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# Open Mathematics

### formerly Central European Journal of Mathematics

Editor-in-Chief: Gianazza, Ugo / Vespri, Vincenzo

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Volume 14, Issue 1

# An S-type upper bound for the largest singular value of nonnegative rectangular tensors

Jianxing Zhao
/ Caili Sang
Published Online: 2016-11-12 | DOI: https://doi.org/10.1515/math-2016-0085

## Abstract

An S-type upper bound for the largest singular value of a nonnegative rectangular tensor is given by breaking N = {1, 2, … n} into disjoint subsets S and its complement. It is shown that the new upper bound is smaller than that provided by Yang and Yang (2011). Numerical examples are given to verify the theoretical results.

MSC 2010: 15A18; 15A42; 15A69

## 1 Introduction

Let ℝ(ℂ) be the real (complex) field, p, q, m, n be positive integers, m, n ≥ 2, N = {1, 2,…, n}, and ${\mathbb{R}}_{+}^{n}$ be the cone {x = (x1, x2, …, xn)T 𲈈 ℝn : xi ≥ 0, iN}. A real (p, q)-th order m × n dimensional rectangular tensor, or simply a real rectangular tensor $\mathcal{A}$ is defined as follows: $A=(ai1⋯ipj1⋯jq);$

where ai1ipj1jq ∈ ℝ for ik = 1,…, m,; k = 1,… p, and jk = 1,… n, k = 1,… q. When p = q = 1, $\mathcal{A}$ is simply a real m × n rectangular matrix. This justifies the word “rectangular”.

For any vector x and any real number α, denote ${x}^{\left[\alpha \right]}=\left({x}_{1}^{\alpha },{x}_{2}^{\alpha },\cdots ,{x}_{n}^{\alpha }{\right)}^{\mathrm{T}}$. Let $\mathcal{A}{x}^{p-1}{y}^{q}$ be a vector in ℝm such that $(Axp−1yq)i=∑i2,⋯,ip=1m∑j1,⋯,jq=1naii2⋯ipj1⋯jqxi2⋯xipyj1⋯yjq,$

where i = 1,…, m. Similarly, let $\mathcal{A}{x}^{p}{y}^{q-1}$ be a vector in ℝn such that $(Axpyq−1)j=∑i1,⋯,ip=1m∑j2,⋯,jq=1nai1⋯ipjj2⋯jqxi1⋯xipyj2yjq,$

where j = 1,…, n. Let l = p + q. If there are a number λ ∈ ℂ, vectors x ∈ λm}\{0}, and y ∈ ℂn\{0} such that $Axp−1yq=λx[l−1],Axpyq−1=λy[l−1],$

then λ is called the singular value of $\mathcal{A}$, and (x, y) is the left and right eigenvectors pair of $\mathcal{A}$, associated with λ, respectively. If λ ∈ℝ; x ∈ ℝm, and y ∈ ℝn, then we say that λ is an ℍ-singular value of $\mathcal{A}$, and (x, y) is the left and right ℍ-eigenvectors pair associated with λ, respectively. If a singular value is not an ℍ-singular value, we call it an ℕ-singular value of $\mathcal{A}$. If p = q = 1, then this is just the usual definition of singular values for a rectangular matrix [1]. We call $λ0=max{|χ|:λisasingularvalueofA}$

is the largest singular value [2].

Note here that the notion of singular values for tensors was first proposed by Lim in [3]. When l is even, the definition in [1] is the same as in [3]. When l is odd, the definition in [1] is slightly different from that in [3], but parallel to the definition of eigenvalues of square matrices [4]; see [1] for details. For the sake of simplicity, the definition of singular values in this paper is the definition in [1].

We recall the weak Perron-Frobenius theorem for nonnegative rectangular tensors, which was given in [2].

#### [2, Theorem 2]

Let $\mathcal{A}$ be a(p, q)-th order m × n dimensional nonnegative tensor Then λ0 is the largest singular value with nonnegative left and right eigenvectors pair $\left(x,y\right)\in {\mathbb{R}}_{+}^{m}\mathrm{\setminus }\left\{0\right\}×{\mathbb{R}}_{+}^{n}\mathrm{\setminus }\left\{0\right\}$ corresponding to it.

The largest singular value of a nonnegative rectangular tensor has a wide range of practical applications in the strong ellipticity condition problem in solid mechanics [5, 6] and the entanglement problem in quantum physics [7, 8]. Recently, there are many results about the properties of square tensors, especially the upper bounds for the ℤ-spectral radius and ℍ-spectral radius of a nonnegative square tensor [913]. However, there are no results about the upper bounds for the largest singular value of a nonnegative rectangular tensor except the following one [2].

