1 引言及预备知识

M-矩阵是一类重要的特殊矩阵, 它在经济学、运筹学等领域有着广泛的应用.M-矩阵是由Ostrowski在1937年首次提出的, 作为矩阵理论的一个研究分支和方法, 关于M-矩阵的研究一直备受关注^[1-3].文献[4-8]讨论了M-矩阵的性质，得出了一些关于M-矩阵的不等式，文献[9-13]讨论了逆M-矩阵的不等式及其相关不等式, 文献[14-16]研究了非负矩阵和M-矩阵之间的关系.文献[17-18]研究了M-矩阵以及逆M-矩阵的主子式的问题.在此基础上, 本文主要讨论M-矩阵的乘法性质及凸组合性质, 得出关于M-矩阵的一些结果.

为了叙述方便, 对符号进行如下约定:M_n(R) 表示实数域R上的所有n×n阶矩阵的集合; A^T表示矩阵A的转置, |A|表示n阶方阵(|a_ij|), ρ(A) 表示矩阵A的谱半径, 其中A=(a_ij)∈M_n(R); 设A=(a_ij)∈M_n(R), 如果a_ij≥0, 则称A为非负矩阵, 记作A≥0;设A=(a_ij)_n, B=(b_ij)_n∈M_n(R), 如果B-A≥0, 称B控制A, 记作A≤B.其他未加说明的符号参见文献[1].

下面是与本论文有关的几个定义及引理.

定义1^[1]  设Z_n={A=(a_ij)∈M_n(R)|a_ij≤0, i≠j, i, j=1, 2, …, n}, 称Z_n中的矩阵A为Z-矩阵.

定义2^[1]  若A∈Z_n, 如果A是正稳定矩阵, 称A是M-矩阵.

定义3^[1]  设A=(a_ij)_n∈M_n(R), 称n阶方阵(m_ij)_n为A的比较矩阵, 记作M(A), 其中m_ij≡ $\left\{ \begin{array}{l} |{a_{ij}}|,j = i,\\ - |{a_{ij}}|,j \ne i. \end{array} \right.$

定义4^[1]  设A∈M_n(R), 如果A的比较矩阵M(A) 是M-矩阵, 称A是H-矩阵.

引理1^[1]  若A∈Z_n, 则下列条件互相等价:

(1) A是M-矩阵;

(2) A=αI-P, P≥0, α > ρ(P);

(3) A是非奇异的, 且A^-1≥0;

(4) 存在正对角矩阵D, 使得DA+A^TD正定;

(5) A的主对角元素是正的且存在正对角矩阵D, 使得AD是行严格对角占优阵;

(6) 存在正向量x > 0, 使得Ax > 0;

(7) A^T是M-矩阵.

引理2^[1]  A, B∈Z_n是给定的矩阵, 假设A=[a_ij]是M-矩阵, 且B≥A.则

(1) B是M-矩阵;

(2) A^-1≥B^-1≥0;

(3) detB≥detA > 0.

引理3^[2]  设A=(a_ij)_n, B=(b_ij)_n∈M_n(R) 且B是非负矩阵, 如果A≤B, 则有ρ(A)≤ρ(B).

2 主要结果

定理1  设A∈M_n是M-矩阵, D∈M_n(R) 是正对角矩阵, 则DA和AD也是M-矩阵.

证明  (1) 首先证DA是M-矩阵.由于A∈M_n是M-矩阵且D是正对角矩阵, 则DA∈Z_n.由于A是M-矩阵, 则由引理1中(4) 知, 存在一个正对角矩阵E, 使得EA+A^TE正定.此外对于正对角矩阵D, 显然ED^-1也是正对角矩阵, 且有(ED^-1)(DA)+(DA)^T(ED^-1)=EA+A^TE正定, 于是由引理1中(4) 可得, DA是M-矩阵.(2) 其次证AD是M-矩阵.设A∈M_n是M-矩阵, 由引理1中(7) 可知, A^T也是M-矩阵.由已证得(1) 可知DA^T是M-矩阵, 因此由引理1中(7) 得, (DA^T)^T=AD^T=AD是M-矩阵.综上可知, DA和AD都是M-矩阵.

