我们知道关于反射矩阵的自反矩阵或反自反矩阵在系统与控制理论、工程、科学计算和其他领域都有广泛的应用^[1-3].近年来, Sylvester复矩阵方程的自反和反自反解的研究非常活跃. Peng和Hu^[4]给出方程$AX=B$的自反和反自反解.文献[5]给出方程${{A}^{H}}XB=C$的自反和反自反解. 2009年, Xu和Li^[6]研究了方程$AX=B$的反问题的Hermitian自反解.文献[7]给出反Hermitian、广义反Hamiltonian矩阵的最佳逼近解, 等等.

上述文献都是考虑方程有解情况下的最佳逼近解, 方程无解情况下的最佳逼近解目前研究很少.不论矩阵方程是否有解, 我们利用复合最速下降法^[8] (the hybrid steepest descent method, HSDM), 求${A_1}Z + Z{B_1} = {C_1}$广义自反最佳逼近解.

首先介绍本文中的符号. ${C}^{m\times n}$表示$m\times n$阶复矩阵的集合, ${R}^{m\times n}$表示$m\times n$阶实矩阵的集合, ${A}^{H}$表示矩阵$A$的共轭转置. $I$表示单位矩阵, 对于矩阵$A$, $B \in {\rm{ }}{C^{m \times n}}$, $A\otimes B$表示$A$与$B$的Kronecker积, $\left\langle A, B \right\rangle={\hbox{trace}}({{B}^{H}}A)$表示$A$与$B$的内积. $\left\| A \right\|$表示矩阵$A$的Frobe-nius范数, ${\left\| A \right\|^2} = \left\langle {A, A} \right\rangle $. ${{\left\| \alpha \right\|}_{2}}$表示向量$\alpha ={{\left( {{x}_{1}}, {{x}_{2}}, \cdots {{x}_{n}} \right)}^{T}}$的2-范数, ${{\left\| \alpha \right\|}_{2}}=\sqrt{{{\alpha }^{T}} \alpha }$.若${P}^{2}=I$且${P}^{H}=P$, 则称矩阵$P$是广义反射矩阵.若$A=PAQ$, 则称矩阵$A$为关于广义反射$P$, $Q$的广义自反矩阵.记${{C}_{r}}^{n\times n}(P, Q)=\{A\in {{C}^{m\times n}}|PAQ=A\}$.定义矩阵$A=({{a}_{ij}})\in {C}^{m\times n}$的拉伸算子为

${\hbox{vec}}(A)={{[{{a}_{11}}, {{a}_{21}}, \cdots, {{a}_{m1}}, {{a}_{12}}, {{a}_{22}}, \cdots, {{a}_{m2}}, \cdots, {{a}_{1n}}, {{a}_{2n}}, \cdots, {{a}_{mn}}]}^{T}}.$

$\nabla $表示梯度算子, 即$\nabla f(\alpha ) = {\left( {\frac{{\partial f}}{{\partial {x_1}}}, \frac{{\partial f}}{{\partial {x_2}}}, \cdots, \frac{{\partial f}}{{\partial {x_n}}}} \right)^T}$, 其中$\alpha = {({x_1}, {x_2}, \cdots, {x_n})^T}$, $H$是希尔伯特空间, 对于映射$T:H\to H$, 所有关于$T$的不动点的集合表示为${\rm{Fix}}(T): = \{ x \in H|T(x) = x$, ${P}_{K}$表示到非空凸集$K$上的投影算子.

本文考虑如下问题:

问题Ⅰ 给定矩阵${A}_{1}$、${P}_{1}\in {C}^{m\times m}$, ${B}_{1}$、${Q}_{1}\in {C}^{n\times n}, $ $C\in {{C}^{m\times n}}, $求$Z\in C_{r}^{m\times n}({{P}_{1}}, {{Q}_{1}})$使得

$\begin{align} \left\|{{A}_{1}}Z+Z{{B}_{1}}-{{C}_{1}}\right\|=\min. \end{align}$

(1.1)

问题Ⅱ 设问题Ⅰ的解集合为${S}_{E}$, 给定${Z}^{*}\in C_{r}^{m\times n}({{P}_{1}}, {{Q}_{1}})$, 求$\hat{Z}\in {S}_{E}$使得

$\begin{align} \left\| \hat{Z}-{{Z}^{*}} \right\|=\underset{Z\in {{S}_{E}}}{\mathop{\min }}\, \left\| Z-{{Z}^{*}} \right\|. \end{align}$

(1.2)

问题Ⅱ是在${S}_{E}$里找一个与矩阵${Z}^{*}\in C_{r}^{m\times n}({{P}_{1}}, {{Q}_{1}})$最接近的矩阵$\hat{Z}$.

本文的结构如下:第2部分介绍问题Ⅰ和问题Ⅱ的一个迭代算法并证明该算法是收敛的; 第3部分用两个数值例子来验证该算法的可行性; 第4部分给出结论.

