1 Introduction

Singular value problems of rectangular tensors have become an important topic in applied mathematics and numerical multilinear algebra, and it has a wide range of practical applications, such as the strong ellipticity condition problem in solid mechanics [1, 2] and the entanglement problem in quantum physics [3, 4].

Let \(\mathbb{R}\) (respectively, \(\mathbb{C}\)) be the real (respectively, complex) field. Assume that \(p,q,m,n\) are positive integers, \(m,n\geq2\), \(l=p+q\), and \(N=\{1,2,\ldots,n\}\). A real \((p,q)\)th order \(m\times n\) dimensional rectangular tensor (or simply a real rectangular tensor) \(\mathcal{A}\) is defined as follows:

$$\mathcal{A}=(a_{i_{1}\cdots i_{p}j_{1}\cdots j_{q}}),\quad a_{i_{1}\cdots i_{p}j_{1}\cdots j_{q}}\in\mathbb{R}, 1\leq i_{1},\ldots,i_{p}\leq m, 1\leq j_{1},\ldots ,j_{q}\leq n. $$

A real rectangular tensor \(\mathcal{A}\) is called nonnegative if \(a_{i_{1}\cdots i_{p}j_{1}\cdots j_{q}}\geq0\) for \(i_{k}=1,\ldots,m, k=1,\ldots, p\), and \(j_{v}=1,\ldots, n,v=1,\ldots,q\).

For vectors \(x=(x_{1},\ldots, x_{m})^{\textrm{T}}\), \(y=(y_{1},\ldots,y_{n})^{\textrm{T}}\) and a real number α, let \(x^{[\alpha]}=(x_{1}^{\alpha},x_{2}^{\alpha},\ldots, x_{m}^{\alpha})^{\textrm{T}}\), \(y^{[\alpha]}=(y_{1}^{\alpha},y_{2}^{\alpha},\ldots,y_{n}^{\alpha})^{\textrm{T}}\), \(\mathcal{A}x^{p-1}y^{q}\) be an m dimension real vector whose ith component is

$$\bigl(\mathcal{A}x^{p-1}y^{q}\bigr)_{i}=\sum _{i_{2},\ldots,i_{p}=1}^{m}\sum_{j_{1},\ldots,j_{q}=1}^{n}a_{ii_{2}\cdots i_{p}j_{1}\cdots j_{q}}x_{i_{2}} \cdots x_{i_{p}}y_{j_{1}}\cdots y_{j_{q}}, $$

and \(\mathcal{A}x^{p}y^{q-1}\) be an n dimension real vector whose jth component is

$$\bigl(\mathcal{A}x^{p}y^{q-1}\bigr)_{j}=\sum _{i_{1},\ldots,i_{p}=1}^{m}\sum_{j_{2},\ldots,j_{q}=1}^{n}a_{i_{1}\cdots i_{p}jj_{2}\cdots j_{q}}x_{i_{1}} \cdots x_{i_{p}}y_{j_{2}}\cdots y_{j_{q}}. $$

If \(\lambda\in\mathbb{C}\), \(x\in\mathbb{C}^{m}\backslash\{0\}\), and \(y\in \mathbb{C}^{n}\backslash\{0\}\) are solutions of

$$ \textstyle\begin{cases} \mathcal{A}x^{p-1}y^{q}=\lambda x^{[l-1]},\\ \mathcal{A}x^{p}y^{q-1}=\lambda y^{[l-1]}, \end{cases} $$

then we say that λ is a singular value of \(\mathcal{A}\), x and y are a left and a right eigenvectors of \(\mathcal{A}\), associated with λ. If \(\lambda\in\mathbb{R}, x\in\mathbb{R}^{m}\), and \(y\in\mathbb{R}^{n}\), then we say that λ is an H-singular value of \(\mathcal{A}\), x and y are a left and a right H-eigenvectors of \(\mathcal{A}\), associated with H-singular value λ [5]. Here,

$$\lambda_{0}=\max\bigl\{ |\lambda|:\lambda \text{ is a singular value of } \mathcal{A}\bigr\} $$

is called the largest singular value [6].

The definition of singular values for tensors was first introduced in [7]. Note here that when l is even, the definitions in [5] is the same as in [7], and when l is odd, the definition in [5] is slightly different from that in [7], but parallel to the definition of eigenvalues of square matrices [8]; see [5] for details.

Recently, many people focus on bounding the largest singular value for nonnegative rectangular tensors [6, 9, 10]. For convenience, we first give some notation. Given a nonempty proper subset S of N, we denote

$$\begin{aligned}& \Delta^{N}:=\bigl\{ (i_{2},\ldots, i_{p},j_{1}, \ldots,j_{q}): i_{2},\ldots, i_{p},j_{1}, \ldots ,j_{q}\in N\bigr\} , \\& \Delta^{S}:=\bigl\{ (i_{2},\ldots, i_{p},j_{1}, \ldots,j_{q}): i_{2},\ldots, i_{p},j_{1}, \ldots ,j_{q}\in S\bigr\} , \\& \Omega^{N}:=\bigl\{ (i_{1},\ldots, i_{p},j_{2}, \ldots,j_{q}): i_{1},\ldots, i_{p},j_{2}, \ldots ,j_{q}\in N\bigr\} , \\& \Omega^{S}:=\bigl\{ (i_{1},\ldots, i_{p},j_{2}, \ldots,j_{q}): i_{1},\ldots, i_{p},j_{2}, \ldots ,j_{q}\in S\bigr\} , \end{aligned}$$

and then

$$\overline{\Delta^{S}}=\Delta^{N}\backslash \Delta^{S},\qquad \overline{\Omega ^{S}}=\Omega^{N} \backslash\Omega^{S}. $$

This implies that, for a nonnegative rectangular tensor \(\mathcal {A}=(a_{i_{1}\cdots i_{p}j_{1}\cdots j_{q}})\), we have, for \(i,j\in S\),

$$\begin{aligned}& r_{i}(\mathcal{A})=\sum_{i_{2},\ldots,i_{p},j_{1},\ldots,j_{q}\in N\atop \delta_{ii_{2}\cdots i_{p}j_{1}\cdots j_{q}}=0}a_{ii_{2}\cdots i_{p}j_{1}\cdots j_{q}}=r_{i}^{\Delta^{S}}( \mathcal{A})+r_{i}^{\overline{\Delta^{S}}}(\mathcal{A}),\quad r_{i}^{j}( \mathcal{A})=r_{i}(\mathcal{A})-a_{ij\cdots jj\cdots j}, \\& c_{j}(\mathcal{A})=\sum_{i_{1},\ldots,i_{p},j_{2},\ldots,j_{q}\in N\atop \delta_{i_{1}\cdots i_{p}jj_{2}\cdots j_{q}}=0}a_{i_{1}\cdots i_{p}jj_{2}\cdots j_{q}}=c_{j}^{\Omega^{S}}( \mathcal{A})+c_{j}^{\overline{\Omega^{S}}}(\mathcal{A}),\quad c_{j}^{i}( \mathcal{A})=c_{j}(\mathcal{A})-a_{i\cdots iji\cdots i}, \end{aligned}$$

where

$$\delta_{i_{1}\cdots i_{p}j_{1}\cdots j_{q}}= \textstyle\begin{cases} 1,& \text{if }i_{1}=\cdots=i_{p}=j_{1}=\cdots=j_{q},\\ 0,&\text{otherwise}, \end{cases} $$

