设$R$、$R^{m\times n}$、$SR^{n\times n}$和$AR^{n\times n}$ 分别表示实数、$m\times n$实矩阵、 $n\times n$实对称矩阵和$n\times n$ 实反对称矩阵集合. $I_{n}$表示$n$阶单位矩阵. $S_{n}(S_{n}=(e_{n},e_{n-1},\cdots,e_1))$ 表示$n\times n$反序单位矩阵($e_{i}$表示$n\times n$ 单位矩阵的第$i$列). $A^T$和tr$(A)$ 分别表示矩阵$A$的转置和迹. 定义矩阵$A$与$B$的内积为$\langle A, B\rangle={\rm tr}(B^TA)$,那么,由这种内积生成的范数, 显然就是Frobenius范数,即$\|A\|^2=\langle A, A\rangle$.

定义1.1     矩阵$A=(a_{ij})\in R^{n\times n}$被称为双对称矩阵,如果$A=A^T=S_nAS_n$,即,$a_{ij}=a_{ji}=a_{n+1-j,n+1-i}$,$( i,j=1,2,\cdots,n)$. 用$BR^{n \times n}$表示$n$阶实双对称矩阵集合.

定义1.2     矩阵$A=(a_{ij})\in R^{n\times n}$被称为双反对称矩阵,如果$A^T=A=-S_nAS_n$,即,$a_{ij}=a_{ji}=-a_{n+1-j,n+1-i}$,$(i,j=1,2,\cdots,n)$. 用$SCR^{n \times n}$表示$n$阶实双反对称矩阵集合.

双对称矩阵在信息理论^[1]、马尔可夫过程^[2]和物理工程问题^[3]等很多领域中有广泛应用,并且已被广泛研究.

线性矩阵方程组已经成为数值计算中的热门课题,已有很多的研究成果. 例如,Yuan和Wang^[4]利用四元数矩阵的复表示和Moore-Penrose广 义逆给出了矩阵方程$AXB+CXD=E$ 带有最小范数的最小二乘$\eta$-Hermitian和$\eta$-anti-Hermitian解 的表达式. Wang^[5]在正则环上研究了矩阵方程组$A_1XB_1=C_1,\; A_2XB_2=C_2$,得到了这个方程组一般解 存在的充分必要条件和表达式. Wang^[6]也在冯$\cdot$诺伊曼正则 环上研究了矩阵方程组. Wang等^[7]给出了矩阵方程组 $A_iX-YB_i=C_i\;(i=1,2,\cdots,s ),\ A_iXB_i-C_iYD_i=E_i \;(i=1,2,\cdots,s)$双(反) 对称解的充分必要条件. Wang和Li^[8]在四元数代数中求出了矩 阵方程组$A_1X=C_1,\;A_2X_2=C_2,\; A_3X_1B_1+A_4X_2B_2=C_3$的最大最小秩解和最小范数解. Wang和Yu^[9]给出了四元数矩阵方程$AX=B$广义双对称和双 反对称解的条件以及解的表达式,并且求出了最大最小秩解. Wang和Yu^[10]也得到了四元数矩阵方程$XA=B$的最小二乘 解的充分必要条件和解的表达式.

迭代方法经常应用于解矩阵方程组. 例如,Lin和Wang^[11]用 迭代法求出了矩阵方程组$A_1X_1B_1+A_2X_2B_2=E,C_1X_1D_1+C_2X_2D_2=F$ 的解. Li和Wang^{[12, 13]} 用迭代法求出了四元数矩阵方程的自反和广义$(P,Q)$自反解. Yin等^{[15, 16]}用迭代法求出了广义Sylvester矩阵方程组的广义自反解. Ding等^[17]用两种方法求解了 矩阵方程组$A_1XB_1=F_1$和$A_2XB_2=F_2$. 在^{[18, 19, 20, 21]}中, Ding等利用基于层次识别原理^{[22, 23]}的迭代法研究了矩阵方程组. Ding和Chen^[24]利用基于 梯度搜索原理的迭代法求解了矩阵方程组 $$\sum\limits_{j=1}^pA_{ij}X_jB_{ij}=M_i,\;\;i=1,2,\cdots,p. (1.1)$$ 值得注意的是,矩阵方程组(1.1)包含了很多的矩阵方程组.

近年来,人们对双对称矩阵、中心对称矩阵等子矩阵约束问题产生 了浓厚兴趣. 然而,由于这些矩阵的特殊结构,在给定的子矩阵约束 条件下讨论 子矩阵约束问题是不适合的,因为这种特殊结构被破坏. 因此, 为了解决这个问题,引入一个定义.

定义1.3    ^[25] 给定$M\in R^{n\times n}$, 如果$n-q$是一个偶数,那么通过删掉$M$的前后$\frac{n-q}{2}$行和列, 得到$M$的一个$q$阶中心主子矩阵, 用$M_c(q)$表示,即,$M_c(q)=[m_{ij}]_{\frac{n-q}{2}\leq i,j\leq n-\frac{n-q}{2}}$.

如果$M\in R^{n\times n}$的中心主子矩阵为$M_{c}(q)=X$, 其他元素为0,那么$M$就被这个中心主子矩阵完全确定. 在这种情况下, 用$\overline{X}_q$表示这种矩阵,即, $$\overline{X}_q=M=\left(\begin{array}{ccc} 0& ~~0~~&0\\ 0&X&0\\ 0&0&0 \end{array}\right) .$$ 显然,奇数(偶数)阶矩阵仅有奇数(偶数)阶中心主子矩阵.

子矩阵约束问题最初来源于子系统扩充问题,并且已被广泛地研究. 例如, Yin等^[14]对一个逆特征值问题给出了顺序主子矩阵约束的$(R,S)$对称解. Liao和Lei^[26] 研究了带有子矩阵约束的双对称解的逆特征值问题. Bai^[27]研究了 带有子矩阵约束的中心对称解的逆特征值问题. Deift和Nanda^[28]讨论了在子矩阵约束下三对角 矩阵的逆特征值问题. Gong等^[29]讨论了矩阵方程$AXA^T=B$中$X$ 带有顺序主子矩阵约束的反对称解. Zhao等^[30]研究了矩阵方程$AX=B$中$X$带有中心主子矩阵约束的 双对称解.

然而,矩阵方程组(1.1)中$X_1,X_2,\cdots,X_p$带有相关的中心主子 矩阵约束的双对称解的问题还没有相关的结论,而且,使用上面文献中的方法,不能解决这个问题. 还应该指出,矩阵方程组(1.1)的系数矩阵经常来自实验, 可能不满足方程组的可解条件. 因此,本文研究以下问题.

问题 I    给定$A_{ij}\in R^{p_i\times n_j}, B_{ij}\in R^{n_j\times m_i}$,$C_i\in R^{p_i\times m_i}$和 $X_{q_j}\in R^{q_j\times q_j}$ $(i=1,2,\cdots,t,$ $j=1,2,\cdots,l)$. 设 $$D_j=\{X_j\in R^{n_j\times n_j}|(X_j)_c(q_j)=X_{q_j},\; X_j-\overline{X}_{q_j}\in BR^{n_j\times n_j}\}. (1.2)$$ 求矩阵组$[X_{1}^{\ast},X_{2}^{\ast},\cdots,X_{l}^{\ast}]\in D_1\times D_2 \times\cdots\times D_l$,使得 $$\sum\limits_{i=1}^t\bigg\|\sum\limits_{j=1}^l A_{ij}X_j^{\ast} B_{ij}-C_i\bigg\|= \min_{[X_1,X_2,\cdots,X_l]\in D_1\times D_2 \times\cdots\times D_l}\sum\limits_{i=1}^t \bigg\|\sum\limits_{j=1}^l A_{ij}X_jB_{ij}-C_i\bigg\|.$$

问题 II    设$S_{E}$表示问题I的解集. 给定矩阵组$[\widehat{X}_1,\widehat{X}_2,\cdots,\widehat{X}_l] \in D_1\times D_2\times\cdots\times D_l$,求$[\hat{X}_1^\ast,\hat{X}_2^\ast,\cdots,\hat{X}_l^\ast]\in S_E$,使得 $$\sum\limits_{j=1}^l\|\hat{X}_j^\ast-\widehat{X}_j\|^2 =\min_{[X_1,X_2\cdots,X_l]\in S_E}\sum\limits_{j=1}^l\|X_j-\widehat{X}_j\|^2.$$

引理2.1     如果矩阵$X\in SR^{n\times n}$,那么 $X+S_nXS_n\in BR^{n\times n}$,$X-S_nXS_n\in SCR^{n\times n}$.

证     由定义(1.1)和定义(1.2)容易得到这个证明.

定义2.1    定义以下两个集合 $$BR^{n \times n}_\ast=\{X|X\in BR^{n\times n},\ \ X_c(q)=0\},$$ $$BR^{n\times n}_\diamond=\{X|X=\left(\begin{array}{ccc} 0& 0&0\\ 0&~X_q~&0\\ 0&0&0 \end{array}\right), \; X\in BR^{n\times n},\ X_q=X_c(q)\in BR^{q\times q}\} .$$ 显然,$BR^{n\times n}_\ast$和$BR^{n\times n}_\diamond$都是$R^{n\times n}$的线性子空间.

引理2.2     (1)~ $R^{n\times n}=SR^{n\times n}\oplus AR^{n\times n}$; (2)~ $SR^{n\times n}=BR^{n\times n}\oplus SCR^{n\times n}$; (3)~ $BR^{n\times n}=BR^{n\times n}_\ast\oplus BR^{n\times n}_\diamond$, \\ 其中 $\oplus$表示直和.

证     (1)式是显然的. 下证(2)式. 证明需要以下3步.

第1步 $\forall X\in SR^{n \times n}$,令 $$X_a=\frac{X+S_nXS_n}{2},\; X_b=\frac{X-S_nXS_n}{2},(2.1)$$ 那么 $X_a\in BR^{n\times n},\ X_b\in SCR^{n\times n}$,以及 $$X=X_a+X_b. (2.2)$$

第2步 如果存在$X' _a\in BR^{n\times n},\ X' _b\in SCR^{n\times n}$满足 $$X=X' _a+X' _b,(2.3)$$ (2.3)式减去(2.2)式得 $$X' _a-X_a=X_b-X' _b. (2.4)$$ (2.4)式两边乘以$S_n$得 $$S_nX' _a S_n-S_n X_a S_n=S_n X_b S_n-S_nX' _b S_n,$$ 即 $$X' _a-X_a=-X_b+X' _b.(2.5)$$ 由(2.4)式和(2.5)式知,$X' _a=X_a,\ X' _b=X_b$.