定理2  设A, B∈M_n是M-矩阵, 则

(1) AB是M-矩阵当且仅当AB∈Z_n;

(2) 如果A, B是2阶方阵, 则AB是M-矩阵.

证明 (1) 若AB是M-矩阵, 显然有AB∈Z_n; 若AB∈Z_n, 因为A, B∈M_n是M-矩阵, 由引理1中(3) 得A和B都是非奇异矩阵, 且A^-1≥0, B^-1≥0.因此, (AB)^-1=B^-1A^-1≥0, 于是由引理1(3) 知, AB是M-矩阵.

(2) 设A=$\left( {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}\\ {{a_{21}}}&{{a_{22}}} \end{array}} \right)$, B=$\left( {\begin{array}{*{20}{c}} {{b_{11}}}&{{b_{12}}}\\ {{b_{21}}}&{{b_{22}}} \end{array}} \right)$是M-矩阵, 则AB=$\left( {\begin{array}{*{20}{c}} {{a_{11}}{b_{11}} + {a_{12}}{b_{21}}}&{{a_{11}}{b_{12}} + {a_{12}}{b_{22}}}\\ {{a_{21}}{b_{11}} + {a_{22}}{b_{22}}}&{{a_{21}}{b_{12}} + {a_{22}}{b_{22}}} \end{array}} \right)$, 由于A, B是M-矩阵, 因此A, B∈Z_n, 且A和B的主对角线元素都是正数.因此有a₁₁ > 0, a₂₂ > 0, b₁₁ > 0, b₂₂ > 0, a₁₂≤0, a₂₁≤0, b₂₁≤0, b₁₂≤0.所以a₁₁b₁₂+a₁₂b₂₂≤0, a₂₁b₁₁+a₂₂b₂₁≤0, AB∈Z_n, 由已证得(1) 可知AB是M-矩阵.

注  一般地, 两个M-矩阵的乘积不一定是M-矩阵, 如

$ \mathit{\boldsymbol{A = }}\left[ {\begin{array}{*{20}{c}} 1&{ - 1}&0\\ { - 1}&3&{ - 2}\\ 0&{ - 1}&4 \end{array}} \right],\mathit{\boldsymbol{B = }}\left( {\begin{array}{*{20}{c}} 1&{ - \frac{1}{2}}&0\\ { - \frac{1}{2}}&1&{ - \frac{1}{3}}\\ 0&{ - \frac{1}{3}}&1 \end{array}} \right),aaa\mathit{\boldsymbol{AB = }}\left( {\begin{array}{*{20}{c}} {\frac{3}{2}}&{ - \frac{3}{2}}&{\frac{1}{3}}\\ { - \frac{5}{2}}&{\frac{{25}}{6}}&{ - 3}\\ {\frac{1}{2}}&{ - \frac{1}{3}}&{\frac{{13}}{3}} \end{array}} \right) \notin {Z_n}. $

定理3  设A, B∈M_n(R) 是M-矩阵, α∈[0, 1], 则

(1) 若αA+(1-α)B是M-矩阵, 则B^-1A没有负的实特征值;

(2) 如果B^-1A是M-矩阵, 则A和B的凸组合是M-矩阵;

(3) B^-1A是M-矩阵当且仅当B^-1A∈Z_n.

证明  (1) 假设B^-1A有负的实特征值-λ, 其中λ > 0, 即

$ \det \left( { - \lambda \mathit{\boldsymbol{I}} - {\mathit{\boldsymbol{B}}^{ - 1}}\mathit{\boldsymbol{A}}} \right) = {\left( { - 1} \right)^n}\det \left( {\lambda \mathit{\boldsymbol{I + }}{\mathit{\boldsymbol{B}}^{ - 1}}\mathit{\boldsymbol{A}}} \right) = 0, $

令α=$\frac{1}{{\lambda + 1}}$, 1-α=$\frac{\lambda }{{\lambda + 1}}$, 则

det (αA+(1-α)B)=det (B(αB^-1A+(1-α)I))=detBdet (αB^-1A+(1-α)I)=detBdet $\left( {\frac{\lambda }{{\lambda + 1}}I + \frac{1}{{\lambda + 1}}{\boldsymbol{B}^{ - 1}}\boldsymbol{A}} \right) = {\left( {\frac{1}{{\lambda + 1}}} \right)^n}$det (λI+B^-1A)=0,

这与αA+(1-α)B是M-矩阵矛盾, 因此B^-1A没有负的实特征值.