and

$$\begin{aligned}& r_{i}^{\Delta^{S}}(\mathcal{A})=\sum_{(i_{2},\ldots,i_{p},j_{1},\ldots ,j_{q})\in\Delta^{S}\atop \delta_{ii_{2}\cdots i_{p}j_{1}\cdots j_{q}}=0}a_{ii_{2}\cdots i_{p}j_{1}\cdots j_{q}},\qquad r_{i}^{\overline{\Delta ^{S}}}( \mathcal{A})=\sum_{(i_{2},\ldots,i_{p},j_{1},\ldots,j_{q})\in \overline{\Delta^{S}}}a_{ii_{2}\cdots i_{p}j_{1}\cdots j_{q}}, \\& c_{j}^{\Omega^{S}}(\mathcal{A})=\sum_{(i_{1},\ldots,i_{p},j_{2},\ldots ,j_{q})\in\Omega^{S}\atop \delta_{i_{1}\cdots i_{p}jj_{2}\cdots j_{q}}=0}a_{i_{1}\cdots i_{p}jj_{2}\cdots j_{q}},\qquad c_{j}^{\overline{\Omega ^{S}}}( \mathcal{A})=\sum_{(i_{1},\ldots,i_{p},j_{2},\ldots,j_{q})\in \overline{\Omega^{S}}}a_{i_{1}\cdots i_{p}jj_{2}\cdots j_{q}}. \end{aligned}$$

In [6], Yang and Yang gave the following bound for the largest singular value of a nonnegative rectangular tensor \(\mathcal{A}\).

Theorem 1

[6], Theorem 4

Let \(\mathcal{A}\) be a \((p,q)\) th order \(m\times n\) dimensional nonnegative rectangular tensor. Then

$$\lambda_{0}\leq\max_{1\leq i\leq m, 1\leq j\leq n}\bigl\{ R_{i}( \mathcal {A}),C_{j}(\mathcal{A})\bigr\} , $$

where

$$R_{i}(\mathcal{A})=\sum_{i_{2},\ldots,i_{p}=1}^{m} \sum_{j_{1},\ldots ,j_{q}=1}^{n}a_{ii_{2}\cdots i_{p}j_{1}\cdots j_{q}}, \qquad C_{j}(\mathcal{A})=\sum_{i_{1},\ldots,i_{p}=1}^{m} \sum_{j_{2},\ldots,j_{q}=1}^{n}a_{i_{1}\cdots i_{p}jj_{2}\cdots j_{q}}. $$

When \(m=n\), He et al. [9] have given an upper bound which is lower than that in Theorem 1.

Theorem 2

[9], Theorem 1.3

Let \(\mathcal{A}\) be a \((p,q)\) th order \(n\times n\) dimensional nonnegative rectangular tensor. Then

$$\lambda_{0}\leq\Phi(\mathcal{A})=\max\bigl\{ \Phi_{1}( \mathcal{A}),\Phi _{2}(\mathcal{A}),\Phi_{3}(\mathcal{A}), \Phi_{4}(\mathcal{A})\bigr\} , $$

where

$$\begin{aligned} \Phi_{1}(\mathcal{A}) =&\max_{i,j\in N,i\neq j}\frac{1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+r_{i}^{j}( \mathcal{A}) \\ &{}+ \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}+r_{i}^{j}( \mathcal {A})\bigr)^{2}+4a_{ij\cdots jj\cdots j}r_{j}(\mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} , \\ \Phi_{2}(\mathcal{A}) =& \max_{i,j\in N,i\neq j}\frac{1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+c_{i}^{j}( \mathcal{A}) \\ &{}+ \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}+c_{i}^{j}( \mathcal {A})\bigr)^{2}+4a_{j\cdots jij\cdots j}c_{j}(\mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} , \\ \Phi_{3}(\mathcal{A}) =& \max_{i,j\in N,i\neq j}\frac{1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+r_{i}^{j}( \mathcal{A}) \\ &{}+ \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}+r_{i}^{j}( \mathcal {A})\bigr)^{2}+4a_{ij\cdots jj\cdots j}c_{j}(\mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} , \\ \Phi_{4}(\mathcal{A}) =& \max_{i,j\in N,i\neq j}\frac{1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+c_{i}^{j}( \mathcal{A}) \\ &{}+ \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}+c_{i}^{j}( \mathcal {A})\bigr)^{2}+4a_{j\cdots jij\cdots j}r_{j}(\mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} . \end{aligned}$$

Similarly, under the condition of \(m=n\), by breaking \(N=\{1,2,\ldots,n\} \) into disjoint subsets S and its complement S̄, Zhao and Sang [10] provided an S-type upper bound for the largest singular value of nonnegative rectangular tensors.

Theorem 3

[10], Theorem 2.2

Let \(\mathcal{A}\) be a \((p,q)\) th order \(n\times n\) dimensional nonnegative rectangular tensor, S be a nonempty proper subset of N, S̄ be the complement of S in N. Then

$$\lambda_{0}\leq U^{S}(\mathcal{A})=\max\bigl\{ U_{1}^{S}(\mathcal{A}),U_{1}^{\bar {S}}( \mathcal{A}),U_{2}^{S}(\mathcal{A}),U_{2}^{\bar{S}}( \mathcal{A})\bigr\} , $$

where

$$\begin{aligned} U_{1}^{S}(\mathcal{A}) =&\max_{i\in S,j\in\bar{S}} \frac{1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+r_{j}^{\overline{\Delta ^{S}}}( \mathcal{A}) \\ &{}+ \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}-r_{j}^{\overline {\Delta^{S}}}( \mathcal{A})\bigr)^{2}+4\max\bigl\{ r_{i}(\mathcal{A}),c_{i}( \mathcal{A})\bigr\} r_{j}^{\Delta^{S}}(\mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} , \\ U_{1}^{\bar{S}}(\mathcal{A}) =&\max_{i\in\bar{S},j\in S} \frac {1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+r_{j}^{\overline {\Delta^{\bar{S}}}}( \mathcal{A}) \\ &{}+ \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}-r_{j}^{\overline {\Delta^{\bar{S}}}}( \mathcal{A})\bigr)^{2}+4\max\bigl\{ r_{i}(\mathcal{A}),c_{i}( \mathcal {A})\bigr\} r_{j}^{\Delta^{\bar{S}}}(\mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} , \\ U_{2}^{S}(\mathcal{A}) =&\max_{i\in S,j\in\bar{S}} \frac{1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+c_{j}^{\overline{\Omega ^{S}}}( \mathcal{A}) \\ &{}+ \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}-c_{j}^{\overline {\Omega^{S}}}( \mathcal{A})\bigr)^{2}+4\max\bigl\{ r_{i}(\mathcal{A}),c_{i}( \mathcal{A})\bigr\} c_{j}^{\Omega^{S}}(\mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} , \\ U_{2}^{\bar{S}}(\mathcal{A}) =&\max_{i\in\bar{S},j\in S} \frac {1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+c_{j}^{\overline {\Omega^{\bar{S}}}}( \mathcal{A}) \\ &{}+ \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}-c_{j}^{\overline {\Omega^{\bar{S}}}}( \mathcal{A})\bigr)^{2}+4\max\bigl\{ r_{i}(\mathcal{A}),c_{i}( \mathcal {A})\bigr\} c_{j}^{\Omega^{\bar{S}}}(\mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} . \end{aligned}$$

In this paper, we continue this research, and give a new S-type upper bound for the largest singular value of nonnegative rectangular tensors. It is proved that the new upper bound is better than those in Theorems 1-3.

