设$ M\in Z^+ $, 记集合$ [M] = \{1, 2, \cdots, M\} $; $ I_j\in Z^+ $, 其中$ j\in [M] $.张量$ {{\cal A}} = ({a}_{{i_1}\cdots{i_M}}) \in {\Bbb C}^{{I_1}\times{I_2}\times\cdots \times{I_M}} $是复数域$ {{\Bbb C}} $上的$ {{I_1} \times{I_2}\times\cdots\times{I_M}} $个元素的多维排列, 其中$ {i_j} \in [{\mathbb I}_j] $, $ {j}\in [{\mathbb M}] $.简记$ {I_1}\times I_2\times\cdots\times{I_M} = {\cal T}^I_M $和$ ({I_1-R_1})\times\cdots \times({I_M-R_M}) = {{\cal T}^{I-R}_M} $.设$ {{\cal A}}\in {\Bbb C}^{{\cal T}^I_M\times {\cal T}^J_M} $和$ {{\cal B}}\in {\Bbb C}^{{{\cal T}^J_M}\times{{\cal T}^K_M}} $, 定义爱因斯坦积$ {\cal A}{*}_{M}{\cal B} $ (参见文献[4])

(1.1) $ \begin{eqnarray} \left({{{\cal A}}{*}_{M}{{\cal B}}}\right)_{{i_1}\cdots{i_M}{k_1}\cdots{k_M}} = \sum\limits_{ {j_r} \in [{\mathbb J}_r], {r} \in [{\mathbb M}] } {a}_{{i_1}\cdots{i_M}{j_1}\cdots{j_M}}{b}_{{j_1}\cdots{j_M}{k_1}\cdots{k_M}}. \end{eqnarray} $

显然, 张量的爱因斯坦积满足结合律.本文记$ {{{\cal A}}{*}_{M}{{\cal B}}} = {{{\cal A}}{*}{{\cal B}}} $; 张量$ {{\cal A}} $的值域为$ {{\cal R(A)}} $和零空间为$ {{\cal N(A)}} $

(1.2) $ \begin{equation} {{\cal R(A)}} = \{ {{\cal Y}} \in {\Bbb C}^{{{\cal T}^I_M}} : \ {{\cal Y}} = {{\cal A}}{*}{{\cal X}}, \ {{\cal X}} \in {\Bbb C}^{{{\cal T}^J_M}}\}, \end{equation} $

(1.3) $ \begin{equation} {{\cal N(A)}} = \{{{\cal X}} \in {\Bbb C}^{{{\cal T}^J_M}}: \ {{\cal A}}{*}{{\cal X}} = {{\cal O}}\}, \end{equation} $

其中$ {{\cal O}} $是零张量; $ {{\cal A}}^{*} $为$ {{\cal A}} $的共轭转置

(1.4) $ \begin{eqnarray} ({{\cal A}}^{*})_{{i_1}\cdots{i_M} {j_1}\cdots{j_M}} = (\bar{{\cal A}})_{{j_1}\cdots{j_M}{i_1}\cdots{i_M}}. \end{eqnarray} $

进一步, 记对角张量为$ {{\cal D}} $

$ \begin{eqnarray} {{\cal D}}_{{j_1}\cdots{j_M}{i_1}\cdots{i_M}} = \left\{\begin{array}{ll}{\sigma_{{j_1}\cdots{j_M}{i_1}\cdots{i_M}}}, \; \; & i_{r} = j_{r}\in [{\mathbb J}_r], \ {r}\in [{\mathbb M}], \\ 0, &\mbox{其它, } \end{array}\right. \end{eqnarray} $

其中$ \sigma_{{j_1}\cdots{j_M}{i_1}\cdots{i_M}}\in {\Bbb C} $; 如果$ {{\cal D}} $的所有对角元素$ {{\sigma}_{{j_1}\cdots{j_M}{i_1}\cdots{i_M}}} = 1 $, 则称$ {{\cal D}} $为单位张量, 记作$ {{\cal I}} $.显然有$ ({{\cal A}} {*}{{\cal B}})^{*} = {{\cal B}}^{*} {*}{{\cal A}}^{*} $以及$ {{\cal I}} {*}{{\cal A}} = {{\cal A}} {*}{{\cal I}} = {{\cal A}} $ (参见文献[18]).

设$ {{\cal A}}\in {\Bbb C}^{{{\cal T}^I_M}\times{{\cal T}^I_M}} $, 若存在张量$ {{\cal X}} \in {\Bbb C}^{{{\cal T}^I_M} \times{{\cal T}^I_M}} $使得$ {{\cal A}} {*}{{\cal X}} = {{\cal X}} {*}{{\cal A}} = {{\cal I}} $, 则$ {{\cal A}} $是可逆的, 且$ {{\cal X}} $是$ {{\cal A}} $的逆, 记作$ {{\cal A}^{-1}} $, 参见文献[2].在文献[18]中, Sun等引入偶数阶张量Moore-Penrose逆的概念:设$ {{\cal A}} \in {\Bbb C}^{{{\cal T}^I_M}\times{{\cal T}^J_M}} $, 则存在唯一的张量$ {{\cal X}} \in {\Bbb C}^{{{\cal T}^J_M}\times{{\cal T}^I_M}} $满足

$ \begin{eqnarray*} (1)\ {{\cal A}} {*}{{\cal X}}{*}{{\cal A}} = {{\cal A}}, \ (2)\ {{\cal X}} {*}{{\cal A}}{*}{{\cal X}} = {{\cal X}}, \ (3)\ ({{\cal A}} {*}{{\cal X}})^{*} = {{\cal A}} {*}{{\cal X}}, \ (4)\ ({{\cal X}} {*}{{\cal A}})^{*} = {{\cal X}} {*}{{\cal A}}. \end{eqnarray*} $

记$ {{\cal X}} = {{\cal A}^{+}} $, 称$ {{\cal A}^{+}} $为$ {{\cal A}} $的Moore-Penrose逆.显然, 若$ {{\cal A}} $是可逆的, 则$ {{\cal A}^{+}} = {{\cal A}^{-1}} $.记

$ {{\cal P}}_{{{\cal A}}} = {{\cal A}}{*}{{\cal A}}^{+}, \ \ {{\cal R}}_{{{\cal A}}} = {{\cal A}}^{+}{*}{{\cal A}}. $

在文献[6]中, Ji和Wei提出偶数阶张量的加权Moore-Penrose逆.

设$ {{\cal A}}\in{\Bbb C}^{{{\cal T}^I_M}\times{{\cal T}^I_M}} $, 则称满足$ {{\cal R}({\cal A}^{p+1})} = {{\cal R}({\cal A}^{p})} $成立的最小正整数$ p $为$ {{\cal A}} $的指标, 记为$ \mbox{Ind}({{\cal A}}) = p $.当指标为1时, 则存在唯一的张量$ {{\cal X}} \in {{{\Bbb C}}^{{{\cal T}^I_M} \times{{\cal T}^I_M}}} $满足

(1.5) $ \begin{eqnarray} (1)\ {{\cal A}}{*}{{\cal X}}{*}{{\cal A}} = {{\cal A}}, \ (2)\ {{\cal X}}{*}{{\cal A}}{*}{{\cal X}} = {{\cal X}}, \ (5)\ {{\cal A}}{*}{{\cal X}} = {{\cal X}}{*}{{\cal A}}. \end{eqnarray} $

记$ {{\cal X}} = {{\cal A}^{\sharp}} $, 称$ {{\cal A}^{\sharp}} $为$ {{\cal A}} $的群逆^[7].

如果方阵$ A $满足$ rank(A^2) = rank(A) $, 则称$ A $的指标为1, 记为$ \mbox{Ind}(A) = 1 $.众所周知, 矩阵$ A $的群逆存在当且仅当$ \mbox{Ind}(A) = 1 $. 2010年, Baksalary和Trenkler^[1]提出core逆的概念并给出若干core逆的性质.矩阵$ A $的core逆存在当且仅当$ \mbox{Ind}(A) = 1 $.随后, 矩阵的core逆问题引起了众多数学工作者的关注. Wang和Liu在文献[20]中给出了一个关于core逆的新的刻画.更多关于矩阵core逆和张量理论、计算及其应用的研究见文献[1, 3, 10-11, 13, 20-22]. Sahoo等在文献[17]中引入张量core逆的定义及其性质.本文在上述研究的基础上, 讨论爱因斯坦积下的张量core逆的性质, 应用张量core逆建立张量core偏序和研究张量方程$ {{\cal A}} {*}{{\cal X}} = {{\cal B}} $的约束最小二乘解等.

定义2.1^[17]  设$ {{\cal A}}\in {\Bbb C}^{{{\cal T}^I_M}\times{{\cal T}^I_M}} $且$ \mbox{Ind}({{\cal A}}) = 1 $, 则存在唯一的矩阵$ {{\cal X}} \in {\Bbb C}^{{{\cal T}^I_M}\times{{\cal T}^I_M}} $满足

$ \begin{eqnarray} {\rm (ⅰ)}\ {{\cal P}}_{{{\cal A}}} = {{\cal A}}{*}{{\cal X}}, \quad {\rm (ⅱ)}\ {\cal R}{({{\cal X}})} \subseteq {{\cal R({{\cal A}})}}. \end{eqnarray} $

记$ {{\cal X}} = {{\cal A}}$, 称$ {{\cal X}} $为$ {{\cal A}} $的core逆.

