3阶对角占优张量的子直和

何建锋

doi:10.13718/j.cnki.xsxb.2020.10.001

3阶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，云南楚雄 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

作者简介:
何建锋(1974-)，男，副教授，主要从事矩阵理论的研究 .

中图分类号: O151.21

On Subdirect Sums of Third-Order Diagonally Dominant Tensors

Jian-feng HE

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

摘要: 利用矩阵与张量之间的联系，将矩阵子直和的概念推广到张量上，定义了张量的子直和.讨论3阶严格对角占优张量的k-子直和仍然是严格对角占优张量，给出3阶S-严格对角占优型张量的k-子直和为S-严格对角占优型张量的条件，并举例说明.
- 张量 /
- 子直和 /
- 严格对角占优张量 /
- 矩阵
Abstract: According to the relation between the matrices and tensors, the subdirect sums of tensors has been introduced, which is the generalization of matrices case. Some conditions have been given to guarantee that the k-subdirect sums of third-order strictly diagonally dominant tensors(SDD) is also SDD. The same situation has been analyzed for S-SDD tensors, and examples been given to illustrate the theoretical results presented.
- tensors /
- subdirect sums /
- strictly diagonally dominant tensors /
- matrix .

方阵的k-子直和是矩阵和的推广，在许多应用中均有出现，例如矩阵完全化问题、区域分解方法中的重叠子域、有限元中的整体刚度矩阵等^[1-3].文献[4]引入了方阵k-子直和的概念，并对其进行了相关研究，证明了正定矩阵的子直和是正定矩阵，对称M-矩阵的子直和是对称M-矩阵等结论.关于方阵k-子直和的一些新的研究成果可参考文献[5-11].

张量是矩阵的高阶推广，在许多科学领域，如信号图像处理^[12]、非线性优化^[13]、高阶统计学^[14]、物理学中的弹性分析^[15-16]和数据挖掘与处理等领域都有重要的应用.有关张量的研究成果可参考文献[17-21]，关于张量的更多文献不在此逐一赘述.

令A=(a_{i₁…i_m})，其中a_{i₁…i_m}∈ $\mathbb{C}$ ( $\mathbb{R}$ )，i_j=1，…，n，j=1，…，m，则称A为一个m阶n维的复(实)张量，记作A∈C^[m，n](R^[m，n]).

鉴于矩阵与张量之间的关系，本文将矩阵子直和的概念推广到张量上，提出张量子直和的概念，并讨论SDD张量、S-SDD-型张量子直和的性质.

为方便讨论，引入以下符号：

${[n] = \{ 1,2, \cdots ,n\} \quad {S_1} = [m - k] = \{ 1,2, \cdots ,m - k\} \quad t = m - k}$

${{S_2} = [m]\backslash {S_1} = \{ m - k + 1, \cdots ,m\} \quad {S_3} = [m + n - k]\backslash [m] = \{ m + 1, \cdots ,m + n - k\} }$

${\Delta ^S} = \{ ({i_2}{i_3} \cdots {i_m}):{i_j} \in S,j = 2, \cdots ,m\} \quad r_i^{{\Delta ^S}}(\mathit{\boldsymbol{A}}) = \sum\limits_{\begin{smallmatrix} ({{i}_{2}}\cdots {{i}_{m}})\in {{\Delta }^{S}} \\ {{\delta }_{i{{i}_{2}}}}\cdots {{i}_{m}}=0 \end{smallmatrix}}{|}{{a}_{i{{i}_{2}}\cdots {{i}_{m}}}}|$

${\delta _{{i_1} \ldots {i_m}}} = \left\{ {\begin{array}{*{20}{l}} 1&{{i_1} = \cdots = {i_m}}\\ 0&{{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\rm{否则}}} \end{array}} \right.$

1. 张量子直和的概念

本节给出张量子直和及相关概念的定义.

定义1^[17] 设张量A=(a_{i₁i₂…i_m})∈R^[m，n]，且满足

${a_{ii}} \ldots i \ge \sum\limits_{{i_2}, \cdots ,{i_m} \in [n]} | {a_{i{i_2} \cdots {i_m}}}| - {a_{ii \cdots i}}\quad i \in [n]$

则称A为对角占优张量.

若对每个i∈[n]，有

${a_{ii \cdots i}} \gt \sum\limits_{{i_2}, \cdots ,{i_m} \in [n]} | {a_{i{i_2} \cdots {i_m}}}| - {a_{ii}}{ \ldots _i}$

则称A为严格对角占优张量，简记为SDD张量.

