1 Introduction

A real order m dimension n tensor \(\mathcal{A}=(a_{i_{1}\cdots i_{m}})\), denoted by \(\mathcal{A}\in R^{[m,n]}\), consists of \(n^{m}\) real entries:

$$a_{i_{1}\cdots i_{m}}\in R, $$

where \(i_{j}=1,\ldots,n\) for \(j=1,\ldots, m\). A tensor \(\mathcal{A}\) is called nonnegative (positive), denoted by \(\mathcal{A}\geq0\) (\(\mathcal{A}>0\)), if every entry \(a_{i_{1}\cdots i_{m}}\geq0\) (\(a_{i_{1}\cdots i_{m}}> 0\), respectively). Given a tensor \(\mathcal{A}=(a_{i_{1}\cdots i_{m}})\in R^{[m,n]}\), if there are a complex number λ and a nonzero complex vector \(x=(x_{1},x_{2},\ldots,x_{n})^{T}\) that are solutions of the following homogeneous polynomial equations:

$$\mathcal{A}x^{m-1}=\lambda x^{[m-1]}, $$

then λ is called an eigenvalue of \(\mathcal{A}\) and x an eigenvector of \(\mathcal{A}\) associated with λ [1–6], where \(\mathcal{A}x^{m-1}\) and \(x^{[m-1]}\) are vectors, whose ith entries are

$$\bigl(\mathcal{A}x^{m-1}\bigr)_{i}=\sum _{i_{2},\ldots,i_{m}\in N} a_{ii_{2}\cdots i_{m}}x_{i_{2}}\cdots x_{i_{m}} \quad \bigl(N=\{1,2,\ldots,n\}\bigr) $$

and \((x^{[m-1]})_{i}=x_{i}^{m-1}\), respectively. Moreover, the spectral radius \(\rho(\mathcal{A})\) [7] of the tensor \(\mathcal{A}\) is defined as

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

Eigenvalues of tensors have become an important topic of study in numerical multilinear algebra, and they have a wide range of practical applications; see [4, 5, 8–21]. Recently, for the largest eigenvalue of a nonnegative tensor, Chang et al. [2] generalized the well-known Perron-Frobenius theorem for irreducible nonnegative matrices to irreducible nonnegative tensors. Here a tensor \(\mathcal{A}=(a_{i_{1}\cdots i_{m}}) \in R^{m,n}\) is called reducible, if there exists a nonempty proper index subset \(I\subset N\) such that

$$a_{i_{1}i_{2}\cdots i_{m} }=0 \quad \mbox{for all } i_{1}\in I, \mbox{for all } i_{2},\ldots,i_{m}\notin I. $$

If \(\mathcal{A}\) is not reducible, then we call \(\mathcal{A}\) irreducible.

Theorem 1

(Theorem 1.4 in [2])

If \(\mathcal{A} \in R^{[m,n]} \) is irreducible nonnegative, then \(\rho(\mathcal{A})\) is a positive eigenvalue with an entrywise positive eigenvector x, i.e., \(x> 0\), corresponding to it.

Subsequently, Yang and Yang [21] extended this theorem to nonnegative tensors.

Theorem 2

(Theorem 2.3 in [21])

If \(\mathcal{A} \in R^{[m,n]} \) is nonnegative, then \(\rho(\mathcal{A})\) is an eigenvalue with an entrywise nonnegative eigenvector x, i.e., \(x\geq0\), \(x\neq0\), corresponding to it.

For the spectral radius of a nonnegative tensor, Yang and Yang [21] provided a lower bound and an upper bound for the spectral radius of a nonnegative tensor.