引理2.1^[17]  设$ {{\cal A}}\in {\Bbb C}^{{{\cal T}^I_M}\times{{\cal T}^I_M}} $且$ \mbox{Ind}({{\cal A}}) = 1 $, 则

(1) ;

(2) .

引理2.2^[17]  设$ {{\cal A}}\in {\Bbb C}^{{{\cal T}^I_M}\times{{\cal T}^I_M}} $且$ \mbox{Ind}({{\cal A}}) = 1 $, 则存在唯一的$ {{\cal X}}\in {\Bbb C}^{{{\cal T}^I_M}\times{{\cal T}^I_M}} $满足

$ \begin{eqnarray} (1)\; {{\cal A}}{*}{{\cal X}}{*}{{\cal A}} = {{\cal A}}, \; (2')\; {{\cal A}}{*}{{\cal X}}^{2} = {{\cal X}}, \; (3)\; ({{\cal A}}{*}{{\cal X}})^{*} = {{\cal A}}{*}{{\cal X}}, \end{eqnarray} $

且$ {{\cal X}} = {{\cal A}}$.

定义2.2^[17]  设$ {{\cal A}} \in {\Bbb C}^{{{\cal T}^I_M}\times{{\cal T}^I_M}} $, 若

(ⅰ)  $ {{\cal A}}{*}{{\cal A}}^{+} = {{\cal A}}^{+}{*}{{\cal A}} $, 则称$ {{\cal A}} $是EP张量;

(ⅱ)$ {{\cal A}}^{2} = {{\cal A}} $, 则称$ {{\cal A}} $是幂等张量;

(ⅲ)  $ {{\cal A}}^{3} = {{\cal A}} $, 则称$ {{\cal A}} $是tripotent张量;

(ⅳ)  $ {{\cal A}}^{2} = {{\cal A}} = {{\cal A}}^{*} $, 则称$ {{\cal A}} $是Hermitian幂等张量;

(ⅴ)  $ {{\cal A}}^{*}{*}{{\cal A}} = {{\cal A}}{*}{{\cal A}}^{*} = {{\cal I}} $, 则称$ {{\cal A}} $是酉张量.

引理2.3^[9,15]  设张量$ {{\cal A}} \in{\Bbb C}^{{{\cal T}^I_M}\times{{\cal T}^K_M}} $, 则在Matlab函数reshape下存在一一映射

$ rsh({{\cal A}}) = A = reshape({{\cal A}}, {{\cal J}}, {{\cal R}}), $

其中$ A\in{\Bbb C}^{{{\cal J}}\times{{\cal R}}} $, $ {{\cal J}} = \prod\limits_{i = 1}^{M}{I_i} $和$ {{\cal R}} = \prod\limits_{i = 1}^{M}{k_i} $.记$ {{\cal A}} $的秩为$ rshrank({{\cal A}}) $

$ rshrank({{\cal A}}) = rank(A). $

注2.1  由文献[15, 引理3.2]有$ rshrank({{\cal A}}{*}{{\cal B}}) = rank(AB)\leq rank(A) = rshrank({{\cal A}}) $, 即$ rshrank({{\cal A}}{*}{{\cal B}})\leq rshrank({{\cal A}}). $

引理2.4^[19]  设$ {{\cal A}} \in {\Bbb C}^{{{\cal T}^I_M}\times{{\cal T}^I_M}} $, 则存在酉张量$ {{\cal U}} \in {\Bbb C}^{{{\cal T}^I_M}\times{{\cal T}^I_M}} $, 使得

(2.1) $ \begin{eqnarray} {{\cal A}} = {{\cal U}}{*} \left[\begin{array}{ccc} {{\cal D}}{*}{{\cal K }} {\quad} & {{\cal D}}{*} {{\cal L}} \\ {{\cal O}}_2 {\quad} & {{\cal O}}_3 \end{array}\right]{*}{{\cal U}}^\ast, \end{eqnarray} $

其中$ {{\cal D}} \in {\Bbb C}^{{{\cal T}^R_M} \times{{\cal T}^R_M}} $是可逆的对角张量, $ {{\cal K}} \in {\Bbb C}^{{{\cal T}^R_M} \times{{\cal T}^R_M}} $和$ {{\cal L}} \in {\Bbb C}^{{{\cal T}^R_M} \times{{\cal T}^{I-R}_M}} $且$ {{\cal K}}{*}{{\cal K}}^\ast +{{\cal L}}{*}{{\cal L}}^\ast = {{\cal I}}_{r} $.

进一步得$ {{\cal A}} $的Moore-Penrose逆

(2.2) $ \begin{eqnarray} {{\cal A}^{+}} = {{\cal U}}{*} \left[\begin{array}{ccc} {{\cal K}}^{*}{*} {{\cal D}}^{-1} {\quad} & {{\cal O}}_1 \\ {{\cal L}}^{*}{*}{{\cal D}} ^{-1} {\quad} & {{\cal O}}_3 \end{array}\right]{*}{{\cal U}}^{*}. \end{eqnarray} $

注2.2  设$ rshrank({{\cal A}}) = rshrank({{\cal A}}^2) $, 由注2.1知

(2.3) $ \begin{eqnarray} rshrank({{\cal D}}) & \leq& rshrank({{\cal A}}) = rshrank\left( {{\cal U}}{*} \left[\begin{array}{ccc} {{\cal D}}{*}{{\cal K}} {\quad} & {{\cal D}}{*} {{\cal L}} \\ {{\cal O}}_2 {\quad} & {{\cal O}}_3 \end{array}\right]{*}{{\cal U}}^\ast\right) \\& = &rshrank\left( {{\cal U}}{*} \left[\begin{array}{ccc} ({{\cal D}}{*}{{\cal K}})^2 {\quad} & {{\cal D}}{*}{{\cal K}}{*}{{\cal D}}{*}{{\cal L}} \\ {{\cal O}}_2 {\quad} & {{\cal O}}_3 \end{array}\right]{*}{{\cal U}}^\ast\right) \\ & \leq& rshrank\left( {{\cal U}}{*} \left[\begin{array}{ccc} {{\cal D}}{*}{{\cal K}} {\quad} & {{\cal O}}_1 \\ {{\cal O}}_2 {\quad} & {{\cal O}}_3 \end{array}\right]{*}{{\cal U}}^\ast\right) \\& = &rshrank({{\cal D}}{*}{{\cal K}}) \leq rshrank({{\cal D}}). \end{eqnarray} $

应用$ {{\cal D}} $是可逆的, 得到$ rshrank({{\cal D}}{*}{{\cal K}}) = rshrank({{\cal D}}) = rshrank({{\cal K}}) $和$ {{\cal K}} $是可逆的.即:若$ rshrank({{\cal A}}) = rshrank\left({{\cal A}}^2\right) $, 则$ {{\cal K}} $是可逆的.

注2.3  如果$ {{\cal A}} $的指标是$ 1 $, $ rshrank({{\cal A}}) = rshrank\left({{\cal A}}^2\right) $, 则$ {{\cal A}} $有群逆, 且

(2.4) $ \begin{eqnarray} {{\cal A}}^{ \sharp} = {{\cal U}}{*} \left[\begin{array}{ccc} {{\cal K}}^{-1}{*}{{\cal D}} ^{-1} {\quad} & {{\cal K}}^{-1}{*}{{\cal D}} ^{-1}{*}{{\cal K}}^{-1}{*}{{\cal L}} \\ {{\cal O}}_2 {\quad} & {{\cal O}}_3 \end{array}\right]{*}{{\cal U}}^{*}. \end{eqnarray} $

引理2.5  设$ {{\cal A}} \in {\Bbb C}^{{{\cal T}^I_M}\times\times{{\cal T}^I_M}} $形如(2.1)式, 则

(ⅰ)  $ {{\cal A}} $是EP张量当且仅当$ {{\cal L}} = {{\cal O}}_{1} $;

(ⅱ)  $ {{\cal A}} $是幂等张量当且仅当$ {{\cal D}}{*}{{\cal K}} = {{\cal I}}_{r} $;

(ⅲ)  $ {{\cal A}} $是Hermitian幂等张量当且仅当$ {{\cal L}} = {{\cal O}}_1, {{\cal D}}{*}{{\cal K}} = {{\cal I}}_{r} $;

(ⅳ)  $ {{\cal A}} $是tripotent张量当且仅当$ ({{\cal D}}{*}{{\cal K}})^{2} = {{\cal I}}_{r} $.

证  应用(2.1)式和$ {{\cal A}}{*}{{\cal A}}^{+} = {{\cal A}}^{+}{*}{{\cal A}} $, 有

$ \begin{eqnarray} {{\cal U}}{*} \left[\begin{array}{ccc} {{\cal D}}{*} {{\cal K}}{*}{{\cal K}}^{*}{*}{{\cal D}}^{-1}+ {{\cal D}}{*} {{\cal L}}{*}{{\cal L}}^{*}{*}{{\cal D}}^{-1}{\quad} & {{\cal O}}_1\\ {{\cal O}}_2 {\quad} & {{\cal O}}_3 \end{array}\right]{*}{{\cal U}}^{*} = {{\cal U}}{*} \left[\begin{array}{ccc} {{\cal K}}^{*}{*}{{\cal K}}{\quad} & {{\cal K}}^{*}{*}{{\cal L}} \\ {{\cal L}}^{*}{*}{{\cal K}} {\quad} & {{\cal L}}^{*}{*}{{\cal L}} \end{array}\right]{*}{{\cal U}}^{*}. \end{eqnarray} $

应用$ {{\cal L}}^{*}{*}{{\cal L}} = {{\cal O}} $, 得到(ⅰ).