第3步 $\forall X_a\in BR^{n\times n},\ X_b\in SCR^{n\times n},$ $$\langle X_a,X_b\rangle= {\rm tr}(X_b^TX_a)={\rm tr}(-S_nX_b^TS_nS_nX_aS_n)=-{\rm tr}(X_b^TX_a)=-\langle X_a,X_b\rangle.$$ 故 $\langle X_a,X_b\rangle=0$.

因此,由以上3步可知,(2)式成立.

(3)式的证明与(2)式的证明类似.

推论2.1     $\forall Z\in R^{n\times n}$,则存在唯一的$Z_1\in BR^{n\times n}_\ast,Z_2\in BR^{n\times n}_\diamond,Z_3\in SCR^{n\times n}$和$Z_4\in AR^{n\times n}$,使得 $Z=Z_1+Z_2+Z_3+Z_4$,其中$\langle Z_i,Z_j\rangle=0,i\neq j,i,j=1,2,3,4.$

定义2.2     定义$\Psi$是从$R^{n\times n}$到$BR^{n\times n}_\ast$的线性投影算子. $\Psi: R^{n\times n} \rightarrow BR^{n\times n}_\ast$ $\Rightarrow$ $Z\rightarrow Z_1$,即,$\Psi(Z)=Z_1$.

$\forall Z\in R^{n\times n},\ W\in BR^{n\times n}_\ast$,则 $$\langle Z ,W\rangle=\langle Z_1+Z_2+Z_3+Z_4,W\rangle=\langle Z_1,W\rangle=\langle \Psi(Z),W\rangle. (2.6)$$

引理2.3    ^[31] 设${\Bbb M}$表示有限维内积空间,${\Bbb N}$是${\Bbb M}$的一个子空间,${\Bbb N}^\perp$ 表示${\Bbb N}$的正交补空间. $\forall x\in {\Bbb M}$,总存在$y_0\in {\Bbb N}$,使得$\|x-y_0\|\leq\|x-y\| ( \forall y\in {\Bbb N})$,其中$\|.\|$ 是定义在${\Bbb M}$上,与内积空间相关的范数. 而且,$y_0$为${\Bbb N}$中唯一最小向量的充分必要条件是$(x-y_0)\perp {\Bbb N}$, 即$(x-y_0)\in {\Bbb N}^\perp$.

问题 A     设$A_{ij},B_{ij}, C_i,X_{q_j}$ $(i=1,2,\cdots,t,j=1,2,\cdots,l)$与问题I中的一致. 求$[\widetilde{X}_1,\widetilde{X}_2,\cdots,\widetilde{X}_l]\in BR^{n_1\times n_1}_\ast \times BR^{n_2\times n_2}_\ast \times \cdots\times BR^{n_l\times n_l}_\ast$,使得 $$\begin{eqnarray*} &&\sum\limits_{i=1}^t \bigg\|\sum\limits_{j=1}^l A_{ij}\widetilde{X}_jB_{ij}-F_i\bigg\| \\ &=& \min_{\scriptsize[X_1,X_2,\cdots,X_l]\in BR^{n_1\times n_1}_\ast \times BR^{n_2\times n_2}_\ast \times \cdots\times BR^{n_l\times n_l}_\ast} \sum\limits_{i=1}^t\bigg\|\sum\limits_{j=1}^lA_{ij}X_jB_{ij}-F_i\bigg\|, \end{eqnarray*}$$ 其中 $F_i=C_i-\sum\limits_{j=1}^lA_{ij}\overline{X}_jB_{ij}$,

$$\begin{equation} \overline{X}_j=\overline{X}_{q_j}=\left(\begin{array}{ccc} 0 & 0 & 0\\ 0 & ~X_{q_j} ~& 0\\ 0 & 0 & 0 \end{array}\right),j=1,2,\cdots,l. \end{equation}$$

(3.1)

引理3.1     问题I的解可表示为$[X_1^\ast,X_2^\ast,\cdots,X_l^\ast]=[\widetilde{X}_1+\overline{X}_1,\widetilde{X}_2+\overline{X}_2,\cdots, \widetilde{X}_l +\overline{X}_l]$,其中 $[\widetilde{X}_1,\widetilde{X}_2,\cdots,\widetilde{X}_l]$是问题A的解.

证     由(1.2)式可得$D_j=\overline{X}_j+BR^{n_j\times n_j}_\ast$. 这表明$D_j$是一个线性流形. $$\begin{eqnarray*} \sum\limits_{i=1}^t \bigg \|\sum\limits_{j=1}^l A_{ij}X_j^\ast B_{ij} -C_i\bigg\| &\Leftrightarrow & \sum\limits_{i=1}^t \bigg\|\sum\limits_{j=1}^l A_{ij}(\overline{X}_j+\widetilde{X}_j)B_{ij}-C_i\bigg\|\\ &\Leftrightarrow &\sum\limits_{i=1}^t \bigg\|\sum\limits_{j=1}^l A_{ij}\widetilde{X}_jB_{ij}-(C_i-\sum\limits_{j=1}^l A_{ij}\overline{X}_jB_{ij}) \bigg\| \\ & \Leftrightarrow &\sum\limits_{i=1}^t\bigg \|\sum\limits_{j=1}^l A_{ij}\widetilde{X}_jB_{ij}-F_i\bigg\|. \end{eqnarray*}$$ 故引理3.1成立.

注1     由于$D_j$是一个线性流形,不能确保得到的解满足给定的子矩阵要求,所以,不能直接构造迭代法解问题I. 因此,把线性流形上的问题I转化为线性子空间上的问题A, 达到求解问题I的目的.

求解问题A,也就是求矩阵方程组

$$\begin{equation} \sum\limits_{j=1}^l A_{ij}X_jB_{ij}=F_i,\;\;i=1,2,\cdots,t. \end{equation}$$

(3.2)

在$BR^{n_1\times n_1}_\ast \times BR^{n_2\times n_2}_\ast \times \cdots\times BR^{n_l\times n_l}_\ast$上的最小二乘解 $[\widetilde{X}_1,\widetilde{X}_2,\cdots,\widetilde{X}_l]$.

引理3.2     设$\widetilde{R}_i$表示方程组(3.2)相应于$[\widetilde{X}_1, \widetilde{X}_2,$ $\cdots,\widetilde{X}_l]$的残量,即, $\widetilde{R}_i=F_i-\sum\limits_{j=1}^l A_{ij} \widetilde{X}_jB_{ij}$. 如果$\Psi(\sum\limits_{i=1}^t A_{ij}^T \widetilde{R}_iB_{ij}^T)=0$ $(j=1,2,\cdots,l)$同时成立, 那么,$[\widetilde{X}_1,\widetilde{X}_2,\cdots,\widetilde{X}_l]$ 就是问题A的解.

证     设${\Bbb L}$ $=\{L|L={\rm diag}(\sum\limits_{j=1}^lA_{1j}X_jB_{1j},\sum\limits_{j=1}^lA_{2j}X_jB_{2j},\cdots,\sum\limits_{j=1}^lA_{tj}X_jB_{tj})\},$ 其中$X_j\in BR^{n_j\times n_j}_\ast$ $(j=1,2,\cdots,l)$. 显然,${\Bbb L}$是一个线性子空间. 对于$[\widetilde{X}_1,\widetilde{X}_2,\cdots,\widetilde{X}_l ]$,记 $$\widetilde{F}={\rm diag} \bigg(\sum\limits_{j=1}^l A_{1j} \widetilde{X}_jB_{1j},\sum\limits_{j=1}^l A_{2j}\widetilde{X}_jB_{2j},\cdots,\sum\limits_{j=1}^l A_{tj}\widetilde{X}_jB_{tj} \bigg),$$ 则 $\widetilde{F}\in {\Bbb L}$. 设$F={\rm diag}(F_1,F_2,\cdots,F_t)$. 由引理2.3知, $[\widetilde{X}_1,\widetilde{X}_2,\cdots,\widetilde{X}_l]$ 是问题A的解,当且仅当$(F-\widetilde{F})\perp {\Bbb L}$,即, $\forall[X_1,X_2,\cdots,X_l]\in BR^{n_1\times n_1}_\ast \times BR^{n_2\times n_2}_\ast \times \cdots\times BR^{n_l\times n_l}_\ast$,

$$\begin{eqnarray} &&\left\langle F-\widetilde{F}, {\rm diag} \bigg(\sum\limits_{j=1}^lA_{1j}X_jB_{1j}, \sum\limits_{j=1}^lA_{2j}X_jB_{2j},\cdots,\sum\limits_{j=1}^lA_{tj}X_jB_{tj} \bigg)\right\rangle \nonumber\\ & =&\left\langle{\rm diag}\left(\widetilde{R}_1,\widetilde{R}_2,\cdots,\widetilde{R}_t\right), {\rm diag}\bigg(\sum\limits_{j=1}^lA_{1j}X_jB_{1j},\sum\limits_{j=1}^lA_{2j}X_jB_{2j},\cdots,\sum\limits_{j=1}^lA_{tj}X_jB_{tj} \bigg)\right\rangle \nonumber\\ & =&0, \end{eqnarray}$$

(3.3)

由(2.6)式知 $$\begin{eqnarray*} &&\left\langle{\rm diag}\left(\widetilde{R}_1,\widetilde{R}_2,\cdots,\widetilde{R}_t\right), {\rm diag}\bigg(\sum\limits_{j=1}^lA_{1j}X_jB_{1j},\sum\limits_{j=1}^lA_{2j}X_jB_{2j},\cdots,\sum\limits_{j=1}^lA_{tj}X_jB_{tj} \bigg)\right\rangle\\ & =&\sum\limits_{i=1}^t \bigg\langle \widetilde{R}_i,\; \sum\limits_{j=1}^lA_{ij}X_jB_{ij}\bigg\rangle =\sum\limits_{i=1}^t \sum\limits_{j=1}^l\langle \widetilde{R}_i,\;A_{ij}X_jB_{ij}\rangle\\ & =& \sum\limits_{j=1}^l \bigg\langle \sum\limits_{i=1}^t A_{ij}^T\widetilde{R}_iB_{ij}^T,\; X_j\bigg\rangle = \sum\limits_{j=1}^l\bigg\langle\Psi \bigg( \sum\limits_{i=1}^t A_{ij}^T\widetilde{R}_iB_{ij}^T\bigg),\; X_j\bigg\rangle. \end{eqnarray*}$$ 因此,如果$\Psi( \sum\limits_{i=1}^t A_{ij}^T\widetilde{R}_iB_{ij}^T)=0$ $(j=1,2,\cdots,l)$同时成立,那么(3.3)式成立,表明 $[\widetilde{X}_1,\widetilde{X}_2,$ $\cdots,\widetilde{X}_l]$是问题A的解.