Theorem 3

(Lemma 5.2 in [21])

Let \(\mathcal{A}=(a_{i_{1}\cdots i_{m}}) \in R^{[m,n]} \) be nonnegative. Then

$$R_{\mathrm{min}}\leq\rho(\mathcal{A})\leq R_{\mathrm{max}}, $$

where \(R_{\mathrm{min}}=\min_{i\in N} R_{i}(\mathcal{A})\), \(R_{\mathrm{max}}= \max_{i\in N} R_{i}(\mathcal{A})\), and \(R_{i}(\mathcal{A}) =\sum_{i_{2},\ldots,i_{m}\in N } a_{ii_{2}\cdots i_{m}}\).

In order to obtain much sharper bounds of the spectral radius of a nonnegative tensor, Li et al. [22] have given an upper bound which estimates the spectral radius more precisely than that in Theorem 3.

Theorem 4

(Theorems 3.3 and 3.5 in [22])

Let \(\mathcal{A}=(a_{i_{1}\cdots i_{m}}) \in R^{[m,n]} \) be nonnegative with \(n \geq2\). Then

$$\rho(\mathcal{A})\leq\Omega_{\mathrm{max}}, $$

where

$$\Omega_{\mathrm{max}}= \max_{\substack{i,j\in N,\\ j\neq i}} \frac{1}{2} \Bigl( a_{i\cdots i}+a_{j\cdots j}+r_{i}^{j}(\mathcal{A})+ \sqrt{ \bigl( a_{i\cdots i}-a_{j\cdots j}+r_{i}^{j}( \mathcal{A}) \bigr)^{2}+ 4a_{ij\cdots j} r_{j}( \mathcal{A})} \Bigr). $$

Furthermore, \(\Omega_{\mathrm{max}}\leq R_{\mathrm{max}}\).

In this paper, we continue this research, and we give a lower bound and an upper bound for \(\rho(\mathcal{A})\) of a nonnegative tensor \(\mathcal{A}\), which all depend only on the entries of \(\mathcal{A}\). It is proved that these bounds are shaper than the corresponding bounds in [21] and [22]. A numerical example is also given to verify the obtained results.

2 New bounds for the spectral radius of nonnegative tensors

In this section, bounds for the spectral radius of a nonnegative tensors are obtained. We first give some notation. Given a nonnegative tensor \(\mathcal{A}=(a_{i_{1}\cdots i_{m}})\in R^{[m,n]}\), we denote

$$\begin{aligned}& \Theta_{i}=\bigl\{ (i_{2},i_{3}, \ldots,i_{m}): i_{j}=i \mbox{ for some } j \in\{2,\ldots,m\}, \mbox{where } i,i_{2},\ldots,i_{m}\in N\bigr\} , \\& \overline{\Theta}_{i}=\bigl\{ (i_{2},i_{3}, \ldots,i_{m}): i_{j}\neq i \mbox{ for any } j \in\{2,\ldots,m \}, \mbox{where } i,i_{2},\ldots,i_{m}\in N\bigr\} , \\& r_{i}(\mathcal{A})=\sum_{\substack{i_{2},\ldots,i_{m}\in N,\\ \delta _{ii_{2}\cdots i_{m}}=0}} a_{ii_{2}\cdots i_{m}}=\sum_{i_{2},\ldots,i_{m}\in N} a_{ii_{2}\cdots i_{m}}-a_{i\cdots i}=R_{i}( \mathcal{A})-a_{i\cdots i}, \\& r_{i}^{j}(\mathcal {A})=\sum_{\substack{\delta_{ii_{2}\ldots i_{m}}=0,\\ \delta_{ji_{2}\cdots i_{m}}=0}} a_{ii_{2}\cdots i_{m}}=\sum_{\substack{i_{2},\ldots,i_{m}\in N,\\ \delta_{ii_{2}\ldots i_{m}}=0}} a_{ii_{2}\cdots i_{m}}-a_{ij\cdots j}=r_{i}( \mathcal{A})-a_{ij\cdots j}, \\& r_{i}^{\Theta_{i}}(\mathcal{A})=\sum_{\substack{(i_{2},\ldots,i_{m})\in\Theta_{i},\\ \delta_{ii_{2}\cdots i_{m}}=0}} |a_{ii_{2}\cdots i_{m}}|,\qquad r_{i}^{\overline{\Theta}_{i}}(\mathcal {A})=\sum _{(i_{2},\ldots,i_{m})\in\overline{\Theta}_{i}} |a_{ii_{2}\cdots i_{m}}|, \end{aligned}$$

where

$$\delta_{i_{1}\cdots i_{m}}=\left \{ \begin{array}{l@{\quad}l} 1, &\mbox{if } i_{1}=\cdots=i_{m}, \\ 0, &\mbox{otherwise}. \end{array} \right . $$

