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弱鏈對角占優B-矩陣線性補問題解的誤差界估計式的改進

周平 李耀堂

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弱鏈對角占優B-矩陣線性補問題解的誤差界估計式的改進

    作者簡介: 周 平(1987?),女,云南人,碩士,講師,主要從事矩陣理論及其應用方面的研究. E-mail:yunpzjy@126.com;
    通訊作者: 李耀堂, liyaotang@ynu.edu.cn
  • 中圖分類號: O151.21

The improvement on the error bound estimation of the solution of the linear supplementary problem for the weak chain diagonally dominant B-matrix

    Corresponding author: LI Yao-tang, liyaotang@ynu.edu.cn
  • CLC number: O151.21

  • 摘要: 利用弱鏈對角占優M-矩陣的逆矩陣的無窮大范數的上界估計式和不等式技巧,給出了弱鏈對角占優B-矩陣線性補問題解的誤差界的2個新估計式,改進了已有結果.
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    [17] 趙仁慶, 劉鵬. 弱鏈對角占優M-矩陣的逆矩陣的無窮大范數的上界估計[J]. 楚雄師范學院學報, 2014, 29(3): 5-11. DOI:  10.3969/j.issn.1671-7406.2014.03.002. Zhao R Q, Liu P. Estimation on upper bounds of the infinity norms of inverses for weakly chained diagonally dominant M-matrices[J]. Journal of Chuxiong Normal University, 2014, 29(3): 5-11.
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出版歷程
  • 收稿日期:  2019-11-28
  • 錄用日期:  2020-05-18
  • 網絡出版日期:  2020-08-05

弱鏈對角占優B-矩陣線性補問題解的誤差界估計式的改進

    作者簡介:周 平(1987?),女,云南人,碩士,講師,主要從事矩陣理論及其應用方面的研究. E-mail:yunpzjy@126.com
    通訊作者: 李耀堂, liyaotang@ynu.edu.cn
  • 1. 文山學院 數學與工程學院,云南 文山 663099
  • 2. 云南大學 數學與統計學院,云南 昆明 650500

摘要: 利用弱鏈對角占優M-矩陣的逆矩陣的無窮大范數的上界估計式和不等式技巧,給出了弱鏈對角占優B-矩陣線性補問題解的誤差界的2個新估計式,改進了已有結果.

English Abstract

  • 線性補問題(${\rm{LCP}}(M,q)$)是在經濟金融學的期權定價問題研究、物理學中的彈性接觸問題和自由邊界問題研究、運籌學中的二次規劃問題和最優停止問題研究等領域[1-5]具有廣泛應用的優化問題,其定義為:求一 ${x^ * } \in {{\bf{R}}^n}$,${x^ * } \geqslant 0$,使其滿足:

    $\qquad {\left( {M{x^ * } + q} \right)^{\rm{T}}}{x^ * } = 0,M{x^ * } + q \geqslant 0,$

    其中系數矩陣 $M = ({m_{ij}}) \in {{\bf{R}}^{n \times n}}$,$q \in {{\bf{R}}^n}$.

    $M$$P$-矩陣(即 $M$ 的所有主子式為正)時,${\rm{LCP}}(M,q)$ 的解 ${x^ * }$ 存在且唯一[1].

    線性補問題 ${\rm{LCP}}(M,q)$ 解的誤差估計在實際應用中具有重要的意義. 2007年陳小君等在文獻[6]中給出如下結果:當 $M$$P$-矩陣時,有

    $\qquad {\left\| {x - {x^ * }} \right\|_\infty } \leqslant \mathop {\max }\limits_{d \in {{[0,1]}^n}} {\left\| {{{\left( {I - D + DM} \right)}^{ - 1}}} \right\|_\infty } \cdot {\left\| {r(x)} \right\|_\infty },$

    其中 $x$${\rm{LCP}}(M,q)$${x^ * }$ 的近似:

    $\qquad r(x) = \min \left\{ {x,Mx + q} \right\},D = {\rm{diag}}({d_i}), 0 \leqslant {d_i} \leqslant 1,i = 1,2, \cdots ,n.$

    ${\;\beta _\infty }(M) = \mathop {\max }\limits_{d \in {{[0,1]}^n}} {\left\| {{{\left( {I - D + DM} \right)}^{ - 1}}D} \right\|_\infty }$. 線性補問題解的誤差界與其系數矩陣 $M$ 密切相關,如何應用 $M$ 的結構和性質對 ${\beta _\infty }(M)$ 進行最小化處理對 ${\rm{LCP}}(M,q)$ 解的誤差估計具有重要意義. 該問題已引起眾多學者的關注和研究,獲得了系數矩陣 $M$ 為某些特殊矩陣時 ${\rm{LCP}}(M,q)$ 解的誤差估計式[7-16]. 本文應用文獻[17]所給弱鏈對角占優 $M$-矩陣的逆矩陣的無窮大范數上界估計式對 ${\rm{LCP}}(M,q)$ 解的誤差估計式進行改進,獲得了2個新估計式. 理論證明和數值算例表明所獲估計式優于文獻[13-15]中的結果.