定义2 设张量A=(a_{i₁…i_m})∈R^[m，n]，n≥2，S是[n]的一个真子集，张量A满足以下两个条件：

(a) |a_i…i|≥(>)r_i^{Δ^S}(A)，∀i∈S；

(b) (|a_i…i|－r_i^{Δ^S}(A))(|a_j…j|－r_j^{Δ^S}(A))≥(>)r_i^{Δ^S}(A)r_j^{Δ^S}(A)，∀i∈S，∀j∈S.

则称张量A为S-对角占优型张量(S-严格对角占优型张量)，简记为S-SDD₀-型张量(S-SDD-型张量).

定义3 设张量A=(a_i₁i₂i₃)∈R^[3，m]，B=(b_i₁i₂i₃)∈R^[3，n]，1≤k≤min{m，n}，称张量

$\mathit{\boldsymbol{C}} = ({c_{{i_1}{i_2}{i_3}}}) \in {R^{[3,m + n - k]}}$

为张量A与张量B的k-阶子直和，记为A $\oplus$ _kB，其中

${c_{{i_1}{i_2}{i_3}}} = \left\{ {\begin{array}{*{20}{l}} {{a_{{i_1}{i_2}{i_3}}}}&{{i_1},{i_2} \in {S_1} \cup {S_2},{i_3} \in {S_1}}\\ {{a_{{i_1}{i_2}{i_3}}}}&{{i_1} \in {S_1},{i_2} \in {S_1} \cup {S_2},{i_3} \in {S_2}}\\ {{a_{{i_1}{i_2}{i_3}}}}&{{i_1} \in {S_2},{i_2} \in {S_1},{i_3} \in {S_2}}\\ {{a_{{i_1}{i_2}{i_3}}} + {b_{({i_1} - t)({i_2} - t)({i_3} - t)}}}&{{i_1},{i_2} \in {S_2},{i_3} \in {S_2}}\\ {{b_{({i_1} - t)({i_2} - t)({i_3} - t)}}}&{{i_1} \in {S_2},{i_2} \in {S_3},{i_3} \in {S_2}}\\ {{b_{({i_1} - t)({i_2} - t)({i_3} - t)}}}&{{i_1} \in {S_3},{i_2} \in {S_2} \cup {S_3},{i_3} \in {S_2}}\\ {{b_{({i_1} - t)({i_2} - t)({i_3} - t)}}}&{{i_1},{i_2} \in {S_2} \cup {S_3},{i_3} \in {S_3}}\\ 0&{{\rm{否则}}} \end{array}} \right.$

例1 设张量A=[A(1，：，：)，A(2，：，：)，A(3，：，：)]，B=[B(1，：，：)，B(2，：，：)，B(3，：，：)]∈R^{[3, 3]}，其中

$A(1{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ;{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ) = \left[ {\begin{array}{*{20}{r}} 1&0&2\\ { - 1}&1&1\\ { - 2}&1&0 \end{array}} \right] \\ A(2{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ) = \left[ {\begin{array}{*{20}{r}} 2&1&{ - 1}\\ 0&1&0\\ 1&0&1 \end{array}} \right]{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} A(3{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ) = \left[ {\begin{array}{*{20}{r}} 1&{ - 1}&1\\ 3&1&{ - 2}\\ 2&2&1 \end{array}} \right]$

$B(1{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ) = \left[ {\begin{array}{*{20}{l}} 3&1&4\\ 0&0&1\\ 2&1&2 \end{array}} \right]\\ B(2{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ) = \left[ {\begin{array}{*{20}{r}} 2&0&{ - 1}\\ 1&5&{ - 1}\\ { - 1}&0&3 \end{array}} \right]{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} B(3{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ) = \left[ {\begin{array}{*{20}{r}} 1&0&1\\ 4&4&{ - 1}\\ { - 3}&{ - 2}&3 \end{array}} \right]$

则张量A与张量B的2-阶子直和为

$\mathit{\boldsymbol{C}} = \mathit{\boldsymbol{A}}{ \oplus _2}\mathit{\boldsymbol{B}} = \\ [C(1{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ),C(2{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ),C(3{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ),C(4{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} )] \in {R^{[3,4]}}$

其中

$C(1{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ) = \left( {\begin{array}{*{20}{r}} 1&0&2&0\\ { - 1}&1&1&0\\ { - 2}&1&0&0\\ 0&0&0&0 \end{array}} \right){\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} C(2{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ) = \left( {\begin{array}{*{20}{l}} 2&1&{ - 1}&0\\ 0&4&1&4\\ 1&0&1&1\\ 0&2&1&2 \end{array}} \right)$

