathcal{H}$. If $k=2$, we obtain Theorem 3.1 and Theorem 4.2 in [7]. If $k\geq3$, we improve some known results about the bounds of $\rho (\mathbb{A}(\mathcal{H}))$ and $\rho(\mathbb{Q}(\mathcal{H}))$ in [4].

Theorem 3.2

Let$k\geq3$, $\mathcal{H}$be ak-uniform hypergraph with degree sequence$d_{1}\geq\cdots\geq d_{n}$, $\mathbb{ A}(\mathcal{H})$be the adjacency tensor of$\mathcal{H}$. Let$A_{1}=\frac{1}{k-1}\binom{n-2}{k-2}$, $\phi_{1}=d_{1}$, and

$$ \phi_{s}=\frac{1}{2} \Biggl\{ d_{s}-A_{1}+ \sqrt {(d_{s}+A_{1})^{2}+4A_{1} \sum_{t=1}^{s-1}(d_{t}-d_{s})} \Biggr\} $$

(3.1)

for$2\leq s\leq n$. Then

$$\begin{aligned} \rho \bigl(\mathbb{A}(\mathcal{H}) \bigr)&\leq\min _{1\leq s\leq n}\phi_{s}. \end{aligned}$$

(3.2)

If$\mathcal{H}$is connected, then the equality in (3.2) holds if and only if$\mathcal{H}$is regular.

Proof

Let $\mathbb{A}=\mathbb{A}(\mathcal{H})$. We apply Theorem 2.1 to $\mathbb{A}(\mathcal{H})$, then we have $M=0$, $N=\frac{1}{(k-1)!}$, $r_{i}=d_{i}$ for $1\leq i\leq n$, $A_{1}=N_{1}$, and (3.1) is from (2.1). Thus (3.2) holds by Theorem 2.1.

If $\mathcal{H}$ is connected, then by Theorem 2.1 the equality in (3.2) holds if and only if $r_{1}(\mathbb{A(\mathcal{H})})=r_{2}(\mathbb{A(\mathcal{H})})=\cdots =r_{n}(\mathbb{A(\mathcal{H})})$, which says exactly that $\mathcal{H}$ is regular, since $r_{i}(\mathbb{A(\mathcal{H})})=d_{i}$ for any $1\leq i\leq n$. □

Theorem 3.3

Let$k\geq3$, $\mathcal{H}$be ak-uniform hypergraph with degree sequence$d_{1}\geq\cdots\geq d_{n}$, $\mathbb{ Q}(\mathcal{H})$be the signless Laplacian tensor of$\mathcal {H}$. Let$A_{1}=\frac{1}{k-1}\binom{n-2}{k-2}$, $\psi_{1}=2d_{1}$, and

$$ \psi_{s}=\frac{1}{2} \Biggl\{ 2d_{s}+d_{1}-A_{1}+ \sqrt {(2d_{s}-d_{1}+A_{1})^{2}+8A_{1} \sum_{t=1}^{s-1}(d_{t}-d_{s})} \Biggr\} $$

(3.3)

for$2\leq s\leq n$. Then

$$ \rho \bigl(\mathbb{Q}(\mathcal{H}) \bigr)\leq\min _{1\leq s\leq n}\psi_{s}. $$

(3.4)

If$\mathcal{H}$is connected, then the equality in (3.4) holds if and only if$\mathcal{H}$is regular.

Proof

Let $\mathbb{A}=\mathbb{Q}(\mathcal{H})$. We apply Theorem 2.1 to $\mathbb{Q}(\mathcal{H})$, then we have $M=d_{1}$, $N=\frac{1}{(k-1)!}$, $r_{i}=2d_{i}$ for $1\leq i\leq n$, $A_{1}=N_{1}$, and (3.3) is from (2.1). Thus (3.4) holds by Theorem 2.1.

If $\mathcal{H}$ is connected, then by Theorem 2.1 the equality in (3.4) holds if and only if $r_{1}(\mathbb{Q(\mathcal{H})})=r_{2}(\mathbb{Q(\mathcal{H})})=\cdots =r_{n}(\mathbb{Q(\mathcal{H})})$, which says exactly that $\mathcal{H}$ is regular, since $r_{i}(\mathbb{Q(\mathcal{H})})=2d_{i}$ for any $1\leq i\leq n$. □

4 Applications to k-uniform directed hypergraph

Directed hypergraphs have found applications in imaging processing [8], optical network communications [15], computer science and combinatorial optimization [10]. However, unlike spectral theory of undirected hypergraphs, there are very few results in spectral theory of directed hypergraphs.