#### see [2, Theorem 4]

Let $\mathcal{A}$be a(p, q)-th order m × n dimensional nonnegative rectangular tensor Then $min1≤i≤m,1≤j≤n{Rj(A),Cj(A)}≤λ0≤max1≤i≤m,1≤j≤n{Rj(A),Cj(A)},$ where $Rj(A)=∑i2,⋯,ip=1m∑j1,⋯,jq=1naii2⋯ipj1⋯jq,Cj(A)=∑i1,⋯,ip=1m∑j2,⋯,jq=1nai1⋯ipjj2⋯jq.$

Throughout this paper, we assume m = n. Our goal in this paper is to give a new upper bound for the largest singular value of a nonnegative rectangular tensor, and prove that the new upper bound is smaller than that in Theorem 1.2.

## 2 Main results

We begin with some notation. Given a nonempty proper subset S of N, we denote $ΔN:={(i2,⋯ip,j1,⋯,jq):i2,⋯ip,j1,⋯,jq∈N},ΔS:={(i2,⋯ip,j1,⋯,jq):i2,⋯ip,j1,⋯,jq∈S},ΩN:={(i1,⋯ip,j2,⋯,jq):i1,⋯ip,j2,⋯,jq∈N},ΩS:={(i1,⋯ip,j2,⋯,jq):i1,⋯ip,j2,⋯,jq∈S},$

and then $△S¯=△N∖△S,ΩS¯=ΩN∖ΩS$

This implies that for a nonnegative rectangular tensor $\mathcal{A}=\left({a}_{{i}_{1}\cdots {i}_{p}{j}_{1}\cdots {j}_{q}}\right)$, we have that for i, jS, $rj(A)=∑i2.⋯.ip.j1,⋯.jq∈Nδii2⋯ipj1⋯jq=0aii2⋯ipj1⋯jq=riΔS(A)+riΔS¯(A),rij(A)=riΔS(A)+riΔS¯(A)−aij⋯jj⋯j,$ $cj(A)=∑i1.⋯,ip,j2…,jq∈Nδi1⋯ipj˙j2⋯jq=0ai1⋯ipjj2⋯jq=cjΩS(A)+cjΩS¯(A),cji(A)=cjΩS(A)+cjΩS¯(A)−aj⋯iji⋯i,$

where $δi1⋯ipj1⋯jq=1,ifi1=⋯=ip=j1=⋯=jq,0,otherwise,$

and $riΔS(A)=∑(i2,⋯,ip,j1.⋯jq)∈ΔSδii2⋯ipj1⋯jq=0aii2⋯ipj1⋯jq,riΔS¯(A)=∑(i2,⋯,ip,j1,⋅⋅,jq)∈ΔS¯aii2⋯ipj1⋯jq,$ $cjΩS(A)=∑(i1⋯,ip.j2…jq)∈ΩSδi1⋯ipjj⋯jq=0ai1⋯ipjj2⋯jq,cjΩS¯(A)=∑(i1,⋯,ip,j2,⋅⋅,jq)∈ΩS¯ai1⋯ipjj2⋯jq.$

By Theorem 1.2, the following lemma is easily obtained.

Let $\mathcal{A}$ be a(p, q)-th order n × n dimensional nonnegative rectangular tensor Then $λ0≥ai⋯ii⋯i,i∈N.$

Let $\mathcal{A}$ be a(p, q)-th order n × n dimensional nonnegative rectangular tensor, S be a nonempty proper subset of N, $\overline{S}$ be the complement of S in N. Then $λ0≤US(A)=max{U1S(A),U1S¯(A),U2S(A),U2S¯(A)},$

where $U1S(A)=maxi∈S,j∈S¯⁡12{ai⋯ii⋯i+aj⋯jj⋯j+rjΔS¯(A)+[(ai⋯ii⋯j−aj⋯jj⋯j−rjΔS¯(A))2+4max{rj(A),ci(A)}rjΔS(A)]12},U1S¯(A)=maxi∈s¯,j∈S⁡12{ai⋯ii⋯i+aj⋯jj⋯j+rjΔS¯¯(A)+[(ai⋯ii⋯j−aj⋯jj⋯j−rjΔS¯¯(A))2+4max{ri(A),ci(A)}rjΔS¯(A)]12},U2S(A)=maxi∈S,j∈S¯⁡12{ai⋯ii⋯i+aj⋯jj⋯j+cjΩS¯(A)+[(ai⋯ii⋯i−aj⋯jj⋯j−cjΩS¯(A))2+4max{ri(A),ci(A)}cjΩS(A)]12},U2S¯(A)=maxi∈s¯,j∈S⁡12{ai⋯ii⋯i+aj⋯jj⋯j+cjΩS¯¯(A)+[(ai⋯ii⋯i−aj⋯jj⋯j−cjΩS¯¯(A))2+4max{ri(A),ci(A)}cjΩS¯(A)]12}.$