本节首先给出解决问题Ⅰ、Ⅱ的迭代算法, 然后证明该算法收敛, 最后研究如何计算到非空凸集$K$上的投影.

算法:

步骤1  输入矩阵${C}_{1}\in {C}^{m\times n}$, ${Z}_{0}\in {C}^{m\times n}$和${Z}^{*}\in C_{r}^{m\times n}({{P}_{1}}, {{Q}_{1}})$.

步骤2  计算

$\begin{eqnarray*}&& A={\hbox{real}}({{A}_{1}}), B={\hbox{imag}}({{A}_{1}}), C={\hbox{real}}({{B}_{1}}), D={\hbox{imag}}({{B}_{1}}), E={\hbox{real}}({{C}_{1}}), F={\hbox{imag}}({{C}_{1}}), \\ && W=I\otimes A+{{C}^{T}}\otimes I, N=I\otimes B+{{D}^{T}}\otimes I, M=2\left( \begin{matrix} {{W}^{T}}W+{{N}^{T}}N&{{N}^{T}}W-{{W}^{T}}N \\ {{W}^{T}}N-{{N}^{T}}W&{{W}^{T}}W+{{N}^{T}}N \\ \end{matrix} \right), \\ && \nu =\frac{2}{\left\| M \right\|}, {{X}^{*}}={\hbox{real}}({{Z}^{*}}), \ \ {{Y}^{*}}={\hbox{imag}}({{Z}^{*}}), k:=1.\end{eqnarray*}$

步骤3

$\begin{eqnarray*}&&{{X}_{k-1}}={\hbox{real}}({{Z}_{k-1}}), {{Y}_{k-1}}={\hbox{imag}}({{Z}_{k-1}}), \\ &&\nabla \psi ({\hbox{vec}}({{Z}_{k-1}}))=M\left( \begin{matrix} {\hbox{ vec}}({{X}_{k-1}}) \\ {\hbox{vec}}({{Y}_{k-1}}) \\ \end{matrix} \right)-2\left( \begin{matrix} {{W}^{T}}&{{N}^{T}} \\ -{{N}^{T}}&{{W}^{T}} \\ \end{matrix} \right)\left( \begin{matrix} {\hbox{ vec}}(E) \\ {\hbox{vec}}(F) \\ \end{matrix} \right), \\ && T\left( \begin{matrix} {\hbox{ vec}}({{X}_{k-1}}) \\ {\hbox{vec}}({{Y}_{k-1}}) \\ \end{matrix} \right)={{P}_{K}}\left\{ \left( \begin{matrix} {\hbox{vec}}({{X}_{k-1}}) \\ {\hbox{vec}}({{Y}_{k-1}}) \\ \end{matrix} \right)-\nu \nabla \psi ({\hbox{vec}}({{Z}_{k-1}})) \right\}, \end{eqnarray*}$

其中

$\begin{eqnarray*}&& K=\left\{ Z|Z={{P}_{1}}Z{{Q}_{1}} \right\}, {\hbox{vec}}({{Z}^{*}})=\left( \begin{matrix} {\hbox{ vec}}({{X}^{*}}) \\ {\hbox{ vec}}({{Y}^{*}}) \\ \end{matrix} \right), \\ && {\hbox{vec}}({{Z}_{k}})=(1-\frac{2}{k+2})T({\hbox{vec}}({{Z}_{k-1}}))+\frac{2}{k+2}({\hbox{vec}}({{Z}^{*}})).\end{eqnarray*}$

步骤4  若$\left\| {{Z}_{k}}-{{Z}_{k-1}} \right\|<\varepsilon $, 则停止; 否则, $k:=k+1$, 转向步骤3.

为了证明该算法的收敛性, 首先给出下面的定义和定理.

定义2.1  设$S\subset H$是非空凸集, $\lambda \in (0, 1)$, $f(x)$是定义在$S$上的函数, 如果对任意${x}_{1}, {{x}_{2}}\in S$, 有

$f(\lambda {{x}_{1}}+(1-\lambda ){{x}_{2}})\le \lambda f({{x}_{1}})+(1-\lambda )f({{x}_{2}}), $

则称函数$f(x)$是$S$上的凸函数.

定义2.2  若$\forall {{x}_{1}}, {{x}_{2}}\in S\subset H$, $\exists \kappa ＞0$, 使$\left\| T(x)-T(y) \right\|\le \kappa \left\| x-y \right\|$成立, 则映射$T:H\to H$称为$\kappa$-Lipschitzian(或$\kappa$-Lipschitzian连续), 特别地, 若对于一切$x, y\in H$, 有$\left\| T(x)-T(y) \right\|\le \left\| x-y \right\|$成立, 则映射$T:H\to H$称为非扩展映射.

定义2.3  对于给定的集合$S\subset H$, 若$\forall x, y\in S$, $\exists \eta ＞0$使

$\langle F(x)-F(y), x-y\rangle \ge \eta {{\left\| x-y \right\|}^{2}}$

成立, 则称映射$F:H\to H$在集合$S$上$\eta $强单调.