引理3.3    ^[32] 设$f(Z)$是子空间${\Bbb Y}$上连续、可微的凸函数. 那么存在$Z_\ast\in {\Bbb Y}$,使得$f(Z_\ast)=\min\limits_{Z \in {\Bbb Y}}f(Z)$,当且仅当梯度 $\nabla f(Z)$在${\Bbb Y}$上的投影等于$0$.

注2     引理3.2与引理3.3是一致的. 事实上, 设$f(X_1,X_2,\cdots,X_l)=\sum\limits_{i=1}^t \|\sum\limits_{j=1}^lA_{ij}X_jB_{ij}-F_i\|^2$,那么$f$是可微的. 构造辅助函数$g(s)=f(X_1+sE_1,X_2+sE_2,\cdots,X_l+sE_l)$, 其中$E_1,E_2,\cdots,E_l$是具有适当大小的任意矩阵. 于是可得

$$\begin{equation} g'(0)=\sum\limits_{j=1}^l\langle \nabla_{X_j} f(X_1,X_2,\cdots,X_l),E_j\rangle. \end{equation}$$

(3.4)

设$R_i=F_i-\sum\limits_{j=1}^lA_{ij}X_jB_{ij}$,那么,又可以得到

$$\begin{equation} g'(0)=-2\sum\limits_{j=1}^l\bigg\langle \sum\limits_{i=1}^t A_{ij}^TR_iB_{ij}^T,E_j\bigg\rangle. \end{equation}$$

(3.5)

由(3.4)式和(3.5)式知

$$\begin{equation} \nabla_{X_j} f(X_1,X_2,\cdots,X_l)=-2\sum\limits_{i=1}^t A_{ij}^TR_iB_{ij}^T,\; j=1,2,\cdots,l. \end{equation}$$

(3.6)

根据(3.6)式,$\Psi(\sum\limits_{i=1}^t A_{ij}^TR_iB_{ij}^T)=0 \ (j=1,2,\cdots,l)$当且仅当 $\nabla_{X_j} f(X_1,X_2,\cdots,X_l)$的投影等于0 $(j=1,2,\cdots,l)$, 这表明引理3.2与引理3.3是一致的.

算法 3.1     任给初始矩阵组$[X_{0,1},X_{0,2}, \cdots,X_{0,l}]\in BR^{n_1\times n_1}_\ast \times BR^{n_2\times n_2}_\ast \times \cdots\times BR^{n_l\times n_l}_\ast$.

1)~ $R_{0,i}=F_i-\sum\limits_{j=1}^l A_{ij}X_{0,j}B_{ij},\; i=1,2,\cdots,t;$ $P_{0,j}=\Psi(\sum\limits_{i=1}^t A_{ij}^TR_{0,i}B_{ij}^T),\ j=1,2,\cdots,l$; $Q_{0,j}=P_{0,j},\ j=1,2,\cdots,l$; $k:=0$;

2)~ 如果$\sum\limits_{i=1}^t \|R_{k,i}\|^2=0$ 或 $\sum\limits_{j=1}^l \|P_{k,j}\|^2=0$,那么停止,否则,计算

3)~ $\alpha_k=\frac{\sum\limits_{j=1}^l \|P_{k,j}\|^2}{ \sum\limits_{i=1}^t\|\sum\limits_{j=1}^l A_{ij}Q_{k,j}B_{ij}\|^2}$; $X_{k+1,j}=X_{k,j}+\alpha_kQ_{k,j},\ j=1,2,\cdots,l$; $$R_{k+1,i}=R_{k,i}-\alpha_k\sum\limits_{j=1}^l A_{ij}Q_{k,j}B_{ij},\ i=1,2,\cdots,t;$$ $$P_{k+1,j}=\Psi(\sum\limits_{i=1}^t A_{ij}^TR_{k+1,i}B_{ij}^T) =P_{k,j}-\alpha_k\Psi(\sum\limits_{i=1}^t A_{ij}^T(\sum\limits_{v=1}^l A_{iv}Q_{k,v}B_{iv})B_{ij}^T),\ j=1,2,\cdots,l;$$ $$\beta_k=\frac{\sum\limits_{j=1}^l\|P_{k+1,j}\|^2}{\sum\limits_{j=1}^l\|P_{k,j}\|^2}; ~~ Q_{k+1,j}=P_{k+1,j}+\beta_kQ_{k,j},\ j=1,2,\cdots,l;$$

4)~ $k:=k+1$,转 2).

注3     1)~ 在算法3.1中,由$P_{k+1,j}=\Psi(\sum\limits_{i=1}^t A_{ij}^TR_{k+1,i}B_{ij}^T)$知,对所有的$k$,$P_{k+1,j} \in BR^{n_j \times n_j}_\ast$. 因为 $Q_{0,j}=P_{0,j}$,所以 对所有的$k$, $Q_{k+1,j}\in BR^{n_j \times n_j}_\ast$,故 $X_{k+1,j}\in BR^{n_j \times n_j}_\ast$.

2)~ 如果$\sum\limits_{i=1}^t \|R_{k,i}\|^2=0$,那么相应的 $[X_{k,1},X_{k,2},\cdots,X_{k,l}]$就是矩阵方程组 $\sum\limits_{j=1}^l A_{ij}X_jB_{ij}=F_i$ $(i=1,2,\cdots,t)$ 在线性子空间$BR^{n_1\times n_1}_\ast \times BR^{n_2\times n_2}_\ast \times \cdots\times BR^{n_l\times n_l}_\ast$上的解. 如果$\sum\limits_{i=1}^t \|R_{k,i}\|^2\neq 0$,而$\sum\limits_{j=1}^l \|P_{k,j}\|^2=0$,那么相应的 $[X_{k,1},X_{k,2},\cdots,X_{k,l}]$就是问题A的解.

引理3.4     假定序列$\{P_{i,v}\}$和$\{Q_{i,v}\}$ $( v=1,2,\cdots,l)$由算法3.1生成. 如果存在一个正整数$k$, 使得对$i=0,1,2,\cdots,k$, $\sum\limits_{v=1}^l\|P_{i,v}\|^2\neq 0,\;\alpha_i\neq 0$以 及$\alpha_i\neq \infty$,那么

(1)~ $\sum\limits_{v=1}^l\langle P_{i,v}\;,\;P_{j,v}\rangle=0,\ i,j=0,1,2,\cdots,k,\;\;i\neq j,$

(2)~ $\sum\limits_{w=1}^t\langle \sum\limits_{u=1}^l A_{wu}Q_{i,u}B_{wu}\;,\; \sum\limits_{u=1}^l A_{wu}Q_{j,u}B_{wu}\rangle=0, \ i,j=0,1,2,\cdots,k,\;\;i\neq j,$

(3)~ $\sum\limits_{v=1}^l\langle Q_{i,v}\;,P_{j,v}\rangle=0, \ i,j=0,1,2,\cdots,k,\;\;i\neq j.$

证     (用数学归纳法) 第一步证明,当$k=1$时, (1)-(3)成立. $$\begin{eqnarray*} \sum\limits_{v=1}^l\langle P_{0,v}\;,P_{1,v}\rangle &=&\sum\limits_{v=1}^l\langle P_{0,v}\;,\;P_{0,v}-\alpha_0\Psi(\sum\limits_{u=1}^t A_{uv}^T(\sum\limits_{w=1}^lA_{uw}Q_{0,w}B_{uw})B_{uv}^T)\rangle\\ &=&\sum\limits_{v=1}^l(\|P_{0,v}\|^2 -\alpha_0\sum\limits_{u=1}^t\langle A_{uv}P_{0,v}B_{uv}\;,\;\sum\limits_{w=1}^lA_{uw}Q_{0,w}B_{uw}\rangle)\\ &=&\sum\limits_{v=1}^l\|P_{0,v}\|^2 -\alpha_0\sum\limits_{u=1}^t\langle \sum\limits_{v=1}^l A_{uv}P_{0,v}B_{uv}\;,\;\sum\limits_{w=1}^lA_{uw}Q_{0,w}B_{uw}\rangle\\ &=&\sum\limits_{v=1}^l\|P_{0,v}\|^2 -\alpha_0\sum\limits_{u=1}^t\langle \sum\limits_{v=1}^l A_{uv}Q_{0,v}B_{uv}\;,\;\sum\limits_{w=1}^lA_{uw}Q_{0,w}B_{uw}\rangle =0. \end{eqnarray*}$$ $$\begin{eqnarray*} &&\sum\limits_{w=1}^t\langle \sum\limits_{u=1}^l A_{wu}Q_{0,u}B_{wu}\;,\; \sum\limits_{u=1}^l A_{wu}Q_{1,u}B_{wu}\rangle \\ &=&\sum\limits_{w=1}^t\langle \sum\limits_{u=1}^l A_{wu}Q_{0,u}B_{wu}\;,\; \sum\limits_{u=1}^l A_{wu}(P_{1,u}+\beta_0Q_{0,u})B_{wu}\rangle \\ &=&\sum\limits_{w=1}^t(\beta_0\|\sum\limits_{u=1}^l A_{wu}Q_{0,u}B_{wu}\|^2+\langle \sum\limits_{u=1}^l A_{wu}Q_{0,u}B_{wu}\;,\; \sum\limits_{u=1}^l A_{wu}P_{1,u}B_{wu}\rangle) \\ &=&\beta_0\sum\limits_{w=1}^t\|\sum\limits_{u=1}^l A_{wu}Q_{0,u}B_{wu}\|^2+\frac{1}{\alpha_0}\sum\limits_{u=1}^l\langle\Psi(\sum\limits_{w=1}^t A_{wu}^T(R_{0,w}-R_{1,w})B_{wu}^T)\;,\; P_{1,u}\rangle\\ &=&\beta_0\sum\limits_{w=1}^t\|\sum\limits_{u=1}^l A_{wu}Q_{0,u}B_{wu}\|^2-\frac{1}{\alpha_0}\sum\limits_{u=1}^l\langle P_{1,u},P_{1,u}\rangle=0. \end{eqnarray*}$$ $$\sum\limits_{v=1}^l\langle Q_{0,v}\;,P_{1,v}\rangle=\sum\limits_{v=1}^l\langle P_{0,v}\;,P_{1,v}\rangle=0.$$ 因此,当$k=1$时,(1)-(3)成立.