$C(3{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ) = \left( {\begin{array}{*{20}{r}} 1&{ - 1}&1&0\\ 3&3&{ - 2}&{ - 1}\\ 2&3&6&{ - 1}\\ 0&{ - 1}&0&3 \end{array}} \right){\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} C(4{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ) = \left( {\begin{array}{*{20}{r}} 0&0&0&0\\ 0&{ - 1}&0&1\\ 0&4&4&{ - 1}\\ 0&{ - 3}&{ - 2}&3 \end{array}} \right)$

2. 对角占优张量的子直和

本节我们讨论(严格)对角占优张量(S-(严格)对角占优型张量)的子直和仍然为(严格)对角占优张量(S-(严格)对角占优型张量)的条件.

定理1 设张量A=(a_i₁i₂i₃)∈R^[3，m]，B=(b_i₁i₂i₃)∈R^[3，n]均为对角占优张量，1≤k≤min{m，n}，且a_iiib_jjj>0，∀i∈S₂，∀j∈[k]，则张量A与张量B的k-阶子直和C=A $\oplus$ _kB也是对角占优张量.

证当i₁∈S₁时，

${c_{{i_1}{i_2}{i_3}}} = \left\{ {\begin{array}{*{20}{l}} {{a_{{i_1}{i_2}{i_3}}}}&{{i_2},{i_3} \in {S_1} \cup {S_2}}\\ 0&{{\rm{否则}}} \end{array}} \right.$

故有

${i_{{i_1}{i_1}{i_1}}} = {a_{{i_1}{i_1}{i_1}}} \ge \sum\limits_{{i_2},{i_3} \in {S_1} \cup {S_2}} | {a_{{i_1}{i_2}{i_3}}}| - {a_{{i_1}{i_1}{i_1}}} = \sum\limits_{{i_2},{i_3} \in {S_1} \cup {S_2} \cup {S_3}} | {c_{{i_1}{i_2}{i_3}}}| - {c_{{i_1}{i_1}{i_1}}}$

当i₁∈S₂时，

$\begin{array}{*{20}{l}} {{c_{{i_1}{i_2}{i_3}}} = \left\{ {\begin{array}{*{20}{l}} {{a_{{i_1}{i_2}{i_3}}}}&{{i_2} \in {S_1} \cup {S_2},{i_3} \in {S_1}}\\ {{a_{{i_1}{i_2}{i_3}}}}&{{i_2} \in {S_1},{i_3} \in {S_2}}\\ {{a_{{i_1}{i_2}{i_3}}} + {b_{({i_1} - t)({i_2} - t)({i_3} - t)}}}&{{i_2},{i_3} \in {S_2}}\\ {{b_{({i_1} - t)({i_2} - t)({i_3} - t)}}}&{{i_2} \in {S_3},{i_3} \in {S_2}}\\ {{b_{({i_1} - t)({i_2} - t)({i_3} - t)}}}&{{i_2} \in {S_2} \cup {S_3},{i_3} \in {S_3}}\\ 0&{{\rm{否则}}} \end{array}} \right.} \end{array}$

故有

$\begin{array}{l} {c_{{i_1}{i_1}{i_1}}} = {a_{{i_1}{i_1}{i_1}}} + {b_{{i_1}{i_1}{i_1}}} \ge (\sum\limits_{{i_2},{i_3} \in {S_1} \cup {S_2}} | {a_{{i_1}{i_2}{i_3}}}| - {a_{{i_1}{i_1}{i_1}}}) + (\sum\limits_{{i_2},{i_3} \in {S_1}} | {b_{({i_1} - t){i_2}{i_3}}}| - {b_{({i_1} - t)({i_1} - t)({i_1} - t)}}) = \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} (\sum\limits_{{i_2},{i_3} \in {S_1} \cup {S_2}} | {a_{{i_1}{i_2}{i_3}}}| + \sum\limits_{{i_2},{i_3} \in {S_1}} | {b_{({i_1} - t){i_2}{i_3}}}|) - ({a_{{i_1}{i_1}{i_1}}} + {b_{({i_1} - t)({i_1} - t)({i_1} - t)}}) = \sum\limits_{{i_2},{i_3} \in {S_2}} | {c_{{i_1}{i_2}{i_3}}}| - {c_{{i_1}{i_1}{i_1}}} \end{array}$

当i₁∈S₃时，

${c_{{i_1}{i_2}{i_3}}} = \left\{ {\begin{array}{*{20}{l}} {{b_{({i_1} - t)({i_2} - t)({i_3} - t)}}}&{{i_2},{i_3} \in {S_2} \cup {S_3}}\\ 0&{{\rm{否则}}} \end{array}} \right.$