Obviously, \(r_{i}(\mathcal{A})= r_{i}^{\Theta_{i}}(\mathcal {A})+r_{i}^{\overline{\Theta}_{i}}(\mathcal{A})\), and \(r_{i}^{j}(\mathcal {A})= r_{i}^{\Theta_{i}}(\mathcal {A})+r_{i}^{\overline{\Theta}_{i}}(\mathcal{A})-|a_{ij\cdots j}|\).

For an irreducible nonnegative tensor, we give the following bounds for the spectral radius.

Lemma 1

Let \(\mathcal{A}=(a_{i_{1}\cdots i_{m}})\in R^{[m,n]}\) be an irreducible nonnegative tensor with \(n \geq2\). Then

$$\Delta_{\mathrm{min}} \leq\rho(\mathcal{A})\leq \Delta_{\mathrm{max}}, $$

where

$$\Delta_{\mathrm{min}}=\min_{\substack{i,j\in N,\\ j\neq i}} \Delta_{i,j}( \mathcal{A}), \qquad \Delta_{\mathrm{max}}=\max_{\substack{i,j\in N, \\ j\neq i}} \Delta_{i,j}(\mathcal{A}) $$

and

$$\Delta_{i,j}(\mathcal{A})=\frac{1}{2} \Bigl( a_{i\cdots i}+ a_{j\cdots j}+r_{i}^{\Theta_{i}}(\mathcal{A}) + \sqrt{ \bigl(a_{i\cdots i}- a_{j\cdots j}+r_{i}^{\Theta_{i}}( \mathcal{A}) \bigr)^{2}+4r_{i}^{\overline{\Theta }_{i}}( \mathcal{A})r_{j}(\mathcal {A})} \Bigr). $$

Proof

Let \(x=(x_{1},x_{2},\ldots,x_{n})^{T}\) be an entrywise positive eigenvector of \(\mathcal{A}\) corresponding to \(\rho(\mathcal{A})\), that is,

$$ \mathcal{A}x^{m-1}=\rho(\mathcal{A})x^{[m-1]}. $$
(1)

Without loss of generality, suppose that

$$x_{t_{n}}\geq x_{t_{n-1}} \geq\cdots\geq x_{t_{2}} \geq x_{t_{1}}>0. $$

(i) We first prove

$$\Delta_{\mathrm{min}}=\min_{\substack{i,j\in N, \\ j\neq i}} \Delta_{i,j}( \mathcal {A})\leq \rho(\mathcal{A}). $$

From (1), we have

$$\sum_{i_{2},\ldots,i_{m}\in N} a_{t_{1}i_{2}\cdots i_{m}}x_{i_{2}}\cdots x_{i_{m}}=\rho(\mathcal{A}) x_{t_{1}}^{m-1}, $$

equivalently,

$$\bigl(\rho(\mathcal{A})-a_{t_{1}\cdots t_{1}}\bigr)x_{t_{1}}^{m-1}= \sum_{\substack{(i_{2},\ldots,i_{m})\in \Theta_{t_{1}},\\ \delta_{t_{1}i_{2}\ldots i_{m}}=0}} a_{t_{1}i_{2}\cdots i_{m}}x_{i_{2}}\cdots x_{i_{m}}+ \sum_{(i_{2},\ldots ,i_{m})\in \overline{\Theta}_{t_{1}} }a_{t_{1}i_{2}\cdots i_{m}}x_{i_{2}} \cdots x_{i_{m}}. $$

