严格对角占优张量的子直和

何建锋

doi:10.6054/j.jscnun.2021048

严格对角占优张量的子直和

何建锋^,

楚雄师范学院数学与统计学院，楚雄 675000

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
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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

The Subdirect Sum of Strictly Diagonally Dominant Tensors

HE Jianfeng^,

School of Mathematics and Statistics, Chuxiong Normal University, Chuxiong 675000, China

摘要

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.
- 子直和 /
- 张量 /
- 严格对角占优
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.
- subdirect sum /
- tensor /
- strictly diagonally dominant

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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.

1. 张量的子直和

对于一个正整数n (≥2)，如果a_{i₁i₂…i_m}∈ $\mathbb{R}$ (i_j=1, 2, …, n; j=1, 2, …, m)，则 $\mathcal{A}$ =(a_{i₁i₂…i_m})称为一个m阶n维实张量，记为 $\mathcal{A}$ ∈ $\mathbb{R}$ ^{[m, n]}.

为了讨论方便，引入下列符号：[n]={1, 2, …, n}；S⊆[n]且S≠Ø, r_i^S(A)= $\sum\limits_{j \ne i, j \in S} {\left| {{a_{ij}}} \right|}$ , Δ^S={(i₂i₃…i_m)|i_j ∈S，j=2，…，m；r_i^{Δ^S}( $\mathcal{A}$ )= $\sum\limits_{\begin{array}{*{20}{c}} {\left( {{i_2} \cdots {i_m}} \right) \in {\mathit{\Delta }^S}}\\ {{\delta _{i{i_2}}}_{ \cdots {i_m}} = 0} \end{array}} {\left| {{a_{i{i_2}}}_{ \cdots {i_m}}} \right|}$ ；当i₁=i₂= …=i_m时，δ_{i₁i₂…i_m}=1，当i₁, i₂, …, i_m不全相等时，δ_{i₁i₂…i_m}=0.

下面先给出几个定义.

定义1^[8] 设方阵A和B的阶分别为n₁和n₂，k为整数且满足1≤k≤min{n₁，n₂}. 如果

$\boldsymbol{A}=\left[\begin{array}{cc} \boldsymbol{A}_{11} & \boldsymbol{A}_{12} \\ \boldsymbol{A}_{21} & \boldsymbol{A}_{22} \end{array}\right], \boldsymbol{B}=\left[\begin{array}{ll} \boldsymbol{B}_{11} & \boldsymbol{B}_{12} \\ \boldsymbol{B}_{21} & \boldsymbol{B}_{22} \end{array}\right],$

其中A₂₂和B₁₁均为k阶方阵，那么

$\boldsymbol{C}=\left[\begin{array}{ccc} \boldsymbol{A}_{11} & \boldsymbol{A}_{12} & \boldsymbol{0} \\ \boldsymbol{A}_{21} & \boldsymbol{A}_{22}+\boldsymbol{B}_{11} & \boldsymbol{B}_{12} \\ \boldsymbol{0} & \boldsymbol{B}_{21} & \boldsymbol{B}_{22} \end{array}\right]$

称为A和B的k-子直和，记为C=A⊕_k B.

定义2^[11] 设矩阵A =(a_ij)∈ $\mathbb{C}$ ^n×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)|a_ii|>r_i^S(A)，∀i∈S；

(ii) (|a_ii|-r_i^S(A))(|a_jj| -r_j^S(A))>r_i^S(A)r_j^S(A)，∀i ∈S，∀j∈ S，

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 $\mathcal{A}$ =(a_{i₁i₂…i_m})∈ $\mathbb{R}$ ^{[m, n]}，如果对于所有的i∈[n]，有

$a_{i i \cdots i}>\sum\limits_{i_{2}, \cdots, i_{m} \in[n]}\left|a_{i i_{2} \cdots i_{m}}\right|-a_{i i \cdots i},$

则称 $\mathcal{A}$ 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 $\mathcal{A}$ =(a_{i₁i₂…i_m})∈ $\mathbb{R}$ ^{[m, n₁]}， $\mathcal{B}$ =(b_{i₁i₂…i_m})∈ $\mathbb{R}$ ^{[m, n₂]}，n₁, n₂≥2，n=n₁+n₂-k，k为整数且1≤k≤min(n₁, n₂)，t=m-k，令