Let λ0 be the largest singular value of $\mathcal{A}$. According to Theorem 1.1, there exist two nonzero nonnegative vectors x = (x1, x2,…, xn)T and y = (y1, y2,…, yn)T such that $Axp−1yq=λ0x[l−1],Axpyq−1=λ0y[l−1].$(1)(2)

Let $xt=max{xi:i∈S},xh=max{xi:i∈S¯};yf=max{yi:i∈S},yg=max{yi:i∈S¯};wi=max{xi,yi},i∈N,ws=max{wi:i∈S},wS¯=max{wi:i∈S¯}.$

Then, at least one of xt and xh is nonzero, and at least of yf and yg is nonzero. Next, we present four cases to prove that.

Case I: Suppose that $ws={x}_{t},{w}_{\overline{S}}={x}_{h}$, then xtyt, xhyh.

(i) If xhxt, then xh = max{wj : iN}. By the h-th equality in (1), we have $(λ0−ah⋯hh⋯h)xhl−1≤λ0xhl−1−ah⋯hh⋯hxhp−1yhq=∑(i2,⋯,ip,j1,⋅⋅,jq)∈ΔSahi2⋯ipj1⋯jqxi2⋯xipyj1⋯yjq+∑(i2,⋯.ip,j1,⋯,jq)∈ΔS¯δhi2⋯ipj1⋯jq=0ahi2⋯ipj1⋯jqxi2⋯xipyj1⋯yjq≤∑(i2,⋯,ip,j1,⋅⋅,jq)∈ΔSahi2⋯ipj1⋯jqxtl−1+∑(i2.⋯,ip,j1,⋯jq)∈ΔS¯δhi2⋯ipj1⋯jq=0ahi2⋯ipj1⋯jqxhl−1=rhΔS(A)xtl−1+rhΔS¯(A)xhl−1.$

Hence, $(λ0−ah⋯hh⋯h−rhΔS¯(A))xhl−1≤rhΔS(A)xtl−1.$(3)

If xt = 0, then ${\lambda }_{0}-{a}_{h\cdots hh\cdots h}-{r}_{h}^{\overline{{\mathrm{\Delta }}^{S}}}\left(\mathcal{A}\right)\le 0$ as xh > 0, and it is obvious that ${\lambda }_{0}\le {U}_{1}^{S}\left(\mathcal{A}\right)$ . Otherwise, xt > 0. From the t-th equality in (1), we have $(λ0−at⋯tt⋯t)xtl−1≤λ0xtl−1−at⋯tt⋯txrp−1ytq=∑i2,⋯,ipj˙.⋅⋅,jq∈Nδti2⋯ipj1⋯jq=0ati2⋯ipj1⋯jqxi2…xjpyj1...yjq≤∑i2.⋯,ip,j.⋯.jq1∈N,δti2⋯ipj1⋯jq=0ati2⋯ipj1⋯jqxhl−1=rt(A)xhl−1,$

i.e., $(λ0−at⋯tt⋯t)xtl−1≤rt(A)xhl−1.$(4)

From Lemma 2.1, we have λ0-atttt ≥ 0. Multiplying (3) with (4), we have $(λ0−at⋯tt⋯t)(λ0−ah⋯hh⋯h−rhΔS¯(A))xtl−1xhl−1≤rt(A)rhΔS(A)xtl−1xhl−1.$

Note that ${x}_{t}^{l-1}{x}_{h}^{l-1}>0$. Then $(λ0−at⋯tt⋯t)(λ0−ah⋯hh⋯h−rhΔS¯(A))≤rt(A)rhΔS(A).$(5)

Solving (5) gives $λ0≤12{at⋯tt⋯t+ah⋯hh⋯h+rhΔS¯(A)+[(at⋯tt⋯t−ah⋯hh⋯h−rhΔS¯(A))2+4rt(A)rhΔS(A)]12}≤maxi∈S,j∈S¯⁡12{ai⋯ii⋯i+aj⋯jj⋯j+rjΔS¯(A)+[(ai⋯ii⋯i−aj⋯jj⋯j−rjΔS¯(A))2+4ri(A)rjΔS(A)]12}≤U1S(A).$