Hence,

$$\begin{aligned} \bigl(\rho(\mathcal{A})-a_{t_{1}\cdots t_{1}}\bigr)x_{t_{1}}^{m-1} \geq& \sum_{\substack{(i_{2},\ldots,i_{m})\in\Theta_{t_{1}}, \\ \delta_{t_{1}i_{2}\ldots i_{m}}=0}} a_{t_{1}i_{2}\cdots i_{m}}x_{t_{1}}^{m-1}+ \sum_{(i_{2},\ldots,i_{m})\in \overline{\Theta}_{t_{1}}} a_{t_{1}i_{2}\cdots i_{m}}x_{t_{2}}^{m-1} \\ =&r_{t_{1}}^{\Theta_{t_{1}}}(\mathcal{A})x_{t_{1}}^{m-1}+ r_{t_{1}}^{\overline{\Theta}_{t_{1}}}(\mathcal{A})x_{t_{2}}^{m-1}, \end{aligned}$$

i.e.,

$$ \bigl(\rho(\mathcal{A})-a_{t_{1}\cdots t_{1}} -r_{t_{1}}^{\Theta_{t_{1}}}( \mathcal{A}) \bigr)x_{t_{1}}^{m-1}\geq r_{t_{1}}^{\overline{\Theta}_{t_{1}}}( \mathcal{A})x_{t_{2}}^{m-1} \geq 0. $$
(2)

Similarly, we have, from (1),

$$\sum_{i_{2},\ldots,i_{m}\in N} a_{t_{2}i_{2}\cdots i_{m}}x_{i_{2}}\cdots x_{i_{m}}=\rho(\mathcal{A}) x_{t_{2}}^{m-1} $$

and

$$ \bigl(\rho(\mathcal{A})-a_{t_{2}\cdots t_{2}} \bigr)x_{t_{2}}^{m-1} \geq r_{t_{2}}(\mathcal{A})x_{t_{1}}^{m-1}\geq 0. $$
(3)

Multiplying inequality (3) with inequality (2) gives

$$\bigl(\rho(\mathcal{A})-a_{t_{1}\cdots t_{1}} -r_{t_{1}}^{\Theta_{t_{1}}}( \mathcal{A}) \bigr) \bigl(\rho(\mathcal {A})-a_{t_{2}\cdots t_{2}} \bigr)x_{t_{1}}^{m-1} x_{t_{2}}^{m-1}\geq r_{t_{2}}(\mathcal{A}) r_{t_{1}}^{\overline{\Theta}_{t_{1}}}(\mathcal{A}) x_{t_{1}}^{m-1}x_{t_{2}}^{m-1}. $$

Note that \(x_{t_{2}}\geq x_{t_{1}}>0\), hence

$$\bigl(\rho(\mathcal{A})-a_{t_{1}\cdots t_{1}} -r_{t_{1}}^{\Theta_{t_{1}}}( \mathcal{A}) \bigr) \bigl(\rho(\mathcal{A})-a_{t_{2}\cdots t_{2}} \bigr) \geq r_{t_{2}}(\mathcal{A}) r_{t_{1}}^{\overline{\Theta}_{t_{1}}}(\mathcal{A}), $$

that is,

$$\rho(\mathcal{A})^{2}- \bigl( a_{t_{1}\cdots t_{1}} +a_{t_{2}\cdots t_{2}} +r_{t_{1}}^{\Theta_{t_{1}}}(\mathcal{A}) \bigr)\rho(\mathcal {A})+a_{t_{2}\cdots t_{2}} \bigl( a_{t_{1}\cdots t_{1}} +r_{t_{1}}^{\Theta_{t_{1}}}( \mathcal{A}) \bigr)\geq r_{t_{2}}(\mathcal{A}) r_{t_{1}}^{\overline{\Theta}_{t_{1}}}( \mathcal{A}). $$