(2) 因为αA+(1-α)B=B(αB^-1A+(1-α)I), 且A, B都是M-矩阵, 所以A∈Z_n, B∈Z_n, αA+(1-α)B∈Z_n.又B^-1A是M-矩阵, 所以αB^-1A+(1-α)I, α∈[0, 1]是M-矩阵.又B是M-矩阵, αA+(1-α)B=B(αB^-1A+(1-α)I)∈Z_n, 由定理2知, αA+(1-α)B是M-矩阵.

(3) 必要性.设B^-1A是M-矩阵, 显然有B^-1A∈Z_n.

充分性.由于A是M-矩阵, 由引理1中(6) 知, 存在正向量x > 0, 使得Ax > 0.再由B是M-矩阵, 知B^-1≥0, 显然B^-1的每一行至少有一个是正数, 所以B^-1Ax > 0, 故由引理1中(6) 知, B^-1A是M-矩阵.

定理4  设A, B∈Z_n, 且A是M-矩阵, 且B≥A, 则

(1) B^-1A≤I且AB^-1≤I,

(2) B^-1A和AB^-1都是M-矩阵,

(3) 对于所有的α∈[0, 1], αA+(1-α)B是M-矩阵,

(4) ∀α∈[0, 1], αC+(1-α)I是M-矩阵, 其中C=B^-1A或C=AB^-1,

(5) (αA+(1-α)B)^-1≤αA^-1+(1-α)B^-1, ∀α∈[0, 1].

证明  (1) 由引理2(1) 知, B是M-矩阵, 则B^-1≥0, 于是由B≥A知, B^-1(B-A)≥0且(B-A)B^-1≥0, 因而有B^-1A≤I, 且AB^-1≤I.

(2) 由已证得(1) 知B^-1A∈Z_n, AB^-1∈Z_n, 此外由于B≥A, 即B-A≥0, 以及A是M-矩阵知, (B-A)A^-1≥0且A^-1(B-A)≥0, 即A^-1B≥I, BA^-1≥I, 因而A^-1B, BA^-1都是非负矩阵, 再由于(B^-1A)^-1=A^-1B, (AB^-1)^-1=BA^-1及由引理1中(3) 知, BA^-1与AB^-1都是M-矩阵.

(3) 因为A∈Z_n, B∈Z_n, 所以∀α∈[0, 1], αA+(1-α)B∈Z_n, 又因为A≤B, 于是有A≤αA+(1-α)B≤B, 由引理2中(1) 知, αA+(1-α)B是M-矩阵.

(4) 因为C是M-矩阵, 所以∀α∈[0, 1], αC+(1-α)I∈Z_n且有αC+(1-α)I≥αC.当α=0时, 显然αC+(1-α)I=I是M-矩阵, 当α≠0时, 由引理2(1) 知, αC+(1-α)I是M-矩阵.