第二步证明,假定当$k=s,\ i\leq s-1$时,(1)-(3)成立,即, $\sum\limits_{v=1}^l\langle P_{i,v},P_{s,v}\rangle=0,$ $\sum\limits_{w=1}^t\langle \sum\limits_{u=1}^l A_{wu}Q_{i,u}B_{wu}, \sum\limits_{u=1}^l A_{wu}Q_{s,u}B_{wu}\rangle=0 ,$ $\sum\limits_{v=1}^l\langle Q_{i,v},P_{s,v}\rangle=0,$ 那么需要证明:当$k=s+1,\ i\leq s$时,(1)-(3)成立. 证明过程如下:

当$k=s+1,\ i\leq s-1$时,

$$\begin{eqnarray*} \sum\limits_{v=1}^l\langle P_{i,v},\;P_{s+1,v}\rangle & =&\sum\limits_{v=1}^l\langle P_{i,v},\;P_{s,v}-\alpha_s \Psi(\sum\limits_{w=1}^tA_{wv}^T(\sum\limits_{u=1}^lA_{wu}Q_{s,u}B_{wu})B_{wv}^T)\rangle\\ & =&-\alpha_s\sum\limits_{w=1}^t\langle\sum\limits_{v=1}^lA_{wv}P_{i,v}B_{wv}\;,\; \sum\limits_{u=1}^lA_{wu}Q_{s,u}B_{wu}\rangle\\ & =&-\alpha_s\sum\limits_{w=1}^t\langle\sum\limits_{v=1}^lA_{wv}(Q_{i,v}-\beta_{i-1}Q_{i-1,v})B_{wv},\; \sum\limits_{u=1}^lA_{wu}Q_{s,u}B_{wu}\rangle = 0. \end{eqnarray*}$$ $$\begin{eqnarray} &&\sum\limits_{w=1}^t\langle \sum\limits_{u=1}^l A_{wu}Q_{i,u}B_{wu},\;\sum\limits_{u=1}^l A_{wu}Q_{s+1,u}B_{wu}\rangle \nonumber\\ &=&\sum\limits_{w=1}^t\langle \sum\limits_{u=1}^l A_{wu}Q_{i,u}B_{wu},\;\sum\limits_{u=1}^l A_{wu}(P_{s+1,u}+\beta_sQ_{s,u})B_{wu}\rangle \nonumber\\ &=&\frac{1}{\alpha_i}\sum\limits_{w=1}^t\langle R_{i,w}-R_{i+1,w},\;\sum\limits_{u=1}^l A_{wu}P_{s+1,u}B_{wu}\rangle \nonumber\\ &=&\frac{1}{\alpha_i}\sum\limits_{u=1}^l\langle\Psi(\sum\limits_{w=1}^t A_{wu}^TR_{i,w}B_{wu}^T)-\Psi(\sum\limits_{w=1}^t A_{wu}^TR_{i+1,w}B_{wu}^T),\; P_{s+1,u}\rangle \nonumber\\ &=&-\frac{1}{\alpha_i}\sum\limits_{u=1}^l \langle P_{i+1,u},\; P_{s+1,u}\rangle. \end{eqnarray}$$

(3.7)

$$\begin{eqnarray*} \sum\limits_{v=1}^l\langle Q_{i,v}\;,\;P_{s+1,v}\rangle &=&\sum\limits_{v=1}^l\langle Q_{i,v}\;,\;P_{s,v}-\alpha_s \Psi(\sum\limits_{w=1}^tA_{wv}^T(\sum\limits_{u=1}^l A_{wu}Q_{s,u}B_{wu})B_{wv}^T)\rangle\\ &=&-\alpha_s\sum\limits_{w=1}^t\langle\sum\limits_{v=1}^l A_{wv} Q_{i,v}B_{wv},\; \sum\limits_{u=1}^l A_{wu}Q_{s,u}B_{wu}\rangle=0. \end{eqnarray*}$$

由算法3.1知,矩阵$Q_{i,v}\ (i\leq s,v=1,2,\cdots,l)$可表示为 $$Q_{i,v}=P_{i,v}+\beta_{i-1}Q_{i-1,v} =P_{i,v}+\beta_{i-1}P_{i-1,v}+\beta_{i-1}\beta_{i-2}P_{i-2,v}+\cdots+\beta_{i-1}\cdots\beta_{0}P_{0,v},$$ 于是可得

$$\begin{equation} \sum\limits_{v=1}^l\langle P_{s,v},\;Q_{s,v}\rangle=\sum\limits_{v=1}^l\|P_{s,v}\|^2. %(3.8)$$ \end{equation}$$

(3.8)

当$k=s+1,\ i=s$时, $$\begin{eqnarray*} \sum\limits_{v=1}^l\langle P_{s,v}\;,P_{s+1,v}\rangle &=&\sum\limits_{v=1}^l\langle P_{s,v}\;,\;P_{s,v}-\alpha_s \Psi(\sum\limits_{w=1}^tA_{wv}^T(\sum\limits_{u=1}^l A_{wu}Q_{s,u}B_{wu})B_{wv}^T)\rangle\\ &=&\sum\limits_{v=1}^l(\|P_{s,v}\|^2 -\alpha_s\langle P_{s,v}\;,\;\Psi(\sum\limits_{w=1}^tA_{wv}^T(\sum\limits_{u=1}^l A_{wu}Q_{s,u}B_{wu})B_{wv}^T)\rangle)\\ &=&\sum\limits_{v=1}^l\|P_{s,v}\|^2 -\alpha_s\sum\limits_{w=1}^t\langle\sum\limits_{v=1}^l A_{wv}P_{s,v}B_{wv}\;,\;\sum\limits_{u=1}^l A_{wu}Q_{s,u}B_{wu}\rangle\\ &=&\sum\limits_{v=1}^l\|P_{s,v}\|^2-\alpha_s\sum\limits_{w=1}^t\langle\sum\limits_{v=1}^l A_{wv}Q_{s,v}B_{wv}\;,\;\sum\limits_{u=1}^l A_{wu}Q_{s,u}B_{wu}\rangle=0. \end{eqnarray*}$$ $$\begin{eqnarray*} &&\sum\limits_{w=1}^t\langle \sum\limits_{u=1}^l A_{wu}Q_{s,u}B_{wu}\;,\; \sum\limits_{u=1}^l A_{wu}Q_{s+1,u}B_{wu}\rangle \\ &=&\sum\limits_{w=1}^t\langle \sum\limits_{u=1}^l A_{wu}Q_{s,u}B_{wu}\;,\; \sum\limits_{u=1}^l A_{wu}(P_{s+1,u}+\beta_sQ_{s,u})B_{wu}\rangle \\ &=&\sum\limits_{w=1}^t(\beta_s\|\sum\limits_{u=1}^l A_{wu}Q_{s,u}B_{wu}\|^2+\frac{1}{\alpha_s}\langle R_{s,w}-R_{s+1,w}\;,\; \sum\limits_{u=1}^l A_{wu}P_{s+1,u}B_{wu}\rangle) \\ &=&\beta_s\sum\limits_{w=1}^t\|\sum\limits_{u=1}^l A_{wu}Q_{s,u}B_{wu}\|^2+\frac{1}{\alpha_s}\sum\limits_{u=1}^l\langle\Psi(\sum\limits_{w=1}^t A_{wu}^T(R_{s,w}-R_{s+1,w})B_{wu}^T)\;, P_{s+1,u}\rangle \\ &=&\beta_s\sum\limits_{w=1}^t\|\sum\limits_{u=1}^l A_{wu}Q_{s,u}B_{wu}\|^2+\frac{1}{\alpha_s}\sum\limits_{u=1}^l\langle P_{s,u}-P_{s+1,u}\;,\; P_{s+1,u}\rangle =0. \end{eqnarray*}$$ $$\begin{eqnarray*} \sum\limits_{v=1}^l\langle Q_{s,v}\;,P_{s+1,v}\rangle &=& \sum\limits_{v=1}^l\langle Q_{s,v}\;,P_{s,v}-\alpha_s \Psi(\sum\limits_{w=1}^tA_{wv}^T(\sum\limits_{u=1}^l A_{wu}Q_{s,u}B_{wu})B_{wv}^T)\rangle\\ &=& \sum\limits_{v=1}^l\|P_{s,v}\|^2-\alpha_s\sum\limits_{v=1}^l\langle Q_{s,v}\;,\; \sum\limits_{w=1}^tA_{wv}^T(\sum\limits_{u=1}^l A_{wu}Q_{s,u}B_{wu})B_{wv}^T\rangle\\ &=& \sum\limits_{v=1}^l\|P_{s,v}\|^2-\alpha_s\sum\limits_{w=1}^t\langle \sum\limits_{v=1}^l A_{wv}Q_{s,v}B_{wv}\;,\; \sum\limits_{u=1}^l A_{wu}Q_{s,u}B_{wu}\rangle\\ &=& \sum\limits_{v=1}^l\|P_{s,v}\|^2-\alpha_s\sum\limits_{w=1}^t\| \sum\limits_{v=1}^l A_{wv}Q_{s,v}B_{wv}\|^2=0. \end{eqnarray*}$$ 由结论$\sum\limits_{v=1}^l\langle P_{s,v}\;,P_{s+1,v}\rangle=0$、 假设$\sum\limits_{w=1}^t\langle \sum\limits_{u=1}^l A_{wu}Q_{i,u}B_{wu}, \sum\limits_{u=1}^l A_{wu}Q_{s,u}B_{wu}\rangle=0 \;(i\leq s-1)$以及(3.7)式可知,当$i\leq s-1$时, $\sum\limits_{w=1}^t\langle \sum\limits_{u=1}^l A_{wu}Q_{i,u}B_{wu},\sum\limits_{u=1}^l A_{wu}Q_{s+1,u}B_{wu}\rangle=0$.

因此,由第一步和第二步知,引理3.4成立.

引理3.4表明,算法3.1生成的矩阵列$\{{\rm diag}( P_{i,1},P_{i,2}, \cdots,P_{i,l})\}$在$R^{(n_1+\cdots+ n_l)\times (n_1+\cdots+ n_l)}$上相互正交. 设$r_j$表示子空间 $BR^{n_j\times n_j}_\ast$的维数,那么存在正整数 $k\leq r_1+r_2+\cdots+r_l$, 使得$\sum\limits_{j=1}^l\|P_{k,j}\|^2=0$,即, 在没有舍入误差的情况下算法3.1至多经过$r_1+r_2+\cdots+r_l$迭代停止.