$\begin{aligned} &c_{i_{1} i_{2} \cdots i_{m}}= \\ &\left\{\begin{array}{ll} a_{i_{1} i_{2} \cdots i_{m}}\ \ \ \ \left(\left(i_{1} i_{2} \cdots i_{m}\right) \in \varDelta^{S_{1} \cup S_{2}} / \varDelta^{S_{2}}\right), \\ a_{i_{1} i_{2} \cdots i_{m}}+b_{i_{1}-t, i_{2}-t, \cdots, i_{m}-t}\ \ \ \ \left(\left(i_{1} i_{2} \cdots i_{m}\right) \in \varDelta^{S_{2}}\right), \\ b_{i_{1}-t, i_{2}-t, \cdots i_{m}-t} \quad\left(\left(i_{1} i_{2} \cdots i_{m}\right) \in \varDelta^{S_{2} \cup S_{3}} / \varDelta^{S_{2}}\right), \\ 0 \quad\left(\left(i_{1} i_{2} \cdots i_{m}\right) \in\left(\varDelta^{[n]} / \varDelta^{S_{1} \cup S_{2}}\right) \cup\left(\varDelta^{[n]} / \varDelta^{S_{2} \cup S_{3}}\right)\right), \end{array}\right. \end{aligned}$

其中，S₁={1, 2, …, n₁-k}，S₂={n₁-k+1, …, n₁}，S₃={n₁+1, …, n}，则张量 $\mathcal{C}$ =(c_{i₁i₂…i_m})∈ $\mathbb{R}$ ^{[m, n]}称为 $\mathcal{A}$ 与 $\mathcal{B}$ 的k-子直和，记为 $\mathcal{C}$ = $\mathcal{A}$ ⊕_k $\mathcal{B}$ .

定义5 设张量 $\mathcal{A}$ =(a_{i₁i₂…i_m})∈ $\mathbb{R}$ ^{[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)|a_i…i|>r_i^{Δ^S}( $\mathcal{A}$ )，∀i∈S；

(ii) (|a_i…i|-r_i^{Δ^S}( $\mathcal{A}$ ))(|a_j…j|-r_j^{Δ^S}( $\mathcal{A}$ ))>r_i^{Δ^S} ( $\mathcal{A}$ )×r_j^{Δ^S}( $\mathcal{A}$ )，∀i∈S，∀j∈ S，

则称 $\mathcal{A}$ 是S-严格对角占优型张量，简记为S-SDD型张量.

2. SDD张量和S-SDD型张量的子直和

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 $\mathcal{A}$ =(a_{i₁i₂…i_m})∈ $\mathbb{R}$ ^{[m, n₁]}， $\mathcal{B}$ =(b_{i₁i₂…i_m})∈ $\mathbb{R}$ ^{[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a_ii…ib_jj…j>0，i∈S₂，j∈[k]，则 $\mathcal{A}$ 与 $\mathcal{B}$ 的k-子直和 $\mathcal{C}$ = $\mathcal{A}$ ⊕_k $\mathcal{B}$ 是SDD张量.

证明分3种情形证明.

情形1. 当i₁ ∈S₁时，有

$c_{i_{1} i_{2} \cdots i_{m}}=\left\{\begin{array}{lc} a_{i_{1} i_{2} \cdots i_{m}}\ \ \ \ \left(\left(i_{2} \cdots i_{m}\right) \in \varDelta^{S_{1} \cup S_{2}}\right), \\ 0\ \ \ \ \left(\left(i_{2} \cdots i_{m}\right) \notin \varDelta^{S_{1} \cup S_{2}}\right), \end{array}\right.$

则

$\begin{aligned} c_{i_{1} i_{1} \cdots i_{1}}&=a_{i_{1} i_{1} \cdots i_{1}}>\sum\limits_{\left(i_{2} \cdots i_{m}\right) \in \varDelta ^{S_{1} \cup S_{2}}}\left|a_{i_{1} i_{2} \cdots i_{m}}\right|-a_{i_{1} i_{1} \cdots i_{1}}= \\ &\sum\limits_{\left(i_{2} \cdots i_{m}\right) \in[n]}\left|c_{i_{1} i_{2} \cdots i_{m}}\right|-c_{i_{1} i_{1} \cdots i_{1}} . \end{aligned}$