    • $N = \left\{ {1,2, \cdots ,n} \right\}$,${{\bf{R}}^{n \times n}}$${{\bf{C}}^{n \times n}}$)是全體 $n \times n$ 階實(復)矩陣構成的集合. 設 $A = ({a_{ij}}) \in {{\bf{R}}^{n \times n}}$,若 ${a_{ij}} \geqslant 0$,$i,j \in {{N}}$,則稱 $A$ 為非負矩陣,記為 $A \geqslant 0$;若 ${a_{ij}} \leqslant 0$,$i \ne j$,$i,j \in N$,則稱 $A$$Z$-矩陣;若 $A$$Z$-矩陣且 ${A^{ - 1}} \geqslant 0$,則稱 $A$$M$-矩陣[18].

      定義 1[19] 設 $A = ({a_{ij}}) \in {{\bf{R}}^{n \times n}}$,若

      $\qquad \left| {{a_{ii}}} \right| > {r_i}(A) = \sum\limits_{j \ne i} {\left| {{a_{ij}}} \right|} ,\forall i \in {{N}},$

      則稱 $A$ 為嚴格對角占優矩陣;若 $\forall i \in N$,$\left| {{a_{ii}}} \right| \geqslant {r_i}(A)$;$J(A) = $$\{ i \in N:|{a_{ii}}| > {r_i}(A)\} \ne \emptyset $ 且對任意一個 $i \notin J(A)$ 存在非零元素鏈 ${a_{i,{i_1}}},\; \; {a_{{i_1},{i_2}}},\; \; \cdots ,{a_{{i_r},j}}$ 使得 $j \in J(A)$,則稱 $A$ 為弱鏈對角占優矩陣.

      定義 2[5] 設 $A = ({a_{ij}}) \in {{\bf{R}}^{n \times n}}$,如果

      $\qquad \sum\limits_{k = 1}^n {{a_{ik}}} > 0,\forall i \in {{N;}}\frac{1}{n}\left( {\sum\limits_{k = 1}^n {{a_{ik}}} } \right) > {a_{ij}},\forall j \in N,\; j \ne i.$

      則稱 $A$$B$-矩陣.

      定義 3[15] 設 $A = ({a_{ij}}) \in {{\bf{R}}^{n \times n}}$,將 $A$ 表示為 $A = {B^ + } + C$,其中

      $\qquad {B^ + } = ({b_{ij}}) = \left( {\begin{array}{*{20}{c}} {{a_{11}} - r_1^ + }& \cdots &{{a_{1n}} - r_1^ + }\\ \vdots & & \vdots \\ {{a_{n1}} - r_n^ + }& \cdots &{{a_{nn}} - r_n^ + } \end{array}} \right),C = ({c_{ij}}) = \left( {\begin{array}{*{20}{c}} {r_1^ + }& \cdots &{r_1^ + }\\ \vdots & & \vdots \\ {r_n^ + }& \cdots &{r_n^ + } \end{array}} \right),$

      $r_i^ + = \max \left\{ {0,{a_{ij}}|j \ne i} \right\}$,若 ${B^ + }$ 為弱鏈對角占優矩陣且主對角元恒為正,則稱矩陣 $A$ 是弱鏈對角占優 $B$-矩陣.

      2016年,文獻[15]給出了 $B$-矩陣與弱鏈對角占優 $B$-矩陣之間的關系,即:若矩陣 $A$$B$-矩陣,則 $A$ 是弱鏈對角占優 $B$-矩陣([15,命題1]).

      文獻[13]給出 $M = ({m_{ij}}) \in {{\bf{R}}^{n \times n}}$$B$-矩陣時 ${\beta _\infty }(M)$ 的如下估計式:

      $\qquad {\beta _\infty }(M) = \mathop {\max }\limits_{d \in {{[0,1]}^n}} {\left\| {{{\left( {I - D + DM} \right)}^{ - 1}}} \right\|_\infty } \leqslant \frac{{n - 1}}{{\min \{ \beta ,1\} }},$

      其中 $\;\beta = \mathop {\min }\limits_{i \in N} \left\{ {{\beta _i}} \right\},\; \; \; {\beta _i} = {b_{ii}} - \displaystyle\sum\limits_{j = 1}^n {\left| {{b_{ij}}} \right|} ,\; \; j \ne i$.