故有

${c_{{i_1}{i_1}{i_1}}} = {b_{({i_1} - t)({i_1} - t)({i_1} - t)}} \ge \sum\limits_{{i_2},{i_3} \in [n]\backslash [k]} {|{b_{({i_1} - t){i_2}{i_3}}}|} - {b_{({i_1} - t)({i_1} - t)({i_1} - t)}} = \sum\limits_{{i_2},{i_3} \in {S_1} \cup {S_2} \cup {S_3}} {|{c_{{i_1}{i_2}{i_3}}}| - {c_{{i_1}{i_1}{i_1}}}}$

综合以上3种情形可知，结论成立.

例2 设张量A=[A(1，：，：)，A(2，：，：)，A(3，：，：)]，B=[B(1，：，：)，B(2，：，：)，B(3，：，：)]，其中

$A(1{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ) = \left[ {\begin{array}{*{20}{c}} {10}&{ - 1}&1\\ 2&0&1\\ 3&1&0 \end{array}} \right]{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} A(2{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ) = \left[ {\begin{array}{*{20}{c}} 2&1&2\\ 1&{11}&1\\ 0&2&1 \end{array}} \right]\\ A(3{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ) = \left[ {\begin{array}{*{20}{l}} 2&2&1\\ 1&2&2\\ 1&1&{12} \end{array}} \right]$

$B(1{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ) = \left[ {\begin{array}{*{20}{c}} {15}&3&0\\ 1&3&3\\ 0&3&1 \end{array}} \right]{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} B(2{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ) = \left[ {\begin{array}{*{20}{r}} 1&4&{ - 2}\\ 0&{13}&1\\ 2&0&2 \end{array}} \right]\\ B(3{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ) = \left[ {\begin{array}{*{20}{l}} 0&1&0\\ 1&0&1\\ 2&2&8 \end{array}} \right]$

易知张量A，B均为对角占优张量，而张量A与张量B的2-阶子直和为

$\mathit{\boldsymbol{C}} = \mathit{\boldsymbol{A}}{ \oplus _2}\mathit{\boldsymbol{B}} =\\ [C(1{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ),C(2{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ),C(3{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ),C(4{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} )]$

其中

$C(1{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ) = \left( {\begin{array}{*{20}{c}} {10}&{ - 1}&1&0\\ 2&0&1&0\\ 3&1&0&0\\ 0&0&0&0 \end{array}} \right){\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} C(2{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ) = \left( {\begin{array}{*{20}{c}} 2&1&2&0\\ 1&{26}&4&0\\ 0&5&4&3\\ 0&0&3&1 \end{array}} \right)$

$C(3{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ) = \left( {\begin{array}{*{20}{c}} 2&1&1&0\\ 2&3&5&{ - 2}\\ 1&2&{25}&1\\ 0&2&0&2 \end{array}} \right){\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} C(4{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ) = \left( {\begin{array}{*{20}{l}} 0&0&0&0\\ 0&0&1&0\\ 0&1&0&1\\ 0&2&2&8 \end{array}} \right)$

计算知

${c_{111}} = 10 \ge 9 = \sum\limits_{{i_2},{i_3} \in [4]} | {c_{1{i_2}{i_3}}}| - {c_{111}}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {c_{222}} = 26 \ge 26 = \sum\limits_{{i_2},{i_3} \in [4]} | {c_{2{i_2}{i_3}}}| - {c_{222}}$

${c_{333}} = 25 \ge 24 = \sum\limits_{{i_2},{i_3} \in [4]} | {c_{3{i_2}{i_3}}}| - {c_{333}}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {c_{444}} = 8 \ge 7 = \sum\limits_{{i_2},{i_3} \in [4]} | {c_{4{i_2}{i_3}}}| - {c_{444}}$

故张量C为对角占优张量.

类似定理1的证明可得如下结论：

定理2 看设张量A=(a_i₁i₂i₃)∈R^[3，m]，B=(b_i₁i₂i₃)∈R^[3，n]均为严格对角占优张量，1≤k≤min{m，n}，且a_iiib_jjj>0，∀i∈S₂，∀j∈[k]，则张量A与张量B的k-阶子直和也是严格对角占优张量.

下面讨论S-SDD-型张量的子直和.先看一个例子.