Furthermore, since

$$\bigl(a_{t_{1}\cdots t_{1}}+a_{t_{2}\cdots t_{2}}+r_{t_{1}}^{\Theta_{t_{1}}}( \mathcal{A}) \bigr)^{2}-4a_{t_{2}\cdots t_{2}} \bigl(a_{t_{1}\cdots t_{1}}+r_{t_{1}}^{\Theta_{t_{1}}}( \mathcal{A}) \bigr)= \bigl(a_{t_{1}\cdots t_{1}}-a_{t_{2}\cdots t_{2}}+r_{t_{1}}^{\Theta_{t_{1}}}( \mathcal{A}) \bigr)^{2}, $$

then solving for \(\rho(\mathcal{A})\) gives

$$\rho(\mathcal{A})\geq\Delta_{t_{1},t_{2}}(\mathcal{A})\geq \min _{\substack{i,j\in N, \\ j\neq i}} \Delta_{i,j}(\mathcal{A})=\Delta_{\mathrm{min}}. $$

(ii) We now prove

$$\rho(\mathcal{A})\leq\max_{\substack{i,j\in N,\\ j\neq i}} \Delta_{i,j}( \mathcal{A})=\Delta_{\mathrm{max}}. $$

From (1), we have

$$\sum_{i_{2},\ldots,i_{m}\in N} a_{t_{n}i_{2}\cdots i_{m}}x_{i_{2}}\cdots x_{i_{m}}=\rho(\mathcal{A}) x_{t_{n}}^{m-1} $$

and

$$\sum_{i_{2},\ldots,i_{m}\in N} a_{t_{n-1}i_{2}\cdots i_{m}}x_{i_{2}}\cdots x_{i_{m}}=\rho(\mathcal{A}) x_{t_{n-1}}^{m-1}. $$

Similar to the proof in (i), we obtain easily

$$\rho(\mathcal{A})\leq \Delta_{t_{n},t_{n-1}}(\mathcal{A})\leq\max _{\substack{i,j\in N,\\ j\neq i}} \Delta_{i,j}(\mathcal{A})=\Delta_{\mathrm{max}}. $$

The conclusion follows from (i) and (ii). □

Now we establish upper and lower bounds for \(\rho(\mathcal{A})\) of a nonnegative tensor \(\mathcal{A}\).

Lemma 2

(Lemma 3.3 in [21])

Suppose \(0\leq \mathcal{A}< \mathcal{C}\). Then \(\rho(\mathcal{A}) \leq\rho (\mathcal{C})\).

Theorem 5

Let \(\mathcal{A}=(a_{i_{1}\cdots i_{m}})\in R^{[m,n]}\) be a nonnegative tensor with \(n \geq2\). Then

$$\Delta_{\mathrm{min}} \leq\rho(\mathcal{A})\leq \Delta_{\mathrm{max}}. $$

Proof

Let \(\mathcal{A}_{k} =\mathcal{A} + \frac{1}{k}\mathcal{E}\), where \(k=1,2,\ldots\) , and \(\mathcal{E}\) denote the tensor with every entry being 1. Then \(\mathcal{A}_{k}\) is a sequence of positive tensors satisfying

$$0 \leq\mathcal{A}< \cdots \mathcal{A}_{k+1} < \mathcal{A}_{k} < \cdots < \mathcal{A}_{1}. $$

By Lemma 2, \(\{\rho(\mathcal{A}_{k})\}_{k=1}^{+\infty}\) is a monotone decreasing sequence with lower bound \(\rho(\mathcal{A})\). From the proof of Theorem 2.3 in [21], we have

$$\lim_{k\rightarrow+\infty} \rho(\mathcal{A}_{k})= \rho( \mathcal{A}). $$

Note that for any \(i,j\in N\), \(j\neq i\),

$$\Delta_{i,j}(\mathcal{A})< \cdots< \Delta_{i,j}( \mathcal{A}_{k+1}) < \Delta_{i,j}(\mathcal{A}_{k}) < \cdots< \Delta_{i,j}(\mathcal{A}_{1}), $$

we obtain easily

$$\lim_{k\rightarrow+\infty} \Delta_{i,j}(\mathcal{A}_{k})= \Delta _{i,j}(\mathcal{A}). $$