定理2.1^[8]  (HSDM)设$T:H\to H$是非扩展映射且Fix$(T)\ne \varnothing $.假设函数$\theta :H\to R\bigcup \infty $存在Gateaux导数$\theta ':H\to H$, 且$\theta '$在$T(H)$上$\kappa$-Lipschitzian且$\eta $强单调.对于任意的${{u}_{0}}\in H$和正实数序列${({{\lambda }_{n}})}_{n\ge 1}$满足:

(L1) $\mathop {\lim }\limits_{n \to \infty } {{\lambda }_{n}}=0$; (L2) $\sum\limits_{n\ge 1}{{{\lambda }_{n}}}=\infty $; (L3) $\mathop {\lim }\limits_{n \to \infty } ({{\lambda }_{n}}-{{\lambda }_{n+1}})\lambda _{_{n+1}}^{-2}=0$;

(或者(W1) $\mathop {\lim }\limits_{n \to \infty } {{\lambda }_{n}}=0$; (W2) $\sum\limits_{n\ge 1}{{{\lambda }_{n}}}=\infty $; (W3) $\sum\limits_{n\ge 1}{\left| {{\lambda }_{n}}-{{\lambda }_{n+1}} \right|}<\infty $), 则由${{u}_{n+1}}:=T({{u}_{n}})-{{\lambda }_{n+1}}{{\theta }^{'}}(T({{u}_{n}}))$生成的序列${{({{u}_{n}})}_{n\ge 0}}$强收敛于唯一的${u}^{*}$, 其中${u}^{*}$满足$\theta ({u^*}) = \inf {\rm{ }}\left\{ {\theta ({\rm{Fix}}(T))} \right\}$ (例如: ${{\lambda }_{n}}={1}/{n}\;$是满足(W1)--(W3) 的序列${({{\lambda }_{n}})}_{n\ge 1}$).

定理2.2 ^[9]  设$K\subset H$是闭凸子集, 假设:

(ⅰ) $\psi :H \to R \cup \left\{ \infty \right\}$是Gateaux可微的凸函数且它的导数$\psi: 'H\to H$在$H$上满足$\gamma$-Lipschitzian;

(ⅱ) $\theta :H \to R \cup \left\{ \infty \right\}$是Gateaux可微的凸函数且它的导数$\theta ': H\to H$在$T(H)$上满足$\kappa$-Lipschitzian和$\eta $强单调, 其中$T:={{P}_{k}}(I-v\psi ')$, $v\in (0, {}^{2}\!\!\diagup\!\!{}_{\gamma }\;]$; 则$\forall {{u}_{0}}\in H$由${u}_{n}:=T({{u}_{n-1}})-{{\lambda }_{n}}\theta '(T({{u}_{n-1}}))$产生的序列${{\left( {{u}_{n}} \right)}_{n\ge 0}}$强收敛于唯一的

${x^*} = \arg \mathop {\inf }\limits_{x \in {K_\psi }} {\mkern 1mu} \theta (x),$

其中${K}_{\psi }:=\underset{x\in K}{\mathop{\arg \inf \psi (x)}}\, \ne \varnothing $.

在集合$K$上定义函数$\psi $和$\theta $, 并证明它们满足定理2.2的条件.

定理2.3  设$\psi ({\hbox{vec}}(Z))={\left\| {{A}_{1}}Z+Z{{B}_{1}}-{{C}_{1}} \right\|}^{2}$, 其中$ Z, {{C}_{1}}\in {{C}^{m\times n}}, $ ${{A}_{1}}\in {{C}^{m\times m}}, $ ${{B}_{1}}\in {{C}^{n\times n}}.$令

$\begin{eqnarray*}&& A={\hbox{real}}({{A}_{1}}), B={\hbox{imag}}({{A}_{1}}), C={\hbox{real}}({{B}_{1}}), D={\hbox{imag}}({{B}_{1}}), E={\hbox{real}}({{C}_{1}}), F={\hbox{imag}}({{C}_{1}}), \\ && X={\hbox{real}}(Z), Y={\hbox{imag}}(Z), W=I\otimes A+{{C}^{T}}\otimes I, N=I\otimes B+{{D}^{T}}\otimes I, \end{eqnarray*}$

则

$\begin{align} \nabla \psi (vec(Z))=\nabla \psi \left( \begin{matrix} {\hbox {vec}}(X) \\ {\hbox {vec}}(Y) \\ \end{matrix} \right) =M\left( \begin{matrix} {\hbox {vec}}(X) \\ {\hbox {vec}}(Y) \\ \end{matrix} \right)-2\left( \begin{matrix} {{W}^{T}}&{{N}^{T}} \\ -{{N}^{T}}&{{W}^{T}} \\ \end{matrix} \right) \left( \begin{matrix} {\hbox {vec}}(E) \\ {\hbox {vec}}(F) \\ \end{matrix} \right). \end{align}$