(5) 令G=AB^-1, ∀α∈[0, 1], 由于

$ \begin{array}{l} \alpha {\mathit{\boldsymbol{A}}^{ - 1}} + \left( {1 - \alpha } \right){\mathit{\boldsymbol{B}}^{ - 1}} - {\left( {\alpha \mathit{\boldsymbol{A + }}\left( {1 - \alpha } \right)\mathit{\boldsymbol{B}}} \right)^{ - 1}} = \\ \alpha {\mathit{\boldsymbol{A}}^{ - 1}} + \left( {1 - \alpha } \right){\mathit{\boldsymbol{B}}^{ - 1}} - {\mathit{\boldsymbol{B}}^{ - 1}}{\left( {\alpha \mathit{\boldsymbol{A}}{\mathit{\boldsymbol{B}}^{ - 1}} + \left( {1 - \alpha } \right)\mathit{\boldsymbol{I}}} \right)^{ - 1}} = \\ {\mathit{\boldsymbol{B}}^{ - 1}}\left( {\alpha \mathit{\boldsymbol{B}}{\mathit{\boldsymbol{A}}^{ - 1}} + \left( {1 - \alpha } \right)\mathit{\boldsymbol{I}}} \right) - {\mathit{\boldsymbol{B}}^{ - 1}}{\left( {\alpha \mathit{\boldsymbol{G + }}\left( {1 - \alpha } \right)\mathit{\boldsymbol{I}}} \right)^{ - 1}} = \\ {\mathit{\boldsymbol{B}}^{ - 1}}\left[ {\alpha {\mathit{\boldsymbol{G}}^{ - 1}} + \left( {1 - \alpha } \right)\mathit{\boldsymbol{I}} - {{\left( {\alpha \mathit{\boldsymbol{G + }}\left( {1 - \alpha } \right)\mathit{\boldsymbol{I}}} \right)}^{ - 1}}} \right] = \\ {\mathit{\boldsymbol{B}}^{ - 1}}\left[ {{\mathit{\boldsymbol{G}}^{ - 1}}\left( {\alpha \mathit{\boldsymbol{I + }}\left( {1 - \alpha } \right)\mathit{\boldsymbol{G}}} \right) - {{\left( {a\mathit{\boldsymbol{C + }}\left( {1 - \alpha } \right)\mathit{\boldsymbol{I}}} \right)}^{ - 1}}} \right] = \\ {\mathit{\boldsymbol{B}}^{ - 1}}\left[ {{\mathit{\boldsymbol{G}}^{ - 1}}\left( {\alpha \mathit{\boldsymbol{I + }}\left( {1 - \alpha } \right)\mathit{\boldsymbol{G}}} \right)\left( {\alpha \mathit{\boldsymbol{G + }}\left( {1 - \alpha } \right)\mathit{\boldsymbol{I}}} \right) - \mathit{\boldsymbol{I}}} \right]{\left( {\alpha \mathit{\boldsymbol{G + }}\left( {1 - \alpha } \right)\mathit{\boldsymbol{I}}} \right)^{ - 1}} = \\ {\mathit{\boldsymbol{B}}^{ - 1}}\left[ {{\mathit{\boldsymbol{G}}^{ - 1}}\left( {{\alpha ^2}\mathit{\boldsymbol{G + }}\alpha \left( {1 - \alpha } \right)\mathit{\boldsymbol{I + }}\alpha \left( {1 - \alpha } \right){\mathit{\boldsymbol{G}}^2} + {{\left( {1 - \alpha } \right)}^2}\mathit{\boldsymbol{G}}} \right) - \mathit{\boldsymbol{I}}} \right]{\left( {\alpha \mathit{\boldsymbol{G + }}\left( {1 - \alpha } \right)\mathit{\boldsymbol{I}}} \right)^{ - 1}} = \\ {\mathit{\boldsymbol{B}}^{ - 1}}\left[ {{\alpha ^2}\mathit{\boldsymbol{I + }}\alpha \left( {1 - \alpha } \right){\mathit{\boldsymbol{G}}^{ - 1}} + \alpha \left( {1 - \alpha } \right)\mathit{\boldsymbol{G}} + {{\left( {1 - \alpha } \right)}^2}\mathit{\boldsymbol{I}} - \mathit{\boldsymbol{I}}} \right]{\left( {\alpha \mathit{\boldsymbol{G + }}\left( {1 - \alpha } \right)\mathit{\boldsymbol{I}}} \right)^{ - 1}} = \\ {\mathit{\boldsymbol{B}}^{ - 1}}\left[ {\alpha \left( {1 - \alpha } \right)\mathit{\boldsymbol{G + }}\alpha \left( {1 - \alpha } \right){\mathit{\boldsymbol{G}}^{ - 1}} + 2\alpha \left( {1 - \alpha } \right)\mathit{\boldsymbol{I}}} \right]{\left( {\alpha \mathit{\boldsymbol{G + }}\left( {1 - \alpha } \right)\mathit{\boldsymbol{I}}} \right)^{ - 1}} = \\ \alpha \left( {1 - \alpha } \right){\mathit{\boldsymbol{B}}^{ - 1}}{\mathit{\boldsymbol{G}}^{ - 1}}{\left( {\mathit{\boldsymbol{I}} - \mathit{\boldsymbol{G}}} \right)^2}{\left( {\alpha \mathit{\boldsymbol{G + }}\left( {1 - \alpha } \right)\mathit{\boldsymbol{I}}} \right)^{ - 1}}, \end{array} $

及B, G, αG+(1-α)I都是M-矩阵且I-G≥0，可知B^-1≥0, C^-1≥0, (αC+(1-α)I)^-1≥0, 于是就有αA^-1+(1-α)B^-1-(αA+(1-α)B)^-1≥0, 因此(αA+(1-α)B)^-1≤αA^-1+(1-α)B^-1.