A directed hypergraph $\overrightarrow{\mathcal{H}}$ is a pair $(V(\overrightarrow{\mathcal{H}}), E(\overrightarrow{\mathcal{H}}))$, where $V(\overrightarrow{\mathcal{H}})=[n]$ is the set of vertices and $E(\overrightarrow{\mathcal{H}})=\{e_{1}, e_{2}, \ldots, e_{m}\}$ is the set of arcs. An arc $e\in E(\overrightarrow{\mathcal{H}})$ is a pair $e=(j_{1},e(j_{1}))$, where $e(j_{1})=\{j_{2},\ldots,j_{t}\}$, $j_{l}\in V(\overrightarrow{\mathcal{H}})$, and $j_{l}\neq j_{h}$ if $l\neq h$ for $l, h\in[t]$ and $t\in[n]$. The vertex $j_{1}$ is called the tail (or out-vertex) and every other vertex $j_{2},\ldots,j_{t}$ is called a head (or in-vertex) of the arc e. The out-degree of a vertex $j\in V(\overrightarrow{\mathcal{H}})$ is defined as $d_{j}^{+}=|E_{j}^{+}|$, where $E_{j}^{+}=\{e\in E(\overrightarrow{\mathcal{H}}): j \mbox{ is the tail of } e\}$. If for any $j\in V(\overrightarrow{\mathcal{H}})$, the degree $d_{j}^{+}$ has the same value d, then $\overrightarrow{\mathcal{H}}$ is called a directed d-out-regular hypergraph.

For a vertex $i\in V(\overrightarrow{\mathcal{H}})$, we denote by $E_{i}$ the set of arcs containing the vertex i, i.e., $E_{i}=\{e\in E(\overrightarrow{\mathcal{H}}): i\in e\}$. Two distinct vertices i and j are weak-connected if there is a sequence of arcs $(e_{1},\ldots,e_{t})$ such that $i\in e_{1}$, $j\in e_{t}$, and $e_{r}\cap e_{r+1}\neq\emptyset$ for all $r\in[t-1]$. Two distinct vertices i and j are strong-connected, denoted by $i\rightarrow j$, if there is a sequence of arcs $(e_{1},\ldots,e_{t})$ such that i is the tail of $e_{1}$, j is a head of $e_{t}$, and a head of $e_{r}$ is the tail of $e_{r+1}$ for all $r\in[t-1]$. A directed hypergraph is called weakly connected if every pair of different vertices of $\overrightarrow{\mathcal{H}}$ is weak-connected. A directed hypergraph is called strongly connected if every pair of different vertices i and j of $\overrightarrow{\mathcal{H}}$ satisfies $i\rightarrow j$ and $j\rightarrow i$.

Similar to the definition of a k-uniform hypergraph, we define a k-uniform directed hypergraph as follows: A directed hypergraph $\overrightarrow{\mathcal{H}}=(V(\overrightarrow {\mathcal{H}}), E(\overrightarrow{\mathcal{H}}))$ is called a k-uniform directed hypergraph if $|e|=k$ for any arc $e\in E(\overrightarrow{\mathcal{H}})$. When $k=2$, then $\overrightarrow{\mathcal{H}}$ is an ordinary digraph.

The following definition for the adjacency tensor and signless Laplacian tensor of a directed hypergraph was proposed by Chen and Qi in [5].

Definition 4.1

([5])

Let $\overrightarrow{\mathcal{H}}=(V(\overrightarrow{\mathcal{H}}), E(\overrightarrow{\mathcal{H}}))$ be a k-uniform directed hypergraph. The adjacency tensor of the directed hypergraph $\overrightarrow {\mathcal{H}}$ is defined as the order k dimension n tensor $\mathbb {A}(\overrightarrow{\mathcal{H}})$, whose $(i_{1}i_{2}\cdots i_{k})$-entry is

$$\bigl(\mathbb{A}(\overrightarrow{\mathcal{H}}) \bigr)_{i_{1}\cdots i_{k}}=\textstyle\begin{cases} \frac{1}{(k-1)!}, & \mbox{if } (i_{1},e(i_{1}))\in E(\overrightarrow{\mathcal{H}}) \mbox{ and } e(i_{1})=(i_{2},\ldots ,i_{k}),\\ 0, & \mbox{otherwise} . \end{cases} $$

Let $\mathbb{D}(\overrightarrow{\mathcal{H}})$ be an order k dimension n diagonal tensor with its diagonal entry $d_{ii\cdots i}$ being $d_{i}^{+}$, the out-degree of vertex i, for all $i\in V(\overrightarrow{\mathcal{H}})=[n]$. Then $\mathbb{Q}(\overrightarrow{\mathcal{H}})=\mathbb{D(\overrightarrow {\mathcal{H}})}+ \mathbb{A(\overrightarrow{\mathcal{H}})}$ is the signless Laplacian tensor of the directed hypergraph $\overrightarrow {\mathcal{H}}$.