(2.1)

证

$\begin{eqnarray*}&&\psi ( {\hbox {vec}}(Z))=\psi \left( \begin{matrix} {\hbox {vec}}(X) \\ {\hbox {vec}}(Y) \\ \end{matrix} \right)\\ &=&{{\left\| {{A}_{1}}Z+Z{{B}_{1}}-{{C}_{1}} \right\|}^{2}}={{\left\| AX-BY+XC-YD-E \right\|}^{2}} \\ && +{{\left\| BX+AY+XD+YC-F \right\|}^{2}} \\ &=&{{\left\| {\hbox {vec}}(AX)- {\hbox {vec}}(BY)+ {\hbox {vec}}(XC)- {\hbox {vec}}(YD)- {\hbox {vec}}(E) \right\|}_{2}^{2}} \\ && +{{\left\| {\hbox {vec}}(BX)+ {\hbox {vec}}(AY)+ {\hbox {vec}}(XD)+ {\hbox {vec}}(YC)- {\hbox {vec}}(F) \right\|}_{2}^{2}} \\ &=&{{\left\| (I\otimes A) {\hbox {vec}}(X)-(I\otimes B) {\hbox {vec}}(Y)+({{C}^{T}}\otimes I) {\hbox {vec}}(X)-({{D}^{T}}\otimes I) {\hbox {vec}}(Y)- {\hbox {vec}}(E) \right\|}_{2}^{2}} \\ && +{{\left\| (I\otimes B) {\hbox {vec}}(X)+(I\otimes A) {\hbox {vec}}(Y)+({{D}^{T}}\otimes I) {\hbox {vec}}(X)+({{C}^{T}}\otimes I) {\hbox {vec}}(Y)- {\hbox {vec}}(F) \right\|}_{2}^{2}} \\ &=&\left\| \left( \begin{matrix} W&-N \\ N&W \\ \end{matrix} \right)\left( \begin{matrix} {\hbox {vec}}(X) \\ {\hbox {vec}}(Y) \\ \end{matrix} \right)-\left( \begin{matrix} {\hbox {vec}}(E) \\ {\hbox {vec}}(F) \\ \end{matrix} \right) \right\|_{2}^{2} \\ &=&\left\langle \left( \begin{matrix} W&-N \\ N&W \\ \end{matrix} \right)\left( \begin{matrix} {\hbox {vec}}(X) \\ {\hbox {vec}}(Y) \\ \end{matrix} \right)-\left( \begin{matrix} {\hbox {vec}}(E) \\ {\hbox {vec}}(F) \\ \end{matrix} \right), \left( \begin{matrix} W&-N \\ N&W \\ \end{matrix} \right)\left( \begin{matrix} {\hbox {vec}}(X) \\ {\hbox {vec}}(Y) \\ \end{matrix} \right)-\left( \begin{matrix} {\hbox {vec}}(E) \\ {\hbox {vec}}(F) \\ \end{matrix} \right) \right\rangle \\ &=&\left\langle \left( \begin{matrix} W&-N \\ N&W \\ \end{matrix} \right)\left( \begin{matrix} {\hbox {vec}}(X) \\ {\hbox {vec}}(Y) \\ \end{matrix} \right), \left( \begin{matrix} W&-N \\ N&W \\ \end{matrix} \right)\left( \begin{matrix} {\hbox {vec}}(X) \\ {\hbox {vec}}(Y) \\ \end{matrix} \right) \right\rangle \\ && -2\left\langle \left( \begin{matrix} W&-N \\ N&W \\ \end{matrix} \right)\left( \begin{matrix} {\hbox {vec}}(X) \\ {\hbox {vec}}(Y) \\ \end{matrix} \right), \left( \begin{matrix} {\hbox {vec}}(E) \\ {\hbox {vec}}(F) \\ \end{matrix} \right) \right\rangle +\left\langle \left( \begin{matrix} {\hbox {vec}}(E) \\ {\hbox {vec}}(F) \\ \end{matrix} \right), \left( \begin{matrix} {\hbox {vec}}(E) \\ {\hbox {vec}}(F) \\ \end{matrix} \right) \right\rangle \\ &=&{{\left( \begin{matrix} {\hbox {vec}}(X) \\ {\hbox {vec}}(Y) \\ \end{matrix} \right)}^{T}}{{\left( \begin{matrix} W&-N \\ N&W \\ \end{matrix} \right)}^{T}}\left( \begin{matrix} W&-N \\ N&W \\ \end{matrix} \right)\left( \begin{matrix} {\hbox {vec}}(X) \\ {\hbox {vec}}(Y) \\ \end{matrix} \right)\\ && -2{{\left( \begin{matrix} {\hbox {vec}}(X) \\ {\hbox {vec}}(Y) \\ \end{matrix} \right)}^{T}}{{\left( \begin{matrix} W&-N \\ N&W \\ \end{matrix} \right)}^{T}}\left( \begin{matrix} {\hbox {vec}}(E) \\ {\hbox {vec}}(F) \\ \end{matrix} \right) \\&& +{{\left( \begin{matrix} {\hbox {vec}}(E) \\ {\hbox {vec}}(F) \\ \end{matrix} \right)}^{T}}\left( \begin{matrix} {\hbox {vec}}(E) \\ {\hbox {vec}}(F) \\ \end{matrix} \right). \end{eqnarray*}$