注  两个M-矩阵的和不一定是M-矩阵.A=$\left( {\begin{array}{*{20}{c}} {\frac{1}{2}}&{ - 1}\\ 0 &{\frac{1}{2}} \end{array}} \right)$, B=$\left( {\begin{array}{*{20}{c}} {\frac{1}{2}}&0\\ { - 1}&{\frac{1}{2}} \end{array}} \right)$是M-矩阵, 由A+B=$\left( {\begin{array}{*{20}{c}} 1 < { - 1}\\ { - 1}< 1 \end{array}} \right)$且det (A+B)=0知, A+B不是M-矩阵.

定理5  设A, B=[b_ij]∈M_n(R), 如果A是M-矩阵, 且M(B)≥A, 则有

(1) B是H-矩阵;

(2) B与|B|都是可逆矩阵;

(3) 0≤|B^-1|≤A^-1;

(4) 0 < detA≤|detB|.

证明  (1) 由于M(B)∈Z_n, 又M(B)≥A, 则由引理2中(1) 知, M(B) 是M-矩阵, 因此B是H-矩阵.

(2) 因为M(B) 是矩阵M-矩阵, 由引理1中(5), 存在正对角矩阵D使得M(B)D是行严格对角占优.由于

$ \begin{array}{l} M\left( \mathit{\boldsymbol{B}} \right)\mathit{\boldsymbol{D = }}\left( {\begin{array}{*{20}{c}} {\left| {{b_{11}}} \right|}&{ - \left| {{b_{12}}} \right|}& \cdots &{ - \left| {{b_{1n}}} \right|}\\ { - \left| {{b_{21}}} \right|}&{ - \left| {{b_{22}}} \right|}& \cdots &{ - \left| {{b_{2n}}} \right|}\\ \cdots & \cdots & \ddots & \cdots \\ { - \left| {{b_{n1}}} \right|}&{ - \left| {{b_{n2}}} \right|}& \cdots &{\left| {{b_{nn}}} \right|} \end{array}} \right)\left( {\begin{array}{*{20}{c}} {{d_1}}&{}&{}&{}\\ {}&{{d_2}}&{}&{}\\ {}&{}& \ddots &{}\\ {}&{}&{}&{{d_n}} \end{array}} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {\begin{array}{*{20}{c}} {\left| {{b_{11}}} \right|{d_1}}&{ - \left| {{b_{12}}} \right|{d_2}}& \cdots &{ - \left| {{b_{1n}}} \right|{d_n}}\\ { - \left| {{b_{21}}} \right|{d_1}}&{ - \left| {{b_{22}}} \right|{d_2}}& \cdots &{ - \left| {{b_{2n}}} \right|{d_n}}\\ \cdots & \cdots & \ddots & \cdots \\ { - \left| {{b_{n1}}} \right|{d_1}}&{ - \left| {{b_{n2}}} \right|{d_2}}& \cdots &{\left| {{b_{nn}}} \right|{d_n}} \end{array}} \right), \end{array} $

则有|b_ii|d_i > $\sum\limits_{j = 1,j \ne i}^n {\left| {{b_{ij}}} \right|} $d_j, 其中i=1, …, n.又