情形2. 当i₁ ∈S₂时，因为 $\mathcal{A}$ 和 $\mathcal{B}$ 是SDD张量，所以

$a_{i_{1} i_{1} \cdots i_{1}}>\sum\limits_{\left(i_{2} \cdots i_{m}\right) \in \varDelta^{ S_{1} \cup S_{2}}}\left|a_{i_{1} i_{2} \cdots i_{m}}\right|-a_{i_{1} i_{1} \cdots i_{1}},$

(1)

$\begin{array}{l} b_{i_{1}-t, i_{1}-t, \cdots, i_{1}-t}>\sum\limits_{\left(i_{2} \cdots i_{m}\right) \in \varDelta^{ S_{2} \cup S_{3}}}\left|b_{i_{1}-t, i_{2}-t, \cdots, i_{m}-t}\right|- \\ \ \ \ \ b_{i_{1}-t, i_{1}-t, \cdots, i_{1}-t} \cdot \end{array}$

(2)

根据定义4，有

$c_{i_{1} i_{1} \cdots i_{1}}=a_{i_{1} i_{1} \cdots i_{1}}+b_{i_{1}-t, i_{1}-t, \cdots, i_{1}-t}.$

(3)

由式(1)~(3)，可知

$\begin{array}{l} c_{i_{1} i_{1} \cdots i_{1}}>\sum\limits_{\left(i_{2} \cdots i_{m}\right) \in \varDelta^{S_{1} \cup S_{2}}}\left|a_{i_{1} i_{2} \cdots i_{m}}\right|-a_{i_{1} i_{1} \cdots i_{1}}+ \\ \ \ \ \ \sum\limits_{\left(i_{2} \cdots i_{m}\right) \in \varDelta^{S_{2} \cup S_{3}}}\left|b_{i_{1}-t, i_{2}-t, \cdots, i_{m}-t}\right|-b_{i_{1}-t, i_{1}-t, \cdots, i_{1}-t}= \\ \ \ \ \ \sum\limits_{\left(i_{2} \cdots i_{m}\right) \in \varDelta^{S_{1} \cup S_{2}}}\left|a_{i_{1} i_{2} \cdots i_{m}}\right|+\sum\limits_{\left(i_{2} \cdots i_{m}\right) \in \varDelta^{S_{2} \cup S_{3}}}\left|b_{i_{1}-t, i_{2}-t, \cdots, i_{m}-t}\right|- \\ \ \ \ \ c_{i_{1} i_{1} \cdots i_{1}}=\sum\limits_{\left(i_{2} \cdots i_{m}\right) \in \varDelta^{[n]}}\left|c_{i_{1} i_{2} \cdots i_{m}}\right|-c_{i_{1} i_{1} \cdots i_{1}}. \end{array}$

情形3. 当i₁ ∈S₃时，有

$c_{i_{1} i_{2} \cdots i_{m}}=\left\{\begin{array}{l} b_{i_{1}-t, i_{2}-t, \cdots, i_{m}-t}\ \ \ \ \left(\left(i_{2} \cdots i_{m}\right) \in \varDelta^{S_{2} \cup S_{3}}\right), \\ 0\ \ \ \ \left(\left(i_{2} \cdots i_{m}\right) \notin \varDelta^{S_{2} \cup S_{3}}\right), \end{array}\right.$

则

$\begin{gathered} c_{i_{1} i_{1} \cdots i_{1}}=b_{i_{1}-t, i_{1}-t, \cdots, i_{1}-t}>\sum\limits_{\left(i_{2} \cdots i_{m}\right) \in \varDelta^{S_{1} \cup S_{2}}}\left|b_{i i_{2} \cdots i_{m}}\right|- \\ b_{i_{1}-t, i_{1}-t, \cdots, i_{1}-t}=\sum\limits_{\left(i_{2} \cdots i_{l}\right) \in[n]}\left|c_{i i_{2} \cdots i_{m}}\right|-c_{i_{1} i_{1} \cdots i_{1}}. \end{gathered}$