故

$\begin{eqnarray*}&& \nabla \psi ( {\hbox {vec}}(Z))=\nabla \psi \left( \begin{matrix} {\hbox {vec}}(X) \\ {\hbox {vec}}(Y) \\ \end{matrix} \right)\\&=&2{{\left( \begin{matrix} W&-N \\ N&W \\ \end{matrix} \right)}^{T}}\left( \begin{matrix} W&-N \\ N&W \\ \end{matrix} \right)\left( \begin{matrix} {\hbox {vec}}(X) \\ {\hbox {vec}}(Y) \\ \end{matrix} \right)-2{{\left( \begin{matrix} W&-N \\ N&W \\ \end{matrix} \right)}^{T}}\left( \begin{matrix} {\hbox {vec}}(E) \\ {\hbox {vec}}(F) \\ \end{matrix} \right), \end{eqnarray*}$

则(2.1) 式得证.

定理2.4  设$\theta ( \hbox {vec}(z))={\left\| z-{{z}^{*}} \right\|}^{2}$, $Z=X+i*Y$, ${Z^*} = {X^*} + i*{Y^*}$, 其中${Z}^{*}$, $Z\in {{C}^{m\times n}}, $ $X, Y\in {R}^{m\times n}$, 则

$\begin{align} \nabla \theta ( {\hbox {vec}}(z))=\nabla \theta \left( \begin{matrix} {\hbox {vec}}(X) \\ {\hbox {vec}}(Y) \\ \end{matrix} \right)=\left( \begin{matrix} 2 {\hbox {vec}}(X-{{X}^{*}}) \\ 2 {\hbox {vec}}(Y-{{Y}^{*}}) \\ \end{matrix} \right). \end{align}$

(2.2)

证

$\begin{eqnarray*}&& \theta ( {\hbox {vec}}(z))={{\left\| z-{{z}^{*}} \right\|}^{2}}={{\left\| (X-{{X}^{*}})+i*(Y-{{Y}^{*}}) \right\|}^{2}}\\&=&\left\langle X-{{X}^{*}}, X-{{X}^{*}}\right\rangle=\left\langle {\hbox {vec}}(X-{{X}^{*}}), {\hbox {vec}}(X-{{X}^{*}})\right\rangle \\&=&{{\left\| X-{{X}^{*}} \right\|}^{2}}+{{\left\| Y-{{Y}^{*}} \right\|}^{2}}=\left\| {\hbox {vec}}(X-{{X}^{*}}) \right\|_{2}^{2}+\left\| {\hbox {vec}}(Y-{{Y}^{*}}) \right\|_{2}^{2} \\&=&\left\langle {\hbox {vec}}(X-{{X}^{*}}), {\hbox {vec}}(X-{{X}^{*}}) \right\rangle +\left\langle {\hbox {vec}}(Y-{{Y}^{*}}), {\hbox {vec}}(Y-{{Y}^{*}}) \right\rangle \\&=&{{[{\hbox {vec}}(X-{{X}^{*}})]}^{T}} {\hbox {vec}}(X-{{X}^{*}})+{{[{\hbox {vec}}(Y-{{Y}^{*}})]}^{T}} {\hbox {vec}}(Y-{{Y}^{*}}), \end{eqnarray*}$

则(2.2) 式得证.

可以证得$\psi '( {\hbox {vec}}(z))=\nabla \psi ( {\hbox {vec}}(z))$和$\theta '( {\hbox {vec}}(z))=\nabla \theta ( {\hbox {vec}}(z))$.因此分别用$\nabla \psi ( {\hbox {vec}}(z))$, $\nabla \theta ( {\hbox {vec}}(z))$来代替$\psi '( {\hbox {vec}}(z))$和$\theta '( {\hbox {vec}}(z))$.

定理2.5  设$\psi ( {\hbox {vec}}(z))$和$\theta ( {\hbox {vec}}(z))$如定理2.3, 2.4中所示, 则有如下的结论:

(a) $\psi ( {\hbox {vec}}(z))$是凸函数;

(b) $\theta ( {\hbox {vec}}(z))$是凸函数;

(c) $\nabla \psi ( {\hbox {vec}}(z))$是$\gamma$-Lipschitizian;

(d) $\nabla \theta ( {\hbox {vec}}(z))$是$\kappa$-Lipschitzian;

(e) $\nabla \theta ( {\hbox {vec}}(z))$是$\eta $强单调的.