Furthermore, since \(\mathcal{A}_{k}\) is positive and also irreducible nonnegative for \(k=1,2,\ldots\) , we have, from Lemma 1,

$$\min_{\substack{i,j\in N, \\ j\neq i}} \Delta_{i,j}(\mathcal{A}_{k}) \leq\rho(\mathcal{A}_{k})\leq \max_{\substack{i,j\in N, \\ j\neq i}} \Delta_{i,j}(\mathcal{A}_{k}). $$

Letting \(k\rightarrow+\infty\), then

$$\Delta_{\mathrm{min}}=\min_{\substack{i,j\in N, \\ j\neq i}} \Delta_{i,j}( \mathcal{A})\leq\rho(\mathcal{A})\leq \max_{\substack{i,j\in N, \\ j\neq i}} \Delta_{i,j}(\mathcal{A})=\Delta_{\mathrm{max}}. $$

The proof is completed. □

We next compare the bounds in Theorem 5 with those in Theorem 3.

Theorem 6

Let \(\mathcal{A}=(a_{i_{1}\cdots i_{m}})\in R^{[m,n]}\) be a nonnegative tensor with \(n \geq2\). Then

$$ R_{\mathrm{min}}\leq\Delta_{\mathrm{min}} \leq \Delta_{\mathrm{max}} \leq R_{\mathrm{max}}. $$
(4)

Proof

We first prove \(R_{\mathrm{min}}\leq\Delta_{\mathrm{min}}\). For any \(i,j\in N\), \(j\neq i\), if \(R_{i}(\mathcal{A})\leq R_{j}(\mathcal{A})\), then

$$a_{ii\cdots i}-a_{jj\cdots j}+r_{i}^{\Theta_{i}}(\mathcal{A}) +r_{i}^{\overline{\Theta}_{i}}(\mathcal{A}) \leq r_{j}(\mathcal{A}). $$

Hence,

$$\begin{aligned}& \bigl(a_{i\cdots i}- a_{j\cdots j}+r_{i}^{\Theta_{i}}( \mathcal{A}) \bigr)^{2}+ 4r_{i}^{\overline{\Theta}_{i}}( \mathcal{A})r_{j}(\mathcal{A}) \\& \quad \geq \bigl(a_{i\cdots i}- a_{j\cdots j}+r_{i}^{\Theta_{i}}(\mathcal{A}) \bigr)^{2} \\& \qquad {}+4r_{i}^{\overline{\Theta}_{i}}(\mathcal{A}) \bigl( a_{ii\cdots i}-a_{jj\cdots j}+r_{i}^{\Theta_{i}}( \mathcal{A}) + r_{i}^{\overline{\Theta}_{i}}(\mathcal {A}) \bigr) \\& \quad = \bigl(a_{i\cdots i}- a_{j\cdots j}+r_{i}^{\Theta_{i}}( \mathcal{A}) \bigr)^{2} \\& \qquad {}+4r_{i}^{\overline{\Theta}_{i}}(\mathcal{A}) \bigl( a_{ii\cdots i}-a_{jj\cdots j}+r_{i}^{\Theta_{i}}( \mathcal{A}) \bigr) +4 \bigl(r_{i}^{\overline{\Theta}_{i}}(\mathcal{A}) \bigr)^{2} \\& \quad = \bigl( a_{i\cdots i}-a_{j\cdots j}+r_{i}^{\Theta_{i}}( \mathcal{A})+2r_{i}^{\overline{\Theta}_{i}}(\mathcal{A}) \bigr)^{2}. \end{aligned}$$