(ii) If xtxh, then xt = max{wj : iN}. Similarly to the proof of (i), we can obtain that $(λ0−ah⋯hh⋯h)(λ0−at⋯tt⋯t−rtΔS¯¯(A))≤rh(A)rtΔS¯(A).$

This gives $λ0≤12{ah⋯hh⋯h+at⋯tt⋯t+rtΔS¯¯(A)+[(ah⋯hh⋯h−at⋯tt⋯t−rtΔS¯¯(A))2+4rh(A)rtΔS¯(A)]12}≤maxi∈s¯,j∈S⁡12{ai⋯ii⋯i+aj⋯jj⋯j+rjΔS¯¯(A)+[(ai⋯ii⋯i−aj⋯jj⋯j−rjΔS¯¯(A))2+4ri(A)rjΔS¯(A)]12}≤U1S¯(A).$

Case II: Suppose that ws = yf, ${w}_{\overline{S}}={y}_{g}$, then yfxf, ygxg. If ygyf, then yg = max{wi : i 𢈈 N}. Similarly to the proof of (i) in Case I, we have $(λ0−af⋯ff⋯f)(λ0−ag⋯gg⋯g−cgΩS¯(A))≤cf(A)cgΩS(A).$

This gives $λ0≤12{af⋯ff⋯f+ag⋯gg⋯g+cgΩS¯(A)+[(af⋯ff⋯f−ag⋯gg⋯g−cgΩS¯(A))2+4cf(A)cgΩS(A)]12}≤maxi∈S,j∈S¯⁡12{ai⋯ii⋯i+aj⋯jj⋯j+cjΩS¯(A)+[(ai⋯ii⋯i−aj⋯jj⋯j−cjΩS¯(A))2+4ci(A)cjΩS(A)]12}≤U2S(A).$

If yfyg, then yf = max{wi : iN}. Similarly to the proof of (ii) in Case I, we have $(λ0−ag⋯gg⋯g)(λ0−af⋯ff⋯f−cfΩS¯¯(A))≤cg(A)cfΩS¯(A).$

This gives $λ0≤12{ag⋯gg⋯g+af⋯ff⋯f+cfΩS¯¯(A)+[(ag⋯g⋯g−af⋯ff⋯f−cfΩS¯¯(A))2+4cg(A)cfΩS¯(A)]12}≤maxi∈s¯,j∈S⁡12{ai⋯ii⋯i+aj⋯jj⋯j+cjΩS¯¯(A)+[(ai⋯ii⋯i−aj⋯jj⋯j−cjΩS¯¯(A))2+4ci(A)cjΩS¯(A)]12}≤U2S¯(A).$

Case III: Suppose that ws = xt, ${w}_{\overline{S}}={y}_{g}$, then xtyt, ygxg. If ygxt, then yg = max{wi : iN}. Similarly to the proof of (i) in Case I, we have $(λ0−at⋯tt⋯t)(λ0−ag⋯gg⋯g−cgΩS¯(A))≤rt(A)cgΩS(A).$

This gives $λ0≤12{at⋯tt⋯t+ag⋯gg⋯g+cgΩS¯(A)+[(at⋯tt⋯t−ag⋯gg⋯g−cgΩS¯(A))2+4rt(A)cgΩS(A)]12}≤maxi∈S,j∈S¯⁡12{ai⋯ii⋯i+aj⋯jj⋯j+cjΩS¯(A)+[(ai⋯ii⋯j−aj⋯jj⋯j−cjΩS¯(A))2+4ri(A)cjΩS(A)]12}≤U2S(A).$

If xtyg, then xt = max{wi : iN}. Similarly to the proof of (ii) in Case I, we can obtain that $(λ0−ag⋯gg⋯g)(λ0−at⋯tt⋯t−rtΔS¯¯(A))≤cg(A)rtΔS¯(A).$

This gives $λ0≤12{ag⋯gg⋯g+at⋯tt⋯t+rtΔS¯¯(A)+[(ag⋯gg⋯g−at⋯tt⋯t−rtΔS¯¯(A))2+4cg(A)rtΔS¯(A)]12}≤maxi∈s¯,j∈S⁡12{ai⋯ii⋯i+aj⋯jj⋯j+rjΔS¯¯(A)+[(ai⋯ii⋯i−aj⋯jj⋯j−rjΔS¯¯(A))2+4ci(A)rjΔS¯(A)]12}≤U1S¯(A).$