      2016年,文獻[14]獲得優于(2)式的如下結果:設 $M = ({m_{ij}}) \in {{\bf{R}}^{n \times n}}$$B$-矩陣,則

      $\qquad {\beta _\infty }(M) = \mathop {\max }\limits_{d \in {{[0,1]}^n}} {\left\| {{{\left( {I - D + DM} \right)}^{ - 1}}} \right\|_\infty } \leqslant \sum\limits_{i = 1}^n {\frac{{n - 1}}{{\min \left\{ {{{\overline \beta }_i},1} \right\}}}} \prod\limits_{j = 1}^{i - 1} {\left( {1 + \frac{1}{{{{\overline \beta }_j}}}\sum\limits_{k = j + 1}^n {\left| {{b_{jk}}} \right|} } \right)} ,$

      其中 $\; \; {\overline \beta _i} = {b_{ii}} - \displaystyle\sum\limits_{j = 1}^n {\left| {{b_{ij}}} \right|} {l_i}({B^ + }),\; \; \; \; {l_i}({B^ + }) = \mathop {\max }\limits_{k \leqslant i \leqslant n} \; \left\{ {\frac{1}{{{b_{ii}}}}\displaystyle\sum\limits_{j = k}^n {\left| {{b_{ij}}} \right|} } \right\},j \ne i$;當 $i = 1$ 時,$\displaystyle\prod\limits_{j = 1}^{i - 1} {\left( {1 + \dfrac{1}{{{{\overline \beta }_j}}}\displaystyle\sum\limits_{k = j + 1}^n {\left| {{b_{jk}}} \right|} } \right)} = 1$.

      文獻[15]中又得到如下結果:設 $M = ({m_{ij}}) \in {{\bf{R}}^{n \times n}}$ 為弱鏈對角占優 $B$-矩陣,則

      $\qquad {\beta _\infty }(M) = \mathop {\max }\limits_{d \in {{[0,1]}^n}} {\left\| {{{\left( {I - D + DM} \right)}^{ - 1}}} \right\|_\infty } \leqslant \sum\limits_{i = 1}^n {\frac{{n - 1}}{{\min \left\{ {{{\tilde \beta }_i},1} \right\}}}} \prod\limits_{j = 1}^{i - 1} {\frac{{{b_{jj}}}}{{{{\tilde \beta }_j}}}} ,$

      其中 $\; {\tilde \beta _i} = {b_{ii}} - \displaystyle\sum\limits_{j = i + 1}^n {\left| {{b_{ij}}} \right|} > 0$,且當 $i = 1$ 時,$\displaystyle\prod\limits_{j = 1}^{i - 1} {\frac{{{b_{jj}}}}{{{{\tilde \beta }_j}}}} = 1$.

      本文繼續對 $M$ 為弱鏈對角占優 $B$-矩陣時 ${\rm{LCP}}(M,q)$ 解的誤差估計問題進行研究. 為敘述方便先給出如下記號:

      $A = ({a_{ij}}) \in {{\bf{R}}^{n \times n}}$,$i,j,k \in N,\; $$i \ne j,\; $$\; \; \; i + 1 \leqslant k \leqslant n$,令

      $\qquad {R_i}\left( A \right) = \frac{{\displaystyle\sum\limits_{j \ne i} {|{a_{ij}}|} }}{{|{a_{ii}}|}},{u_i}\left( A \right) = \frac{{\displaystyle\sum\limits_{j = i + 1} {|{a_{ij}}|} }}{{|{a_{ii}}|}},{s_k}\left( A \right) = \max \left\{ {\frac{{\displaystyle\sum\limits_{i + k \ne j,k \leqslant j \leqslant n} {|{a_{i + k,j}}|} }}{{|{a_{i + k,i + k}}|}}} \right\},$

      $\qquad {t_{ki}} =\left\{ {\begin{array}{*{20}{l}} { \dfrac{{|{a_{ki}}|}}{{|{a_{kk}}| - \displaystyle\sum\limits_{j = i + 1,k \ne j}^n {{a_{kj}}} }},}&{ |{a_{ki}}| \ne 0,}\\ {0,}&{|{a_{ki}}| = 0,} \end{array}} \right.{t_i}\left( A \right) = \left\{ {\begin{array}{*{20}{l}} {\mathop {\max }\limits_{i + 1 \leqslant k \leqslant n} \{ {t_{ki}}\} ,}&{1 \leqslant i \leqslant n - 1,}\\ {0,}&{i = n}, \end{array}} \right.$

      $\qquad {p_k}\left( A \right) = \max \left\{ {\frac{{|{a_{i + k,k}}| + \displaystyle\sum\limits_{h = k + 1,h \ne i + k}^n {|{a_{i + k,h}}|} {s_k}}}{{|{a_{i + k,i + k}}|}},i = 1,2, \cdots ,n - k} \right\}.$

    • 本節我們應用文獻[17]所給的如下引理 1,研究 $M$ 為弱鏈對角占優 $B$-矩陣時 $LCP(M,q)$ 解的誤差估計問題.