Clearly, the adjacency tensor and the signless Laplacian tensor of a k-uniform directed hypergraph $\overrightarrow{\mathcal{H}}$ are nonnegative k-uniform tensors, but not symmetric in general. For any $1\leq i\leq n$, we have

$$r_{i} \bigl(\mathbb{A(\overrightarrow{\mathcal{H}})} \bigr)=\sum _{i_{2},\ldots, i_{k}=1}^{n} \bigl(\mathbb{A(\overrightarrow{ \mathcal{H}})} \bigr)_{ii_{2}\cdots i_{k}} =d_{i}^{+} $$

and

$$r_{i} \bigl(\mathbb{Q(\overrightarrow{\mathcal{H}})} \bigr)=\sum _{i_{2},\ldots, i_{k}=1}^{n} \bigl(\mathbb{Q(\overrightarrow{ \mathcal{H}})} \bigr)_{ii_{2}\cdots i_{k}} =2d_{i}^{+}. $$

The following statement is an alternative explanation of weak irreducibility.

Definition 4.2

([9, 12])

Suppose that $\mathbb{A}=(a_{i_{1}i_{2}\ldots i_{k}})_{1\le i_{j}\le n (j=1, \ldots, k)}$ is a nonnegative tensor of order k and dimension n. We call a nonnegative matrix $G(\mathbb{A})$ the representation associated matrix to the nonnegative tensor $\mathbb{A}$ if the $(i, j)$th entry of $G(\mathbb{A})$ is defined to be the summation of $a_{ii_{2}\ldots i_{k}}$ with indices $\{i_{2},\ldots, i_{k}\}\ni j$. We call the tensor $\mathbb {A}$ weakly reducible if its representation $G(\mathbb{A})$ is a reducible matrix.

Let $A=(a_{ij})$ be a nonnegative square matrix of order n. The associated digraph $D(A)=(V,E)$ of A (possibly with loops) is defined to be the digraph with vertex set $V=\{1,2,\ldots,n\}$ and arc set $E=\{(i,j)\mid a_{ij}>0\}$.

Now we give a characterization of a strongly connected k-uniform directed hypergraph.

Theorem 4.3

Let$\overrightarrow{\mathcal{H}}$be ak-uniform directed hypergraph, $\mathbb{A}=\mathbb{A}(\overrightarrow{\mathcal {H}})=(a_{i_{1}i_{2}\cdots i_{k}})$be the adjacency tensor of$\overrightarrow{\mathcal{H}}$, $G(\mathbb{A})$be the representation associated matrix of$\mathbb{A}$, and$D(G(\mathbb{A}))$be the associated directed graph of$G(\mathbb{A})$. Then the following four conditions are equivalent:

• (i) $\mathbb{A}$is weakly irreducible.

• (ii) $G(\mathbb{A})$is irreducible.

• (iii) $D(G(\mathbb{A}))$is strongly connected.

• (iv) $\overrightarrow{\mathcal{H}}$is strongly connected.

Proof

By Proposition 15 in [27] and $\mathbb{A}=\mathbb {A}(\overrightarrow{\mathcal{H}})$ is a nonnegative tensor, we have (i) ⇔ (ii) ⇔ (iii). Now we show (iii) ⇔ (iv).

(iii) ⇒ (iv): Let $D(G(\mathbb{A}))$ is strongly connected, now we show $\overrightarrow{\mathcal{H}}$ is strongly connected.