应用(2.1)式和$ {{\cal A}}^{2} = {{\cal A}} $, 有

$ \begin{eqnarray} {{\cal U}}{*} \left[\begin{array}{ccc} ({{\cal D}}{*}{{\cal K}})^{2} {\quad} & {{\cal D}}{*}{{\cal K}}{*}{{\cal D}}{*}{{\cal L}} \\ {{\cal O}}_2 {\quad} & {{\cal O}}_3 \end{array}\right]{*}{{\cal U}}^{*} = {{\cal U}}{*} \left[\begin{array}{ccc} {{\cal D}}{*}{{\cal K}} {\quad} & {{\cal D}}{*}{{\cal L}} \\ {{\cal O}}_2 {\quad} & {{\cal O}}_3 \end{array}\right]{*}{{\cal U}}^{*}, \end{eqnarray} $

以及$ \mbox{Ind}({{\cal A}}) = 1 $.因此$ {{\cal K}} $和$ {{\cal D}}{*}{{\cal K}} $是可逆的.应用$ ({{\cal D}}{*}{{\cal K}})^{2} = {{\cal D}}{*}{{\cal K}} $, 得到(ⅱ).

应用(2.1)式和$ {{\cal A}}^{2} = {{\cal A}} = {{\cal A}}^{*} $, 有

$ \begin{eqnarray} {{\cal U}}{*} \left[\begin{array}{ccc} ({{\cal D}}{*}{{\cal K}})^{2} {\quad} & {{\cal D}}{*}{{\cal K}}{*}{{\cal D}}{*}{{\cal L}}\\ {{\cal O}}_2 {\quad} & {{\cal O}}_3 \end{array}\right]{*}{{\cal U}}^{*} = {{\cal U}}{*} \left[\begin{array}{ccc} {{\cal D}}{*}{{\cal K}}{\quad} & {{\cal D}}{*}{{\cal L}} \\ {{\cal O}}_2 {\quad} & {{\cal O}}_3 \end{array}\right]{*}{{\cal U}}^{*} = {{\cal U}}{*} \left[\begin{array}{ccc} ({{\cal D}}{*}{{\cal K}})^{*} {\quad} & {{\cal O}}_1\\ ({{\cal D}}{*}{{\cal L}})^{*} {\quad} & {{\cal O}}_3 \end{array}\right]{*}{{\cal U}}^{*}. \end{eqnarray} $

应用$ {{\cal D}}{*}{{\cal L}} = {{\cal O}} $, $ {{\cal D}}{*}{{\cal K}} = ({{\cal D}}{*}{{\cal K}})^{*} $, $ ({{\cal D}}{*}{{\cal K}})^{2} = {{\cal D}}{*}{{\cal K}} $和$ {{\cal K}}{*}{{\cal K}}^\ast +{{\cal L}}{*}{{\cal L}}^\ast = {{\cal I}}_{r} $, 得到(ⅲ).

应用(2.1)式和$ {{\cal A}}^{3} = {{\cal A}} $, 有

$ \begin{eqnarray} {{\cal U}}{*} \left[\begin{array}{ccc} ({{\cal D}}{*}{{\cal K}})^{3}{\quad} & ({{\cal D}}{*}{{\cal K}})^{2}{*}{{\cal D}}{*}{{\cal L}} \\ {{\cal O}}_2 {\quad} & {{\cal O}}_3 \end{array}\right]{*}{{\cal U}}^{*} = {{\cal U}}{*} \left[\begin{array}{ccc} {{\cal D}}{*}{{\cal K}} {\quad} & {{\cal D}}{*}{{\cal L}} \\ {{\cal O}}_2 {\quad} & {{\cal O}}_3 \end{array}\right]{*}{{\cal U}}^{*}, \end{eqnarray} $

以及$ \mbox{Ind}({{\cal A}}) = 1 $.应用$ ({{\cal D}}{*}{{\cal K}})^{3} = {{\cal D}}{*}{{\cal K}} $, 得到(ⅳ).

定理3.1  设$ {{\cal A}} \in {\Bbb C}^{{{\cal T}^I_M}\times\times{{\cal T}^I_M}} $形如(1.1)式且$ rshrank({{\cal A}}) = rshrank({{\cal A}}^2) $, 则

(3.1)

证  设$ {{\cal A}^{+}} $如(2.2)式给出, 则

$ \begin{eqnarray} {{\cal P}}_{{{\cal A}}} & = &{{\cal A}}{*}{{\cal A}^{+}} = {{\cal U}}{*} \left[\begin{array}{ccc} {{\cal D}} {*}{{\cal K}} {\quad} & {{\cal D}} {*}{{\cal L}} \\ {{\cal O}}_2 {\quad}& {{\cal O}}_3 \end{array}\right]{*}{{\cal U}}^{*} {*} {{\cal U}}{*} \left[\begin{array}{ccc} {{\cal K}}^{*}{*} {{\cal D}}^{-1} {\quad} & {{\cal O}}_1 \\ {{\cal L}}^{*}{*} {{\cal D}}^{-1} {\quad} & {{\cal O}}_3 \end{array}\right]{*}{{\cal U}}^{*} \\ & = &{{\cal U}}{*} \left[\begin{array}{ccc} {{\cal I}}_{r} {\quad} & {{\cal O}}_1 \\ {{\cal O}}_2 {\quad} & {{\cal O}}_3 \end{array}\right]{*}{{\cal U}}^{*}. \end{eqnarray} $

设$ {{\cal B}}\in {\Bbb C}^{{{\cal T}^I_M}\times{{\cal T}^I_M}} $是$ {{\cal A}} $的core逆, 记

(3.2) $ \begin{eqnarray} {{\cal B}} = {{\cal U}}{*} \left[\begin{array}{ccc} {{\cal W}} {\quad} & {{\cal X}} \\ {{\cal Y}} {\quad} & {{\cal Z}} \end{array}\right]{*}{{\cal U}}^{*}, \end{eqnarray} $

其中$ {{\cal W}} \in {\Bbb C}^{{{\cal T}^R_M} \times{{\cal T}^R_M}} $.由定义2.1的条件(ⅰ)成立当且仅当$ {{\cal D}} {*} {{\cal K}}{*}{{\cal W}} + {{\cal D}} {*}{{\cal L}}{*}{{\cal Y}} = {{\cal I}}_r $和$ {{\cal K}}{*}{{\cal X}} +{{\cal L}}{*}{{\cal Z}} = {{\cal O}} $.根据引理2.2, 我们知道 , 由定义2.1的条件(ⅱ)成立当且仅当$ {{\cal Y}} = {{\cal O}} $和$ {{\cal Z}} = {{\cal O}} $.因此, $ {{\cal D}} {*}{{\cal K}}{*}{{\cal W}} = {{\cal I}}_r $和$ {{\cal K}}{*}{{\cal X}} = {{\cal O}} $.由于$ {{\cal K}} $是非奇异的, 因此条件(ⅱ)等价于$ {{\cal X}} = {{\cal O}} $.定理3.1得证.

定理3.2  设$ {{\cal A}} $, $ {{\cal A}^{+}} $, $ {{\cal A}}^\sharp $, 和$ {{\cal A}} $分别形如(2.1), (2.2), (2.4)和(3.1)式, 则

(ⅰ) $ {{\cal P}}_{{{\cal A}}} = {{\cal A}}{*}{{\cal A}}^{+} = {{\cal A}}{*}{{\cal A}} $;

(ⅱ)

证  由于

$ {{\cal A}}{*}{{\cal A}}^{+} = {{\cal U}}{*} \left[\begin{array}{ccc} {{\cal D}}{*}{{\cal K }} {\quad} & {{\cal D}}{*}{{\cal L}} \\ {{\cal O}}_2 {\quad} & {{\cal O}}_3 \end{array}\right]{*}{{\cal U}}^\ast{*} {{\cal U}}{*} \left[\begin{array}{ccc} {{\cal K}}^{*}{*} {{\cal D}}^{-1} {\quad} & {{\cal O}}_1 \\ {{\cal L}}^{*}{*} {{\cal D}}^{-1} {\quad} & {{\cal O}}_3 \end{array}\right]{*}{{\cal U}}^{*} \nonumber \\ \;\;\;\;\;\;\;\;\;\; = {{\cal U}}{*} \left[\begin{array}{ccc} {{\cal I}}_{r} {\quad} & {{\cal O}}_1 \\ {{\cal O}}_2 {\quad} & {{\cal O}}_3 \end{array}\right]{*}{{\cal U}}^{*}, $

得到$ {{\cal P}}_{{{\cal A}}} = {{\cal A}}{*}{{\cal A}}^{+} = {{\cal A}}{*}{{\cal A}} $.