在算法3.1中,如果$\alpha_i= 0$,那么 $\sum\limits_{v=1}^l\|P_{i,v}\|^2=0$. 如果$\alpha_i= \infty$, 那么 $\sum\limits_{w=1}^t\|\sum\limits_{v=1}^l A_{wv}Q_{i,v}$ $B_{wv}\|^2=0.$ 于是 $\sum\limits_{v=1}^l A_{wv}Q_{i,v}B_{wv}=0,w=1,2,\cdots,t$. 在上式两边用$R_{i,w}$做内积可得 $\langle \sum\limits_{v=1}^l A_{wv}Q_{i,v}B_{wv},R_{i,w}\rangle=0.$ 从而, $$\begin{eqnarray*} \sum\limits_{w=1}^t\langle\sum\limits_{v=1}^lA_{wv}Q_{i,v}B_{wv},\;R_{i,w}\rangle &=&\sum\limits_{v=1}^l\langle Q_{i,v},\;\sum\limits_{w=1}^t A_{wv}^TR_{i,w}B_{wv}^T\rangle\\ &=&\sum\limits_{v=1}^l\langle Q_{i,v},\;\Psi(\sum\limits_{w=1}^t A_{wv}^TR_{i,w}B_{wv}^T)\rangle\\ &=& \sum\limits_{v=1}^l \langle Q_{i,v},\;P_{i,v}\rangle=\sum\limits_{v=1}^l\|P_{i,v}\|^2=0. \end{eqnarray*}$$ 因此,如果存在一个正整数$i$使得$\alpha_i= 0$或$\alpha_i= \infty$, 那么相应的$[X_1,X_2,\cdots,X_l]$就是问题A的解.

由引理3.4和以上讨论,可得以下定理.

定理3.1     任给初始矩阵组$[X_{0,1},X_{0,2},\cdots,X_{0,l}]\in BR^{n_1\times n_1}_\ast \times BR^{n_2\times n_2}_\ast \times \cdots\times BR^{n_l\times n_l}_\ast$,通过算法3.1, 至多经过$r_1+r_2+\cdots+r_l$迭代可得问题A的一个解.

注4     理论上,任给初始矩阵组,通过算法3.1,至多经过$r_1+r_2+\cdots+r_l$ 迭代可得问题A的一个解. 但实际上,舍入误差是不可避免的,因而问题A的解有时要超过 $r_1+r_2+\cdots+r_l$迭代才能得到.

设${\Bbb W}=\{[M_1,M_2,\cdots,M_l]|M_j= \Psi(\sum\limits_{i=1}^tA_{ij}^TH_iB_{ij}^T),\;j=1,2,\cdots,l\},$ 其中$H_i\in R^{p_i\times m_i}$. 显然,${\Bbb W}$是 $BR^{n_1\times n_1}_\ast \times BR^{n_2\times n_2}_\ast \times \cdots\times BR^{n_l\times n_l}_\ast$ 的一个线性子空间.

引理3.5     假设$[\widetilde{X}_1,\widetilde{X}_2,\cdots,\widetilde{X}_l]$ 是问题A的解. 那么问题A的任意解可表示为 $[\widetilde{X}_1+M_1^\ast,\widetilde{X}_2+M_2^\ast,\cdots, \widetilde{X}_l+M_l^\ast]$,当且仅当

$$\begin{equation} \sum\limits_{j=1}^lA_{ij}M_j^\ast B_{ij}=0,\;\;i=1,2,\cdots,t, \end{equation}$$

(3.9)

其中$[M_1^\ast,M_2^\ast,\cdots,M_l^\ast]\in BR^{n_1\times n_1}_\ast \times BR^{n_2\times n_2}_\ast \times \cdots\times BR^{n_l\times n_l}_\ast$.

证     设$[M_1^\ast,M_2^\ast,\cdots,M_l^\ast]\in BR^{n_1\times n_1}_\ast \times BR^{n_2\times n_2}_\ast \times \cdots\times BR^{n_l\times n_l}_\ast$. 如果$[\widetilde{X}_1+M_1^\ast,\widetilde{X}_2+M_2^\ast,\cdots,\widetilde{X}_l+M_l^\ast]$ 是问题A的一个解,那么由引理3.2的证明可知

$$\begin{eqnarray} &&\sum\limits_{i=1}^t \bigg\|\sum\limits_{j=1}^lA_{ij}\widetilde{X}_jB_{ij}-F_i\bigg\|^2 \nonumber\\ &=& \sum\limits_{i=1}^t\bigg\|\sum\limits_{j=1}^lA_{ij}(\widetilde{X}_j+M_j^\ast)B_{ij}-F_i\bigg\|^2 =\sum\limits_{i=1}^t \bigg\|\sum\limits_{j=1}^lA_{ij}M_j^{\ast}B_{ij}-\widetilde{R}_i\bigg\|^2 \nonumber\\ &=&\bigg\|{\rm diag}\bigg(\sum\limits_{j=1}^lA_{1j}M_j^{\ast}B_{1j}-\widetilde{R}_1,\sum\limits_{j=1}^lA_{2j}M_j^{\ast}B_{2j}-\widetilde{R}_2,\cdots, \sum\limits_{j=1}^lA_{tj}M_j^{\ast}B_{tj}-\widetilde{R}_t\bigg) \bigg\|^2 \nonumber\\ &=&\bigg\|{\rm diag}\bigg(\sum\limits_{j=1}^lA_{1j}M_j^{\ast}B_{1j}, \sum\limits_{j=1}^lA_{2j}M_j^{\ast}B_{2j},\cdots, \sum\limits_{j=1}^lA_{tj}M_j^{\ast}B_{tj}\bigg) -\left(\widetilde{R}_1,\widetilde{R}_2,\cdots,\widetilde{R}_t \right) \bigg\|^2 \nonumber\\ &=&\bigg\|\bigg(\sum\limits_{j=1}^lA_{1j}M_j^{\ast}B_{1j},\sum\limits_{j=1}^lA_{2j}M_j^{\ast}B_{2j},\cdots,\sum\limits_{j=1}^lA_{tj}M_j^{\ast}B_{tj} \bigg)\bigg\|^2 +\left\|\left(\widetilde{R}_1,\widetilde{R}_2,\cdots, \widetilde{R}_t\right)\right\|^2 \nonumber\\ &=&\sum\limits_{i=1}^t \bigg\|\sum\limits_{j=1}^lA_{ij}M_j^{\ast}B_{ij} \bigg\|^2+\sum\limits_{i=1}^t\|\widetilde{R}_i\|^2. \end{eqnarray}$$ %

(3.10)

于是可得$\sum\limits_{j=1}^lA_{ij}M_j^{\ast}B_{ij}=0$.

反过来,如果$[\widetilde{X}_1+M_1^\ast,\widetilde{X}_2+M_2^\ast,\cdots,\widetilde{X}_l+M_l^\ast]\in BR^{n_1\times n_1}_\ast \times BR^{n_2\times n_2}_\ast \times \cdots\times BR^{n_l\times n_l}_\ast$,$\sum\limits_{j=1}^lA_{ij}M_j^{\ast}B_{ij}=0,\; (i=1,2,\cdots,t)$,那么

$$\sum\limits_{i=1}^t\bigg\|\sum\limits_{j=1}^lA_{ij} (\widetilde{X}_j+M_j^\ast)B_{ij}-F_i\bigg\|^2 =\sum\limits_{i=1}^t\bigg\|\sum\limits_{j=1}^lA_{ij}\widetilde{X}_jB_{ij}-F_i\bigg\|^2,$$

(3.11)

表明$[\widetilde{X}_1+M_1^\ast,\widetilde{X}_2+M_2^\ast,\cdots, \widetilde{X}_l+M_l^\ast]$是问题A的一个解.

定理3.2     如果选取初始的矩阵组$[X_{0,1},X_{0,2},\cdots,X_{0,l}]\in {\Bbb W}$,那么,通过算法3.1可得问题A的最小范数解.

证     由算法3.1和定理3.1知,如果选取初始矩阵组$[X_{0,1},X_{0,2},\cdots,X_{0,l}]\in {\Bbb W}$,可得问题A的一个解$[\widetilde{X}_1^{\ast},\widetilde{X}_2^{\ast}, \cdots,\widetilde{X}_l^{\ast}]$,并且存在$\widetilde{H}_i\in R^{p_i\times m_i}$,使得$\widetilde{X}_j^{\ast}=\Psi(\sum\limits_{i=1}^tA_{ij}^T\widetilde{H}_iB_{ij}^T), \;j=1,2,\cdots,l.$ 由引理3.5知,问题A的任意解可表示为$[\widetilde{X}_1^{\ast}+M_1^\ast,\widetilde{X}_2^{\ast}+M_2^\ast,\cdots,\widetilde{X}_l^{\ast}+M_l^\ast]$, 其中 $[M_1^\ast,M_2^\ast,\cdots,M_l^\ast]\in BR^{n_1\times n_1}_\ast \times BR^{n_2\times n_2}_\ast \times \cdots\times BR^{n_l\times n_l}_\ast$,并且满足(3.9)式,而且 $$\sum\limits_{j=1}^l\langle\widetilde{X}_j^{\ast},M_j^{\ast}\rangle =\sum\limits_{j=1}^l\langle\Psi(\sum\limits_{i=1}^tA_{ij}^T\widetilde{H}_iB_{ij}^T),\;M_j^{\ast}\rangle =\sum\limits_{i=1}^t\langle\widetilde{H}_i,\;\sum\limits_{j=1}^lA_{ij}M_j^{\ast}B_{ij}\rangle=0.$$ 因此 $$\sum\limits_{j=1}^l\|\widetilde{X}_j^{\ast}+M_j^{\ast}\|^2=\sum\limits_{j=1}^l\langle\widetilde{X}_j^{\ast}+M_j^{\ast}\;,\; \widetilde{X}_j^{\ast}+M_j^{\ast}\rangle =\sum\limits_{j=1}^l\|\widetilde{X}_j^{\ast}\|^2+\sum\limits_{j=1}^l\|M_j^{\ast}\|^2 \geq \sum\limits_{j=1}^l\|\widetilde{X}_j^{\ast}\|^2.$$ 故$[\widetilde{X}_1^{\ast},\widetilde{X}_2^{\ast},\cdots,\widetilde{X}_l^{\ast}]$是问题A的最小范数解.

注5     问题A的解集是非空的,由引理3.5知,这个解集是一个闭凸锥. 因此,问题A存在唯一解. 如果$[\widetilde{X}_1,\widetilde{X}_2,\cdots,\widetilde{X}_l]\in {\Bbb W}$ 是问题A的解,那么它就是唯一最小范数解.