例3 设张量A=(a_i₁i₂i₃)∈R^{[3, 4]}，其中

$A(3{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ) = \left( {\begin{array}{*{20}{c}} {0.6}&{0.2}&0&{0.1}\\ {0.2}&{0.3}&0&{0.1}\\ 0&0&{2.2}&{0.2}\\ {0.1}&{0.1}&{0.2}&{0.2} \end{array}} \right)\\ A(4{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ) = \left( {\begin{array}{*{20}{c}} {0.8}&{0.2}&{0.1}&0\\ {0.2}&{0.3}&{0.1}&0\\ {0.1}&{0.1}&{0.1}&0\\ 0&0&0&{2.2} \end{array}} \right)$

取S={1，2}，由于

${|{a_{111}}| = 2.6 \gt 0.4 = r_1^{{\Delta ^S}}(\mathit{\boldsymbol{A}})\quad |{a_{222}}| = 2.6 \gt 0.4 = r_2^{{\Delta ^S}}(\mathit{\boldsymbol{A}})}$

${(|{a_{111}}| - r_1^{{\Delta ^S}}(\mathit{\boldsymbol{A}}))(|{a_{333}}| - r_3^{{\Delta ^{\bar S}}}(\mathit{\boldsymbol{A}})) = 3.25 \gt 0.39 = r_1^{{\Delta ^{\bar S}}}(\mathit{\boldsymbol{A}})r_3^{{\Delta ^S}}(\mathit{\boldsymbol{A}})}$

${|{a_{111}}| - r_1^{{\Delta ^S}}(\mathit{\boldsymbol{A}}))(|{a_{444}}| - r_4^{{\Delta ^{\bar S}}}(\mathit{\boldsymbol{A}})) = 4.62 \gt 0.45 = r_1^{{\Delta ^{\bar S}}}(\mathit{\boldsymbol{A}})r_4^{{\Delta ^S}}(\mathit{\boldsymbol{A}})}$

${|{a_{222}}| - r_2^{{\Delta ^S}}(\mathit{\boldsymbol{A}}))(|{a_{333}}| - r_3^{{\Delta ^S}}(\mathit{\boldsymbol{A}})) = 3.52 \gt 0.52 = r_2^{{\Delta ^{\bar S}}}(\mathit{\boldsymbol{A}})r_3^{{\Delta ^S}}(\mathit{\boldsymbol{A}})}$

${|{a_{222}}| - r_2^{{\Delta ^S}}(\mathit{\boldsymbol{A}}))(|{a_{444}}| - r_4^{{\Delta ^{\bar S}}}(\mathit{\boldsymbol{A}})) = 4.62 \gt 0.46 = r_2^{{\Delta ^{\bar S}}}(\mathit{\boldsymbol{A}})r_4^{{\Delta ^S}}(\mathit{\boldsymbol{A}})}$

即A为{1，2}-SDD-型张量.但是对于张量C=A $\oplus$ ₂A，其中

$C(1{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ) = \left( {\begin{array}{*{20}{c}} {2.6}&{0.1}&{0.1}&{0.02}&0&0\\ {0.1}&{0.2}&{0.1}&{0.02}&0&0\\ {0.1}&{0.1}&{0.2}&{0.03}&0&0\\ {0.02}&{0.03}&{0.03}&{0.04}&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0 \end{array}} \right)\\ C(2{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ) = \left( {\begin{array}{*{20}{c}} {0.4}&0&{0.1}&{0.1}&0&0\\ 0&{2.6}&{0.1}&{0.1}&0&0\\ {0.1}&{0.1}&{0.1}&{0.1}&0&0\\ {0.1}&{0.1}&{0.1}&{0.1}&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0 \end{array}} \right)$

$C(3{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ) = \left( {\begin{array}{*{20}{c}} {0.6}&{0.2}&0&{0.1}&0&0\\ {0.2}&{0.3}&0&{0.1}&0&0\\ 0&0&{4.8}&{0.3}&{0.1}&{0.02}\\ {0.1}&{0.1}&{0.3}&{0.4}&{0.1}&{0.02}\\ 0&0&{0.1}&{0.1}&{0.2}&{0.03}\\ 0&0&{0.02}&{0.03}&{0.03}&{0.04} \end{array}} \right)\\ C(4{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ) = \left( {\begin{array}{*{20}{c}} {0.8}&{0.2}&{0.1}&0&0&0\\ {0.2}&{0.3}&{0.1}&0&0&0\\ {0.1}&{0.1}&{0.5}&0&{0.1}&{0.1}\\ 0&0&0&{4.8}&{0.1}&{0.1}\\ 0&0&{0.1}&{0.1}&{0.1}&{0.1}\\ 0&0&{0.1}&{0.1}&{0.1}&{0.1} \end{array}} \right)$