由于 , 得到 .因此有

定理3.2得证.

定理3.3  设$ {{\cal A}} \in{\Bbb C}^{{{\cal T}^I_M}\times{{\cal T}^I_M}} $形如(2.1)式且$ rshrank({{\cal A}}) = rshrank({{\cal A}}^2) $.则

$ rshrank({{\cal A}}^{*}{*}{{\cal A}}) = rshrank({{\cal A}}{*}{{\cal A}^{*}}) = rshrank({{\cal A}}) = r, r>0, $

则$ {{\cal A}}^{*}{*}{{\cal A}}^2 $群可逆.

证  由定理3.1, 有$ {{\cal D}}{*}{{\cal K}} $是可逆的, 又由注2.1, 得到

$ \begin{eqnarray} {{\cal A}}^{*}{*}{{\cal A}}^2 & = & {{\cal U}}{*} \left[\begin{array}{ccc} ({{\cal D}}{*}{{\cal K}})^{*} {\quad} & {{\cal O}}_1 \\ ({{\cal D}} {*}{{\cal L}})^{*} {\quad} & {{\cal O}}_3 \end{array}\right]{*}{{\cal U}}^\ast{*} {{\cal U}}{*}\left[\begin{array}{ccc} {{\cal D}}{*}{{\cal K }} {\quad} & {{\cal D}}{*} {{\cal L}} \\ {{\cal O}}_2 {\quad} & {{\cal O}}_3 \end{array}\right]{{\cal U}}^\ast{*} {{\cal U}}{*}\left[\begin{array}{ccc} {{\cal D}}{*}{{\cal K }}{\quad} & {{\cal D}}{*} {{\cal L}} \\ {{\cal O}}_2 {\quad} & {{\cal O}}_3 \end{array}\right]{*}{{\cal U}}^\ast \\& = &{{\cal U}}{*} \left[\begin{array}{ccc} ({{\cal D}}{*}{{\cal K}})^{*} {\quad} & {{\cal O}}_1 \\ ({{\cal D}}{*} {{\cal L}})^{*} {\quad}& {{\cal O}}_3 \end{array}\right]{*}{{\cal U}}^\ast{*} {{\cal U}}{*}\left[\begin{array}{ccc} ({{\cal D}}{*}{{\cal K}})^2 {\quad} & {{\cal D}}{*}{{\cal K}}{*}{{\cal D}}{*} {{\cal L}} \\ {{\cal O}}_2 {\quad} & {{\cal O}}_3 \end{array}\right]{*}{{\cal U}}^\ast, \end{eqnarray} $

$ \begin{eqnarray*} \nonumber rshrank({{\cal A}}^{*}{*}{{\cal A}}^2) & = &rshrank\left({{\cal U}}{*} \left[\begin{array}{ccc} ({{\cal D}}{*}{{\cal K}})^{*} {\quad} & {{\cal O}}_1 \\ ({{\cal D}}{*} {{\cal L}})^{*} {\quad} & {{\cal O}}_3 \end{array}\right]{*} \left[\begin{array}{ccc} ({{\cal D}}{*}{{\cal K}})^2 {\quad} & {{\cal D}}{*}{{\cal K}}{*}{{\cal D}}{*} {{\cal L}} \\ {{\cal O}}_2 {\quad} & {{\cal O}}_3 \end{array}\right]{{\cal U}}^\ast\right) \nonumber \\ & \geq & rshrank\left( \left[\begin{array}{ccc} ({{\cal D}}{*}{{\cal K}})^{*} {\quad} & {{\cal O}}_1 \\ ({{\cal D}}{*} {{\cal L}})^{*} {\quad}& {{\cal O}}_3 \end{array}\right]{*} \left[\begin{array}{ccc} ({{\cal D}}{*}{{\cal K}})^2 {\quad}& {{\cal D}}{*}{{\cal K}}{*}{{\cal D}}{*} {{\cal L}} \\ {{\cal O}}_2 {\quad} & {{\cal O}}_3 \end{array}\right]\right. {\nonumber}\\ &&\left. {*} \left[\begin{array}{ccc} ({{\cal D}}{*}{{\cal K}})^{-1}{\quad} & ({{\cal D}}{*}{{\cal K}})^{-2}{*}{{\cal D}}{*}{{\cal L}} \\ {{\cal O}}_2 {\quad} & {{\cal O}}_3 \end{array}\right] \right) \nonumber \\ & = & rshrank({{\cal A}}^\ast{*}{{\cal A}}) = r \nonumber \end{eqnarray*} $

和$ rshrank({{\cal A}}^{*}{*}{{\cal A}}^2)\leq rshrank({{\cal A}}^\ast{*}{{\cal A}}) = r $, 因此$ rshrank({{\cal A}}^{*}{*}{{\cal A}}^2) = rshrank({{\cal A}}^{*}{*}{{\cal A}}) = r $.同理有$ rshrank({{\cal A}}{*}{{\cal A}}^{*}) = r $.

由于

$ \begin{eqnarray*} && rshrank({{\cal A}}^{*}{*}{{\cal A}}^2{*}{{\cal A}}^{*}{*}{{\cal A}}^2) \\ & = &rshrank\bigg({{\cal U}}{*} \left[\begin{array}{ccc} ({{\cal D}}{*}{{\cal K}})^{*}{\quad} & {{\cal O}}_1 \\ ({{\cal D}}{*} {{\cal L}})^{*} {\quad} & {{\cal O}}_3 \end{array}\right] {*}\left[\begin{array}{ccc} ({{\cal D}}{*}{{\cal K}})^2 {\quad} & {{\cal D}}{*}{{\cal K}}{*}{{\cal D}}{*} {{\cal L}} \\ {{\cal O}}_2 {\quad} & {{\cal O}}_3 \end{array}\right] \nonumber {*}\left[\begin{array}{ccc} ({{\cal D}}{*}{{\cal K}})^{*}{\quad} & {{\cal O}}_1 \\ ({{\cal D}}{*} {{\cal L}})^{*} {\quad} & {{\cal O}}_3 \end{array}\right]\\ \nonumber &&{*} \left[\begin{array}{ccc} ({{\cal D}}{*}{{\cal K}})^2{\quad} & {{\cal D}}{*}{{\cal K}}{*}{{\cal D}}{*}{{\cal L}} \\ {{\cal O}}_2 {\quad} & {{\cal O}}_3 \end{array}\right]{*}{{\cal U}}^\ast\bigg) \nonumber \\ & \geq& rshrank\left({{\cal U}}{*} \left[\begin{array}{ccc} (({{\cal D}}{*}{{\cal K}})^{*} {{\cal D}}{*}{{\cal K}})^{-1} {\quad} & {{\cal O}}_1 \\ {{\cal O}}_1 {\quad} & {{\cal O}}_3 \end{array}\right]{*} \left[\begin{array}{ccc} ({{\cal D}}{*}{{\cal K}})^{*} {\quad} & {{\cal O}}_1 \\ ({{\cal D}}{*} {{\cal L}})^{*} {\quad}& {{\cal O}}_3 \end{array}\right] {*} \left[\begin{array}{ccc} ({{\cal D}}{*}{{\cal K}})^2 {\quad} & {{\cal D}}{*}{{\cal K}}{*}{{\cal D}}{*} {{\cal L}} \\ {{\cal O}}_2 {\quad} & {{\cal O}}_3 \end{array}\right] \right. \nonumber \\&&\left. \nonumber {*} \left[\begin{array}{ccc} ({{\cal D}}{*}{{\cal K}})^{*} {\quad} & {{\cal O}}_1 \\ ({{\cal D}}{*} {{\cal L}})^{*} {\quad}& {{\cal O}}_3 \end{array}\right]{*} \left[\begin{array}{ccc} ({{\cal D}}{*}{{\cal K}})^2 {\quad} & {{\cal D}}{*}{{\cal K}}{*}{{\cal D}}{*} {{\cal L}} \\ {{\cal O}}_2 {\quad} & {{\cal O}}_3 \end{array}\right] {*} \left[\begin{array}{ccc} ({{\cal D}}{*}{{\cal K}})^{-2}{\quad} & {{\cal O}}_1 \\ {{\cal O}}_2 {\quad} & {{\cal O}}_3 \end{array}\right]{*}{{\cal U}}^\ast\right) \nonumber \\ & = & rshrank({{\cal A}}{*}{{\cal A}}^\ast) = r, \nonumber \end{eqnarray*} $

以及$ rshrank({{\cal A}}^{*}{*} {{\cal A}}^2{*} {{\cal A}}^{*}{*}{{\cal A}}^2) \leq rshrank({{\cal A}}^{*}{*}{{\cal A}}^2) = r $, 我们得到$ rshrank({{\cal A}}^{*}{*}{{\cal A}}^2{*} {{\cal A}}^{*}{*}{{\cal A}}^2) \ = \ r $.即$ rshrank({{\cal A}}^{*}{*}{{\cal A}}^2) = rshrank ({{\cal A}}^{*}{*}{{\cal A}}^2{*} {{\cal A}}^{*}{*}{{\cal A}}^2) = r $, 所以$ ({{\cal A}}^{*}{*}{{\cal A}}^2)^{\sharp} $存在.