定理3.3     对任意初始矩阵组$[X_{0,1},X_{0,2},\cdots,X_{0,l}]\in BR^{n_1\times n_1}_\ast \times BR^{n_2\times n_2}_\ast \times \cdots\times BR^{n_l\times n_l}_\ast$,算法3.1第$k$步生成的矩阵组 $[X_{k,1},X_{k,2},\cdots,X_{k,l}]$满足以下优化问题 $$\sum\limits_{i=1}^t \bigg\|\sum\limits_{j=1}^l A_{ij}X_{k,j}B_{ij}-F_i\bigg\|^2 =\min_{[X_1,X_2,\cdots,X_l]\in {\Bbb U}} \sum\limits_{i=1}^t\bigg\|\sum\limits_{j=1}^lA_{ij}X_jB_{ij}-F_i \bigg\|^2,$$ 其中${\Bbb U}$表示仿射子空间,具有如下形式 $${\Bbb U}=[X_{0,1},X_{0,2},\cdots,X_{0,l}]+ {\rm span}\{[Q_{0,1},Q_{0,2},\cdots,Q_{0,l}],\cdots,[Q_{k-1,1},Q_{k-1,2},\cdots,Q_{k-1,l}]\}.$$

证     对任意初始矩阵组$[X_1,X_2,\cdots,X_l]\in {\Bbb U}$,存在实数列$\{t_i\}_0^{k-1}$,使得 $$[X_1,X_2,\cdots,X_l]=[X_{0,1},X_{0,2},\cdots,X_{0,l}]+\sum\limits_{i=0}^{k-1}t_i[Q_{i,1},Q_{i,2},\cdots,Q_{i,l}].$$ 令$g(t_o,t_1,\cdots,t_{k-1})=\sum\limits_{w=1}^t\|\sum\limits_{j=1}^lA_{wj}(X_{0,j}+\sum\limits_{i=0}^{k-1}t_iQ_{i,j})B_{wj}-F_w\|^2.$ 由引理3.4知 $$\begin{eqnarray*} && g(t_o,t_1,\cdots,t_{k-1}) \\ &=&\sum\limits_{w=1}^t \bigg\|\sum\limits_{j=1}^lA_{wj}X_{0,j}B_{wj}-F_w +\sum\limits_{i=0}^{k-1}t_i\sum\limits_{j=1}^lA_{wj}Q_{i,j}B_{wj} \bigg\|^2\\ &=&\sum\limits_{w=1}^t\bigg\|-R_{0,w}+\sum\limits_{i=0}^{k-1 }t_i\sum\limits_{j=1}^lA_{wj}Q_{i,j}B_{wj}\bigg\|^2\\ &=&\sum\limits_{w=1}^t \bigg(\|R_{0,w}\|^2+\sum\limits_{i=0}^{k-1}t_i^2 \bigg\|\sum\limits_{j=1}^lA_{wj}Q_{i,j}B_{wj}\bigg\|^2 -2\sum\limits_{i=0}^{k-1}t_i\bigg\langle\sum\limits_{j=1}^lA_{wj} Q_{i,j}B_{wj},\;R_{0,w}\bigg\rangle\bigg), \end{eqnarray*}$$ 其中$R_{0,w}\;(w=1,2,\cdots,t)$是初始矩阵组$[X_{0,1},X_{0,2}, \cdots,X_{0,l}]$对应的残量. 由算法3.1知,$R_{0,w}$可表示为

$$R_{0,w}=R_{i,w}+\alpha_{i-1}\sum\limits_{j=1}^lA_{wj}Q_{i-1,j}B_{wj}+\cdots+\alpha_{0}\sum\limits_{j=1}^lA_{wj}Q_{0,j}B_{wj}.$$

(3.12)

由于$g(t_o,t_1,\cdots,t_{k-1})$是连续可微函数, 所以使$g(t_o,t_1,\cdots,t_{k-1})$最小,当且仅当 $$\frac{\partial g(t_o,t_1,\cdots,t_{k-1})}{\partial t_i}=0.$$ 由引理3.4、(3.8)式以及(3.12)式知 $$\begin{eqnarray*} t_i&=&\frac{\sum\limits_{w=1}^t\langle\sum\limits_{j=1}^lA_{wj}Q_{i,j}B_{wj},\;R_{0,w}\rangle}{\sum\limits_{w=1}^t\|\sum\limits_{j=1}^lA_{wj}Q_{i,j}B_{wj}\|^2} =\frac{\sum\limits_{w=1}^t\langle\sum\limits_{j=1}^lA_{wj}Q_{i,j}B_{wj},\;R_{i,w}\rangle}{\sum\limits_{w=1}^t\|\sum\limits_{j=1}^lA_{wj}Q_{i,j}B_{wj}\|^2}\\ &=&\frac{\sum\limits_{j=1}^l\langle Q_{i,j},\;\Psi(\sum\limits_{w=1}^tA_{wj}^TR_{i,w}B_{wj}^T)\rangle}{\sum\limits_{w=1}^t\|\sum\limits_{j=1}^lA_{wj}Q_{i,j}B_{wj}\|^2} =\frac{\sum\limits_{j=1}^l\langle Q_{i,j},\;P_{i,j}\rangle}{\sum\limits_{w=1}^t\|\sum\limits_{j=1}^lA_{wj}Q_{i,j}B_{wj}\|^2}\\ &=&\frac{\sum\limits_{j=1}^l\|P_{i,j}\|^2}{\sum\limits_{w=1}^t\|\sum\limits_{j=1}^lA_{wj}Q_{i,j}B_{wj}\|^2} =\alpha_i. \end{eqnarray*}$$ 因为 $$\min_{t_i}g(t_o,t_1,\cdots,t_{k-1})= \min_{[X_1,X_2,\cdots,X_l]\in {\Bbb U}} \sum\limits_{i=1}^t\bigg\|\sum\limits_{j=1}^lA_{ij}X_jB_{ij}-F_i\bigg\|^2,$$ 所以定理3.3成立.

定理3.3表明,如果$[X_{k-1,1},X_{k-1,2},\cdots,X_{k-1,l}] \in {\Bbb U}$,$[X_{k,1},X_{k,2},\cdots,X_{k,l}]\in {\Bbb U}$,那么 $$\sum\limits_{i=1}^t\bigg\|\sum\limits_{j=1}^l A_{ij}X_{k,j}B_{ij} -F_i\bigg\|\leq\sum\limits_{i=1}^t \bigg\|\sum\limits_{j=1}^l A_{ij}X_{k-1,j}B_{ij}-F_i\bigg\|,$$ 表明残量序列$\{\sum\limits_{i=1}^t\|\sum\limits_{j=1}^l A_{ij}X_{k,j}B_{ij}-F_i\|\}(k=1,2,\cdots)$单调递减. 这个单调递减的性质确保 算法3.1快速的收敛.

由于问题I的解集是一个非空的闭凸锥,所以问题II存在唯一解. 给定矩阵组$[\widehat{X}_1,\widehat{X}_2,$ $\cdots,\widehat{X}_l] \in D_1\times D_2\times\cdots\times D_l$,那么 $$\min\sum\limits_{i=1}^t\bigg\|\sum\limits_{j=1}^l A_{ij}X_jB_{ij}-C_i\bigg\| =\min\sum\limits_{i=1}^t\bigg\|\sum\limits_{j=1}^l A_{ij}(X_j-\widehat{X}_j)B_{ij} -(C_i-\sum\limits_{j=1}^lA_{ij}\widehat{X}_jB_{ij})\bigg\|.$$

设$N_j=X_j-\widehat{X}_j\;(j=1,2,\cdots,l),\ \widehat{F}_i=C_i-\sum\limits_{j=1}^lA_{ij}\widehat{X}_jB_{ij}$,则 求问题II的唯一解等价于求下述最小二乘问题的最小Frobenius范数解 $$\min_{[N_1,N_2,\cdots,N_l]}\sum\limits_{i=1}^t \bigg\|\sum\limits_{j=1}^l A_{ij}N_jB_{ij}-\widehat{F}_i\bigg\|. (4.1)$$ 应用算法3.1,设初始矩阵组$[N_{0,1},N_{0,2},\cdots,N_{0,l}]=[0,0,\cdots,0]$,可得问题(4.1)唯一最小Frobenius范数解$[N_1^*,N_2^*,\cdots,N_l^*]$, 从而得到问题II的唯一解$[\hat{X}_1^\ast,\hat{X}_2^\ast,\cdots,\hat{X}_l^\ast]$. 在这种情况下, $\hat{X}_1^\ast,\hat{X}_2^\ast,\cdots,\hat{X}_l^\ast$可表示为$\hat{X}_1^\ast=N_1^*+\widehat{X}_1,\;\hat{X}_2^\ast =N_2^*+\widehat{X}_2,\;\cdots,\;\hat{X}_l^\ast=N_l^*+\widehat{X}_l$.