证  (a)设$\psi( {\hbox {vec}}(Z))=\left\| {{A}_{1}}Z+Z{{B}_{1}}-{{C}_{1}} \right\|$, 得

$\begin{eqnarray*} &&\psi (\lambda {\hbox {vec}}({{Z}_{1}})+(1-\lambda ) {\hbox {vec}}({{Z}_{2}}))\\&=&\left\| {{A}_{1}}[\lambda {{Z}_{1}}+(1-\lambda ){{Z}_{2}}]+[\lambda {{Z}_{1}}+(1-\lambda ){{Z}_{2}}]{{B}_{1}}-{{C}_{1}} \right\| \\&=&\left\| \lambda ({{A}_{1}}{{Z}_{1}}+{{Z}_{1}}{{B}_{1}}-{{C}_{1}})+(1-\lambda )({{A}_{1}}{{Z}_{2}}+{{Z}_{2}}{{B}_{1}}-{{C}_{1}}) \right\| \\ &\le &\lambda \left\| {{A}_{1}}{{Z}_{1}}+{{Z}_{1}}{{B}_{1}}-{{C}_{1}} \right\|+(1-\lambda )(\left\| {{A}_{1}}{{Z}_{2}}+{{Z}_{2}}{{B}_{1}}-{{C}_{1}} \right\| \\ &=&\lambda \psi ( {\hbox {vec}}({{Z}_{1}}))+(1-\lambda )\psi ( {\hbox {vec}}({{Z}_{2}})), \lambda \in (0, \ 1). \end{eqnarray*}$

所以由凸集定义可得, $\psi ( {\hbox {vec}}(Z))$是凸函数.

(b)设$f( {\hbox {vec}}(Z))=\left\| Z-{{Z}^{*}} \right\|$, 则

$\begin{eqnarray*}&& f[\lambda {\hbox {vec}}({{Z}_{1}})+(1-\lambda ) {\hbox {vec}}({{Z}_{2}})]=\left\| \lambda {{Z}_{1}}+(1-\lambda ){{Z}_{2}}-{{Z}^{*}} \right\| \\&=&\left\| \lambda {{Z}_{1}}-\lambda {{Z}^{*}}+\lambda {{Z}^{*}}+(1-\lambda ){{Z}_{2}}-{{Z}^{*}} \right\| \\& =&\left\| \lambda ({{Z}_{1}}-{{Z}^{*}})+(1-\lambda )({{Z}_{2}}-{{Z}^{*}}) \right\|\le \left\| \lambda ({{Z}_{1}}-{{Z}^{*}}) \right\|+\left\| (1-\lambda )({{Z}_{2}}-{{Z}^{*}}) \right\| \\& =&\lambda \left\| {{Z}_{1}}-{{Z}^{*}} \right\|+(1-\lambda )\left\| {{Z}_{2}}-{{Z}^{*}} \right\|=\lambda f( {\hbox {vec}}({{Z}_{1}}))+(1-\lambda )f( {\hbox {vec}}({{Z}_{2}})), \end{eqnarray*}$

其中$\lambda \in (0, \ 1).$所以, 由凸集定义可得(b)成立.

(c)令

$\begin{eqnarray*}&& W=I\otimes A+{{C}^{T}}\otimes I, N=I\otimes B+{{D}^{T}}\otimes I, \\ && M=2\left( \begin{matrix} {{W}^{T}}W+{{N}^{T}}N&{{N}^{T}}W-{{W}^{T}}N \\ {{W}^{T}}N-{{N}^{T}}W&{{W}^{T}}W+{{N}^{T}}N \\ \end{matrix} \right), \end{eqnarray*}$

根据(2.1) 式有

$\begin{eqnarray*} && \left\| \nabla \psi (vec({{Z}_{1}}))-\nabla \psi ( {\hbox {vec}}({{Z}_{2}})) \right\| ={{\left\| M\left( \begin{matrix} {\hbox {vec}}({{X}_{1}}) \\ {\hbox {vec}}({{Y}_{1}}) \\ \end{matrix} \right)-M\left( \begin{matrix} {\hbox {vec}}({{X}_{2}}) \\ {\hbox {vec}}({{Y}_{2}}) \\ \end{matrix} \right) \right\|}_{2}} \\ &\le& \left\| M \right\|{{\left\| \left( \begin{matrix} {\hbox {vec}}({{X}_{1}}) \\ {\hbox {vec}}({{Y}_{1}}) \\ \end{matrix} \right)-\left( \begin{matrix} {\hbox {vec}}({{X}_{2}}) \\ {\hbox {vec}}({{Y}_{2}}) \\ \end{matrix} \right) \right\|}_{2}}=\left\| M \right\|{{\left\| {\hbox {vec}}({{Z}_{1}})- {\hbox {vec}}({{Z}_{2}}) \right\|}_{2}}. \end{eqnarray*}$

因此, $\nabla \psi ( {\hbox {vec}}(Z))$是$\gamma {\hbox-Lipschitizian}$, 其中$\gamma =\left\| M \right\|$.