引理3.1  设$ {{\tilde{\cal A}}} = {\tilde{\cal {B}}}{*}{ \tilde{\cal{C}}} $, $ rshrank({\tilde{\cal A}}) = rshrank({\tilde{\cal A}}^2) $, $ rshrank({\tilde{\cal {B}}}) = rshrank({\tilde{\cal {C}}}) = r $, 其中

$ {\tilde{\cal {B}}} = \left[\begin{array}{ccc} ({{\cal D}}{*} {{\cal K}})^{*} \\ ({{\cal D}}{*}{{\cal L}})^{*} \end{array}\right], {\quad} {\tilde{\cal {C}}} = \left[\begin{array}{ccc} ({{\cal D}}{*} {{\cal K}})^2 {\quad} & {{\cal D}}{*}{{\cal K}}{*}{{\cal D}}{*}{{\cal L}} \end{array}\right], $

则$ {\tilde{\cal {C}}}{*}{ \tilde{\cal{B}}} $是可逆的.

证  由于$ {\tilde{\cal {A}}}^{2} = { \tilde{\cal{B}}}{*}{\tilde{\cal {C}}} {*}{\tilde{\cal {B}}}{*}{\tilde{\cal {C}}} $, 若$ {\tilde{\cal {C}}}{*}{\tilde{\cal {B}}} $不可逆, 则$ rshrank({\tilde{\cal {A}}}^{2}) < rshrank({\tilde{\cal {B}}}{*}{\tilde{\cal {C}}}) = rshrank({\tilde{\cal {A}}}) $与$ \mbox{Ind}({\tilde{\cal {A}}}) = 1 $矛盾.因此, $ {\tilde{\cal {C}}}{*}{\tilde{\cal {B}}} $是可逆的.

定理3.4  设$ {{\cal A}}\in{\Bbb C}^{{{\cal T}^I_M}\times{{\cal T}^I_M}} $形如(2.1)式, $ rshrank({{\cal A}}^{2}) = rshrank({{\cal A}}) = r $, 则

证  设$ {{\cal A}} $有形如(2.1)式的形式, 由定理3.1和定理3.3, 有$ {{\cal D}}{*}{{\cal K}} $是可逆的, $ {{\cal A}}^{*}{*}{{\cal A}}^2 $是群可逆的, 以及

$ \begin{eqnarray} {{\cal A}}^{*}{*}{{\cal A}}^{2} & = &{{\cal U}}{*} \left[\begin{array}{ccc} ({{\cal D}}{*} {{\cal K}})^{*} {\quad} & {{\cal O}}_1 \\ ({{\cal D}}{*}{{\cal L}})^{*} {\quad}& {{\cal O}}_3 \end{array}\right]{*} \left[\begin{array}{ccc} ({{\cal D}}{*} {{\cal K}})^2 {\quad} & {{\cal D}}{*}{{\cal K}}{*}{{\cal D}}{*}{{\cal L}} \\ {{\cal O}}_2 {\quad} & {{\cal O}}_3 \end{array}\right]{*}{{\cal U}}^{*} \\& = &{{\cal U}}{*} \left[\begin{array}{ccc} ({{\cal D}}{*} {{\cal K}})^{*} \\ ({{\cal D}}{*}{{\cal L}})^{*} \end{array}\right]{*} \left[\begin{array}{ccc} ({{\cal D}}{*} {{\cal K}})^2 {\quad} & {{\cal D}}{*}{{\cal K}}{*}{{\cal D}}{*}{{\cal L}} \end{array}\right]{*}{{\cal U}}^{*}. \end{eqnarray} $

应用引理3.1, 我们知道

$ \left[\begin{array}{ccc} ({{\cal D}}{*} {{\cal K}})^2 {\quad} & {{\cal D}}{*}{{\cal K}}{*}{{\cal D}}{*}{{\cal L}} \end{array}\right]{*} \left[\begin{array}{ccc} ({{\cal D}}{*} {{\cal K}})^{*} \\ ({{\cal D}}{*}{{\cal L}})^{*} \end{array}\right] $

是可逆的.因此得到$ ({{\cal A}}^{*}{*}{{\cal A}}^2)^\sharp $的表达式

$ \begin{eqnarray} ({{\cal A}}^{*}{*}{{\cal A}}^2)^\sharp & = & {{\cal U}}{*}\left[\begin{array}{ccc} ({{\cal D}}{*} {{\cal K}})^{*} \\ ({{\cal D}}{*} {{\cal L}})^{*} \end{array}\right]{*} \left(\begin{array}{ccc} \left[\begin{array}{ccc} ({{\cal D}}{*} {{\cal K}})^2 {\quad} & {{\cal D}}{*}{{\cal K}}{*}{{\cal D}}{*}{{\cal L}} \end{array}\right]{*} \left[\begin{array}{ccc} ({{\cal D}}{*} {{\cal K}})^{*} \\ ({{\cal D}}{*}{{\cal L}})^{*} \end{array}\right]\end{array}\right)^{-2}\\ &&{*} \left[\begin{array}{ccc} ({{\cal D}}{*} {{\cal K}})^2 {\quad} & {{\cal D}}{*}{{\cal K}}{*}{{\cal D}}{*}{{\cal L}} \end{array}\right]{*}{{\cal U}}^{*}. \end{eqnarray} $

应用(2.1)式和$ ({{\cal A}}^{*}{*}{{\cal A}}^2)^\sharp $的表达式, 我们得到

即 .

矩阵偏序是满足自反性、传递性和反对称性的二元关系.以下是三种重要的矩阵偏序.

第一种偏序是由Hartwing在文献[5]提出的减序

(4.1) $ \begin{eqnarray} A\leq B \Leftrightarrow rank(B-A) = rank(B)-rank(A). \end{eqnarray} $

第二种偏序是由Mitra在文献[14]中提出的sharp序

(4.2) $ \begin{eqnarray} A\mathop \le \limits ^{\#}B \Leftrightarrow A^{\#}A = A^{\#}B , AA^{\#} = BA^{\#}. \end{eqnarray} $

第三种偏序是由Baksalary和Trenkler在文献[1]提出的core序

(4.3)

类似于矩阵偏序, 我们称在张量上满足下述三个条件的二元关系$ "\preceq" $为张量偏序.

(1) 自反性, $ {{\cal A}}\preceq{{\cal A}} $;

(2) 对称性, 若$ {{\cal A}}\preceq{{\cal B}} $且$ {{\cal B}}\preceq{{\cal A}} $, 则$ {{\cal A}} = {{\cal B}} $;

(3) 传递性, 若$ {{\cal A}}\preceq{{\cal B}} $且$ {{\cal B}}\preceq{{\cal C}} $, 则$ {{\cal A}}\preceq{{\cal C}} $.

在本节, 我们考虑张量的偏序并研究张量偏序的性质和刻画等.

定义4.1  设$ {{\cal A}}, {{\cal B}} \in {\Bbb C}^{{{\cal T}^I_M} \times{{\cal T}^I_M}} $.我们称满足

(4.4) $ \begin{eqnarray} {{\cal A}}\preceq {{\cal B}} \Leftrightarrow rshrank({{\cal B-A}}) = rshrank({{\cal B}})-rshrank({{\cal A}}) \end{eqnarray} $

的二元关系$ {{\cal A}}\preceq {{\cal B}} $为$ {{\cal T}} $-减序.

以下定理证明$ {{\cal T}} $-减序是一个张量偏序.

定理4.1  $ {{\cal T}} $减序是一个张量偏序.

证  自反性显然成立.

设$ {{\cal A}}\mathop \preceq{{\cal B}} $, $ {{\cal B}}\mathop \preceq{{\cal A}} $, 即有$ rshrank({{\cal B-A}}) = rshrank({{\cal B}})-rshrank({{\cal A}}) $, $ rshrank({{\cal A-B}}) = rshrank({{\cal A}})-rshrank({{\cal B}}) $.由于$ rshrank({{\cal A-B}}) = rshrank({{\cal B-A}}) $, 则$ rshrank({{\cal A}}) = rshrank({{\cal B}}) $, 则$ rshrank({{\cal A-B}}) = 0 $, 也就是说$ {{\cal A}} = {{\cal B}} $.

设$ {{\cal A}}\mathop \preceq{{\cal B}} $, $ {{\cal B}}\mathop \preceq{{\cal C}} $, 即有$ rshrank({{\cal B-A}}) = rshrank({{\cal B}})- rshrank({{\cal A}}) $, $ rshrank({{\cal B-C}}) = rshrank({{\cal B}})-rshrank({{\cal C}}) $, 则

$ \begin{eqnarray*} rshrank({{\cal C}})-rshrank({{\cal A}}) &\leq & rshrank({{\cal C-A}}) = rshrank({{\cal (B-A)+(C-B)}})\nonumber\\ &\leq& rshrank{{\cal (B-A)}}+rshrank{{\cal (C-B)}}\nonumber\\ & = &rshrank({{\cal B}})-rshrank({{\cal A}}) +rshrank({{\cal C}})- rshrank({{\cal B}})\nonumber\\ & = &rshrank({{\cal C}})-rshrank({{\cal A}}), \end{eqnarray*} $

即$ rshrank({{\cal C-A}}) = rshrank({{\cal C}})-rshrank({{\cal A}}) $.因此有$ {{\cal A}}\mathop \preceq{{\cal C}} $.