例 5.1     求解矩阵方程组 $$\label{eqn 5.1} \left\{ \begin{array}{ll} A_{11}X_1B_{11}+A_{12}X_2B_{12}=C_1,\\ A_{21}X_1B_{21}+A_{22}X_2B_{22}=C_2, \end{array} \right. (5.1)$$ 其中 $$A_{11} = \left(\begin{array}{rrrrrrrr} 1 & 3 & -2 & 5 & -1 & 4 & 2 & 2\\ 3 & -7 & 1 & -8 & 2 & -9 & 4 & -8\\ 3 & -2 & 4 & -4 & 5 & -3 & 5 & -4\\ 11 & 6 & 12 & 7 & 10 & 4 & 12 & 7\\ -5 & 1 & -5 & 2 & -3 & 3 & -6 & 2\\ 9 & 4 & 6 & 3 & 7 & 5 & 8 & 3\\ 8 & -3 & 9 & -3 & 5 & -6 & 7 & -2 \end{array}\right), B_{11} = \left(\begin{array}{rrrrrrr} -1 & 4 & -1 & 4 & -1 & 4 & -1\\ 5 & -1 & 5 & -1 & 5 & -1 & 5\\ -1 & -2 & -1 & -2 & -1 & -2 & -1\\ 3 & 9 & 3 & 9 & 3 & 9 & 3\\ 7 & -8 & 7 & -8 & 7 & -8 & 7\\ -3 & 5 & -3 & 5 & -3 & 5 & -3\\ 4 & -6 & 4 & -6 & 4 & -6 & 4\\ 7 & 2 & 7 & 2 & 7 & 2 & 7 \end{array}\right),$$ $$A_{12} = \left(\begin{array}{rrrrrrrrr} 3 & -4 & 1 & -3 & 7 & -8 & 2 & -1 & 4\\ -1 & 3 & -2 & 4 & -2 & 6 & -1 & 2 & -2\\ 3 & -5 & 4 & -6 & 6 & -4 & 4 & -3 & 1\\ 3 & -4 & 1 & -7 & 5 & -2 & 3 & -5 & 6\\ -1 & 3 & -3 & 6 & -2 & 1 & -5 & -6 & -4\\ 3 & -5 & 2 & -8 & 1 & -2 & 3 & -7 & 1\\ 1 & 2 & 5 & 2 & 3 & 5 & 1 & 1 & 4 \end{array}\right), B_{12} = \left(\begin{array}{rrrrrrr} -9 & 4 & -9 & 4 & -9 & 4 & -9\\ -2 & 3 & -2 & 3 & -2 & 3 & -2\\ 3 & 5 & 3 & 5 & 3 & 5 & 3\\ 2 & -6 & 2 & -6 & 2 & -6 & 2\\ 9 & 3 & 9 & 3 & 9 & 3 & 9\\ 4 & -13 & 4 & -13 & 4 & -13 & 4\\ 11 & -5 & 11 & -5 & 11 & -5 & 11\\ -3 & 7 & -3 & 7 & -3 & 7 & -3\\ 6 & 1 & 6 & 1 & 6 & 1 & 6 \end{array}\right),$$ $$A_{21} =\left(\begin{array}{rrrrrrrrrr} 3 & -1 & 0 & -8 & 2 & -9 & 4 & -7\\ 3 & -2 & 4 & -4 & 5 & -3 & 5 & -4\\ 11 & 6 & 12 & 7 & 10 & 4 & 12 & 7\\ -5 & 1 & -5 & 2 & -3 & 4 & -6 & 2\\ 7 & 4 & 6 & 3 & 7 & 5 & 8 & 3\\ 1 & -3 & 9 & -3 & 5 & -6 & 7 & -4 \end{array}\right), B_{21} =\left(\begin{array}{rrrrrrr} 3 & -1 & 4 & -1 & 4 & -2\\ -1 & 5 & -1 & 5 & -1 & 5\\ -3 & -1 & -2 & -1 & -2 & -1\\ 8 & 3 & 6 & 3 & 8 & 2\\ -8 & 7 & -4 & 7 & -4 & 7\\ 4 & -3 & 5 & -3 & 5 & -2\\ -6 & 4 & -6 & 4 & -6 & 4\\ 1 & 3 & 2 & 6 & 2 & 6 \end{array}\right),$$ $$A_{22} =\left(\begin{array}{rrrrrrrrrr} 3 & -1 & 3 & -1 & 3 & -1 & 5 & -2 & 2\\ -2 & 3 & -2 & 3 & -2 & 3 & -4 & 1 & -1\\ -3 & 5 & -3 & 5 & -3 & 5 & -1 & 2 & -3\\ -1 & 3 & -1 & 3 & -1 & 3 & -2 & 2 & -1\\ 3 & -1 & 3 & -1 & 3 & -1 & 1 & -1 & 2\\ -3 & 5 & -3 & 5 & -3 & 5 & -4 & 3 & -3 \end{array}\right), B_{22} =\left(\begin{array}{rrrrrrr} 5 & -3 & 5 & -3 & 6 & -1\\ -6 & 2 & -6 & 2 & -6 & 2\\ -8 & 4 & -8 & 4 & -8 & 4\\ -1 & -8 & -1 & -8 & -1 & -8\\ -3 & 2 & -3 & 2 & -3 & 2\\ -1 & -2 & -1 & -2 & -1 & -2\\ 4 & -7 & 4 & -7 & 4 & -7\\ 3 & 12 & 3 & 12 & 3 & 12\\ 2 & -5 & 2 & -5 & 2 & -5 \end{array}\right),$$ $$C_1 =\left(\begin{array}{rrrrrrr} 262 & -1446 & 262 & -1446 & 262 & -1446 & 262\\ 152 & 1246 & 152 & 1246 & 152 & 1246 & 152\\ 1224 & -450 & 1224 & -450 & 1224 & -450 & 1224\\ 2110 & -2092 & 2110 & -2092 & 2110 & -2092 & 2110\\ -770 & 800 & -770 & 800 & -770 & 800 & -770\\ 1876 & -1106 & 1876 & -1106 & 1876 & -1106 & 1876\\ 1696 & -520 & 1696 & -520 & 1696 & -520 & 1696\\ \end{array}\right),$$ $$C_2 =\left(\begin{array}{rrrrrrr} -110 & -410 & -502 & -590 & -642 & -628\\ 68 & 282 & -210 & 348 & -144 & 398\\ -1952 & 852 & -1704 & 954 & -1690 & 1132\\ -64 & -406 & 156 & -418 & 178 & -430\\ -422 & 308 & -246 & 380 & -218 & 450\\ -154 & 418 & -476 & 514 & -418 & 550 \end{array}\right).$$ toeplitz$(1:n)$表示第1行为$(1,2,\cdots,n)$的$n$阶Toeplitz矩阵. hilb$(n)$表示$n$阶Hilbert矩阵. 设中心主子矩阵为 $X_{q_1}={\rm toeplitz}(1:4)$,$X_{q_2}={\rm hilb}(5)$. 首先计算$F_i=C_i-A_{i1}\overline{X}_1B_{i1}-A_{i2} \overline{X}_2B_{i2}\;(i=1,2)$,其中$\overline{X}_j$ $(j=1,2)$为形如(3.1)式具有适当大小的矩阵. 于是可得以下方程组 $$\label{eqn 5.2} \left\{ \begin{array}{ll} A_{11}X_1B_{11}+A_{12}X_2B_{12}=F_1,\\ A_{21}X_1B_{21}+A_{22}X_2B_{22}=F_2 . \end{array} \right. (5.2).$$ 设初始矩阵为$[X_{0,1},X_{0,2}]=[{\rm zeros}(8), {\rm zeros}(9)]$. 应用算法3.1, 迭代69步可得方程组(5.2)在子空间$BR^{8\times 8}_\ast \times BR^{9\times 9}_\ast$ 上的最小Frobenius范数最小二乘解$(\widetilde{X}_1,\widetilde{X}_2)$ {\scriptsize $$\widetilde{X}_1 =\left(\begin{array}{rrrrrrrr} -3.4116& 20.1512& 13.8336 & 8.8217 & 2.9464 & -13.3935 & -26.0629 & -10.8094\\ 20.1512& 22.1981& 6.1023 & -9.7545 & -7.9694 & 4.1094 & 8.8559 & -26.0629\\ 13.8336& 6.1023& 0 & 0 & 0 & 0 & 4.1094 & -13.3935\\ 8.8217& -9.7545& 0 & 0 & 0 & 0 & -7.9694 & 2.9464\\ 2.9464& -7.9694& 0 & 0 & 0 & 0 & -9.7545 & 8.8217\\ -13.3935& 4.1094& 0 & 0 & 0 & 0 & 6.1023 & 13.8336\\ -26.0629& 8.8559& 4.1094 & -7.9694 & -9.7545 & 6.1023 & 22.1981 & 20.1512\\ -10.8094& -26.0629& -13.3935 & 2.9464 & 8.8217 & 13.8336 & 20.1512 & -3.4116 \end{array}\right),$$ $$\widetilde{X}_2 = \left(\begin{array}{rrrrrrrrr} 25.1095 & -6.9018 & -26.6546 & 11.8188 & 15.9120 & 1.6025& -7.5709& -2.5952& -15.4706\\ -6.9018 & 32.4696 & -4.2254 & 7.1416 & -11.2847 & 8.8638& 11.2418& -20.1741& -2.5952\\ -26.6546 & -4.2254 & 0 & 0 & 0 & 0& 0& 11.2418& -7.5709\\ 11.8188 & 7.1416 & 0 & 0 & 0 & 0& 0& 8.8638& 1.6025\\ 15.9120 & -11.2847 & 0 & 0 & 0 & 0& 0& -11.2847& 15.9120\\ 1.6025 & 8.8638 & 0 & 0 & 0 & 0& 0& 7.1416& 11.8188\\ -7.5709 & 11.2418 & 0 & 0 & 0 & 0& 0& -4.2254& -26.6546\\ -2.5952 & -20.1741 & 11.2418 & 8.8638 & -11.2847 & 7.1416& -4.2254& 32.4696& -6.9018\\ -15.4706 & -2.5952 & -7.5709 & 1.6025 & 15.9120 & 11.8188& -26.6546& -6.9018& 25.1095 \end{array}\right),$$ } 残量范数为 $$\|F_1-A_{11}\widetilde{X}_1 B_{11}-A_{12}\widetilde{X}_2 B_{12}\|+\|F_2-A_{21}\widetilde{X}_1 B_{21}-A_{22}\widetilde{X}_2 B_{22}\|= 709.9595.$$ 同样的,应用算法3.1,可以计算序列$\{(X_{k,1},X_{k,2})\}$. 设 $\delta_k=\frac{\|(X_{k,1},X_{k,2})-(\widetilde{X}_1,\widetilde{X}_2)\|}{\|(\widetilde{X}_1,\widetilde{X}_2)\|}$ 表示解的相对误差,$r_k=\sum\limits_{i=1}^2||F_i-A_{i1}X_{k,1}B_{i1}-A_{i2}X_{k,2}B_{i2}||$表示残量范数. 求得的结果如图 1. 从图 1上可见,随着迭代次数的增加,$r_k$逐渐下降,$\delta_k$逐渐下降并趋向于0.

矩阵方程组(5.1)的最小Frobenius范数最小二乘解为 {\scriptsize $$X_1^\ast =\left(\begin{array}{rrrrrrrr} -3.4116& 20.1512& 13.8336 & 8.8217& 2.9464 & -13.3935 & -26.0629 & -10.8094\\ 20.1512& 22.1981& 6.1023 & -9.7545& -7.9694 & 4.1094 & 8.8559 & -26.0629\\ 13.8336& 6.1023& 1 & 2& 3 & 4 & 4.1094 & -13.3935\\ 8.8217& -9.7545& 2 & 1& 2 & 3 & -7.9694 & 2.9464\\ 2.9464& -7.9694& 3 & 2& 1 & 2 & -9.7545 & 8.8217\\ -13.3935& 4.1094& 4 & 3& 2 & 1 & 6.1023 & 13.8336\\ -26.0629& 8.8559& 4.1094 & -7.9694& -9.7545 & 6.1023 & 22.1981 & 20.1512\\ -10.8094& -26.0629& -13.3935 & 2.9464& 8.8217 & 13.8336 & 20.1512 & -3.4116 \end{array}\right),$$ $$X_2^\ast = \left(\begin{array}{rrrrrrrrr} 25.1095 & -6.9018 & -26.6546 & 11.8188& 15.9120& 1.6025& -7.5709& -2.5952& -15.4706\\ -6.9018 & 32.4696 & -4.2254 & 7.1416& -11.2847& 8.8638& 11.2418& -20.1741& -2.5952\\ -26.6546 & -4.2254 & 1.0000 & 0.5000& 0.3333& 0.2500& 0.2000& 11.2418& -7.5709\\ 11.8188 & 7.1416 & 0.5000 & 0.3333& 0.2500& 0.2000& 0.1667& 8.8638& 1.6025\\ 15.9120 & -11.2847 & 0.3333 & 0.2500& 0.2000& 0.1667& 0.1429& -11.2847& 15.9120\\ 1.6025 & 8.8638 & 0.2500 & 0.2000& 0.1667& 0.1429& 0.1250& 7.1416& 11.8188\\ -7.5709 & 11.2418 & 0.2000 & 0.1667& 0.1429& 0.1250& 0.1111& -4.2254& -26.6546\\ -2.5952 & -20.1741 & 11.2418 & 8.8638& -11.2847& 7.1416& -4.2254& 32.4696& -6.9018\\ -15.4706 & -2.5952 & -7.5709 & 1.6025& 15.9120& 11.8188& -26.6546& -6.9018& 25.1095 \end{array}\right).$$ }