Case IV: Suppose that $ws=yf,{w}_{\overline{S}}={x}_{h}$, then yfxf, xhyh. If xhyf, then xh = max{wj : iN}. Similarly to the proof of (i) in Case I, we have $(λ0−af⋯ff⋯f)(λ0−ah⋯hh⋯h−rhΔS¯(A))≤cf(A)rhΔS(A).$

This gives $λ0≤12{af⋯ff⋯f+ah⋯hh⋯h+rhΔS¯(A)+[(af⋯ff⋯f−ah⋯hh⋯h−rhΔS¯(A))2+4cf(A)rhΔS(A)]12}≤maxi∈S,j∈S¯⁡12{ai⋯ii⋯i+aj⋯jj⋯j+rjΔS¯(A)+[(ai⋯ii⋯i−aj⋯jj⋯j−rjΔS¯(A))2+4ci(A)rjΔS(A)]12}≤U1S(A).$

If yfxh, then yf = max{wi} : iN}. Similarly to the proof of (ii) in Case I, we can obtain that $(λ0−ah⋯hh⋯h)(λ0−af⋯ff⋯f−cfΩS¯¯(A))≤rh(A)cfΩS¯(A).$

This gives $λ0≤12{ah⋯hh⋯h+af⋯ff⋯f+cfΩS¯¯(A)+[(ah⋯hh⋯h−af⋯ff⋯f−cfΩS¯¯(A))2+4rh(A)cfΩS¯(A)]12}≤maxi∈s¯,j∈S⁡12{ai⋯ii⋯i+aj⋯jj⋯j+cjΩS¯¯(A)+[(ai⋯ii⋯i−aj⋯jj⋯j−cjΩS¯¯(A))2+4ri(A)cjΩS¯(A)]12}≤U2S¯(A).$

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

Next, we compare the upper bound in Theorem 2.2 with that in Theorem 1.2.

Let $\mathcal{A}$ be a(p, q)-th order n × n dimensional nonnegative rectangular tensor, S be a nonempty proper subset of $N,\overline{S}$ be the complement of S in N. Then $US(A)≤maxi,j∈N⁡{Ri(A),Cj(A)}.$

## 3 Numerical examples

In this section, two numerical examples are given to verify the theoretical results.

Let $\mathcal{A}=\left({a}_{ijkl}\right)$ be a (2, 2)-th order 3 × 3 dimensional nonnegative rectangular tensor with entries defined as follows: $A(:,:,1,1)=210000212,A(:,:,2,1)=110200221,A(:,:,3,1)=9101001122,A(:,:,1,2)=021002220,A(:,:,2,2)=111120222,A(:,:,3,2)=100002221,A(:,:,1,3)=101110121,A(:,:,2,3)=001101111,A(:,:,3,3)=110201222.$

Let S ={1, 2}. Obviously $\overline{S}=\left\{3\right\}$. By Theorem 1.2, we have $λ0≤45.$

By Theorem 2.2, we have $λ0≤38.4165.$

In fact, λ0 = 31.3781. This example shows that the upper bound in Theorem 2.2 is smaller than that in Theorem 1.2.

Let $\mathcal{A}=\left({a}_{ijkl}\right)$ be a(2, 2)-th order 2 × 2 dimensional nonnegative rectangular tensor with entries defined as follows: $a1111=a1112=a1222=a2112=a2121=a2221=1,$

other aijkl = 0. By Theorem 2.2, we have $λ0≤3.$

In fact, λ0 = 3. This example shows that the upper bound in Theorem 2.2 is sharp.

## 4 Conclusions

In this paper, by breaking N into disjoint subsets S and its complement, an S-type upper bound ${U}^{S}\left(\mathcal{A}\right)$ for the largest singular value of a nonnegative rectangular tensor $\mathcal{A}$ with m = n is obtained, which improves the upper bound in [2]. Then an interesting problem is how to pick S to make ${U}^{S}\left(\mathcal{A}\right)$ as small as possible. But this is difficult when n is large. In the future, we will research this problem.

## Acknowledgement

The authors are very indebted to the reviewers for their valuable comments and corrections, which improved the original manuscript of this paper. This work is supported by the National Natural Science Foundation of China (Nos.11361074,11501141), Foundation of Guizhou Science and Technology Department (Grant No.[2015]2073) and Natural Science Programs of Education Department of Guizhou Province (Grant No. [2016]066).

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Accepted: 2016-10-11

Published Online: 2016-11-12

Published in Print: 2016-01-01

Citation Information: Open Mathematics, Volume 14, Issue 1, Pages 925–933, ISSN (Online) 2391-5455,

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