When

$$a_{i\cdots i}-a_{j\cdots j}+r_{i}^{\Theta_{i}} ( \mathcal{A})+2r_{i}^{\overline{\Theta}_{i}}(\mathcal{A}) > 0, $$

we have

$$\begin{aligned}& a_{i\cdots i}+ a_{j\cdots j}+r_{i}^{\Theta_{i}}(\mathcal{A}) + \sqrt{ \bigl(a_{i\cdots i}- a_{j\cdots j}+r_{i}^{\Theta_{i}}( \mathcal{A}) \bigr)^{2}+4r_{i}^{\overline{\Theta}_{i}} ( \mathcal{A})r_{j}(\mathcal{A})} \\& \quad \geq a_{i\cdots i}+ a_{j\cdots j}+r_{i}^{\Theta_{i}}( \mathcal{A})+ \bigl( a_{i\cdots i}-a_{j\cdots j}+r_{i}^{\Theta_{i}}( \mathcal{A})+2r_{i}^{\overline {\Theta}_{i}}(\mathcal{A}) \bigr) \\& \quad = 2 \bigl( a_{i\cdots i}+r_{i}^{\Theta_{i}}( \mathcal{A})+r_{i}^{\overline {\Theta}_{i}}(\mathcal{A}) \bigr) \\& \quad = 2R_{i}(\mathcal{A}). \end{aligned}$$

When

$$a_{i\cdots i}-a_{j\cdots j}+r_{i}^{\Theta_{i}}( \mathcal{A})+2r_{i}^{\overline{\Theta}_{i}} (\mathcal{A})\leq0, $$

that is,

$$a_{i\cdots i}+r_{i}^{\Theta_{i}}(\mathcal{A})+2r_{i}^{\overline{\Theta }_{i}}( \mathcal{A})\leq a_{j\cdots j}, $$

we have

$$\begin{aligned}& a_{i\cdots i}+ a_{j\cdots j}+r_{i}^{\Theta_{i}}(\mathcal{A}) + \sqrt{ \bigl(a_{i\cdots i}- a_{j\cdots j}+r_{i}^{\Theta_{i}}( \mathcal {A}) \bigr)^{2}+4r_{i}^{\overline{\Theta}_{i}}( \mathcal{A})r_{j}(\mathcal {A})} \\& \quad \geq a_{i\cdots i}+ a_{j\cdots j}+r_{i}^{\Theta_{i}}( \mathcal{A}) + \sqrt{ \bigl(a_{i\cdots i}- a_{j\cdots j}+r_{i}^{\Theta_{i}}( \mathcal{A}) \bigr)^{2}} \\& \quad = a_{i\cdots i}+ a_{j\cdots j}+r_{i}^{\Theta_{i}}( \mathcal{A})- \bigl( a_{i\cdots i}-a_{j\cdots j}+r_{i}^{\Theta_{i}}( \mathcal{A}) \bigr) \\& \quad = 2a_{j\cdots j} \\& \quad \geq 2 \bigl(a_{i\cdots i}+r_{i}^{\Theta_{i}}( \mathcal{A})+2r_{i}^{\overline{\Theta}_{i}}(\mathcal {A}) \bigr) \\& \quad \geq 2 \bigl(a_{i\cdots i}+r_{i}^{\Theta_{i}}( \mathcal{A})+r_{i}^{\overline{\Theta}_{i}} (\mathcal{A}) \bigr) \\& \quad = 2R_{i}(\mathcal{A}) . \end{aligned}$$