(d)设$\theta ( {\hbox {vec}}(Z))={\left\| Z-{{Z}^{*}} \right\|}^{2}$, 由(2.2) 式得

$\begin{eqnarray*}&&\left\| \nabla \theta ( {\hbox {vec}}({{Z}_{1}})-\nabla \theta ( {\hbox {vec}}({{Z}_{2}}) \right\|={{\left\| 2 {\hbox {vec}}({{Z}_{1}}-{{Z}^{*}})-2 {\hbox {vec}}({{Z}_{2}}-{{Z}^{*}}) \right\|}_{2}}\\&=&\left\| 2[{\hbox {vec}}(X)-{\hbox {vec}}({{X}^{*}})]-2[{\hbox {vec}}(Y)-{\hbox {vec}}({{X}^{*}})] \right\|_{2} \\&=&2\left\| {\hbox {vec}}(X)- {\hbox {vec}}(Y) \right\|_{2} =2{{\left\| {\hbox {vec}}({{Z}_{1}}-{{Z}_{2}}) \right\|}_{2}}=2\left\| {{Z}_{1}}-{{Z}_{2}} \right\|, \end{eqnarray*}$

因此$\nabla \theta ( {\hbox {vec}}(z))$是$\kappa$-Lipschitzian, 其中$\kappa =2$.

(e)再由(2.2) 式可以得到

$\begin{array}{l} \left\langle {\nabla \theta \left( {\begin{array}{*{20}{c}} {{\rm{ vec(real}}({Z_1}))}\\ {{\rm{vec(imag}}({Z_1}))} \end{array}} \right) - \nabla \theta \left( {\begin{array}{*{20}{c}} {{\rm{vec(real}}({Z_2}))}\\ {{\rm{vec(imag}}({Z_2}))} \end{array}} \right),} \right.\\ \left. {\left( {\begin{array}{*{20}{c}} {{\rm{ vec(real}}({Z_1}))}\\ {{\rm{ vec(imag}}({Z_1}))} \end{array}} \right) - \left( {\begin{array}{*{20}{c}} {{\rm{ vec(real}}({Z_2}))}\\ {{\rm{vec(imag}}({Z_2}))} \end{array}} \right)} \right\rangle \\ = 2\left\langle {\left( {\begin{array}{*{20}{c}} {{\rm{vec(real}}({Z_1}))}\\ {{\rm{vec(imag}}({Z_1}))} \end{array}} \right) - \left( {\begin{array}{*{20}{c}} {{\rm{vec(real}}({Z_2}))}\\ {{\rm{ vec(imag}}({Z_2}))} \end{array}} \right),} \right.\\ \left. {\left( {\begin{array}{*{20}{c}} {{\rm{vec(real}}({Z_1}))}\\ {{\rm{ vec(imag}}({Z_1}))} \end{array}} \right) - \left( {\begin{array}{*{20}{c}} {{\rm{vec(real}}({Z_2}))}\\ {{\rm{ vec(imag}}({Z_2}))} \end{array}} \right)} \right\rangle \\ = 2\left\| {\left( {\begin{array}{*{20}{c}} {{\rm{vec(real}}({Z_1}))}\\ {{\rm{ vec(imag}}({Z_1}))} \end{array}} \right) - \left( {\begin{array}{*{20}{c}} {{\rm{vec(real}}({Z_2}))}\\ {{\rm{ vec(imag}}({Z_2}))} \end{array}} \right)} \right\|_2^2. \end{array}$

根据定义2.2, 可以推出$\nabla \theta ({\hbox{vec}}(z))$是$\eta $强单调的, 其中$\eta =2$.

由定理2.5可知, $\psi ({\hbox{vec}}(z))$和$\theta ({\hbox{vec}}(z))$满足定理2.2的条件, 由定理2.2可以推出该算法是收敛的.

在算法中, 需要计算到凸集$K$上的投影.为此我们给出如下定理.

定理2.6  设${P}_{1}\in {{C}^{m\times m}}, {{Q}_{1}}\in {C}^{n\times n}$是广义自反矩阵, $Z\in {C}^{m\times n}$是关于矩阵${P}_{1}$和${Q}_{1}$的自反矩阵, $X={\hbox{real}}(Z), Y={\hbox{imag}}(Z), K=\left\{ Z|Z={{P}_{1}}Z{{Q}_{1}} \right\}, {{\alpha }_{1}}, {{\alpha }_{2}}, \cdots, {\alpha }_{s}$是矩阵$\left( \begin{matrix} P&-Q \\ Q&P \\ \end{matrix} \right)-I$零空间的标准正交基, 其中$P={\hbox{real}}({{Q}_{1}}^{T}\otimes {{P}_{1}}), Q={\hbox{imag}}({{Q}_{1}}^{T}\otimes {{P}_{1}})$, 则到凸集$K$上的投影为