综上所述，当i₁ ∈S₁∪S₂∪S₃时，有

$c_{i_{1} i_{1} \cdots i_{1}}>\sum\limits_{\left(i_{2} \cdots i_{m}\right) \in[n]}\left|c_{i i_{2} \cdots i_{m}}\right|-c_{i_{1} i_{1} \cdots i_{1}},$

因此， $\mathcal{C}$ = $\mathcal{A}$ ⊕_k $\mathcal{B}$ 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 $\mathcal{A}$ = (a_ijk) ∈ $\mathbb{R}$ ^{[3, 4]}，其中

$\begin{array}{c} \mathcal{A}(:,:, 1)=\left(\begin{array}{cccc} 2.6 & 0.1 & 0.1 & 0.02 \\ 0.1 & 0.2 & 0.1 & 0.02 \\ 0.1 & 0.1 & 0.2 & 0.03 \\ 0.02 & 0.03 & 0.03 & 0.04 \end{array}\right), \\ \mathcal{A}(:,:, 2)=\left(\begin{array}{cccc} 0.4 & 0 & 0.1 & 0.1 \\ 0 & 2.6 & 0.1 & 0.1 \\ 0.1 & 0.1 & 0.1 & 0.1 \\ 0.1 & 0.1 & 0.1 & 0.1 \end{array}\right), \\ \mathcal{A}(:,:, 3)=\left(\begin{array}{cccc} 0.6 & 0.2 & 0 & 0.1 \\ 0.2 & 0.3 & 0 & 0.1 \\ 0 & 0 & 2.2 & 0.2 \\ 0.1 & 0.1 & 0.2 & 0.2 \end{array}\right), \\ \mathcal{A}(:,:, 4)=\left(\begin{array}{cccc} 0.8 & 0.2 & 0.1 & 0 \\ 0.2 & 0.3 & 0.1 & 0 \\ 0.1 & 0.1 & 0.1 & 0 \\ 0 & 0 & 0 & 2.2 \end{array}\right). \end{array}$

令S={1, 2}，则张量 $\mathcal{A}$ 是S-SDD型张量. 但是2-子直和 $\mathcal{C}$ = $\mathcal{A}$ ⊕₂ $\mathcal{A}$ 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

$\begin{array}{l} \left(\left|c_{111}\right|-r_{1}^{\varDelta^{S}}(\mathcal{C})\right)\left(\left|c_{555}\right|-r_{5}^{\varDelta^{\bar{S}}}(\mathcal{C})\right)=-0.22<0= \\ \ \ \ \ \ \ \ r_{1}^{\varDelta^{\bar{S}}}(\mathcal{C}) r_{5}^{\varDelta^{S}}(\mathcal{C}). \end{array}$

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 $\mathcal{A}$ =(a_{i₁i₂…i_m})∈ $\mathbb{R}$ ^{[m, n₁]}是S-SDD型张量，张量 $\mathcal{B}$ =(b_{i₁i₂…i_m})∈ $\mathbb{R}$ ^{[m, n₂]}是SDD张量，S是S₁的一个非空子集，n₁, n₂≥2，k为整数且1≤k≤min(m, n)，集合S₁、S₂、S₃与定义4中的相同. 如果a_ii…ib_jj…j>0，i∈S₂，j∈[k]，则k-子直和 $\mathcal{C}$ = $\mathcal{A}$ ⊕_k $\mathcal{B}$ 是S-SDD型张量.

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

(i) |c_i…i| ≥r_i^{Δ^S}( $\mathcal{C}$ )，∀i∈S；

(ii) (|c_i…i|-r_i^{Δ^S}( $\mathcal{C}$ ))(|c_j…j|-r_j^{Δ^S}( $\mathcal{C}$ ))≥r_i^{Δ^S}( $\mathcal{C}$ )r_j^{Δ^S}( $\mathcal{C}$ )，∀i∈S，∀j∈ S.

当S=S₁时，S在[n₁+n₂-k]中的补集为S₂∪S₃.