$C(5{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ) = \left( {\begin{array}{*{20}{c}} 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&{0.6}&{0.2}&0&{0.1}\\ 0&0&{0.2}&{0.3}&0&{0.1}\\ 0&0&0&0&{2.2}&{0.2}\\ 0&0&{0.1}&{0.1}&{0.2}&{0.2} \end{array}} \right)\\ C(6{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ) = \left( {\begin{array}{*{20}{c}} 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&{0.8}&{0.2}&{0.1}&0\\ 0&0&{0.2}&{0.3}&{0.1}&0\\ 0&0&{0.1}&{0.1}&{0.1}&0\\ 0&0&0&0&0&{2.2} \end{array}} \right)$

当i=1，j=5时，有

$(|{c_{111}}| - r_1^{{\Delta ^S}}(\mathit{\boldsymbol{C}}))(|{c_{555}}| - r_5^{{\Delta ^{\bar S}}}(\mathit{\boldsymbol{C}})) = - 0.22 \lt 0 = r_1^{{\Delta ^{\bar S}}}(\mathit{\boldsymbol{C}})r_5^{{\Delta ^S}}(\mathit{\boldsymbol{C}})$

即张量C不是{1，2}-SDD-型张量.

例3表明，S-SDD-型张量的子直和不一定是S-SDD-型张量.因此，寻找S-SDD-型张量的子直和仍然为S-SDD-型张量的条件是有意义的.

定理3 设张量A=(a_i₁i₂i₃)∈R^[3，m]为S-SDD-型张量，S为S₁的子集，B=(b_i₁i₂i₃)∈R^[3，n]为SDD张量，1≤k≤min{m，n}，a_iii>0，∀i∈S₂，b_jjj>0，j∈[k]，则张量A与张量B的k-阶子直和C=A $\oplus$ _kB也是S-SDD-型张量.

证先证明S=S₁时的情形.

要证明C是S-SDD-型张量，需证明：

(i) |c_iii|>r_i^{Δ^S}(C)，∀i∈S；

(ii) (|c_iii|－r_i^{Δ^S}(C))(|c_jjj|－r_j^{Δ^S}(C))>r_i^{Δ^S}(C)r_j^{Δ^S}(C)，∀i∈S，∀j∈S.

此时，由于S=S₁，则S=S₂∪S₃.下面逐一证明这两个条件成立.

(i) 由张量A为S-SDD-型张量知|a_iii|>r_i^{Δ^S}(A)，∀i∈S₁，从而有

$|{c_{iii}}| = |{a_{iii}}| \gt r_i^{{\Delta ^S}}(\mathit{\boldsymbol{A}}) = r_i^{{\Delta ^S}}(\mathit{\boldsymbol{C}})\quad \forall i \in {S_1}$

(ii) 当j∈S₂时，由i∈S=S₁，有

${\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} r_i^{{\Delta ^S}}(\mathit{\boldsymbol{C}}) = r_i^{{\Delta ^S}}(\mathit{\boldsymbol{A}})\quad r_j^{{\Delta ^S}}(\mathit{\boldsymbol{C}}) = r_j^{{\Delta ^S}}(\mathit{\boldsymbol{A}})$

$r_j^{{\Delta ^{\bar S}}}(\mathit{\boldsymbol{C}}) = r_j^{{\Delta ^{\bar S}}}(\mathit{\boldsymbol{A}}) + {r_{j - t}}(\mathit{\boldsymbol{B}})\quad r_i^{{\Delta ^{\bar S}}}(\mathit{\boldsymbol{C}}) = r_i^{{\Delta ^{\bar S}}}(\mathit{\boldsymbol{A}})$

由张量A为S-SDD-型张量得

$(|{a_{iii}}| - r_i^{{\Delta ^S}}(\mathit{\boldsymbol{A}}))(|{a_{jji}}| - r_j^{{\Delta ^{\bar S}}}(\mathit{\boldsymbol{A}})) \gt r_i^{{\Delta ^{\bar S}}}(\mathit{\boldsymbol{A}})r_j^{{\Delta ^S}}(\mathit{\boldsymbol{A}})\quad \forall i \in S,\quad \forall j \in \bar S$