      引理 1[17] 設 $A = ({a_{ij}}) \in {{\bf{R}}^{n \times n}}$ 為弱鏈對角占優 $B$-矩陣,則

      $\qquad \begin{split} {\left\| {{A^{ - 1}}} \right\|_\infty } \leqslant & \max \left\{ {\sum\limits_{i = 1}^n {\left[ {\frac{1}{{{a_{ii}}\left( {1 - {u_i}\left( A \right){t_i}\left( A \right)} \right)}}} \right]} } \right.\prod\limits_{j = 1}^{i - 1} {\frac{{{u_j}\left( A \right)}}{{1 - {u_j}\left( A \right){t_j}\left( A \right)}}} ,\\ &\qquad \;\left. {\sum\limits_{i = 1}^n {\left[ {\frac{{{p_i}\left( A \right)}}{{{a_{ii}}\left( {1 - {u_i}\left( A \right){t_i}\left( A \right)} \right)}}} \right]\prod\limits_{j = 1}^{i - 1} {\left( {1 + \frac{{{p_j}\left( A \right){u_j}\left( A \right)}}{{1 - {u_j}\left( A \right){t_j}\left( A \right)}}} \right)} } } \right\}. \end{split}$

      其中,當$i = 1$時,有

      $\qquad \prod\limits_{j = 1}^{i - 1} {\frac{{{u_j}\left( A \right)}}{{1 - {u_j}\left( A \right){t_j}\left( A \right)}}} = 1,\prod\limits_{j = 1}^{i - 1} {\left( {1 + \frac{{{p_j}\left( A \right){u_j}\left( A \right)}}{{1 - {u_j}\left( A \right){t_j}\left( A \right)}}} \right)} = 1.$

      引理 2[13] 若 $\gamma > 0,\; \; \eta \geqslant 0$,則對任意的 $x \in [0,1]$,有

      $\qquad \frac{1}{{1 - x + \gamma x}} \leqslant \frac{1}{{\min \left\{ {\gamma ,1} \right\}}},\frac{{\eta x}}{{1 - x + \gamma x}} \leqslant \frac{\eta }{\gamma }.$

      定理 1 設 $M = ({m_{ij}}) \in {{\bf{R}}^{n \times n}}$ 為弱鏈對角占優 $B$-矩陣,令 $M = {B^ + } + C$,${B^ + } = ({b_{ij}})$$C$ 同(1)式,則

      $\qquad \begin{split} {\beta _\infty }(M) =& \mathop {\max }\limits_{d \in {{[0,1]}^n}} {\left\| {{{\left( {I - D + DM} \right)}^{ - 1}}} \right\|_\infty } \leqslant \\ &\max \left\{ {\sum\limits_{i = 1}^n {\frac{{n - 1}}{{\min \left\{ {{{\overset{\frown} \beta }_i},1} \right\}}}} \prod\limits_{j = 1}^{i - 1} {\frac{1}{{{{\overset{\frown} \beta }_j}}}\sum\limits_{k = j + 1}^n {\left| {{b_{jk}}} \right|} } } \right.,\left. {\sum\limits_{i = 1}^n {\frac{{(n - 1){p_i}({B^ + })}}{{\min \left\{ {{{\overset{\frown} \beta }_i},1} \right\}}}} \prod\limits_{j = 1}^{i - 1} {\left( {1 + \frac{{{p_j}({B^ + })}}{{{{\overset{\frown} \beta }_j}}}\sum\limits_{k = j + 1}^n {\left| {{b_{jk}}} \right|} } \right)} } \right\}. \end{split}$

      其中 ${{\overset{\frown} \beta } _i} = {b_{ii}} - \displaystyle\sum\limits_{j = i + 1}^n {\left| {{b_{ij}}} \right|} {t_i}({B^ + })$;$i = 1$ 時,$\displaystyle\prod\limits_{j = 1}^{i - 1} {\frac{1}{{{{{\overset{\frown} \beta } }_j}}}\displaystyle\sum\limits_{k = j + 1}^n {\left| {{b_{jk}}} \right|} } = 1$,$\displaystyle\prod\limits_{j = 1}^{i - 1} {\left( {1 + \frac{{{p_j}({B^ + })}}{{{{{\overset{\frown} \beta } }_j}}}\sum\limits_{k = j + 1}^n {\left| {{b_{jk}}} \right|} } \right)} = 1$.

      證明 令 ${M_D} = I - D + DM$,則

      $\qquad {M_D} = I - D + D({B^ + } + C)\; \; = (I - D + D{B^ + }) + DC\; ,$

      $I - D + D{B^ + } = B_D^ + ,\;DC = {C_D}$,則 ${M_D}$ 可表示為 ${M_D} = B_D^ + + {C_D}$,由文獻[15]中的定理2知,$B_D^ + $ 是弱鏈對角占優 $M$-矩陣,且

      $\qquad {\left\| {M_D^{ - 1}} \right\|_\infty } \leqslant {\left\| {{{\left( {I + {{(B_D^ + )}^{ - 1}}{C_D}} \right)}^{ - 1}}} \right\|_\infty } \cdot {\left\| {{{(B_D^ + )}^{ - 1}}} \right\|_\infty } \leqslant (n - 1){\left\| {{{(B_D^ + )}^{ - 1}}} \right\|_\infty }.$

