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严格对角占优张量的子直和

何建锋

何建锋. 严格对角占优张量的子直和[J]. 华南师范大学学报(自然科学版), 2021, 53(3): 102-105. DOI: 10.6054/j.jscnun.2021048
引用本文: 何建锋. 严格对角占优张量的子直和[J]. 华南师范大学学报(自然科学版), 2021, 53(3): 102-105. DOI: 10.6054/j.jscnun.2021048
HE Jianfeng. The Subdirect Sum of Strictly Diagonally Dominant Tensors[J]. Journal of South China Normal University (Natural Science Edition), 2021, 53(3): 102-105. DOI: 10.6054/j.jscnun.2021048
Citation: HE Jianfeng. The Subdirect Sum of Strictly Diagonally Dominant Tensors[J]. Journal of South China Normal University (Natural Science Edition), 2021, 53(3): 102-105. DOI: 10.6054/j.jscnun.2021048

严格对角占优张量的子直和

基金项目: 

国家自然科学基金项目 61463002

云南省教育厅科学研究基金项目 2019J0399

详细信息
    通讯作者:

    何建锋, Email: hjf@cxtc.edu.cn

  • 中图分类号: O151.21

The Subdirect Sum of Strictly Diagonally Dominant Tensors

  • 摘要: 根据张量与矩阵之间的联系,将方阵子直和及S-严格对角占优矩阵的概念推广到张量上,给出了张量子直和与S-严格对角占优型张量的定义,用分类讨论的方法证明2个严格对角占优张量的子直和仍然为严格对角占优张量,并讨论了S-严格对角占优型张量的情形,给出2个张量的子直和为S-严格对角占优型张量的条件.
    Abstract: The concept of subdirect sum of square matrices is extended to tensors according to the relation between matrix and tensor. The definitions of subdirect sum of tensors and S-strictly diagonally dominant tensors are given. It is proved, with the method of classification, that the subdirect sum of two strictly diagonally dominant tensors is also a strictly diagonally dominant tensor. Moreover, the condition ensuring that the subdirect sum of two tensors is the S-strictly diagonally dominant tensor is also given.
  • 矩阵作为代数学中的一个基本内容,已在科学和工程的各个方面得到广泛应用. 随着大数据分析的发展,人们对张量的研究日益增加. 目前,有关张量的研究成果已较为丰富[1-5]. 此处所提的张量也可以称作超矩阵,相比于矩阵元素有2个下标,张量元素的下标个数可以大于2个. 鉴于矩阵与张量之间的联系,许多矩阵理论中的内容已被推广到张量上进行研究,如严格对角占优矩阵[6]、特征值[1]、正定性[1]和Perron-Frobenius定理[7]等.

    方阵的子直和是矩阵和的一种推广,FALLAT和JOHNSON[8]给出了方阵子直和的定义,并对其性质进行研究. 方阵子直和在矩阵的完备化[9]、区域分解方法中的重叠子域[10]等问题中均有涉及. 目前关于方阵子直和的研究已有许多结果[8, 11-17].

    本文利用矩阵与张量之间的维数关系,定义了张量子直和与S-严格对角占优型张量,证明了严格对角占优张量的k-子直和是严格对角占优张量,并给出了张量的子直和为S-严格对角占优型张量的一个充分条件.

    对于一个正整数n (≥2),如果ai1i2imR (ij=1, 2, …, n; j=1, 2, …, m),则A=(ai1i2im)称为一个mn维实张量,记为AR[m, n].

    为了讨论方便,引入下列符号:[n]={1, 2, …, n};S⊆[n]且S≠Ø, riS(A)=ji,jS|aij|, ΔS={(i2i3im)|ijSj=2,…,mriΔS(A)=(i2im)ΔSδii2im=0|aii2im|;当i1=i2= …=im时,δi1i2im=1,当i1, i2, …, im不全相等时,δi1i2im=0.

    下面先给出几个定义.

    定义1[8]  设方阵AB的阶分别为n1n2k为整数且满足1≤k≤min{n1n2}. 如果

    A=[A11A12A21A22],B=[B11B12B21B22],

    其中A22B11均为k阶方阵,那么

    C=[A11A120A21A22+B11B120B21B22]

    称为ABk-子直和,记为C=Ak B.