2 Main results

Theorem 4

Let \(\mathcal{A}\) be a \((p,q)\) th order \(n\times n\) dimensional nonnegative rectangular tensor, S be a nonempty proper subset of N, S̄ be the complement of S in N. Then

$$\lambda_{0}\leq\Psi^{S}(\mathcal{A})=\max\bigl\{ \Psi_{1}^{S}(\mathcal{A}),\Psi _{1}^{\bar{S}}( \mathcal{A}),\Psi_{2}^{S}(\mathcal{A}),\Psi_{2}^{\bar {S}}( \mathcal{A}),\Psi_{3}^{S}(\mathcal{A}),\Psi_{3}^{\bar{S}}( \mathcal {A}),\Psi_{4}^{S}(\mathcal{A}),\Psi_{4}^{\bar{S}}( \mathcal{A})\bigr\} , $$

where

$$\begin{aligned} \Psi_{1}^{S}(\mathcal{A}) =&\max_{i\in S,j\in\bar{S}} \frac{1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+r_{i}^{\Delta^{S}}( \mathcal {A})+r_{j}^{\overline{\Delta^{S}}}(\mathcal{A}) \\ &{}+ \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}+r_{i}^{\Delta ^{S}}( \mathcal{A})-r_{j}^{\overline{\Delta^{S}}}(\mathcal {A})\bigr)^{2}+4r_{i}^{\overline{\Delta^{S}}}( \mathcal{A})r_{j}^{\Delta^{S}}(\mathcal {A}) \bigr]^{\frac{1}{2}} \bigr\} , \\ \Psi_{1}^{\bar{S}}(\mathcal{A}) =&\max_{i\in\bar{S},j\in S} \frac {1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+r_{i}^{\Delta ^{\bar{S}}}( \mathcal{A})+r_{j}^{\overline{\Delta^{\bar{S}}}}(\mathcal {A}) \\ &{}+ \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}+r_{i}^{\Delta^{\bar {S}}}( \mathcal{A})-r_{j}^{\overline{\Delta^{\bar{S}}}}(\mathcal {A})\bigr)^{2}+4r_{i}^{\overline{\Delta^{\bar{S}}}}( \mathcal{A})r_{j}^{\Delta^{\bar {S}}}(\mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} , \\ \Psi_{2}^{S}(\mathcal{A}) =&\max_{i\in S,j\in\bar{S}} \frac{1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+c_{i}^{\Omega ^{S}}( \mathcal{A})+c_{j}^{\overline{\Omega^{S}}}(\mathcal{A}) \\ &{}+ \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}+c_{i}^{\Omega ^{S}}( \mathcal{A})-c_{j}^{\overline{\Omega^{S}}}(\mathcal {A})\bigr)^{2}+4c_{i}^{\overline{\Omega^{S}}}( \mathcal{A})c_{j}^{\Omega^{S}}(\mathcal {A}) \bigr]^{\frac{1}{2}} \bigr\} , \\ \Psi_{2}^{\bar{S}}(\mathcal{A}) =&\max_{i\in\bar{S},j\in S} \frac {1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+c_{i}^{\Omega ^{\bar{S}}}( \mathcal{A})+c_{j}^{\overline{\Omega^{\bar{S}}}}(\mathcal {A}) \\ &{}+ \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}+c_{i}^{\Omega^{\bar {S}}}( \mathcal{A})-c_{j}^{\overline{\Omega^{\bar{S}}}}(\mathcal {A})\bigr)^{2}+4c_{i}^{\overline{\Omega^{\bar{S}}}}( \mathcal{A})c_{j}^{\Omega^{\bar {S}}}(\mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} , \\ \Psi_{3}^{S}(\mathcal{A}) =&\max_{i\in S,j\in\bar{S}} \frac{1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+r_{i}^{\Delta^{S}}( \mathcal {A})+c_{j}^{\overline{\Omega^{S}}}(\mathcal{A}) \\ &{}+ \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}+r_{i}^{\Delta ^{S}}( \mathcal{A})-c_{j}^{\overline{\Omega^{S}}}(\mathcal {A})\bigr)^{2}+4r_{i}^{\overline{{\Delta^{S}}}}( \mathcal{A})c_{j}^{\Omega^{S}}(\mathcal {A}) \bigr]^{\frac{1}{2}} \bigr\} , \\ \Psi_{3}^{\bar{S}}(\mathcal{A}) =&\max_{i\in\bar{S},j\in S} \frac {1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+r_{j}^{\overline {\Delta^{\bar{S}}}}( \mathcal{A})+c_{i}^{\Omega^{\bar{S}}}(\mathcal{A}) \\ &{}+ \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}-r_{j}^{\overline {\Delta^{\bar{S}}}}( \mathcal{A})+c_{i}^{\Omega^{\bar{S}}}(\mathcal {A})\bigr)^{2}+4r_{j}^{\Delta^{\bar{S}}}( \mathcal{A})c_{i}^{{\overline{\Omega ^{\bar{S}}}}}(\mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} , \\ \Psi_{4}^{S}(\mathcal{A}) =&\max_{i\in S,j\in\bar{S}} \frac{1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+r_{j}^{\overline{\Delta ^{S}}}( \mathcal{A})+c_{i}^{\Omega^{S}}(\mathcal{A}) \\ &{}+ \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}-r_{j}^{\overline {\Delta^{S}}}( \mathcal{A})+c_{i}^{\Omega^{S}}(\mathcal{A})\bigr)^{2}+4r_{j}^{\Delta ^{S}}( \mathcal{A})c_{i}^{{\overline{\Omega^{S}}}}(\mathcal{A}) \bigr]^{\frac {1}{2}} \bigr\} , \\ \Psi_{4}^{\bar{S}}(\mathcal{A}) =&\max_{i\in\bar{S},j\in S} \frac {1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+r_{i}^{\Delta ^{\bar{S}}}( \mathcal{A})+c_{j}^{\overline{\Omega^{\bar{S}}}}(\mathcal{A}) \\ &{}+ \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}+r_{i}^{\Delta ^{\bar{S}}}( \mathcal{A})-c_{j}^{\overline{\Omega^{\bar{S}}}}(\mathcal {A})\bigr)^{2}+4r_{i}^{{\overline{\Delta^{\bar{S}}}}}( \mathcal{A})c_{j}^{\Omega ^{\bar{S}}}(\mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} . \end{aligned}$$

Proof

Because \(\lambda_{0}\) is the largest singular value of \(\mathcal{A}\), from Theorem 2 in [6], there are nonnegative nonzero vectors \(x=(x_{1},x_{2},\ldots,x_{n})^{\mathrm{T}}\) and \(y=(y_{1},y_{2},\ldots,y_{n})^{\mathrm{T}}\), such that

$$\begin{aligned}& \mathcal{A}x^{p-1}y^{q}=\lambda_{0} x^{[l-1]}, \end{aligned}$$
(1)
$$\begin{aligned}& \mathcal{A}x^{p}y^{q-1}=\lambda_{0} y^{[l-1]}. \end{aligned}$$
(2)

Let

$$\begin{aligned}& x_{t}=\max\{x_{i}:i\in S\},\qquad x_{h}=\max \{x_{i}:i\in\bar{S}\};\\& y_{f}=\max\{y_{i}:i\in S\},\qquad y_{g}=\max\{y_{i}:i\in\bar{S}\}; \\& w_{i}=\max\{x_{i},y_{i}\}, \quad i\in N,\qquad w_{S}=\max\{w_{i}:i\in S\},\qquad w_{\bar{S}}=\max\{ w_{i}:i\in\bar{S}\}. \end{aligned}$$

Then at least one of \(x_{t}\) and \(x_{h}\) is nonzero, and at least one of \(y_{f}\) and \(y_{g}\) is nonzero. We next divide into four cases to prove.