      由引理1得

      $\qquad \begin{split} {\left\| {{{(B_D^ + )}^{ - 1}}} \right\|_\infty } \leqslant & \max \left\{ {\sum\limits_{i = 1}^n {\left[ {\frac{1}{{(1 - {d_i} + {b_{ii}}{d_i})\left( {1 - {u_i}(B_D^ + ){t_i}(B_D^ + )} \right)}}} \right]} } \right.\prod\limits_{j = 1}^{i - 1} {\frac{{{u_j}(B_D^ + )}}{{1 - {u_j}(B_D^ + ){t_j}(B_D^ + )}}} ,\\ &\qquad\; \left. {\sum\limits_{i = 1}^n {\left[ {\frac{{{p_i}(B_D^ + )}}{{(1 - {d_i} + {b_{ii}}{d_i})\left( {1 - {u_i}(B_D^ + ){t_i}(B_D^ + )} \right)}}} \right]\prod\limits_{j = 1}^{i - 1} {\left( {1 + \frac{{{p_j}(B_D^ + ){u_j}(B_D^ + )}}{{1 - {u_j}(B_D^ + ){t_j}(B_D^ + )}}} \right)} } } \right\}, \end{split}$

      再由引理2知,對任意的 $i,j,k \in N$$j \ne i$,有

      $\qquad {u_i}(B_D^ + ) = \frac{{\displaystyle\sum\limits_{j = i + 1}^n {\left| {{b_{ij}}} \right|{d_i}} }}{{1 - {d_i} + {b_{ii}}{d_i}}} \leqslant \frac{{\displaystyle\sum\limits_{j = i + 1}^n {\left| {{b_{ij}}} \right|} }}{{{b_{ii}}}} = {u_i}({B^ + }),$

      $\qquad {t_{ki}}(B_D^ + ) = \frac{{|{b_{ki}}|{d_k}}}{{1 - {d_k} + {b_{kk}}{d_k} - \displaystyle\sum\limits_{j = i + 1,k \ne j}^n {|{b_{kj}}|{d_k}} }} \leqslant \frac{{|{b_{ki}}|}}{{{b_{kk}} - \displaystyle\sum\limits_{j = i + 1,k \ne j}^n {|{b_{kj}}|} }} = {t_{ki}}({B^ + }),$

      $\qquad {t_k}(B_D^ + ) \leqslant \mathop {\max }\limits_{i + 1 \leqslant k \leqslant n} \left\{ {\frac{{|{b_{ki}}|}}{{{b_{kk}} - \displaystyle\sum\limits_{j = i + 1,k \ne j}^n {|{b_{kj}}|} }}} \right\} = {t_k}({B^ + }) < 1,$

      $\qquad {s_k}(B_D^ + ) = \max \left\{ {\frac{{\displaystyle\sum\limits_{i + k \ne j} {|{b_{i + k,j}}|{d_{i + k}}} }}{{1 - {d_{i + k}} + {b_{i + k,i + k}}{d_{i + k}}}}} \right\} \leqslant \max \left\{ {\frac{{\displaystyle\sum\limits_{i + k \ne j} {|{b_{i + k,j}}|} }}{{{b_{i + k,i + k}}}}} \right\} = {s_k}({B^ + }),$

      $\qquad {p_k}(B_D^ + ) = \max \left\{ {\frac{{|{a_{i + k,k}}| + \displaystyle\sum\limits_{h = k + 1,h \ne i + k}^n {|{a_{i + k,h}}|} {s_k}}}{{|{a_{i + k,i + k}}|}}} \right\} \leqslant \max \left\{ {\frac{{|{b_{i + k,k}}| + \displaystyle\sum\limits_{h = k + 1,h \ne i + k}^n {|{b_{i + k,h}}|} {s_k}({B^ + })}}{{{b_{i + k,i + k}}}}} \right\} = {p_k}({B^ + }). $

      于是由(8)~(12)式與引理2得

      $\qquad \frac{{{u_i}(B_D^ + )}}{{1 - {u_i}(B_D^ + ){t_i}(B_D^ + )}} = \frac{{\displaystyle\sum\limits_{j = i + 1}^n {\left| {{b_{ij}}} \right|{d_i}} }}{{1 - {d_i} + {b_{ii}}{d_i} - \displaystyle\sum\limits_{j = i + 1}^n {\left| {{b_{ij}}} \right|{d_i}{t_i}(B_D^ + )} }} \leqslant \frac{{\displaystyle\sum\limits_{j = i + 1}^n {\left| {{b_{ij}}} \right|} }}{{{b_{ii}} - \displaystyle\sum\limits_{j = i + 1}^n {\left| {{b_{ij}}} \right|} {t_i}({B^ + })}} = \frac{1}{{{{\overset{\frown} \beta }_i}}}\displaystyle\sum\limits_{j = i + 1}^n {\left| {{b_{ij}}} \right|} ,$

      $\qquad \begin{split} \frac{1}{{1 - {d_i} + {b_{ii}}{d_i} - \left( {1 - {u_i}(B_D^ + ){t_i}(B_D^ + )} \right)}} =& \frac{1}{{1 - {d_i} + {b_{ii}}{d_i} - \displaystyle\sum\limits_{j = i + 1}^n {|{b_{ij}}|} {d_i}{t_i}(B_D^ + )}}\leqslant\\ & \frac{1}{{\min \left\{ {{b_{ii}} - \displaystyle\sum\limits_{j = i + 1}^n {|{b_{ij}}|} {t_i}({B^ + }),1} \right\}}} = \frac{1}{{\min \left\{ {{{\overset{\frown} \beta }_i},1} \right\}}}. \end{split}$