又因B为SDD张量，故|b_iii|>r_i(B)，∀i∈[n].从而有

$\begin{array}{l} \begin{array}{*{20}{l}} {(|{c_{iii}}| - r_i^{{\Delta ^S}}(\mathit{\boldsymbol{C}}))(|{c_{jjj}}| - r_j^{{\Delta ^{\bar S}}}(\mathit{\boldsymbol{C}})) = }\\ {(|{a_{iii}}| - r_i^{{\Delta ^S}}(\mathit{\boldsymbol{A}}))(|{a_{jjj}} + {b_{(j - t)(j - t)(j - t)}}| - r_j^{{\Delta ^{\bar S}}}(\mathit{\boldsymbol{A}}) - {r_{j - t}}(\mathit{\boldsymbol{B}})) = } \end{array}\\ \begin{array}{*{20}{l}} {(|{a_{iii}}| - r_i^{{\Delta ^S}}(\mathit{\boldsymbol{A}}))((|{a_{jjj}}| - r_j^{{\Delta ^{\bar S}}}(\mathit{\boldsymbol{A}})) + (|{b_{(j - t)(j - t)(j - t)}}| - {r_{j - t}}(\mathit{\boldsymbol{B}}))) \gt }\\ {(|{a_{iii}}| - r_i^{{\Delta ^S}}(\mathit{\boldsymbol{A}}))(|{a_{jji}}| - r_j^{{\Delta ^{\bar S}}}(\mathit{\boldsymbol{A}})) \gt } \end{array}\\ r_i^{{\Delta ^{\bar S}}}(\mathit{\boldsymbol{A}})r_j^{{\Delta ^S}}(\mathit{\boldsymbol{A}}) = r_i^{{\Delta ^{\bar S}}}(\mathit{\boldsymbol{C}})r_j^{{\Delta ^S}}(\mathit{\boldsymbol{C}}) \end{array}$

当j∈S₃时，由i∈S=S₁知r_j^{Δ^S}(C)=0，r_j^{Δ^S}(C)=r_j－t^{Δ^S}(B)，故

$\begin{array}{*{20}{c}} {(|{c_{iii}}| - r_i^{{\Delta ^S}}(\mathit{\boldsymbol{C}}))(|{c_{jjj}}| - r_j^{{\Delta ^{\bar S}}}(\mathit{\boldsymbol{C}})) = (|{a_{iii}}| - r_i^{{\Delta ^S}}(\mathit{\boldsymbol{A}}))(|{b_{(j - t)(j - t)(j - t)}}| - r_j^{{\Delta ^{\bar S}}}(\mathit{\boldsymbol{B}})) \gt }\\ {0 = r_i^{{\Delta ^{\bar S}}}(\mathit{\boldsymbol{C}})r_j^{{\Delta ^S}}(\mathit{\boldsymbol{C}})} \end{array}$

综上所述，可知结论成立.

当|S| < |S₁|时，可类似证明结论成立.

例4 取例3中的S-SDD-型张量A，例2中的SDD张量B，S={1，2}，则C=A $\oplus$ ₂B=[C(1，：，：)，C(2，：，：)，C(3，：，：)，C(4，：，：)，C(5，：，：)]，其中

$C(1{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ) = \left[ {\begin{array}{*{20}{c}} {2.6}&{0.1}&{0.1}&{0.02}&0\\ {0.1}&{0.2}&{0.1}&{0.02}&0\\ {0.1}&{0.1}&{0.2}&{0.03}&0\\ {0.02}&{0.03}&{0.03}&{0.04}&0\\ 0&0&0&0&0 \end{array}} \right]\\ C(2{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ) = \left[ {\begin{array}{*{20}{c}} {0.4}&0&{0.1}&{0.1}&0\\ 0&{2.6}&{0.1}&{0.1}&0\\ {0.1}&{0.1}&{0.1}&{0.1}&0\\ {0.1}&{0.1}&{0.1}&{0.1}&0\\ 0&0&0&0&0 \end{array}} \right]$

$C(3{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ) = \left[ {\begin{array}{*{20}{c}} {0.6}&{0.2}&0&{0.1}&0\\ {0.2}&{0.3}&0&{0.1}&0\\ 0&0&{17.2}&{3.2}&1\\ {0.1}&{0.1}&{1.2}&{3.2}&3\\ 0&0&0&3&0 \end{array}} \right]\\ C(4{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} :{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ) = \left[ {\begin{array}{*{20}{c}} {0.8}&{0.2}&{0.1}&0&0\\ {0.2}&{0.3}&{0.1}&0&0\\ {0.1}&{0.1}&{1.1}&0&{ - 2}\\ 0&0&0&{15.2}&1\\ 0&0&2&4&2 \end{array}} \right]$

由于

$|{c_{111}}| = 2.6 \gt 0.4 = r_1^{{\Delta ^S}}(\mathit{\boldsymbol{C}}){\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} |{c_{222}}| = 2.6 \gt 0.4 = r_2^{{\Delta ^S}}(\mathit{\boldsymbol{C}})$