    定义2[11]  设矩阵A =(aij)∈Cn×nn≥2,S是集合[n]的一个非空子集,如果以下2个条件成立:

    (i)|aii|>riS(A),∀iS

    (ii) (|aii|-riS(A))(|ajj| -rjS(A))>riS(A)rjS(A),∀iS,∀jS

    则称A是S-严格对角占优矩阵,简记为S-SDD矩阵.

    定义3[18]  设张量A=(ai1i2im)∈R[m, n],如果对于所有的i∈[n],有

    aiii>i2,,im[n]|aii2im|aiii,

    则称A是严格对角占优张量,简记为SDD张量.

    由定义1和定义2,利用矩阵与张量之间的联系,可定义张量子直和与S-SDD型张量:

    定义4  设张量A=(ai1i2im)∈R[m, n1]B=(bi1i2im)∈R[m, n2]n1, n2≥2,n=n1+n2-kk为整数且1≤k≤min(n1, n2),t=m-k,令

    ci1i2im={ai1i2im    ((i1i2im)ΔS1S2/ΔS2),ai1i2im+bi1t,i2t,,imt    ((i1i2im)ΔS2),bi1t,i2t,imt((i1i2im)ΔS2S3/ΔS2),0((i1i2im)(Δ[n]/ΔS1S2)(Δ[n]/ΔS2S3)),

    其中,S1={1, 2, …, n1-k},S2={n1-k+1, …, n1},S3={n1+1, …, n},则张量C=(ci1i2im)∈R[m, n]称为ABk-子直和,记为C=Ak B.

    定义5  设张量A=(ai1i2im)∈R[m, n]n≥2,S是集合[n]的一个非空子集,如果以下2个条件成立:

    (i)|aii|>riΔS(A),∀iS

    (ii) (|aii|-riΔS(A))(|ajj|-rjΔS(A))>riΔS (ArjΔS(A),∀iS,∀jS

    则称AS-严格对角占优型张量,简记为S-SDD型张量.

    本节证明2个SDD张量的子直和仍然是SDD张量,以及1个SDD张量与1个S-SDD型张量的子直和是S-SDD型张量.

    定理 1  设张量A=(ai1i2im)∈R[m, n1]B=(bi1i2im)∈R[m, n2]都是SDD张量,k为整数,且1≤ k≤min(n1, n2),n1, n2≥2,n=n1+n2-k. 如果aiiibjjj>0,iS2j∈[k],则ABk-子直和C=Ak B是SDD张量.

    证明  分3种情形证明.

    情形1. 当i1S1时,有

    ci1i2im={ai1i2im    ((i2im)ΔS1S2),0    ((i2im)ΔS1S2),

    ci1i1i1=ai1i1i1>(i2im)ΔS1S2|ai1i2im|ai1i1i1=(i2im)[n]|ci1i2im|ci1i1i1.

    情形2. 当i1S2时,因为AB是SDD张量,所以

    ai1i1i1>(i2im)ΔS1S2|ai1i2im|ai1i1i1, (1)
    bi1t,i1t,,i1t>(i2im)ΔS2S3|bi1t,i2t,,imt|    bi1t,i1t,,i1t (2)

    根据定义4,有

    ci1i1i1=ai1i1i1+bi1t,i1t,,i1t. (3)

    由式(1)~(3),可知

    ci1i1i1>(i2im)ΔS1S2|ai1i2im|ai1i1i1+    (i2im)ΔS2S3|bi1t,i2t,,imt|bi1t,i1t,,i1t=    (i2im)ΔS1S2|ai1i2im|+(i2im)ΔS2S3|bi1t,i2t,,imt|    ci1i1i1=(i2im)Δ[n]|ci1i2im|ci1i1i1.

    情形3. 当i1S3时,有

    ci1i2im={bi1t,i2t,,imt    ((i2im)ΔS2S3),0    ((i2im)ΔS2S3),

    ci1i1i1=bi1t,i1t,,i1t>(i2im)ΔS1S2|bii2im|bi1t,i1t,,i1t=(i2il)[n]|cii2im|ci1i1i1.