定义4.2  设$ {{\cal A}}, {{\cal B}} \in {\Bbb C}^{{{\cal T}^I_M} \times{{\cal T}^I_M}} $且$ \mbox{Ind}({{\cal A}}) = \mbox{Ind}({{\cal B}}) = 1 $.我们称满足

(4.5)

的二元关系为$ {{\cal T}} $-core序.

为了证明$ {{\cal T}} $-core序是张量偏序, 我们首先要证明两个引理.

引理4.1  设$ {{\cal A}} \in {\Bbb C}^{{{\cal T}^I_M}\times{{\cal T}^I_M}} $和$ {{\cal B}} \in {\Bbb C}^{{{\cal T}^I_M}\times{{\cal T}^I_M}} $且有$ \mbox{Ind}({{\cal A}}) = \mbox{Ind}({{\cal B}}) = 1 $, 此外, $ {{\cal A}} $形如(2.1)式, 则 当且仅当

(4.6) $ \begin{eqnarray} {{\cal B}} = {{\cal U}}{*} \left [\begin{array}{ccc} {{\cal D}}{*}{{\cal K}} {\quad} & {{\cal D}}{*}{{\cal L}} \\ {{\cal O}}_2 {\quad} & {{\cal Z}} \end{array}\right] {*} {{\cal U}}^\ast, \end{eqnarray} $

其中$ {{\cal D}}{*} {{\cal K}} $非奇异并且$ {{\cal Z}} \in {\Bbb C}^{{{\cal T}^{I-R}_M}\times{{{\cal T}^{I-R}_M}}} $是指标为1的张量.

证  设$ {{\cal B}} $的分解形如(3.2)式, 从(2.1), (3.1)和(3.2)式, 得到

(4.7)

因此, 成立当且仅当$ {{{\cal W}}} = {{\cal D}}{*}{{\cal K}} $, $ {{\cal X}} = {{\cal D}}{*}{{\cal L}} $.此外, 由

(4.8)

知道, 成立当且仅当$ {{\cal W}} = {{\cal D}}{*}{{\cal K}} $, $ {{\cal Y}} = {{\cal O}} $.即有(4.6)式成立, 其中$ {{\cal D}}{*}{{\cal K}} \in {\Bbb C}^{{R_1}\times\cdots\times{R_M} \times{R_1} \times\cdots\times{R_M}} $是可逆的.由$ \mbox{Ind}({{\cal B}}) = 1 $, 得到$ {{\cal Z}} \in {\Bbb C}^{{\cal T}^{I-R}_M\times {\cal T}^{I-R}_M} $的指标为1.

引理4.2  设$ {{\cal B}} $形如(4.6)式且$ {\mbox{Ind}}({{\cal A}}) = 1 $, 则

(4.9) $ \begin{eqnarray} {{\cal B}}^{\sharp} = {{\cal U}}{*} \left [\begin{array}{ccc} ({{\cal D}}{*}{{\cal K}})^{-1}{\quad} & {{\cal X}}_0 \\ {{\cal O}}_2 {\quad} & {{\cal Z}}^{\sharp} \end{array}\right]{*} {{\cal U}}^\ast, \end{eqnarray} $

其中$ {{\cal X}}_0 = ({{\cal D}}{*}{{\cal K}})^{-1}{*}{{\cal K}}^{-1}{*}{{\cal L}}- ({{\cal D}}{*}{{\cal K}})^{-2}{*}{{\cal D}}{*}{{\cal L}}{*}{{\cal Z}}^{\#}{*}{{\cal Z}}- {{\cal K}}^{-1}{*}{{\cal L}}{*}{{\cal Z}}^{\#} $.

证  假设$ {{\cal X}} $是$ {{\cal B}} $的群逆, 将其分解为

(4.10) $ \begin{eqnarray} {{\cal X}} = {{\cal U}}{*} \left [\begin{array}{ccc} {{\cal X}}_1 {\quad} & {{\cal X}}_2 \\ {{\cal X}}_3 {\quad} & {{\cal X}}_4 \end{array}\right]{*} {{\cal U}}^\ast. \end{eqnarray} $

由$ {{\cal B}}{*}{{\cal X}}{*}{{\cal B}} = {{\cal B}} $, 得到

(4.11) $ \cal U*\left[ {\begin{array}{*{20}{c}}{\cal D*K*{X}_1*\cal D*K + D*L*{X}_3*\cal D*K}\\{\cal Z*{X}_3*\cal D*K}\end{array}\;\;\begin{array}{*{20}{c}}{\cal D*K*{X}_1*\cal D*L + D*L*{X}_3*\cal D*L}\\{ + \cal D*K*{X}_2*\cal Z + D*L*{X}_4*\cal Z}\\{\cal Z*{X}_3*\cal D*L + Z*{X}_4*\cal Z}\end{array}} \right]*{\cal U^ * }\\ = \cal U*\left[ {\begin{array}{*{20}{c}}{\cal D*K\quad }&{\cal D*L}\\{{\cal O}_2\quad }&\cal Z\end{array}} \right]*{\cal U^ * }, $

即

(4.12) $ \begin{equation} {{\cal D}}{*}{{\cal K}}{*}{{\cal X}}_1{*} {{\cal D}}{*}{{\cal K}}+{{\cal D}}{*}{{\cal L}}{*}{{\cal X}}_3{*} {{\cal D}}{*}{{\cal K}} = {{\cal D}}{*}{{\cal K}} , \end{equation} $

(4.13) $ \begin{equation} {{\cal D}}{*}{{\cal K}}{*}{{\cal X}}_1{*} {{\cal D}}{*}{{\cal L}}+{{\cal D}}{*}{{\cal L}} {*}{{\cal X}}_3{*} {{\cal D}}{*}{{\cal L}} +{{\cal D}}{*}{{\cal K}}{*}{{\cal X}}_2{*} {{\cal Z}}+{{\cal D}} {*}{{\cal L}} {*}{{\cal X}}_4{*} {{\cal Z}} = {{\cal D}}{*}{{\cal L}}, \end{equation} $

(4.14) $ \begin{equation} {{\cal Z}}{*}{{\cal X}}_3{*}{{\cal D}}{*}{{\cal K}} = {{\cal O}}_2, \end{equation} $

(4.15) $ \begin{equation} {{\cal Z}}{*}{{\cal X}}_3{*}{{\cal D}}{*}{{\cal L}}+{{\cal Z}}{*}{{\cal X}}_4{*}{{\cal Z}} = {{\cal Z}}. \end{equation} $

由$ {{\cal D}}{*}{{\cal K}} $可逆, (3.16)式可等价表达为$ {{\cal Z}}{*}{{\cal X}}_3 = {{\cal O}}_2 $.

由$ {{\cal X}}{*}{{\cal B}}{*}{{\cal X}} = {{\cal X}} $, 有

(4.16) $ \begin{eqnarray} &&{{\cal U}}{*} \left [\begin{array}{ccc} {{\cal X}}_1{*}{{\cal D}}{*}{{\cal K}}{*} {{\cal X}}_1+{{\cal X}}_1{*}{{\cal D}}{*}{{\cal L}}{*} {{\cal X}}_3+ {{\cal X}}_2{*}{{\cal Z}}{*}{{\cal X}}_3 \ & {{\cal X}}_1{*}{{\cal D}}{*}{{\cal K}}{*} {{\cal X}}_2+{{\cal X}}_1{*}{{\cal D}}{*}{{\cal L}}{*} {{\cal X}}_4 +{{\cal X}}_2{*}{{\cal Z}}{*}{{\cal X}}_4 \\ {{\cal X}}_3{*}{{\cal D}}{*}{{\cal K}}{*} {{\cal X}}_1+{{\cal X}}_3{*}{{\cal D}}{*}{{\cal L}}{*} {{\cal X}}_3+ {{\cal X}}_4{*}{{\cal Z}}{*}{{\cal X}}_3 \ & {{\cal X}}_3{*}{{\cal D}}{*}{{\cal K}}{*} {{\cal X}}_2+{{\cal X}}_3{*}{{\cal D}}{*}{{\cal L}}{*} {{\cal X}}_4 +{{\cal X}}_4{*}{{\cal Z}}{*}{{\cal X}}_4 \end{array}\right]{*} {{\cal U}}^\ast \\ && = {{\cal U}}{*} \left [\begin{array}{ccc} {{\cal X}}_1 {\quad} & {{\cal X}}_2 \\ {{\cal X}}_3 {\quad} & {{\cal X}}_4 \end{array}\right]{*} {{\cal U}}^\ast. \end{eqnarray} $

即

(4.17) $ \begin{equation} {{\cal X}}_1{*}{{\cal D}}{*}{{\cal K}}{*} {{\cal X}}_1+{{\cal X}}_1{*}{{\cal D}}{*}{{\cal L}}{*} {{\cal X}}_3+ {{\cal X}}_2{*}{{\cal Z}}{*}{{\cal X}}_3 = {{\cal X}}_1 , \end{equation} $

(4.18) $ \begin{equation} {{\cal X}}_1{*}{{\cal D}}{*}{{\cal K}}{*} {{\cal X}}_2+{{\cal X}}_1{*}{{\cal D}}{*}{{\cal L}}{*} {{\cal X}}_4 +{{\cal X}}_2{*}{{\cal Z}}{*}{{\cal X}}_4 = {{\cal X}}_2 , \end{equation} $