Case I: If \(w_{S}=x_{t}, w_{\bar{S}}=x_{h}\), then \(x_{t}\geq y_{t}, x_{h}\geq y_{h}\).

(i) If \(x_{h}\geq x_{t}\), then \(x_{h}=\max\{w_{i}:i\in N\}\). From (3) of Theorem 2.2 in [10], we have

$$\begin{aligned} & \bigl(\lambda_{0} -a_{h\cdots hh\cdots h}-r_{h}^{\overline{\Delta^{S}}}( \mathcal {A})\bigr)x_{h}^{l-1}\leq r_{h}^{\Delta^{S}}( \mathcal{A})x_{t}^{l-1}. \end{aligned}$$
(3)

If \(x_{t}=0\), by \(x_{h}>0\), we have \(\lambda_{0} -a_{h\cdots hh\cdots h}-r_{h}^{\overline{\Delta^{S}}}(\mathcal{A})\leq0\). Then \(\lambda_{0}\leq a_{h\cdots hh\cdots h}+r_{h}^{\overline{\Delta^{S}}}(\mathcal {A})\leq\Psi_{1}^{S}(\mathcal{A})\). Otherwise, \(x_{t}>0\). From (1), we have

$$\begin{aligned} (\lambda_{0} -a_{t\cdots tt\cdots t})x_{t}^{l-1} \leq& \lambda_{0} x_{t}^{l-1}-a_{t\cdots tt\cdots t}x_{t}^{p-1}y_{t}^{q} \\ =&\sum_{(i_{2},\ldots, i_{p},j_{1},\ldots, j_{q})\in\Delta^{S} \atop \delta_{ti_{2}\cdots i_{p}j_{1}\cdots j_{q}}=0}a_{ti_{2}\cdots i_{p}j_{1}\cdots j_{q}}x_{i_{2}} \cdots x_{i_{p}}y_{j_{1}}\cdots y_{j_{q}} \\ &{}+\sum_{(i_{2},\ldots, i_{p},j_{1},\ldots, j_{q})\in\overline{\Delta ^{S}}}a_{ti_{2}\cdots i_{p}j_{1}\cdots j_{q}}x_{i_{2}} \cdots x_{i_{p}}y_{j_{1}}\cdots y_{j_{q}} \\ \leq&\sum_{(i_{2},\ldots, i_{p},j_{1},\ldots, j_{q})\in\Delta^{S} \atop \delta_{ti_{2}\cdots i_{p}j_{1}\cdots j_{q}}=0}a_{ti_{2}\cdots i_{p}j_{1}\cdots j_{q}}x_{t}^{l-1} +\sum_{(i_{2},\ldots, i_{p},j_{1},\ldots, j_{q})\in\overline{\Delta^{S}}} a_{ti_{2}\cdots i_{p}j_{1}\cdots j_{q}}x_{h}^{l-1} \\ =& r_{t}^{\Delta^{S}}(\mathcal{A})x_{t}^{l-1}+r_{t}^{\overline{\Delta ^{S}}}( \mathcal{A})x_{h}^{l-1}, \end{aligned}$$

i.e.,

$$\begin{aligned} & \bigl(\lambda_{0} -a_{t\cdots tt\cdots t}-r_{t}^{\Delta^{S}}( \mathcal {A})\bigr)x_{t}^{l-1}\leq r_{t}^{\overline{\Delta^{S}}}( \mathcal{A})x_{h}^{l-1}. \end{aligned}$$
(4)

If \(\lambda_{0} -a_{t\cdots tt\cdots t}-r_{t}^{\Delta^{S}}(\mathcal{A})\leq 0\), then \(\lambda_{0}\leq a_{t\cdots tt\cdots t}+r_{t}^{\Delta^{S}}(\mathcal{A})\leq \Psi_{1}^{S}(\mathcal{A})\). If \(\lambda_{0} -a_{t\cdots tt\cdots t}-r_{t}^{\Delta^{S}}(\mathcal{A})> 0\), multiplying (3) with (4) and noting that \(x_{t}^{l-1}x_{h}^{l-1}>0\), we have

$$\begin{aligned} & \bigl(\lambda_{0}-a_{t\cdots tt\cdots t}-r_{t}^{\Delta^{S}}( \mathcal{A})\bigr) \bigl(\lambda _{0} -a_{h\cdots hh\cdots h}-r_{h}^{\overline{\Delta^{S}}}( \mathcal{A})\bigr) \leq r_{t}^{\overline{\Delta^{S}}}(\mathcal{A})r_{h}^{\Delta^{S}}( \mathcal{A}). \end{aligned}$$
(5)

Solving \(\lambda_{0}\) in (5) gives

$$\begin{aligned} \lambda_{0} \leq& \frac{1}{2} \bigl\{ a_{t\cdots tt\cdots t}+a_{h\cdots hh\cdots h}+r_{t}^{\Delta^{S}}( \mathcal{A})+r_{h}^{\overline{\Delta ^{S}}}(\mathcal{A}) \\ &{}+ \bigl[\bigl(a_{t\cdots tt\cdots t}+r_{t}^{\Delta^{S}}( \mathcal{A})-a_{h\cdots hh\cdots h}-r_{h}^{\overline{\Delta^{S}}}(\mathcal{A}) \bigr)^{2}+4r_{t}^{\overline {\Delta^{S}}}(\mathcal{A})r_{h}^{\Delta^{S}}( \mathcal{A}) \bigr]^{\frac {1}{2}} \bigr\} \\ \leq&\max_{i\in S,j\in\bar{S}}\frac{1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+r_{i}^{\Delta^{S}}( \mathcal {A})+r_{j}^{\overline{\Delta^{S}}}(\mathcal{A}) \\ &{}+ \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}+r_{i}^{\Delta ^{S}}( \mathcal{A})-r_{j}^{\overline{\Delta^{S}}}(\mathcal {A})\bigr)^{2}+4r_{i}^{\overline{\Delta^{S}}}( \mathcal{A})r_{j}^{\Delta^{S}}(\mathcal {A}) \bigr]^{\frac{1}{2}} \bigr\} \\ =&\Psi_{1}^{S}(\mathcal{A}). \end{aligned}$$

(ii) If \(x_{t}\geq x_{h}\), similar to the proof of (i), we have

$$\begin{aligned} & \bigl(\lambda_{0} -a_{h\cdots hh\cdots h}-r_{h}^{\Delta^{\bar{S}}}( \mathcal {A})\bigr) \bigl(\lambda_{0} -a_{t\cdots tt\cdots t}-r_{t}^{\overline{\Delta^{\bar {S}}}}( \mathcal{A})\bigr)\leq r_{h}^{\overline{\Delta^{\bar{S}}}}(\mathcal {A})r_{t}^{\Delta^{\bar{S}}}(\mathcal{A}), \end{aligned}$$

and

$$\begin{aligned} \lambda_{0} \leq& \frac{1}{2} \bigl\{ a_{h\cdots hh\cdots h}+a_{t\cdots tt\cdots t}+r_{h}^{\Delta^{\bar{S}}}( \mathcal{A})+r_{t}^{\overline{\Delta ^{\bar{S}}}}(\mathcal{A}) \\ &{}+ \bigl[\bigl(a_{h\cdots hh\cdots h}-a_{t\cdots tt\cdots t}+r_{h}^{\Delta^{\bar {S}}}( \mathcal{A})-r_{t}^{\overline{\Delta^{\bar{S}}}}(\mathcal {A})\bigr)^{2}+4r_{h}^{\overline{\Delta^{\bar{S}}}}( \mathcal{A})r_{t}^{\Delta^{\bar {S}}}(\mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} \\ \leq&\max_{i\in\bar{S},j\in S}\frac{1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+r_{i}^{\Delta^{\bar{S}}}( \mathcal {A})+r_{j}^{\overline{\Delta^{\bar{S}}}}(\mathcal{A}) \\ &{}+ \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}+r_{i}^{\Delta^{\bar {S}}}( \mathcal{A})-r_{j}^{\overline{\Delta^{\bar{S}}}}(\mathcal {A})\bigr)^{2}+4r_{i}^{\overline{\Delta^{\bar{S}}}}( \mathcal{A})r_{j}^{\Delta^{\bar {S}}}(\mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} \\ =&\Psi_{1}^{\bar{S}}(\mathcal{A}). \end{aligned}$$