      再由(13)~(14)式,有

      $\qquad {\left\| {{{(B_D^ + )}^{ - 1}}} \right\|_\infty } \leqslant \max \left\{ {\sum\limits_{i = 1}^n {\frac{1}{{\min \left\{ {{{\overset{\frown} \beta }_i},1} \right\}}}} \prod\limits_{j = 1}^{i - 1} {\left( {1 + \frac{1}{{{{\overset{\frown} \beta }_j}}}\sum\limits_{j = i + 1}^n {\left| {{b_{ij}}} \right|} } \right)} } \right.,\left. {\sum\limits_{i = 1}^n {\frac{{{p_i}({B^ + })}}{{\min \left\{ {{{\overset{\frown} \beta }_i},1} \right\}}}} \prod\limits_{j = 1}^{i - 1} {\left( {1 + \frac{{{p_j}({B^ + })}}{{{{\overset{\frown} \beta }_j}}}\sum\limits_{k = j + 1}^n {\left| {{b_{jk}}} \right|} } \right)} } \right\}.$

      由(6),(15)式即得定理1成立. 證畢.

      $B$-矩陣和弱鏈對角占優 $B$-矩陣的關系可直接得如下推論 1.

      推論 1 設 $M = ({m_{ij}}) \in {{\bf{R}}^{n \times n}}$$B$-矩陣,令 $M = {B^ + } + C$,${B^ + } = ({b_{ij}})$$C$ 同(1)式,則

      $\qquad \begin{split} {\beta _\infty }(M) = \mathop {\max }\limits_{d \in {{[0,1]}^n}} {\left\| {{{\left( {I - D + DM} \right)}^{ - 1}}} \right\|_\infty } \leqslant &\max \left\{ {\sum\limits_{i = 1}^n {\frac{{n - 1}}{{\min \left\{ {{{\overset{\frown} \beta }_i},1} \right\}}}} \prod\limits_{j = 1}^{i - 1} {\frac{1}{{{{\overset{\frown} \beta }_j}}}\sum\limits_{k = j + 1}^n {\left| {{b_{jk}}} \right|} } } \right.,\\ &\qquad \;\left. {\sum\limits_{i = 1}^n {\frac{{(n - 1){p_i}({B^ + })}}{{\min \left\{ {{{\overset{\frown} \beta }_i},1} \right\}}}} \prod\limits_{j = 1}^{i - 1} {\left( {1 + \frac{{{p_j}({B^ + })}}{{{{\overset{\frown} \beta }_j}}}\sum\limits_{k = j + 1}^n {\left| {{b_{jk}}} \right|} } \right)} } \right\}. \end{split}$

      下面對估計式(3)~(5)進行比較.

      定理 2 如果 $M = ({m_{ij}}) \in {{\bf{R}}^{n \times n}}$ 是弱鏈對角占優 $B$-矩陣,令 $M = {B^ + } + C$,${B^ + } = ({b_{ij}})$$C$ 同(1)式,則

      $\qquad \begin{split} &\max \left\{ {\sum\limits_{i = 1}^n {\frac{{n - 1}}{{\min \left\{ {{{\overset{\frown} \beta }_i},1} \right\}}}} \prod\limits_{j = 1}^{i - 1} {\frac{1}{{{{\overset{\frown} \beta }_j}}}\sum\limits_{k = j + 1}^n {\left| {{b_{jk}}} \right|} } } \right.,\left. {\sum\limits_{i = 1}^n {\frac{{(n - 1){p_i}({B^ + })}}{{\min \left\{ {{{\overset{\frown} \beta }_i},1} \right\}}}} \prod\limits_{j = 1}^{i - 1} {\left( {1 + \frac{{{p_j}({B^ + })}}{{{{\overset{\frown} \beta }_j}}}\sum\limits_{k = j + 1}^n {\left| {{b_{jk}}} \right|} } \right)} } \right\}\leqslant\\ & \qquad\; \sum\limits_{i = 1}^n {\frac{{n - 1}}{{\min \left\{ {{{\overline \beta }_i},1} \right\}}}} \prod\limits_{j = 1}^{i - 1} {\left( {1 + \frac{1}{{{{\overline \beta }_j}}}\sum\limits_{k = j + 1}^n {\left| {{b_{jk}}} \right|} } \right)} \leqslant \sum\limits_{i = 1}^n {\frac{{n - 1}}{{\min \left\{ {{{\tilde \beta }_i},1} \right\}}}} \prod\limits_{j = 1}^{i - 1} {\frac{{{b_{jj}}}}{{{{\tilde \beta }_j}}}} . \end{split}$

      其中 ${\overline \beta _i}$,${\tilde \beta _i}$${{\overset{\frown} \beta } _i}$ 分別由(3)式,(4)式和(5)式定義.