(4.19) $ \begin{equation} {{\cal X}}_3{*}{{\cal D}}{*}{{\cal K}}{*} {{\cal X}}_1+{{\cal X}}_3{*}{{\cal D}}{*}{{\cal L}}{*} {{\cal X}}_3+ {{\cal X}}_4{*}{{\cal Z}}{*}{{\cal X}}_3 = {{\cal X}}_3, \end{equation} $

(4.20) $ \begin{equation} {{\cal X}}_3{*}{{\cal D}}{*}{{\cal K}}{*} {{\cal X}}_2+{{\cal X}}_3{*}{{\cal D}}{*}{{\cal L}}{*} {{\cal X}}_4 +{{\cal X}}_4{*}{{\cal Z}}{*}{{\cal X}}_4 = {{\cal X}}_4. \end{equation} $

由$ {{\cal B}}{*}{{\cal X}} = {{\cal X}}{*}{{\cal B}} $, 有

(4.21) $ \begin{eqnarray} &&{{\cal U}}{*} \left [\begin{array}{ccc} {{\cal D}}{*}{{\cal K}}{*}{{\cal X}}_1+{{\cal D}}{*}{{\cal L}}{*}{{\cal X}}_3 {\quad} & {{\cal D}}{*}{{\cal K}}{*}{{\cal X}}_2+{{\cal D}}{*}{{\cal L}}{*}{{\cal X}}_4 \\ {{\cal Z}}{*}{{\cal X}}_3 {\quad} & {{\cal Z}}{*}{{\cal X}}_4 \end{array}\right]{*} {{\cal U}}^\ast \\ & = &{{\cal U}}{*} \left [\begin{array}{ccc} {{\cal X}}_1{*}{{\cal D}}{*}{{\cal K}} {\quad} & {{\cal X}}_1{*}{{\cal D}}{*}{{\cal L}}+{{\cal X}}_2{*} {{\cal Z}} \\ {{\cal X}}_3{*}{{\cal D}}{*}{{\cal K}} {\quad}& {{\cal X}}_3{*}{{\cal D}}{*}{{\cal L}}+ {{\cal X}}_4{*} {{\cal Z}} \end{array}\right]{*} {{\cal U}}^\ast, \end{eqnarray} $

即

(4.22) $ \begin{equation} {{\cal D}}{*}{{\cal K}}{*}{{\cal X}}_1+{{\cal D}}{*}{{\cal L}}{*}{{\cal X}}_3 = {{\cal X}}_1{*}{{\cal D}}{*}{{\cal K}} , \end{equation} $

(4.23) $ \begin{equation} {{\cal D}}{*}{{\cal K}}{*}{{\cal X}}_2+{{\cal D}}{*}{{\cal L}}{*}{{\cal X}}_4 = {{\cal X}}_1{*}{{\cal D}}{*}{{\cal L}}+{{\cal X}}_2{*} {{\cal Z}} , \end{equation} $

(4.24) $ \begin{equation} {{\cal Z}}{*}{{\cal X}}_3 = {{\cal X}}_3{*}{{\cal D}}{*}{{\cal K}}, \end{equation} $

(4.25) $ \begin{equation} {{\cal Z}}{*}{{\cal X}}_4 = {{\cal X}}_3{*}{{\cal D}}{*}{{\cal L}}+{{\cal X}}_4{*} {{\cal Z}}. \end{equation} $

由$ {{\cal Z}}{*}{{\cal X}}_3 = {{\cal O}}_2 $, 得到$ {{\cal X}}_3{*}{{\cal D}}{*}{{\cal K}} = {{\cal Z}}{*}{{\cal X}}_3 = {{\cal O}}_2 $, 又$ {{\cal D}}{*}{{\cal K}} $可逆, 得到$ {{\cal X}}_3 = {{\cal O}}_2 $.应用$ {{\cal X}}_3 = {{\cal O}}_2 $和(4.12)式, 得到$ {{\cal X}}_1 = ({{\cal D}}{*}{{\cal K}})^{-1} $.将$ {{\cal X}}_3 = {{\cal O}}_2 $和$ {{\cal X}}_1 = ({{\cal D}}{*}{{\cal K}})^{-1} $代入(4.13), (4.15), (4.18), (4.20), (4.23)和(4.25)式, 得到

(4.26) $ \begin{equation} {{\cal D}}{*}{{\cal K}}{*}{{\cal X}}_2{*}{{\cal Z}}+{{\cal D}} {*}{{\cal L}} {*}{{\cal X}}_4{*} {{\cal Z}} = {{\cal O}}_1, \end{equation} $

(4.27) $ \begin{equation} {{\cal Z}}{*}{{\cal X}}_4{*}{{\cal Z}} = {{\cal Z}}, \end{equation} $

(4.28) $ \begin{equation} {{\cal K}}^{-1} {{\cal L}}{*}{{\cal X}}_4+{{\cal X}}_2{*}{{\cal Z}}{*}{{\cal X}}_4 = {{\cal O}}_1, \end{equation} $

(4.29) $ \begin{equation} {{\cal X}}_4{*}{{\cal Z}}{*}{{\cal X}}_4 = {{\cal X}}_4, \end{equation} $

(4.30) $ \begin{equation} {{\cal D}}{*}{{\cal K}}{*}{{\cal X}}_2+{{\cal D}} {*}{{\cal L}}{*}{{\cal X}}_4 = {{\cal K}}^{-1}{*}{{\cal L}}+{{\cal X}}_2{*}{{\cal Z}}, \end{equation} $

(4.31) $ \begin{equation} {{\cal Z}}{*}{{\cal X}}_4 = {{\cal X}}_4{*}{{\cal Z}}. \end{equation} $

由(4.27), (4.29)和(4.31)式, 得到$ {{\cal X}}_4 = {{\cal Z}}^{\#} $.

由(4.30)式, $ {{\cal X}}_4 = {{\cal Z}}^{\#} $和$ {{\cal D}}{*}{{\cal K}} $可逆, 知道

$ \begin{eqnarray} {{\cal X}}_2& = &({{\cal D}}{*}{{\cal K}})^{-1}{*}{{\cal K}}^{-1}{*}{{\cal L}}+ ({{\cal D}}{*}{{\cal K}})^{-1}{*}{{\cal X}}_2{*}{{\cal Z}}- ({{\cal D}}{*}{{\cal K}})^{-1}{*}{{\cal D}}{*}{{\cal L}}{*}{{\cal Z}}^{\#} \\& = &({{\cal D}}{*}{{\cal K}})^{-1}{*}{{\cal K}}^{-1}{*}{{\cal L}}+ ({{\cal D}}{*}{{\cal K}})^{-2}{*}{{\cal D}}{*}{{\cal K}}{*}{{\cal X}}_2{*}{{\cal Z}}- ({{\cal D}}{*}{{\cal K}})^{-1}{*}{{\cal D}}{*}{{\cal L}}{*}{{\cal Z}}^{\#}, \end{eqnarray} $

进而由(4.26)式, 得到

$ \begin{eqnarray} {{\cal X}}_2& = &({{\cal D}}{*}{{\cal K}})^{-1}{*}{{\cal K}}^{-1}{*}{{\cal L}}- ({{\cal D}}{*}{{\cal K}})^{-2}{*}{{\cal D}}{*}{{\cal L}}{*}{{\cal Z}}^{\#}{*}{{\cal Z}}- ({{\cal D}}{*}{{\cal K}})^{-1}{*}{{\cal D}}{*}{{\cal L}}{*}{{\cal Z}}^{\#} \\& = &({{\cal D}}{*}{{\cal K}})^{-1}{*}{{\cal K}}^{-1}{*}{{\cal L}}- ({{\cal D}}{*}{{\cal K}})^{-2}{*}{{\cal D}}{*}{{\cal L}}{*}{{\cal Z}}^{\#}{*}{{\cal Z}}- {{\cal K}}^{-1}{*}{{\cal L}}{*}{{\cal Z}}^{\#}. \end{eqnarray} $

证毕.

定理4.2  T-core序是一个张量偏序.

证  为了证明$ {{\cal T}} $-core序是一个张量偏序.我们要需要证明$ {{\cal T}} $-core序满足(ⅰ)自反性, 即 ; (ⅱ)反身性, 即 和 表明$ {{\cal A}} = {{\cal B}} $; (ⅲ)传递性, 即 和 表明 , 对于$ {{\cal A}}, {{\cal B}} $, $ {{\cal C}} \in {\Bbb C}^{{{\cal T}^I_M}\times{{\cal T}^I_M}} $.由于自反性显然满足, 我们开始证明(ⅱ).设 且 , 得到

(4.32)

由张量core满足{2}逆和(4.32)式第二个等式的右边, 得到 .因此, 应用引理2.1得到$ {{\cal B}} = {{\cal A}}{*}{{\cal B}}^ {\sharp}{*}{{\cal B}} $.应用(2.1), (4.6)和(4.9)式, 得到

(4.33) $ \begin{eqnarray} {{\cal B}} = {{\cal A}}{*}{{\cal B}}^ {\sharp}{*}{{\cal B}} = {{\cal U}}{*} \left [\begin{array}{ccc} {{\cal D}}{*}{{\cal K}} {\quad} & {{\cal D}}{*}{{\cal L}}+{{\cal D}}{*}{{\cal K}}{*}{{\cal X}}_{0}{*}{{\cal Z}}+ {{\cal D}}{*}{{\cal L}}{*}{{\cal Z}}^{\sharp}{*}{{\cal Z}} \\ {{\cal O}}_2 {\quad} & {{\cal O}}_3 \end{array}\right]{*} {{\cal U}}^\ast. \end{eqnarray} $

通过比较(4.6)和(4.33)式, 得到了$ {{\cal B}} = {{\cal A}}{*}{{\cal B}}^ {\sharp}{*}{{\cal B}} $成立的充要条件就是$ {{\cal Z}} = 0 $.但是若$ {{\cal Z}} = 0 $, 则$ {{\cal B}} $与$ {{\cal A}} $在(2.1)式的形式一致, 即$ {{\cal T}} $-core偏序具有反身性.