Case II: Assume that \(w_{S}=y_{f}, w_{\bar{S}}=y_{g}\). If \(y_{g}\geq y_{f}\), similar to the proof of (i), we have

$$\begin{aligned} & \bigl(\lambda_{0} -a_{f\cdots ff\cdots f}-c_{f}^{\Omega^{S}}( \mathcal {A})\bigr) \bigl(\lambda_{0} -a_{g\cdots gg\cdots g}-c_{g}^{\overline{\Omega ^{S}}}( \mathcal{A})\bigr)\leq c_{f}^{\overline{\Omega^{S}}}(\mathcal {A})c_{g}^{\Omega^{S}}(\mathcal{A}), \end{aligned}$$

and

$$\begin{aligned} \lambda_{0} \leq& \frac{1}{2} \bigl\{ a_{f\cdots ff\cdots f}+a_{g\cdots gg\cdots g}+c_{f}^{\Omega^{S}}( \mathcal{A})+c_{g}^{\overline{\Omega ^{S}}}(\mathcal{A}) \\ &{}+ \bigl[\bigl(a_{f\cdots ff\cdots f}-a_{g\cdots gg\cdots g}+c_{f}^{\Omega ^{S}}( \mathcal{A})-c_{g}^{\overline{\Omega^{S}}}(\mathcal {A})\bigr)^{2}+4c_{f}^{\overline{\Omega^{S}}}( \mathcal{A})c_{g}^{\Omega^{S}}(\mathcal {A}) \bigr]^{\frac{1}{2}} \bigr\} \\ \leq&\max_{i\in S,j\in\bar{S}}\frac{1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+c_{i}^{\Omega^{S}}( \mathcal {A})+c_{j}^{\overline{\Omega^{S}}}(\mathcal{A}) \\ &{}+ \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}+c_{i}^{\Omega ^{S}}( \mathcal{A})-c_{j}^{\overline{\Omega^{S}}}(\mathcal {A})\bigr)^{2}+4c_{i}^{\overline{\Omega^{S}}}( \mathcal{A})c_{j}^{\Omega^{S}}(\mathcal {A}) \bigr]^{\frac{1}{2}} \bigr\} \\ =&\Psi_{2}^{S}(\mathcal{A}). \end{aligned}$$

If \(y_{f}\geq y_{g}\), similarly, we have

$$\begin{aligned} & \bigl(\lambda_{0} -a_{g\cdots gg\cdots g}-c_{g}^{\Omega^{\bar{S}}}( \mathcal {A})\bigr) \bigl(\lambda_{0} -a_{f\cdots ff\cdots f}-c_{f}^{\overline{\Omega^{\bar {S}}}}( \mathcal{A})\bigr)\leq c_{g}^{\overline{\Omega^{\bar{S}}}}(\mathcal {A})c_{f}^{\Omega^{\bar{S}}}(\mathcal{A}) \end{aligned}$$

and

$$\begin{aligned} \lambda_{0} \leq& \frac{1}{2} \bigl\{ a_{g\cdots gg\cdots g}+a_{f\cdots ff\cdots f}+c_{g}^{\Omega^{\bar{S}}}( \mathcal{A})+c_{f}^{\overline{\Omega ^{\bar{S}}}}(\mathcal{A}) \\ &{}+ \bigl[\bigl(a_{g\cdots g\cdots g}-a_{f\cdots ff\cdots f}+c_{g}^{\Omega^{\bar {S}}}( \mathcal{A})-c_{f}^{\overline{\Omega^{\bar{S}}}}(\mathcal {A})\bigr)^{2}+4c_{g}^{\overline{\Omega^{\bar{S}}}}( \mathcal{A})c_{f}^{\Omega^{\bar {S}}}(\mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} \\ \leq&\max_{i\in\bar{S},j\in S}\frac{1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+c_{i}^{\Omega^{\bar{S}}}( \mathcal {A})+c_{j}^{\overline{\Omega^{\bar{S}}}}(\mathcal{A}) \\ &{}+ \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}+c_{i}^{\Omega^{\bar {S}}}( \mathcal{A})-c_{j}^{\overline{\Omega^{\bar{S}}}}(\mathcal {A})\bigr)^{2}+4c_{i}^{\overline{\Omega^{\bar{S}}}}( \mathcal{A})c_{j}^{\Omega^{\bar {S}}}(\mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} \\ =&\Psi_{2}^{\bar{S}}(\mathcal{A}). \end{aligned}$$

Case III: Assume that \(w_{S}=x_{t}, w_{\bar{S}}=y_{g}\). If \(y_{g}\geq x_{t}\), similar to the proof of (i), we have

$$\begin{aligned} & \bigl(\lambda_{0}-a_{t\cdots tt\cdots t}-r_{t}^{\Delta^{S}}( \mathcal{A})\bigr) \bigl(\lambda _{0} -a_{g\cdots gg\cdots g}-c_{g}^{\overline{\Omega^{S}}}( \mathcal{A})\bigr) \leq r_{t}^{\overline{{\Delta^{S}}}}(\mathcal{A})c_{g}^{\Omega^{S}}( \mathcal{A}) \end{aligned}$$

and

$$\begin{aligned} \lambda_{0} \leq& \frac{1}{2} \bigl\{ a_{t\cdots tt\cdots t}+a_{g\cdots gg\cdots g}+r_{t}^{\Delta^{S}}( \mathcal{A})+c_{g}^{\overline{\Omega ^{S}}}(\mathcal{A}) \\ &{}+ \bigl[\bigl(a_{t\cdots tt\cdots t}-a_{g\cdots gg\cdots g}+r_{t}^{\Delta ^{S}}( \mathcal{A})-c_{g}^{\overline{\Omega^{S}}}(\mathcal {A})\bigr)^{2}+4r_{t}^{\overline{{\Delta^{S}}}}( \mathcal{A})c_{g}^{\Omega^{S}}(\mathcal {A}) \bigr]^{\frac{1}{2}} \bigr\} \\ \leq&\max_{i\in S,j\in\bar{S}}\frac{1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+r_{i}^{\Delta^{S}}( \mathcal {A})+c_{j}^{\overline{\Omega^{S}}}(\mathcal{A}) \\ &{}+ \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}+r_{i}^{\Delta ^{S}}( \mathcal{A})-c_{j}^{\overline{\Omega^{S}}}(\mathcal {A})\bigr)^{2}+4r_{i}^{\overline{{\Delta^{S}}}}( \mathcal{A})c_{j}^{\Omega^{S}}(\mathcal {A}) \bigr]^{\frac{1}{2}} \bigr\} \\ =&\Psi_{3}^{S}(\mathcal{A}). \end{aligned}$$