For any $i,j\in V(\overrightarrow{\mathcal{H}})=V(D(G(\mathbb{A})))$, there exists a directed path P from i to j in $D(G(\mathbb{A}))$ by $D(G(\mathbb{A}))$ being strongly connected. We assume $P=ij_{1}j_{2}\cdots j_{t}j$, then $(i,j_{1}),(j_{1},j_{2}),\ldots , (j_{t},j)\in E(D(G(\mathbb{A})))$, which implies $\sum_{j_{1}\in\{i_{2},\ldots, i_{k}\}} a_{ii_{2}\cdots i_{k}}>0$, $\sum_{j_{2}\in\{i_{2},\ldots, i_{k}\}} a_{j_{1}i_{2}\cdots i_{k}}>0$, …, $\sum_{j_{t}\in\{i_{2},\ldots, i_{k}\}} a_{j_{t-1}i_{2}\cdots i_{k}}>0$, and $\sum_{j\in\{i_{2},\ldots, i_{k}\}} a_{j_{t}i_{2}\cdots i_{k}}>0$, thus there exists a sequence of arcs ($e_{1},e_{2},\ldots,e_{t},e_{t+1}$), where $e_{l} \in\overrightarrow{\mathcal{H}}$ and $l\in[t+1]$, such that i is the tail of $e_{1}$, $j_{1}$ is a head of $e_{1}$, $j_{l}$ is the tail of $e_{l+1}$, $j_{l+1}$ is a head of $e_{l+1}$ for $1\leq l\leq t-1$, $j_{t}$ is the tail of $e_{t+1}$, j is a head of $e_{t+1}$, say, $i\rightarrow j$ in ${\mathcal{H}}$. Therefore $\overrightarrow{\mathcal {H}}$ is strongly connected.

(iv) ⇒ (iii): Let $\overrightarrow{\mathcal{H}}$ be strongly connected. Now we show that $D(G(\mathbb{A}))$ is strongly connected.

For any $i,j\in V(D(G(\mathbb{A})))=V(\overrightarrow{\mathcal{H}})$, $i\rightarrow j$ in $\overrightarrow{\mathcal{H}}$ by $\overrightarrow {\mathcal{H}}$ being strongly connected, say, there exists a sequence of arcs ($e_{1},e_{2},\ldots,e_{t},e_{t+1}$), where $e_{l} \in\overrightarrow {\mathcal{H}}$ for $l\in[t+1]$, such that i is the tail of $e_{1}$, j is a head of $e_{t+1}$, and a head of $e_{r}$ is the tail of $e_{r+1}$ for all $r\in[t]$. We assume that $j_{r}$ is the tail of $e_{r+1}$ and a head of $e_{r}$ for all $r\in[t]$, then $\sum_{j_{1}\in\{i_{2},\ldots, i_{k}\}} a_{ii_{2}\cdots i_{k}}>0$, $\sum_{j_{r+1}\in\{i_{2}, \ldots, i_{k}\}} a_{j_{r}i_{2}\cdots i_{k}}>0$ for $1\leq r\leq t-1$, and $\sum_{j\in\{i_{2},\ldots, i_{k}\}} a_{j_{t}i_{2}\cdots i_{k}}>0$. Thus $(i,j_{1})\in E(D(G(\mathbb{A})))$, $(j_{r},j_{r+1})\in E(D(G(\mathbb {A})))$ for $1\leq r\leq t-1$ and $(j_{t},j)\in E(D(G(\mathbb{A})))$, which implies that there exists a walk $ij_{1}j_{2}\cdots j_{t}j$ in $D(G(\mathbb{A}))$. Therefore $D(G(\mathbb{A}))$ is strongly connected. □

Recently, several papers studied the spectral radii of the adjacency tensor $\mathbb{A}(\overrightarrow{\mathcal{H}})$ and the signless Laplacian tensor $\mathbb{Q}(\overrightarrow{\mathcal {H}})$ of a k-uniform directed hypergraph $\overrightarrow{\mathcal{H}}$ (see [5, 24] and so on).

Let $\overrightarrow{\mathcal{H}}$ be a k-uniform directed hypergraph. If $\overrightarrow{\mathcal{H}}$ is strongly connected, then by Theorem 4.3 and the above definitions, $\mathbb{A}(\overrightarrow{\mathcal{H}})$ and thus $\mathbb {Q}(\overrightarrow{\mathcal{H}})$ are weakly irreducible. Thus we can apply Theorem 2.1 to the adjacency tensor $\mathbb {A}(\overrightarrow{\mathcal{H}})$ and the signless Laplacian tensor $\mathbb{Q}(\overrightarrow{\mathcal{H}})$ of a (strongly connected) k-uniform directed hypergraph $\overrightarrow{\mathcal{H}}$. If $k= 2$, we obtain Theorem 2.7 in [11]. If $k\geq3$, we obtain some new results about the bounds of $\rho (\mathbb{A}(\overrightarrow{\mathcal{H}}))$ and $\rho(\mathbb {Q}(\overrightarrow{\mathcal{H}}))$ as follows.