$ \begin{array}{l} \mathit{\boldsymbol{BD = }}\left( {\begin{array}{*{20}{c}} {{b_{11}}}&{{b_{12}}}& \cdots &{{b_{1n}}}\\ {{b_{21}}}&{{b_{22}}}& \cdots &{{b_{2n}}}\\ \cdots & \cdots & \ddots & \cdots \\ {{b_{n1}}}&{{b_{n2}}}& \cdots &{{b_{nn}}} \end{array}} \right)\left( {\begin{array}{*{20}{c}} {{d_1}}&{}&{}&{}\\ {}&{{d_2}}&{}&{}\\ {}&{}& \ddots &{}\\ {}&{}&{}&{{d_n}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {{b_{11}}{d_1}}&{{b_{12}}{d_2}}& \cdots &{{b_{1n}}{d_n}}\\ {{b_{21}}{d_1}}&{{b_{22}}{d_2}}& \cdots &{{b_{2n}}{d_n}}\\ \cdots & \cdots & \ddots & \cdots \\ {{b_{n1}}{d_1}}&{{b_{n2}}{d_2}}& \cdots &{{b_{nn}}{d_n}} \end{array}} \right),\\ \left| \mathit{\boldsymbol{B}} \right|\mathit{\boldsymbol{D = }}\left( {\begin{array}{*{20}{c}} {\left| {{b_{11}}} \right|{d_1}}&{\left| {{b_{12}}} \right|{d_2}}& \cdots &{\left| {{b_{1n}}} \right|{d_n}}\\ {\left| {{b_{21}}} \right|{d_1}}&{\left| {{b_{22}}} \right|{d_2}}& \cdots &{\left| {{b_{2n}}} \right|{d_n}}\\ \cdots & \cdots & \ddots & \cdots \\ {\left| {{b_{n1}}} \right|{d_1}}&{ - \left| {{b_{n2}}} \right|{d_2}}& \cdots &{\left| {{b_{nn}}} \right|{d_n}} \end{array}} \right), \end{array} $

所以BD与|B|D都是行严格对角占优, 因而BD与|B|D是可逆矩阵.再由D是可逆阵知, B与|B|都是可逆矩阵.

(3) 取α > $\mathop {\max }\limits_{1 \le i \le n} ${|b_ii|}, 令P=αI-A.因为A是M-矩阵且M(B)≥A知P≥0, 又由引理1中(2) 知α > ρ(P), 再构造对角矩阵D=(d_ij), 其中d_ii=$\left\{ \begin{array}{l} 1,{b_{ij}} > 0\\ - 1,{b_{ij}} < 0 \end{array} \right.$, 则DB的主对角线元素大于0, 易知|αI-DB|≤P.令R=αI-DB, 即|R|≤P, 由引理3知ρ(R)≤ρ(|R|)≤ρ(P) < α, 因此|(DB)^-1|=|α^-1(1-α^-1R)^-1|=$\left| {\sum\limits_{k = 0}^\infty {{\alpha ^{ - k - 1}}} {\boldsymbol{R}^k}} \right| \le \sum\limits_{k = 0}^\infty {{\alpha ^{ - k - 1}}} {\left| \boldsymbol{R} \right|^k} \le \sum\limits_{k = 0}^\infty {{\alpha ^{ - k - 1}}} {\boldsymbol{P}^k} = {\boldsymbol{A}^{ - 1}}$, 再由D的构造知, |(DB)^-1|=|B^-1|, 故综上可得A^-1≥|B^-1|≥0.

(4) 对n作数学归纳法证明.当n=1时, 由已证得(3) 知a₁₁^-1≥|b₁₁^-1|, 因此a₁₁≤|b₁₁|, 即|detB|≥detA.假设n-1时成立, 将A与B进行分块, A=$\left( {\begin{array}{*{20}{c}} {{A_{11}}}&{{A_{12}}}\\ {{A_{21}}}&{{A_{nn}}} \end{array}} \right)$, B=$\left( {\begin{array}{*{20}{c}} {{B_{11}}}&{{B_{12}}}\\ {{B_{21}}}&{{B_{nn}}} \end{array}} \right)$, 其中A₁₁, B₁₁∈M_n-1.由归纳假设知|detB₁₁| > detA₁₁, 再由(3) 知A^-1-|B^-1|的(n, n) 元素$\frac{{{\rm{det}}{\boldsymbol{A}_{11}}}}{{{\rm{det}}\boldsymbol{A}}} - \frac{{|{\rm{det}}{\boldsymbol{B}_{11}}|}}{{|{\rm{det}}\boldsymbol{B}|}}$≥0, 进而由A是M-矩阵及B是可逆矩阵知, detA > 0, |detB| > 0, |detB| > $\frac{{|{\rm{det}}{\boldsymbol{B}_{11}}|}}{{{\rm{det}}{\boldsymbol{A}_{11}}}}$detA > detA > 0.综上得证.