(1) 因为 $\mathcal{A}$ 是S-SDD型张量，有|a_i…i|>r_i^{Δ^S}( $\mathcal{A}$ ) (i∈S)，则

$\left|c_{i \cdots i}\right|=\left|a_{i \cdots i}\right|>r_{i}^{\varDelta^{S}}(\mathcal{A})=r_{i}^{\varDelta^{S}}(\mathcal{C}) \quad(i \in S) .$

从而条件(i)成立.

(2) 下面分2种情形证明条件(ii)成立.

① 当j∈S₂时，由i∈S=S₁，有

$\begin{gathered} r_{i}^{\varDelta^{S}}(\mathcal{C})=r_{i}^{\varDelta^{S}}(\mathcal{A}), r_{j}^{\varDelta^{S}}(\mathcal{C})=r_{j}^{\varDelta^{S}}(\mathcal{A}), r_{i}^{\varDelta^{\bar{S}}}(\mathcal{C})=r_{i}^{\varDelta^{\bar{S}}}(\mathcal{A}) ,\\ r_{j}^{\varDelta^{\bar{S}}}(\mathcal{C})=r_{j}^{\varDelta^{\bar{S}}}(\mathcal{A})+r_{j-t}(\mathcal{B}), t=n_{1}-k. \end{gathered}$

因为 $\mathcal{A}$ 是S-SDD型张量， $\mathcal{B}$ 是SDD张量，则∀i∈S，∀j∈ S，有

$\left(\left|a_{i \cdots i}\right|-r_{i}^{\varDelta^{S}}(\mathcal{A})\right)\left(\left|a_{j \cdots j}\right|-r_{j}^{\varDelta^{\bar{S}}}(\mathcal{A})\right)>r_{i}^{\varDelta^{\bar{S}}}(\mathcal{A}) r_{j}^{\varDelta^{S}}(\mathcal{A}),$

且|b_i…i|>r_i( $\mathcal{B}$ ) (i∈[n₂]). 从而

$\begin{array}{l} \left(\left|c_{i \cdots i}\right|-r_{i}^{\varDelta^{S}}(\mathcal{C})\right)\left(\left|c_{j \cdots j}\right|-r_{j}^{\varDelta^{\bar{S}}}(\mathcal{C})\right)=\left(\left|a_{i \cdots i}\right|-r_{i}^{\varDelta^{S}}(\mathcal{A})\right) \times \\ \ \ \ \ \left(\left|a_{j \cdots j}+b_{j-t, \cdots, j-t}\right|-r_{j}^{\varDelta^{\bar{S}}}(\mathcal{A})-r_{j-t}(\mathcal{B})\right)= \\ \ \ \ \ \left(\left|a_{i \cdots i}\right|-r_{i}^{\varDelta^S}(\mathcal{A})\right)\left(\left(\left|a_{j \cdots j}\right|-r_{j}^{\varDelta^{\bar{S}}}(\mathcal{A})\right)+\right. \\ \ \ \ \ \left.\left(\left|b_{j-t, \cdots, j-t}\right|-r_{j-t}(\mathcal{B})\right)\right)>\left(\left|a_{i \cdots i}\right|-r_{i}^{\varDelta^{S}}(\mathcal{A})\right) \times \\ \ \ \ \ \left(\left|a_{j \cdots j}\right|-r_{j}^{\varDelta^{\bar{S}}}(\mathcal{A})\right)>r_{i}^{\varDelta^{\bar{S}}}(\mathcal{A}) r_{j}^{\varDelta^{S}}(\mathcal{A})=r_{i}^{\varDelta^{\bar{S}}}(\mathcal{C}) r_{j}^{\varDelta^{S}}(\mathcal{C}). \end{array}$

由此可知当j∈S₂时，条件(ii)成立.

② 当j∈S₃时，由i ∈S=S₁，有r_j^{Δ^S}( $\mathcal{C}$ )=0，r_j^{Δ^S}( $\mathcal{C}$ )=r_j-t^{Δ^S}( $\mathcal{B}$ )，从而

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， $\mathcal{C}$ 是S-SDD型张量.

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