Theorem 4.4

Let$k\geq3$, $\overrightarrow{\mathcal{H}}$be ak-uniform directed hypergraph with out-degree sequence$d_{1}^{+}\geq\cdots\geq d_{n}^{+}$, $\mathbb{ A}(\overrightarrow{\mathcal{H}})$be the adjacency tensor of$\overrightarrow{\mathcal{H}}$. Let$A_{1}=\frac{1}{k-1}\binom {n-2}{k-2}$, $\phi_{1}=d_{1}^{+}$, and

$$ \phi_{s}=\frac{1}{2} \Biggl\{ d_{s}^{+}-A_{1}+ \sqrt { \bigl(d_{s}^{+}+A_{1} \bigr)^{2}+4A_{1} \sum_{t=1}^{s-1} \bigl(d_{t}^{+}-d_{s}^{+} \bigr)} \Biggr\} $$

(4.1)

for$2\leq s\leq n$. Then

$$\begin{aligned} \rho \bigl(\mathbb{A}(\overrightarrow{\mathcal{H}}) \bigr)&\leq \min_{1\leq s\leq n}\phi_{s}. \end{aligned}$$

(4.2)

Moreover, if$\overrightarrow{\mathcal{H}}$is a strongly connectedk-uniform directed hypergraph, then the equality in (4.2) holds if and only if$d_{1}^{+}=d_{2}^{+}=\cdots=d_{n}^{+}$.

Proof

Let $\mathbb{A}=\mathbb{A}(\overrightarrow{\mathcal{H}})$. We apply Theorem 2.1 to $\mathbb{A}(\overrightarrow{\mathcal{H}})$, then we have $M=0$, $N=\frac{1}{(k-1)!}$, $r_{i}=d^{+}_{i}$ for $1\leq i\leq n$, $A_{1}=N_{1}$, and (4.1) is from (2.1). Thus (4.2) holds by Theorem 2.1, and the equality in (4.2) holds if and only if $d_{1}^{+}=d_{2}^{+}=\cdots=d_{n}^{+}$ by Theorem 2.1 and Theorem 4.3. □

Theorem 4.5

Let$k\geq3$, $\overrightarrow{\mathcal{H}}$be ak-uniform directed hypergraph with out-degree sequence$d_{1}^{+}\geq\cdots\geq d_{n}^{+}$, $\mathbb{Q}(\overrightarrow{\mathcal{H}})$be the signless Laplacian tensor of$\overrightarrow{\mathcal{H}}$. Let$A_{1}=\frac {1}{k-1}\binom{n-2}{k-2}$, $\psi_{1}=2d_{1}^{+}$, and

$$ \psi_{s}=\frac{1}{2} \Biggl\{ 2d_{s}^{+}+d_{1}^{+}-A_{1}+ \sqrt { \bigl(2d_{s}^{+}-d_{1}^{+}+A_{1} \bigr)^{2}+8A_{1}\sum_{t=1}^{s-1} \bigl(d_{t}^{+}-d_{s}^{+} \bigr)} \Biggr\} $$

(4.3)

for$2\leq s\leq n$. Then

$$ \rho \bigl(\mathbb{Q}(\overrightarrow{\mathcal{H}}) \bigr)\leq \min_{1\leq s\leq n}\psi_{s}. $$

(4.4)

Moreover, if$\overrightarrow{\mathcal{H}}$is a strongly connectedk-uniform directed hypergraph, then the equality in (4.4) holds if and only if$d_{1}^{+}=d_{2}^{+}=\cdots=d_{n}^{+}$.

Proof

Let $\mathbb{A}=\mathbb{Q}(\overrightarrow{\mathcal{H}})$. We apply Theorem 2.1 to $\mathbb{Q}(\overrightarrow{\mathcal{H}})$, then we have $M=d_{1}^{+}$, $N=\frac{1}{(k-1)!}$, $r_{i}=2d_{i}^{+}$ for $1\leq i\leq n$, $A_{1}=N_{1}$, and (4.3) is from (2.1). Thus (4.4) holds by Theorem 2.1, and the equality in (4.4) holds if and only if $d_{1}^{+}=d_{2}^{+}=\cdots=d_{n}^{+}$ by Theorem 2.1 and Theorem 4.3. □