例 5.2     矩阵$A_{11},B_{11},A_{12},B_{12},A_{21}, B_{21},A_{22},B_{22},C_1$和$C_2$分别为 $$A_{11} = \left(\begin{array}{cc} {\rm hilb}(n/2)~ & {\rm ones}(n/2) \\ {\rm hankel}(1:n/2)~ & {\rm zeros}(n/2) \end{array}\right),\;\;\;\;\;\; B_{11}={\rm eye}(n),$$ $$A_{12} = \left(\begin{array}{cc} {\rm toeplitz}(1:n/2) ~& {\rm ones}(n/2)\\ {\rm zeros}(n/2) ~ & {\rm ones}(n/2) \end{array}\right),\;\;\;\;\;\;B_{12} = {\rm ones}(n),$$ $$A_{21} = \left(\begin{array}{cc} {\rm hankel}(1:n/2)~ & {\rm ones}(n/2)\\ {\rm toeplitz}(1:n/2)~ & {\rm zeros}(n/2) \end{array}\right),\;\;\;\;\;\;\;B_{21}=-{\rm eye}(n),$$ $$A_{22} = {\rm hankel}(1:n), \qquad B_{22}={\rm hadamard}(n),$$ $$C_1={\rm tridiag}([1,5,-1],n),\; \; C_2={\rm toeplitz}(1:n)\times{\rm hankel}(1:n),$$ 其中 hankel$(1:n)$表示第1行为$(1,2,\cdots,n)$的$n$阶Hankel矩阵. $C_1$是由向量$[1,5,-1]$生成的三对角矩阵. $X_{q_1}={\rm toeplitz}(1:8)$和$X_{q_2}={\rm hilb}(8)$为给定的中心主子 矩阵. 应用算法3.1,分别求解当$n=12,24,48,96$时所对应的矩阵方程组(5.1). 首先计算$F_i=C_i-A_{i1}\overline{X}_1B_{i1} -A_{i2}\overline{X}_2B_{i2}\;(i=1,2)$, 其中$\overline{X}_j \ (j=1,2)$是形如(3.1)式具有适当大小的矩阵. 设$[X_{0,1},X_{0,2}]=[{\rm zeros}(n),{\rm zeros}(n)]$,应用算法(3.1), 得到矩阵方程组 (5.2)在$BR^{n\times n}_\ast \times BR^{n\times n}_\ast$上的 最小Frobenius范数最小二乘解. 表 1列出了有关的数值结果, 即列出了当$n=12,24,48,96$时矩阵$A_{11},A_{12},$ $A_{21},A_{22}$的秩,以及在停止标准为 $\sum\limits_{j=1}^2\|P_{k,j}\|^2\leq 10e-010$的条件下的迭代次数 (IT),停止条件$\sum\limits_{j=1}^2\|P_{k,j}\|^2$(TER), 残量范数$\sum\limits_{i=1}^2\|F_i-A_{i1}X_{k,1}B_{i1}- A_{i2}X_{k,2}B_{i2}\|$(RES)和CPU时间.

尽管$A_{11},A_{12}$和$A_{21}$的条件数很大,但是从表 1可以看出,算法3.1的收敛速度是很快的. 为了节省篇幅,我们仅给出当$n=12$时方程组(5.1)最小范数最小 二乘解. {\tiny $$X_1^\ast =\left(\begin{array}{rrrrrrrrrrrr} 7.1818& 10.0101 & -5.3302 & -7.1262 & -8.6368 & -8.8701 & -9.2279 & -9.4798& -8.4007& -6.1948 & 10.2040 & 8.5820 \\ 10.0101& 17.4332 & -10.2940 & -7.5255 & -7.9086 & -7.6042 & -7.4557 & -9.2785& -10.4503& -15.6862 & 23.8558 & 10.2040 \\ -5.3302& -10.2940 & 1 & 2 & 3 & 4 & 5 & 6& 7& 8 & -15.6862 & -6.1948 \\ -7.1262& -7.5255 & 2 & 1 & 2 & 3 & 4 & 5& 6& 7 & -10.4503 & -8.4007 \\ -8.6368& -7.9086 & 3 & 2 & 1 & 2 & 3 & 4& 5& 6 & -9.2785 & -9.4798 \\ -8.8701& -7.6042 & 4 & 3 & 2 & 1 & 2 & 3& 4& 5 & -7.4557 & -9.2279 \\ -9.2279& -7.4557 & 5 & 4 & 3 & 2 & 1 & 2& 3& 4 & -7.6042 & -8.8701 \\ -9.4798& -9.2785 & 6 & 5 & 4 & 3 & 2 & 1& 2& 3 & -7.9086 & -8.6368 \\ -8.4007& -10.4503 & 7 & 6 & 5 & 4 & 3 & 2& 1& 2 & -7.5255 & -7.1262 \\ -6.1948& -15.6862 & 8 & 7 & 6 & 5 & 4 & 3& 2& 1 & -10.2940 & -5.3302 \\ 10.2040& 23.8558 & -15.6862 & -10.4503 & -9.2785 & -7.4557 & -7.6042 & -7.9086& -7.5255& -10.2940 & 17.4332 & 10.0101 \\ 8.5820& 10.2040 & -6.1948 & -8.4007 & -9.4798 & -9.2279 & -8.8701 & -8.6368& -7.1262& -5.3302 & 10.0101 & 7.1818 \end{array}\right),$$ } {\tiny $$X_2^\ast = \left(\begin{array}{rrrrrrrrrrrrr} 6.6141& 0.9170& -0.5848 & 1.5692 & 0.7367 & -0.7189 & -3.2931 & -2.4996 & -0.4744 & 1.6602 & -0.0941 & 3.5430 \\ 0.9170& -0.3455& 0.9143 & 1.4465 & 0.4298 & -1.0287 & -0.7414 & 0.2572 & 0.6862 & -0.4000 & 0.1754 & -0.0941 \\ -0.5848& 0.9143& 1.0000 & 0.5000 & 0.3333 & 0.2500 & 0.2000 & 0.1667 & 0.1429 & 0.1250 & -0.4000 & 1.6602 \\ 1.5692& 1.4465& 0.5000 & 0.3333 & 0.2500 & 0.2000 & 0.1667 & 0.1429 & 0.1250 & 0.1111 & 0.6862 & -0.4744 \\ 0.7367& 0.4298& 0.3333 & 0.2500 & 0.2000 & 0.1667 & 0.1429 & 0.1250 & 0.1111 & 0.1000 & 0.2572 & -2.4996 \\ -0.7189& -1.0287& 0.2500 & 0.2000 & 0.1667 & 0.1429 & 0.1250 & 0.1111 & 0.1000 & 0.0909 & -0.7414 & -3.2931 \\ -3.2931& -0.7414& 0.2000 & 0.1667 & 0.1429 & 0.1250 & 0.1111 & 0.1000 & 0.0909 & 0.0833 & -1.0287 & -0.7189 \\ -2.4996& 0.2572& 0.1667 & 0.1429 & 0.1250 & 0.1111 & 0.1000 & 0.0909 & 0.0833 & 0.0769 & 0.4298 & 0.7367 \\ -0.4744& 0.6862& 0.1429 & 0.1250 & 0.1111 & 0.1000 & 0.0909 & 0.0833 & 0.0769 & 0.0714 & 1.4465 & 1.5692 \\ 1.6602& -0.4000& 0.1250 & 0.1111 & 0.1000 & 0.0909 & 0.0833 & 0.0769 & 0.0714 & 0.0667 & 0.9143 & -0.5848 \\ -0.0941& 0.1754& -0.4000 & 0.6862 & 0.2572 & -0.7414 & -1.0287 & 0.4298 & 1.4465 & 0.9143 & -0.3455 & 0.9170 \\ 3.5430& -0.0941& 1.6602 & -0.4744 & -2.4996 & -3.2931 & -0.7189 & 0.7367 & 1.5692 & -0.5848 & 0.9170 & 6.6141 \end{array}\right).$$ }

图 2描述了当$n=12,24,48,96$时停止条件$\sum\limits_{j=1}^2\|P_{k,j}\|^2$的收敛曲线. 从图 2中可以看到, 随着迭代次数的增加,$\sum\limits_{j=1}^2\|P_{k,j}\|^2$逐渐趋向于0.

图 3描述了当$n=24$时残量范数的收敛曲线. 图 3表明残量范数是平稳下降的.

下面求解问题II. 不失一般性,令$n=24$. 设$\widehat{X}_1$是 ones$(24)$带有中心主子矩阵为toeplitz$(1:8)$的矩阵, $\widehat{X}_2$是eye$(24)$带有中心主子矩阵为hilb(8)的矩阵. 为了求与$[\widehat{X}_1,\widehat{X}_2]$相关的最佳逼近解 $[\widehat{X}_1^\ast, \widehat{X}_2^\ast]$,设$N_j=X_j-\widehat{X}_j\;(j=1,2)$, $\widehat{F_i}=C_i-A_{i1}\widehat{X}_1B_{i1}-A_{i2} \widehat{X}_2B_{i2}$, 初始矩阵为$[N_{0,1},N_{0,2}]=[{\rm zeros}(24),{\rm zeros}(24)]$, 应用算法3.1,迭代910步,可得方程组$A_{i1}N_1B_{i1} +A_{i2}N_2B_{i2}=\widehat{F_i}\;(i=1,2)$ 的最小范数解$[N_1^\ast,N_2^\ast]$. 因而与 $[\widehat{X}_1,\widehat{X}_2]$相关的最佳逼近解为 $\hat{X}_1^\ast=N_1^*+\widehat{X}_1,\hat{X}_2^\ast =N_2^*+\widehat{X}_2$.

本文提出了一种算法,即,算法3.1求解问题I. 首先,把问题I转化为等价的问题A. 然后,构造了算法3.1求解问题A,并且证明了此算法的收敛性. 本文还应用此算法求解问题II. 最后,给出了数值试验说明此算法是有效的.