(ⅲ), 设 且 , 即

(4.34)

(4.35)

通过应用(4.34)和(4.35)式的第二个等式和引理2.1, 有 .由(3.1), (4.6)和(4.9)式, 得到

(4.36)

由(3.1)式, 得到了 .因此, .

类似地, 应用(4.34)和(4.35)式的第一个等式和引理2.1, 得到

此外, 引理2.1和引理2.2, 得到

(4.37)

由(3.1), (4.6)和(4.37)式, 得到

即 .因此, 有 .因此, 得到 .

定理4.3  若$ {{\cal A}}, \ {{\cal B}}\in {\Bbb C}^{{{\cal T}^I_M}\times{{\cal T}^I_M}} $并且$ rshrank({{\cal A}}^{2}) = rshrank({{\cal A}}) $和$ rshrank({{\cal B}}^{2}) = rshrank({{\cal B}}) $, 则下列条件等价

$ \rm(ⅰ) $;

$ \rm(ⅱ) $$ {{\cal A}}^{+}{*}{{\cal A}} = {{\cal A}}^{+}{*}{{\cal B}} $和$ {{\cal A}}^2 = {{\cal B}}{*}{{\cal A}} $.

证  由引理2.1, (4.5)式的第一个条件可写为 .在等式 的两边左乘$ {{\cal A}}^{+}{*}{{\cal A}} $, 得到$ {{\cal A}}^ {+}{*}{{\cal A}} = {{\cal A}}^ {+}{*}{{\cal B}} $.在等式$ {{\cal A}}^ {+}{*}{{\cal A}} = {{\cal A}}^ {+}{*}{{\cal B}} $两边左乘$ {{\cal A}}^{\sharp}{*}{{\cal A}} $, 得到 .因此, .类似地, 由引理2.1, (3.5)式的第一个条件可写为$ {{\cal A}}{*} {{\cal A}}^ {+} = {{\cal B}}{*}{{\cal A}}^ {\sharp}{*} {{\cal A}}{*}{{\cal A}}^ {+} $.在等式 两边右乘$ {{\cal A}}^ {2} $, 得到$ {{\cal A}}^2 = {{\cal B}}{*} {{\cal A}} $.此外, 在等式$ {{\cal A}}^2 = {{\cal B}}{*} {{\cal A}} $两边右乘$ {{\cal A}}^ {\sharp}{*}{{\cal A}}^ {+} $, 得到 .即 .

设$ {{\cal A}} \in{\Bbb C}^{{{\cal T}^I_M}\times{I{\cal T}^I_M}} $和$ {{\cal B}} \in{\Bbb C}^{{{\cal T}^I_M}} $.若方程

(5.1) $ \begin{eqnarray} {{\cal A}}{*}{{\cal X}} = {{\cal B}} \end{eqnarray} $

有解, 则通解为

$ {{\cal X}} = {{\cal A}}^{(1)}{*}{{\cal B}} +({{\cal I}}-{{\cal A}}^{(1)}{{\cal A}}){*}{{\cal Y}}, $

其中$ {{\cal A}}^{(1)} $是$ {{\cal A}} $的$ \left\{1\right\} $逆并且$ {{\cal Y}} \in{\Bbb C}^{{{\cal T}^I_M}} $是任意张量.当方程(5.1)无解时, 则 是其极小范数最小二乘解^[18].在文献[17]中, Sahoo等研究了方程(5.1)在约束条件$ {{\cal B}}\in{\Bbb {{\cal R({{\cal A}})}}} $, $ \mbox{Ind}({{\cal A}}) = 1 $和$ {{\cal X}}\in{\Bbb {{\cal R({{\cal A}})}}} $下的解, 并证明 是唯一解.当$ {{\cal B}}\notin{\Bbb {{\cal R({{\cal A}})}}} $, 易证方程(5.1)是无解的.本节我们考虑方程(5.1)在$ {x \in {\Bbb {\cal R}}({\cal A})} $下的最小二乘解.

定理5.1  设$ {{\cal A}} $和$ {{\cal B}} $形如(5.1)式且$ {\mbox{Ind}}({{\cal A}}) = 1 $, 则 是$ \|{{\cal A}}{*}{{\cal X}} -{{\cal B}}\|_F^2 = \min, $$ {{{\cal X}} \in {\Bbb {\cal R}}({\cal A})} $的唯一解.

证  由于$ {{\cal X}}\in{\Bbb {\cal R(A)}} $, 则存在$ {{\cal Y}} \in{\Bbb C}^{{{\cal T}^I_M}} $使得$ {{\cal X}} = {{\cal A{*}Y}} $.设$ {{\cal A}} $形如(2.1)式.记

则

$ \begin{eqnarray*} \nonumber \left\| {{\cal A{*}X}-{{\cal B}}} \right\|_F^2 & = & \left\| {{\cal A{*}A{*}Y}-{{\cal B}}} \right\|_F^2 \\ \nonumber & = &\left\|{{\cal U}}{*}\left[\begin{array}{ccc} {{\cal D}} {*}{{\cal K}}{\quad} & {{\cal D}} {*}{{\cal L}} \\ {{\cal O}}_2 {\quad} & {{\cal O}}_3 \end{array}\right]{*}{{\cal U}}^\ast{*}{{\cal U}}{*}\left[\begin{array}{ccc} {{\cal D}}{*}{{\cal K}}{\quad} & {{\cal D}} {*}{{\cal L}}\\ {{\cal O}}_2 {\quad} & {{\cal O}}_3 \end{array}\right] {*}{{\cal U}}^\ast{*}{{\cal Y}}-{{\cal U}}{*} {{\cal U}}^\ast{*}{{\cal B}}\right\|_F^2 \\ \nonumber & = & \left\|{{\cal U}}{*}\left[\begin{array}{ccc} ({{\cal D}} {*}{{\cal K}})^2 {\quad} & {{\cal D}} {*}{{\cal K}}{*}{{\cal D}} {*}{{\cal L}} \\ {{\cal O}}_2 {\quad} & {{\cal O}}_3 \end{array}\right]{*} \left[\begin{array}{ccc} y_1 \\ y_2 \end{array}\right] -{{\cal U}}{*}\left[\begin{array}{ccc} b_1 \\ b_2 \end{array}\right] \right\|_F^2 \\ \nonumber & = & \left\|\left[\begin{array}{ccc} ({{{\cal D}}{*}{\cal K}})^2{*}y_1+{{\cal D}} {*}{{\cal K}}{*}{{\cal D}} {*}{{\cal L}}{*}y_2-b_1 \\ b_2 \end{array}\right]\right\|_F^2 \\ \nonumber & = &\left\| ({{{\cal D}}{*}{\cal K}})^2{*}y_1+{{\cal D}} {*}{{\cal K}}{*}{{\cal D}} {*}{{\cal L}}{*}y_2-b_1 \right\|_F^2+\left\| b_2 \right\|_F^2, \end{eqnarray*} $

其中$ {{\cal A}}, \ {{\cal U}} \in{\Bbb C}^{{{\cal T}^I_M}\times{{\cal T}^I_M}} $, $ {y_1} \in{\Bbb C}^{{{\cal T}^R_M}} $, $ {y_2} \in{\Bbb C}^{{{\cal T}^{I-R}_M}} $, $ {b_1 } \in{\Bbb C}^{{{\cal T}^R_M}} $, $ { b_2} \in{\Bbb C}^{{{\cal T}^{I-R}_M}} $, 和$ {{{\cal O}}_4} \in{\Bbb C}^{{{\cal T}^{I-R}_M}} $.由于$ {{\cal D}}{*}{{\cal K}} $可逆, 取$ y_1 = ({{\cal D}}{*}{\cal K})^{-2}{*} b_1 - {{\cal K}}^{-1}{*} {{\cal L}}{*}y_2 $, 则有

$ \min\limits_{y_1, y_2} \left\| ({{{\cal D}}{*}{\cal K}})^2{*}y_1 +{{\cal D}} {*}{{\cal K}}{*}{{\cal D}} {*}{{\cal L}}{*}y_2-b_1\right\|_F^2 = 0. $

因而$ \| {{\cal A}}{*} {{\cal X}}- {{\cal B}}\|_F^2 = \|b_2\|_F^2 $, 即得

定理5.1证毕.