If \(x_{t}\geq y_{g}\), similarly, we have

$$\begin{aligned} & \bigl(\lambda_{0} -a_{g\cdots gg\cdots g}-c_{g}^{\Omega^{\bar{S}}}( \mathcal {A})\bigr) \bigl(\lambda_{0} -a_{t\cdots tt\cdots t}-r_{t}^{\overline{\Delta^{\bar {S}}}}( \mathcal{A})\bigr)\leq c_{g}^{{\overline{\Omega^{\bar{S}}}}}(\mathcal {A})r_{t}^{\Delta^{\bar{S}}}(\mathcal{A}) \end{aligned}$$

and

$$\begin{aligned} \lambda_{0} \leq& \frac{1}{2} \bigl\{ a_{t\cdots tt\cdots t}+a_{g\cdots gg\cdots g}+r_{t}^{\overline{\Delta^{\bar{S}}}}( \mathcal{A})+c_{g}^{\Omega ^{\bar{S}}}(\mathcal{A}) \\ &{}+ \bigl[\bigl(a_{t\cdots tt\cdots t}-a_{g\cdots gg\cdots g}+r_{t}^{\overline {\Delta^{\bar{S}}}}( \mathcal{A})-c_{g}^{\Omega^{\bar{S}}}(\mathcal {A})\bigr)^{2}+4r_{t}^{\Delta^{\bar{S}}}( \mathcal{A})c_{g}^{{\overline{\Omega ^{\bar{S}}}}}(\mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} \\ \leq&\max_{i\in\bar{S},j\in S}\frac{1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+r_{j}^{\overline{\Delta^{\bar {S}}}}( \mathcal{A})+c_{i}^{\Omega^{\bar{S}}}(\mathcal{A}) \\ &{}+ \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}-r_{j}^{\overline {\Delta^{\bar{S}}}}( \mathcal{A})+c_{i}^{\Omega^{\bar{S}}}(\mathcal {A})\bigr)^{2}+4r_{j}^{\Delta^{\bar{S}}}( \mathcal{A})c_{i}^{{\overline{\Omega ^{\bar{S}}}}}(\mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} \\ =&\Psi_{3}^{\bar{S}}(\mathcal{A}). \end{aligned}$$

Case IV: Assume that \(w_{S}=y_{f}, w_{\bar{S}}=x_{h}\). If \(x_{h}\geq y_{f}\), similar to the proof of (i), we have

$$\begin{aligned} & \bigl(\lambda_{0}-a_{f\cdots ff\cdots f}-c_{f}^{\Omega^{S}}( \mathcal{A})\bigr) \bigl(\lambda _{0} -a_{h\cdots hh\cdots h}-r_{h}^{\overline{\Delta^{S}}}( \mathcal{A})\bigr) \leq c_{f}^{{\overline{\Omega^{S}}}}(\mathcal{A})r_{h}^{\Delta^{S}}( \mathcal{A}) \end{aligned}$$

and

$$\begin{aligned} \lambda_{0} \leq& \frac{1}{2} \bigl\{ a_{f\cdots ff\cdots f}+a_{h\cdots hh\cdots h}+r_{h}^{\overline{\Delta^{S}}}( \mathcal{A})+c_{f}^{\Omega ^{S}}(\mathcal{A}) \\ &{}+ \bigl[\bigl(a_{f\cdots ff\cdots f}-a_{h\cdots hh\cdots h}-r_{h}^{\overline {\Delta^{S}}}( \mathcal{A})+c_{f}^{\Omega^{S}}(\mathcal{A})\bigr)^{2}+4c_{f}^{{\overline {\Omega^{S}}}}( \mathcal{A})r_{h}^{\Delta^{S}}(\mathcal{A}) \bigr]^{\frac {1}{2}} \bigr\} \\ \leq&\max_{i\in S,j\in\bar{S}}\frac{1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+r_{j}^{\overline{\Delta^{S}}}( \mathcal {A})+c_{i}^{\Omega^{S}}(\mathcal{A}) \\ &{}+ \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}-r_{j}^{\overline {\Delta^{S}}}( \mathcal{A})+c_{i}^{\Omega^{S}}(\mathcal{A})\bigr)^{2}+4c_{i}^{{\overline {\Omega^{S}}}}( \mathcal{A})r_{j}^{\Delta^{S}}(\mathcal{A}) \bigr]^{\frac {1}{2}} \bigr\} \\ =&\Psi_{4}^{S}(\mathcal{A}). \end{aligned}$$

If \(y_{f}\geq x_{h}\), similarly, we have

$$\begin{aligned} & \bigl(\lambda_{0} -a_{h\cdots hh\cdots h}-r_{h}^{\Delta^{\bar{S}}}( \mathcal {A})\bigr) \bigl(\lambda_{0} -a_{f\cdots ff\cdots f}-c_{f}^{\overline{\Omega^{\bar {S}}}}( \mathcal{A})\bigr)\leq r_{h}^{{\overline{\Delta^{\bar{S}}}}}(\mathcal {A})c_{f}^{\Omega^{\bar{S}}}(\mathcal{A}) \end{aligned}$$

and

$$\begin{aligned} \lambda_{0} \leq& \frac{1}{2} \bigl\{ a_{h\cdots hh\cdots h}+a_{f\cdots ff\cdots f}+r_{h}^{\Delta^{\bar{S}}}( \mathcal{A})+c_{f}^{\overline{\Omega ^{\bar{S}}}}(\mathcal{A}) \\ &{}+ \bigl[\bigl(a_{h\cdots hh\cdots h}-a_{f\cdots ff\cdots f}+r_{h}^{\Delta ^{\bar{S}}}( \mathcal{A})-c_{f}^{\overline{\Omega^{\bar{S}}}}(\mathcal {A})\bigr)^{2}+4r_{h}^{{\overline{\Delta^{\bar{S}}}}}( \mathcal{A})c_{f}^{\Omega ^{\bar{S}}}(\mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} \\ \leq&\max_{i\in\bar{S},j\in S}\frac{1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+r_{i}^{\Delta^{\bar{S}}}( \mathcal {A})+c_{j}^{\overline{\Omega^{\bar{S}}}}(\mathcal{A}) \\ &{}+ \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}+r_{i}^{\Delta ^{\bar{S}}}( \mathcal{A})-c_{j}^{\overline{\Omega^{\bar{S}}}}(\mathcal {A})\bigr)^{2}+4r_{i}^{{\overline{\Delta^{\bar{S}}}}}( \mathcal{A})c_{j}^{\Omega ^{\bar{S}}}(\mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} \\ =&\Psi_{4}^{\bar{S}}(\mathcal{A}). \end{aligned}$$

The conclusion follows from Cases I, II, III and IV. □

We next give the following comparison theorem for these upper bounds in Theorems 1-4.

Theorem 5

Let \(\mathcal{A}\) be a \((p,q)\) th order \(n\times n\) dimensional nonnegative rectangular tensor, S be a nonempty proper subset of N, S̄ be the complement of S in N. Then

$$\Psi^{S}(\mathcal{A})\leq U^{S}(\mathcal{A})\leq\Phi( \mathcal{A})\leq\max_{i,j\in N}\bigl\{ R_{i}( \mathcal{A}),C_{j}(\mathcal{A})\bigr\} . $$

Proof

I. By Remark 2.2 in [9], \(\Phi(\mathcal{A})\leq\max_{i,j\in N}\{R_{i}(\mathcal{A}),C_{j}(\mathcal{A})\}\) holds.

II. Next, we prove \(U^{S}(\mathcal{A})\leq\Phi(\mathcal{A})\). Here, we only prove \(U_{1}^{S}(\mathcal{A})\leq\Phi(\mathcal{A})\). Similarly, we can prove \(U_{1}^{\bar{S}}(\mathcal{A}),U_{2}^{S}(\mathcal{A}), U_{2}^{\bar{S}}(\mathcal{A})\leq\Phi(\mathcal{A})\), respectively.