${(|{c_{111}}| - r_1^{{\Delta ^S}}(\mathit{\boldsymbol{C}}))(|{c_{333}}| - r_3^{{\Delta ^{\bar S}}}(\mathit{\boldsymbol{C}})) = 5.72 \gt 0.39 = r_1^{{\Delta ^{\bar S}}}(\mathit{\boldsymbol{C}})r_3^{{\Delta ^S}}(\mathit{\boldsymbol{C}})}$

${(|{c_{111}}| - r_1^{{\Delta ^S}}(\mathit{\boldsymbol{C}}))(|{c_{444}}| - r_4^{{\Delta ^{\bar S}}}(\mathit{\boldsymbol{C}})) = 6.82 \gt 0.45 = r_1^{{\Delta ^{\bar S}}}(\mathit{\boldsymbol{C}})r_4^{{\Delta ^S}}(\mathit{\boldsymbol{C}})}$

${(|{c_{111}}| - r_1^{{\Delta ^S}}(\mathit{\boldsymbol{C}}))(|{c_{555}}| - r_5^{{\Delta ^{\bar S}}}(\mathit{\boldsymbol{C}})) = 2.2 \gt 0 = r_1^{{\Delta ^{\bar S}}}(\mathit{\boldsymbol{C}})r_5^{{\Delta ^S}}(\mathit{\boldsymbol{C}})}$

${(|{c_{222}}| - r_2^{{\Delta ^S}}(\mathit{\boldsymbol{C}}))(|{c_{333}}| - r_3^{{\Delta ^{\bar S}}}(\mathit{\boldsymbol{C}})) = 5.72 \gt 0.52 = r_2^{{\Delta ^{\bar S}}}(\mathit{\boldsymbol{C}})r_3^{{\Delta ^S}}(\mathit{\boldsymbol{C}})}$

${(|{c_{222}}| - r_2^{{\Delta ^S}}(\mathit{\boldsymbol{C}}))(|{c_{444}}| - r_4^{{\Delta ^{\bar S}}}(\mathit{\boldsymbol{C}})) = 6.82 \gt 0.6 = r_2^{{\Delta ^{\bar S}}}(\mathit{\boldsymbol{C}})r_4^{{\Delta ^S}}(\mathit{\boldsymbol{C}})}$

${(|{c_{222}}| - r_2^{{\Delta ^S}}(\mathit{\boldsymbol{C}}))(|{c_{555}}| - r_5^{{\Delta ^{\bar S}}}(\mathit{\boldsymbol{C}})) = 2.2 \gt 0 = r_2^{{\Delta ^{\bar S}}}(\mathit{\boldsymbol{C}})r_5^{{\Delta ^S}}(\mathit{\boldsymbol{C}})}$

所以，张量C为S-SDD-型张量.

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3阶对角占优张量的子直和

楚雄师范学院数学与统计学院，云南楚雄 675000

作者简介:
何建锋(1974-)，男，副教授，主要从事矩阵理论的研究 .

On Subdirect Sums of Third-Order Diagonally Dominant Tensors

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

1. 张量子直和的概念

2. 对角占优张量的子直和

计量

3阶对角占优张量的子直和

作者简介: 何建锋(1974-)，男，副教授，主要从事矩阵理论的研究
楚雄师范学院数学与统计学院，云南楚雄 675000

English Abstract

On Subdirect Sums of Third-Order Diagonally Dominant Tensors

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目录

全文HTML

留言板

3阶对角占优张量的子直和

楚雄师范学院 数学与统计学院，云南 楚雄 675000

作者简介: 何建锋(1974-)，男，副教授，主要从事矩阵理论的研究 .

On Subdirect Sums of Third-Order Diagonally Dominant Tensors

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

1. 张量子直和的概念

2. 对角占优张量的子直和

计量

出版历程

3阶对角占优张量的子直和

作者简介: 何建锋(1974-)，男，副教授，主要从事矩阵理论的研究 楚雄师范学院 数学与统计学院，云南 楚雄 675000

English Abstract

On Subdirect Sums of Third-Order Diagonally Dominant Tensors

全文HTML

目录

全文HTML

楚雄师范学院数学与统计学院，云南楚雄 675000

作者简介:
何建锋(1974-)，男，副教授，主要从事矩阵理论的研究 .

作者简介: 何建锋(1974-)，男，副教授，主要从事矩阵理论的研究
楚雄师范学院数学与统计学院，云南楚雄 675000