      證明 因為矩陣 ${B^ + }$ 是弱鏈對角占優的,且 ${b_{ii}} > 0$. 由 ${p_i}$ 的定義知,對任意的 $i \in N$,有 $0 \leqslant {p_i} \leqslant 1$;對任意的 $k \in N$,$i + 1 \leqslant k \leqslant n$,$1 \leqslant i \leqslant n - 1$,有

      $\qquad \begin{split} {l_{ki}}({B^ + }) - {t_{ki}}({B^ + }) =& \frac{{|{b_{ki}}| + \displaystyle\sum\limits_{j = i + 1,j \ne k}^n {|{b_{kj}}|} }}{{|{b_{kk}}|}} - \frac{{|{b_{ki}}|}}{{|{b_{kk}}| - \displaystyle\sum\limits_{j = i + 1,j \ne k}^n {|{b_{kj}}|} }}= \\ & \frac{{\displaystyle\sum\limits_{j = i + 1,j \ne k}^n {|{b_{kj}}|} \left( {|{b_{kk}}| - |{b_{ki}}| - \displaystyle\sum\limits_{j = i + 1,j \ne k}^n {|{b_{kj}}|} } \right)}}{{|{b_{kk}}|\left( {|{b_{kk}}| - \displaystyle\sum\limits_{j = i + 1,j \ne k}^n {|{b_{kj}}|} } \right)}} \geqslant 0. \end{split}$

      從而 ${l_{ki}}({B^ + }) \geqslant {t_{ki}}({B^ + })$,進而 $0 \leqslant {t_i}({B^ + }) \leqslant {l_i}({B^ + }) < 1$. 根據(18)式,$\forall i \in N$

      $\qquad \; {\overline \beta _i} = {b_{ii}} - \sum\limits_{j = 1}^n {\left| {{b_{ij}}} \right|} {l_i}({B^ + }) \leqslant {b_{ii}} - \sum\limits_{j = 1}^n {\left| {{b_{ij}}} \right|} {t_i}({B^ + }) = {{\overset{\frown} \beta } _i},$

      $\qquad \frac{1}{{\min \left\{ {{{{\overset{\frown} \beta } }_i},1} \right\}}} \leqslant \frac{1}{{\min \left\{ {{{\overline \beta }_i},1} \right\}}}.$

      對任意的 $1 \leqslant j \leqslant n$,有

      $\qquad \frac{1}{{{{{\overset{\frown} \beta } }_j}}}\sum\limits_{k = j + 1}^n {\left| {{b_{jk}}} \right|} \leqslant 1 + \frac{1}{{{{\overline \beta }_j}}}\sum\limits_{k = j + 1}^n {\left| {{b_{jk}}} \right|} .$

      由(19)式和(20)式,知

      $\qquad \sum\limits_{i = 1}^n {\frac{{n - 1}}{{\min \left\{ {{{{\overset{\frown} \beta } }_i},1} \right\}}}} \prod\limits_{j = 1}^{i - 1} {\frac{1}{{{{{\overset{\frown} \beta } }_j}}}\sum\limits_{k = j + 1}^n {\left| {{b_{jk}}} \right|} } \leqslant \sum\limits_{i = 1}^n {\frac{{n - 1}}{{\min \left\{ {{{\overline \beta }_i},1} \right\}}}} \prod\limits_{j = 1}^{i - 1} {\left( {1 + \frac{1}{{{{\overline \beta }_j}}}\sum\limits_{k = j + 1}^n {\left| {{b_{jk}}} \right|} } \right)} ,$

      $\qquad \sum\limits_{i = 1}^n {\frac{{(n - 1){p_i}({B^ + })}}{{\min \left\{ {{{{\overset{\frown} \beta } }_i},1} \right\}}}} \prod\limits_{j = 1}^{i - 1} {\left( {1 + \frac{{{p_j}({B^ + })}}{{{{{\overset{\frown} \beta } }_j}}}\sum\limits_{k = j + 1}^n {\left| {{b_{jk}}} \right|} } \right)} \leqslant \sum\limits_{i = 1}^n {\frac{{n - 1}}{{\min \left\{ {{{\overline \beta }_i},1} \right\}}}} \prod\limits_{j = 1}^{i - 1} {\left( {1 + \frac{1}{{{{\overline \beta }_j}}}\sum\limits_{k = j + 1}^n {\left| {{b_{jk}}} \right|} } \right)} .$

      另外,注意到

      $\qquad \; {\overline \beta _i} = {b_{ii}} - \sum\limits_{j = 1}^n {\left| {{b_{ij}}} \right|} {l_i}({B^ + }) \leqslant {b_{ii}} - \sum\limits_{j = i + 1}^n {\left| {{b_{ij}}} \right|} = {\tilde \beta _i},$

      $\qquad \frac{1}{{\min \left\{ {{{\tilde \beta }_i},1} \right\}}} \leqslant \frac{1}{{\min \left\{ {{{\overline \beta }_i},1} \right\}}}.$