(i) Suppose that

$$\begin{aligned} U_{1}^{S}(\mathcal{A}) =& \max_{i\in S,j\in\bar{S}} \frac{1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+r_{j}^{\overline{\Delta ^{S}}}( \mathcal{A})\\ &{}+ \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}-r_{j}^{\overline{\Delta^{S}}}( \mathcal{A})\bigr)^{2}+4r_{i}(\mathcal {A})r_{j}^{\Delta^{S}}( \mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} . \end{aligned}$$

From the proof of Theorem 2.2 in [10], we can see that the bound \(U_{1}^{S}(\mathcal{A})\) is obtained by solving \(\lambda_{0}\) from

$$\begin{aligned} & (\lambda_{0}-a_{i\cdots ii\cdots i}) \bigl( \lambda_{0} -a_{j\cdots jj\cdots j}-r_{j}^{\overline{\Delta^{S}}}( \mathcal{A})\bigr) \leq r_{i}(\mathcal{A})r_{j}^{\Delta^{S}}( \mathcal{A}). \end{aligned}$$
(6)

From the proof of Theorem 1.3 in [9], we can see that the bound

$$\begin{aligned} \Phi_{1}(\mathcal{A}) =& \max_{i,j\in N,i\neq j}\frac{1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+r_{j}^{i}( \mathcal{A})\\ &{} + \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}-r_{j}^{i}( \mathcal {A})\bigr)^{2}+4a_{ji\cdots ii\cdots i}r_{i}(\mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} \end{aligned}$$

is obtained by solving \(\lambda_{0}\) from

$$\begin{aligned} & (\lambda_{0}-a_{i\cdots ii\cdots i}) \bigl( \lambda_{0} -a_{j\cdots jj\cdots j}-r_{j}^{i}( \mathcal{A})\bigr)\leq a_{ji\cdots ii\cdots i}r_{i}(\mathcal{A}). \end{aligned}$$
(7)

Taking \(i\in S\), \(j\in\bar{S}\) in (7), by the proof of Theorem 6 in [11], we know that if \(\lambda_{0}\) satisfies (6), then \(\lambda _{0}\) satisfies (7), which implies that

$$\begin{aligned} \Phi_{1}(\mathcal{A}) \geq&\max_{i\in S,j\in\bar{S}} \frac {1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+r_{j}^{i}( \mathcal{A})\\ &{} + \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}-r_{j}^{i}( \mathcal {A})\bigr)^{2}+4a_{ji\cdots ii\cdots i}r_{i}(\mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} \\ \geq&U_{1}^{S}(\mathcal{A}). \end{aligned}$$

Obviously, \(U_{1}^{S}(\mathcal{A})\leq\Phi(\mathcal{A})\).

(ii) Suppose that

$$\begin{aligned} U_{1}^{S}(\mathcal{A}) =& \max_{i\in S,j\in\bar{S}} \frac{1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+r_{j}^{\overline{\Delta ^{S}}}( \mathcal{A}) \\ &{}+ \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}-r_{j}^{\overline {\Delta^{S}}}( \mathcal{A})\bigr)^{2}+4c_{i}(\mathcal{A})r_{j}^{\Delta^{S}}( \mathcal {A}) \bigr]^{\frac{1}{2}} \bigr\} . \end{aligned}$$

Similar to the proof of (i), we can obtain \(U_{1}^{S}(\mathcal{A})\leq\Phi _{3}(\mathcal{A})\leq\Phi(\mathcal{A})\).

III. Finally, we prove that \(\Psi^{S}(\mathcal{A})\leq U^{S}(\mathcal{A})\). Here, we only prove \(\Psi_{1}^{S}(\mathcal{A})\leq U^{S}(\mathcal{A})\). Similarly, we can prove \(\Psi_{1}^{\bar{S}}(\mathcal{A}),\Psi_{2}^{S}(\mathcal {A}),\Psi_{2}^{\bar{S}}(\mathcal{A}),\Psi_{3}^{S}(\mathcal{A}),\Psi_{3}^{\bar {S}}(\mathcal{A}),\Psi_{4}^{S}(\mathcal{A}),\Psi_{4}^{\bar{S}}(\mathcal {A})\leq U^{S}(\mathcal{A})\), respectively.

Let \(i\in S\) and \(j\in\bar{S}\). From the proof of Theorem 4, we can see that the bound \(\Psi_{1}^{S}(\mathcal{A})\) is obtained by solving \(\lambda_{0}\) from

$$\begin{aligned} & \bigl(\lambda_{0}-a_{i\cdots ii\cdots i}-r_{i}^{\Delta^{S}}( \mathcal{A})\bigr) \bigl(\lambda _{0} -a_{j\cdots jj\cdots j}-r_{j}^{\overline{\Delta^{S}}}( \mathcal{A})\bigr) \leq r_{i}^{\overline{\Delta^{S}}}(\mathcal{A})r_{j}^{\Delta^{S}}( \mathcal{A}). \end{aligned}$$
(8)

(i) Suppose that \(r_{i}^{\overline{\Delta^{S}}}(\mathcal{A})r_{j}^{\Delta ^{S}}(\mathcal{A})=0\). If \(\lambda_{0}-a_{i\cdots ii\cdots i}-r_{i}^{\Delta^{S}}(\mathcal{A})>0\), i.e., \(\lambda_{0}>a_{i\cdots ii\cdots i}+r_{i}^{\Delta^{S}}(\mathcal{A})\), then \(\lambda_{0} -a_{j\cdots jj\cdots j}-r_{j}^{\overline{\Delta^{S}}}(\mathcal {A})\leq0\), and for any \(i\in S\),

$$(\lambda_{0}-a_{i\cdots ii\cdots i}) \bigl(\lambda_{0} -a_{j\cdots jj\cdots j}-r_{j}^{\overline{\Delta^{S}}}(\mathcal{A})\bigr)\leq0\leq r_{i}(\mathcal {A})r_{j}^{\Delta^{S}}(\mathcal{A}). $$

That is to say, if \(\lambda_{0}\) satisfies (8), then \(\lambda _{0}\) satisfies (6), which implies that \(\Psi_{1}^{S}(\mathcal{A})\leq U_{1}^{S}(\mathcal{A})\leq U^{S}(\mathcal{A})\).

If \(\lambda_{0}-a_{i\cdots ii\cdots i}-r_{i}^{\Delta^{S}}(\mathcal{A})\leq0\), then \(\lambda_{0} -a_{j\cdots jj\cdots j}-r_{j}^{\overline{\Delta ^{S}}}(\mathcal{A})\geq0\), i.e., \(\lambda_{0} \geq a_{j\cdots jj\cdots j}+r_{j}^{\overline {\Delta^{S}}}(\mathcal{A})\). From (3), we can obtain \(\lambda_{0} -a_{j\cdots jj\cdots j}-r_{j}^{\overline{\Delta^{S}}}(\mathcal {A})\leq r_{j}^{\Delta^{S}}(\mathcal{A})\), i.e.,

$$\begin{aligned} & \lambda_{0} -a_{j\cdots jj\cdots j}\leq r_{j}( \mathcal{A}). \end{aligned}$$
(9)