    综上所述,当i1S1S2S3时,有

    ci1i1i1>(i2im)[n]|cii2im|ci1i1i1,

    因此,C= Ak B是SDD张量.

    定理1表明2个SDD张量的子直和仍然是SDD张量,但是,2个S-SDD型张量的子直和不一定是S-SDD型张量.

    例 1  张量A= (aijk) ∈R[3, 4],其中

    A(:,:,1)=(2.60.10.10.020.10.20.10.020.10.10.20.030.020.030.030.04),A(:,:,2)=(0.400.10.102.60.10.10.10.10.10.10.10.10.10.1),A(:,:,3)=(0.60.200.10.20.300.1002.20.20.10.10.20.2),A(:,:,4)=(0.80.20.100.20.30.100.10.10.100002.2).

    S={1, 2},则张量A是S-SDD型张量. 但是2-子直和C=A2 A不是S-SDD型张量,此时定义5中的条件(ii)不成立,因为,当i=1,j=5时,有

    (|c111|rΔS1(C))(|c555|rΔˉS5(C))=0.22<0=       rΔˉS1(C)rΔS5(C).

    下面给出2个张量的子直和为S-SDD型张量的一个条件.

    定理 2  设张量A=(ai1i2im)∈R[m, n1]是S-SDD型张量,张量B=(bi1i2im)∈R[m, n2]是SDD张量,SS1的一个非空子集,n1, n2≥2,k为整数且1≤k≤min(m, n),集合S1S2S3与定义4中的相同. 如果aiiibjjj>0,iS2j∈[k],则k-子直和C=Ak B是S-SDD型张量.

    证明  先证明S=S1时的情形. 要证明张量C是S-SDD型张量,需证明以下2个条件成立:

    (i) |cii| ≥riΔS(C),∀iS

    (ii) (|cii|-riΔS(C))(|cjj|-rjΔS(C))≥riΔS(C)rjΔS(C),∀iS,∀jS.

    S=S1时,S在[n1+n2-k]中的补集为S2S3.

    (1) 因为A是S-SDD型张量,有|aii|>riΔS(A) (iS),则

    |cii|=|aii|>rΔSi(A)=rΔSi(C)(iS).

    从而条件(i)成立.

    (2) 下面分2种情形证明条件(ii)成立.

    ① 当jS2时,由iS=S1,有

    rΔSi(C)=rΔSi(A),rΔSj(C)=rΔSj(A),rΔˉSi(C)=rΔˉSi(A),rΔˉSj(C)=rΔˉSj(A)+rjt(B),t=n1k.

    因为A是S-SDD型张量,B是SDD张量,则∀iS,∀jS,有

    (|aii|rΔSi(A))(|ajj|rΔˉSj(A))>rΔˉSi(A)rΔSj(A),

    且|bii|>ri(B) (i∈[n2]). 从而

    (|cii|rΔSi(C))(|cjj|rΔˉSj(C))=(|aii|rΔSi(A))×    (|ajj+bjt,,jt|rΔˉSj(A)rjt(B))=    (|aii|rΔSi(A))((|ajj|rΔˉSj(A))+    (|bjt,,jt|rjt(B)))>(|aii|rΔSi(A))×    (|ajj|rΔˉSj(A))>rΔˉSi(A)rΔSj(A)=rΔˉSi(C)rΔSj(C).

    由此可知当jS2时,条件(ii)成立.

    ② 当jS3时,由iS=S1,有rjΔS(C)=0,rjΔS(C)=rj-tΔS(B),从而

    (|cii|rΔSi(C))(|cjj|rΔˉSj(C))=    (|aii|rΔSi(A))(|bjt,,jt|rΔˉSjt(B))>    rΔˉSi(C)rΔSj(C).

    综上所述,当S=S1时,结论成立.

    当|S| < |S1|时,证明过程类似. 由此可知,C是S-SDD型张量.

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出版历程
  • 收稿日期:  2020-09-20
  • 网络出版日期:  2021-07-05
  • 刊出日期:  2021-06-24

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