Therefore,

$$\frac{1}{2} \Bigl(a_{i\cdots i}+ a_{j\cdots j}+r_{i}^{\Theta_{i}}( \mathcal{A}) + \sqrt{ \bigl(a_{i\cdots i}- a_{j\cdots j}+r_{i}^{\Theta_{i}}( \mathcal{A}) \bigr)^{2}+4r_{i}^{\overline{\Theta }_{i}}( \mathcal{A})r_{j}(\mathcal {A})} \Bigr)\geq R_{i}( \mathcal{A}), $$

which implies

$$\begin{aligned}& \min_{\substack{i,j\in N, \\ j\neq i}}\frac{1}{2} \Bigl(a_{i\cdots i}+ a_{j\cdots j}+r_{i}^{\Theta_{i}}(\mathcal{A}) + \sqrt{ \bigl(a_{i\cdots i}- a_{j\cdots j}+r_{i}^{\Theta_{i}}( \mathcal{A}) \bigr)^{2}+4r_{i}^{\overline{\Theta }_{i}}(\mathcal{A}) r_{j}(\mathcal{A})} \Bigr) \\& \quad \geq \min_{i\in N}R_{i}(\mathcal{A}), \end{aligned}$$

i.e., \(R_{\mathrm{min}} \leq\Delta_{\mathrm{min}}\).

On the other hand, if for any \(i,j\in N\), \(j\neq i\),

$$R_{j}(\mathcal{A}) \leq R_{i}(\mathcal{A}), $$

then

$$a_{jj\cdots j}-a_{ii\cdots i}-r_{i}^{\Theta_{i}}(\mathcal{A}) +r_{j}(\mathcal{A}) \leq r_{i}^{\overline{\Theta}_{i}}(\mathcal{A}). $$

Similarly, we can also obtain

$$\frac{1}{2} \Bigl(a_{i\cdots i}+ a_{j\cdots j}+r_{i}^{\Theta_{i}}( \mathcal{A}) + \sqrt{ \bigl(a_{i\cdots i}- a_{j\cdots j}+r_{i}^{\Theta_{i}}( \mathcal{A}) \bigr)^{2}+4r_{i}^{\overline{\Theta }_{i}}( \mathcal{A})r_{j}(\mathcal {A})} \Bigr)\geq R_{j}( \mathcal{A}), $$

and that \(R_{\mathrm{min}}\leq\Delta_{\mathrm{min}}\). Hence, the first inequality in (4) holds. In a similar way, we can prove that the last inequality in (4) also holds. The conclusion follows. □

Example 1

Consider the nonnegative tensor

$$\mathcal{A}=\bigl[A(:,:,1),A(:,:,2),A(:,:,3)\bigr]\in R^{[3,3]}, $$

where

$$\begin{aligned}& A(:,:,1)=\left ( \begin{array}{@{}c@{\quad}c@{\quad}c@{}} 0.2192 &0.4411& 0.5232\\ 0.7637 &0.5239& 0.8330\\ 0.7993 &0.3710& 0.5328 \end{array} \right ), \\& A(:,:,2)= \left ( \begin{array}{@{}c@{\quad}c@{\quad}c@{}} 0.4380& 0.0482& 0.1325\\ 0.1803& 0.6729& 0.1809 \\ 0.3773& 0.1079& 0.8965 \end{array} \right ), \\& A(:,:,3)=\left ( \begin{array}{@{}c@{\quad}c@{\quad}c@{}} 0.0779 & 0.1982 & 0.4691\\ 0.5135 & 0.8284 & 0.7352\\ 0.1135 & 0.1163 & 0.8645 \end{array} \right ). \end{aligned}$$

We now compute the bounds for \(\rho(\mathcal{A})\). By Theorem 3, we have

$$2.5474 \leq\rho(\mathcal{A}) \leq5.2318. $$

By Theorem 4, we have

$$\rho(\mathcal{A})\leq5.0753. $$

By Theorem 5, we have

$$3.0097 \leq\rho(\mathcal{A})\leq 4.7894. $$

It is easy to see that the bounds in Theorem 5 are sharper than those in Theorem 3 (Lemma 5.2 of [21]), and that the upper bound in Theorem 5 is sharper than that in Theorem 4 (Theorem 3.3 of [22]) in some cases.

3 Conclusions

In this paper, we obtain a lower and an upper bound for the spectral radius of a nonnegative tensor, which improved the known bounds obtained by Yang and Yang [21], and Li et al. [22].