By \(\lambda_{0}-a_{i\cdots ii\cdots i}-r_{i}^{\Delta^{S}}(\mathcal{A})\leq 0\leq r_{i}^{\overline{\Delta^{S}}}(\mathcal{A})\), i.e., \(\lambda_{0}-a_{i\cdots ii\cdots i}\leq r_{i}(\mathcal{A})\), we have

$$\begin{aligned} & \lambda_{0}-a_{i\cdots ii\cdots i}-r_{i}^{\overline{\Delta^{\bar {S}}}}( \mathcal{A})\leq r_{i}^{\Delta^{\bar{S}}}(\mathcal{A}). \end{aligned}$$
(10)

Multiplying (9) with (10), we can obtain

$$\begin{aligned} & (\lambda_{0} -a_{j\cdots jj\cdots j}) \bigl( \lambda_{0}-a_{i\cdots ii\cdots i}-r_{i}^{\overline{\Delta^{\bar{S}}}}( \mathcal{A})\bigr)\leq r_{i}^{\Delta^{\bar {S}}}(\mathcal{A})r_{j}( \mathcal{A}), \end{aligned}$$
(11)

which implies that if \(\lambda_{0}\) satisfies (8), then \(\lambda_{0}\) satisfies (6), consequently, \(\Psi_{1}^{S}(\mathcal{A})\leq U_{1}^{\bar{S}}(\mathcal{A})\leq U^{S}(\mathcal{A})\).

(ii) Suppose that \(r_{i}^{\overline{\Delta^{S}}}(\mathcal{A})r_{j}^{\Delta ^{S}}(\mathcal{A})>0\). Then dividing (8) by \(r_{i}^{\overline{\Delta^{S}}}(\mathcal {A})r_{j}^{\Delta^{S}}(\mathcal{A})\), we have

$$\begin{aligned} & \frac{(\lambda_{0}-a_{i\cdots i}-r_{i}^{\Delta^{S}}(\mathcal {A}))}{r_{i}^{\overline{\Delta^{S}}}(\mathcal{A})} \frac{(\lambda_{0}-a_{j\cdots j}-r_{j}^{\overline{\Delta^{S}}}(\mathcal {A}))}{r_{j}^{\Delta^{S}}(\mathcal{A})}\leq1. \end{aligned}$$
(12)

Furthermore, if \(\frac{\lambda_{0}-a_{i\cdots i}-r_{i}^{\Delta^{S}}(\mathcal {A})}{r_{i}^{\overline{\Delta^{S}}}(\mathcal{A})}\geq1\), then by Lemma 2.3 in [12] and (12), we have

$$\begin{aligned} & \frac{(\lambda_{0}-a_{i\cdots i})}{r_{i}(\mathcal{A})} \frac{(\lambda_{0}-a_{j\cdots j}-r_{j}^{\overline{\Delta^{S}}}(\mathcal {A}))}{r_{j}^{\Delta^{S}}(\mathcal{A})} \leq \frac{(\lambda_{0}-a_{i\cdots i}-r_{i}^{\Delta^{S}}(\mathcal {A}))}{r_{i}^{\overline{\Delta^{S}}}(\mathcal{A})} \frac{(\lambda_{0}-a_{j\cdots j}-r_{j}^{\overline{\Delta^{S}}}(\mathcal {A}))}{r_{j}^{\Delta^{S}}(\mathcal{A})} \leq1. \end{aligned}$$

Thus, (6) holds, which implies that if \(\lambda_{0}\) satisfies (8), then \(\lambda_{0}\) satisfies (6), consequently, \(\Psi_{1}^{S}(\mathcal{A})\leq U_{1}^{S}(\mathcal{A})\). And if \(\frac{\lambda_{0}-a_{i\cdots i}-r_{i}^{\Delta^{S}}(\mathcal {A})}{r_{i}^{\overline{\Delta^{S}}}(\mathcal{A})}\leq1\), then (10) holds, which leads to (11) from (9). This implies that if \(\lambda_{0}\) satisfies (8), then \(\lambda_{0}\) satisfies (6), consequently, \(\Psi_{1}^{S}(\mathcal{A})\leq U_{1}^{\bar{S}}(\mathcal{A})\leq U^{S}(\mathcal{A})\). The conclusion follows immediately from what we have proved. □

3 Numerical examples

Example 1

Let \(\mathcal{A}=(a_{ijkl})\) be a \((2,2)\)th order \(3\times3\) dimensional nonnegative rectangular tensor with entries defined as follows:

$$\begin{aligned}& A(:,:,1,1)= \begin{bmatrix} 6& 1& 1\\ 1& 0& 0\\ 1& 1& 1 \end{bmatrix},\qquad A(:,:,2,1)= \begin{bmatrix} 2& 2& 0\\ 1& 1& 0\\ 2& 2& 1 \end{bmatrix},\\& A(:,:,3,1)= \begin{bmatrix} 3& 0& 3\\ 3& 2& 0\\ 1& 1& 1 \end{bmatrix}, \\& A(:,:,1,2)= \begin{bmatrix} 1& 0& 0\\ 1& 1& 1\\ 1& 2& 2 \end{bmatrix},\qquad A(:,:,2,2)= \begin{bmatrix} 2& 2& 0\\ 0& 3& 2\\ 1& 2& 0 \end{bmatrix},\\& A(:,:,3,2)= \begin{bmatrix} 1& 1& 0\\ 0& 1& 2\\ 2& 1& 2 \end{bmatrix}, \\& A(:,:,1,3)= \begin{bmatrix} 0& 1& 0\\ 1& 2& 2\\ 1& 1& 1 \end{bmatrix},\qquad A(:,:,2,3)= \begin{bmatrix} 0& 0& 1\\ 0& 0& 1\\ 0& 1& 2 \end{bmatrix},\\& A(:,:,3,3)= \begin{bmatrix} 1& 1& 1\\ 1& 0& 1\\ 2& 1& 0 \end{bmatrix}. \end{aligned}$$

By Theorem 1, we have

$$\lambda_{0}\leq33. $$

By Theorem 2, we have

$$\lambda_{0}\leq 32.8924. $$

Taking \(S=\{1,2\},\bar{S}=\{3\}\), by Theorem 3, we have

$$\lambda_{0}\leq32.0540; $$

by Theorem 4, we have

$$\lambda_{0}\leq30.0965. $$

In fact, \(\lambda_{0}=29.8830\). This example shows that the upper bound in Theorem 4 is smaller than those in Theorems 1-3.

Example 2

Let \(\mathcal{A}=(a_{ijkl})\) be a \((2,2)\)th order \(2\times2\) dimensional nonnegative rectangular tensor with entries defined as follows:

$$a_{1111}=a_{1112}=a_{1222}=a_{2112}=a_{2121}=a_{2221}=1, $$

the other \(a_{ijkl}=0\). By Theorem 4, we have

$$\lambda_{0}\leq3. $$

In fact, \(\lambda_{0}=3\). This example shows that the upper bound in Theorem 4 is sharp.

4 Conclusions

In this paper, a new S-type upper bound \(\Psi^{S}(\mathcal{A})\) of the largest singular value for a nonnegative rectangular tensor \(\mathcal {A}\) with \(m=n\) is obtained by breaking N into disjoint subsets S and its complement. It is proved that the bound \(\Psi^{S}(\mathcal{A})\) is better than those in [6, 9, 10].

Note here that when \(n=2\), \(\Phi(\mathcal{A})=U^{S}(\mathcal{A})=\Psi ^{S}(\mathcal{A})\), and when \(n\geq3\), \(\Phi(\mathcal{A})\geq U^{S}(\mathcal{A})\geq\Psi ^{S}(\mathcal{A})\) always holds. How to pick S to make \(\Psi^{S}(\mathcal{A})\) as small as possible is an interesting problem, but difficult when n is large. We will research this problem in the future.