      對任意的 $1 \leqslant j \leqslant n$,有

      $\qquad 1 + \frac{1}{{{{\overline \beta }_j}}}\sum\limits_{k = j + 1}^n {\left| {{b_{jk}}} \right|} \leqslant 1 + \frac{1}{{{{\tilde \beta }_j}}}\sum\limits_{k = j + 1}^n {\left| {{b_{jk}}} \right|} = \frac{{{{\tilde \beta }_j} + \displaystyle\sum\limits_{k = j + 1}^n {\left| {{b_{jk}}} \right|} }}{{{{\tilde \beta }_j}}} = \frac{{{b_{jj}}}}{{{{\tilde \beta }_j}}}.$

      由(23)式和(24)式得

      $\qquad \sum\limits_{i = 1}^n {\frac{{n - 1}}{{\min \left\{ {{{\overline \beta }_i},1} \right\}}}} \prod\limits_{j = 1}^{i - 1} {\left( {1 + \frac{1}{{{{\overline \beta }_j}}}\sum\limits_{k = j + 1}^n {\left| {{b_{jk}}} \right|} } \right)} \leqslant \sum\limits_{i = 1}^n {\frac{{n - 1}}{{\min \left\{ {{{\tilde \beta }_i},1} \right\}}}} \prod\limits_{j = 1}^{i - 1} {\frac{{{b_{jj}}}}{{{{\tilde \beta }_j}}}} .$

      綜合(21)~(25)式知定理 2結論成立. 證畢.

      注 1 定理2表明本文中所獲估計式比(3)式和(4)式更優.

    • 例 1 考慮如下弱鏈對角占優 $B$-矩陣[15]

      $\qquad M = \left[ {\begin{array}{*{20}{c}} {1.5}&{0.2}&{0.4}&{0.5} \\ { - 0.1}&{1.5}&{0.5}&{0.1} \\ {0.5}&{ - 0.1}&{1.5}&{0.1} \\ {0.4}&{0.4}&{0.8}&{1.8} \end{array}} \right],$

      $M = {B^ + } + C$,且

      $\qquad {B^ + } = \left[ {\begin{array}{*{20}{c}} 1&{ - 0.3}&{ - 0.1}&0 \\ { - 0.6}&1&0&{ - 0.4} \\ 0&{ - 0.6}&1&{ - 0.4} \\ { - 0.4}&{ - 0.4}&0&1 \end{array}} \right].$

      由(4)式計算得

      $\qquad \mathop {\max }\limits_{d \in {{[0,1]}^4}} {\left\| {{{\left( {I - D + DM} \right)}^{ - 1}}} \right\|_\infty } \leqslant 41.111\;1,$

      由(5)式計算得

      $\qquad \mathop {\max }\limits_{d \in {{[0,1]}^4}} {\left\| {{{\left( {I - D + DM} \right)}^{ - 1}}} \right\|_\infty } \leqslant 10.275\;1.$

      可見(5)式優于(4)式.

      例 2 考慮 $B$-矩陣[13]

      $\qquad {M_k} = \left[ {\begin{array}{*{20}{c}} {1.5}&{0.5}&{0.4}&{0.5} \\ { - 0.1}&{1.7}&{0.7}&{0.6} \\ {0.8}&{ - 0.1\dfrac{k}{{k + 1}}}&{1.8}&{0.7} \\ 0&{0.7}&{0.8}&{1.8} \end{array}} \right],$

      ${M_k}$ 可表示為 ${M_k} = B_k^ + + C_k^ + $$k \geqslant 1$),此時

      $\qquad B_k^ + = \left[ {\begin{array}{*{20}{c}} 1&0&{ - 0.1}&0 \\ { - 0.8}&1&0&{ - 0.1} \\ 0&{ - 0.1\dfrac{k}{{k + 1}} - 0.8}&1&{ - 0.1} \\ { - 0.8}&{ - 0.1}&0&1 \end{array}} \right].$

      由(2)式計算得

      $\qquad \mathop {\max }\limits_{d \in {{[0,1]}^4}} {\left\| {{{\left( {I - D + DM} \right)}^{ - 1}}} \right\|_\infty } \leqslant 30(k + 1),{\text{當}}k \to \infty {\text{時}},30(k + 1) \to \infty .$

      由(3)式計算得

      $\qquad \mathop {\max }\limits_{d \in {{[0,1]}^4}} {\left\| {{{\left( {I - D + DM} \right)}^{ - 1}}} \right\|_\infty } \leqslant 14.104\;4,$

      由(4)式計算得

      $\qquad \mathop {\max }\limits_{d \in {{[0,1]}^4}} {\left\| {{{\left( {I - D + DM} \right)}^{ - 1}}} \right\|_\infty } \leqslant 15.267\;5,$

      由(16)式計算得

      $\qquad \mathop {\max }\limits_{d \in {{[0,1]}^4}} {\left\| {{{\left( {I - D + DM} \right)}^{ - 1}}} \right\|_\infty } \leqslant 9.502\;7.$

      可見(16)式優于(2)~(4)式.

參考文獻 (19)

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