$\begin{array}{l} \;{P_K}\left( {\begin{array}{*{20}{c}} {{\rm{vec}}(X)}\\ {{\rm{vec}}(Y)} \end{array}} \right)\\ = \left\langle {\left( {\begin{array}{*{20}{c}} {{\rm{vec}}(X)}\\ {{\rm{vec}}(Y)} \end{array}} \right),{\alpha _1}} \right\rangle {\alpha _1} + \left\langle {\left( {\begin{array}{*{20}{c}} {{\rm{vec}}(X)}\\ {{\rm{vec}}(Y)} \end{array}} \right),{\alpha _2}} \right\rangle {\alpha _2}\; + \cdots + \left\langle {\left( {\begin{array}{*{20}{c}} {{\rm{vec}}(X)}\\ {{\rm{vec}}(Y)} \end{array}} \right),{\alpha _s}} \right\rangle {\alpha _s}. \end{array}$

(2.3)

证

$\begin{eqnarray}&&Z={{P}_{1}}Z{{Q}_{1}}\Leftrightarrow {\hbox{vec}}(Z)={\hbox{vec}}({{P}_{1}}Z{{Q}_{1}})\nonumber\\ &\Leftrightarrow& {\hbox{vec}}(Z)={{Q}_{1}}^{T}\otimes {{P}_{1}}{\hbox{vec}}(Z) \nonumber\\ &\Leftrightarrow& {\hbox{vec}}(X)+i*{\hbox{vec}}(Y)=(P+i*Q)({\hbox{vec}}(X)+i*{\hbox{vec}}(Y)) \nonumber\\ &\Leftrightarrow& {\hbox{vec}}(X)+i*{\hbox{vec}}(Y)=(P{\hbox{vec}}(X)-Q{\hbox{vec}}(Y))+i*(Q{\hbox{vec}}(X)+P{\hbox{vec}}(Y)) \nonumber\\ &\Leftrightarrow&\left( \begin{matrix} {\hbox{vec}}(X) \\ {\hbox{vec}}(Y) \\ \end{matrix} \right)=\left( \begin{matrix} P{\hbox{vec}}(X)-Q{\hbox{vec}}(Y) \\ Q{\hbox{vec}}(X)+P{\hbox{vec}}(Y) \\ \end{matrix} \right)\nonumber\\ &\Leftrightarrow&\left[\left( \begin{matrix} P &-Q \\ Q&P \\ \end{matrix} \right)-I \right]\left( \begin{matrix} {\hbox{vec}}(X) \\ {\hbox{vec}}(Y) \\ \end{matrix} \right)=0 \Leftrightarrow\left[\left( \begin{matrix} P &-Q \\ Q&P \\ \end{matrix} \right)-I \right]\left( \begin{matrix} {\hbox{vec}}(X) \\ {\hbox{vec}}(Y) \\ \end{matrix} \right)=0. \end{eqnarray}$

(2.4)

凸集$K$中的元素就是线性方程(2.4) 的解.因此只要给定(2.4) 式解集合的标准正交基, 就可以利用(2.3) 式计算出vec$({{Z}_{0}})$在$K$上的投影.

下面用两个数值例子来验证上述算法的可行性, 所有的实验都是在MATLAB2007R中进行的.

例1 考虑矩阵方程${A_1}Z + Z{B_1} = {C_1}$, 其中

$\begin{eqnarray*}&& {{A}_{1}}=\left( \begin{matrix} 2&1-i&6 \\ 5&4+2i&-3 \\ -1+i&4&8 \\ \end{matrix}\right), {{B}_{1}}=\left( \begin{matrix} 5-3i&2&-6 \\ 6&-7&0 \\ 2+4i&4&-3 \\ \end{matrix} \right), \\ &&{{C}_{1}}=\left( \begin{matrix} -26-2i&-52+37i&35+24i \\ -21+75i&28+31i&-55+33i \\ 80+76i&12+29i&-7-i \\ \end{matrix} \right), \\ &&{{P}_{1}}=\left( \begin{matrix} 0&-i&0 \\ i&0&0 \\ 0&0&1 \\ \end{matrix} \right), {{Q}_{1}}=\left( \begin{matrix} 0&i&0 \\ -i&0&0 \\ 0&0&-1 \\ \end{matrix} \right), {{Z}^{*}}=\left( \begin{matrix} 0&0&0 \\ 0&0&0 \\ 0&0&0 \\ \end{matrix} \right), {{Z}_{0}}=\left( \begin{matrix} 10&10&10 \\ 10&10&10 \\ 10&10&10 \\ \end{matrix} \right). \end{eqnarray*}$

可以验证该矩阵方程是相容的, 且有唯一如下广义自反解:

$Z=\left( \begin{matrix} -2-5i&1+4i&-8+7i \\ 1+4i&2+5i&7+8i \\ 6+3i&-3+6i&0 \\ \end{matrix} \right)\in C_{r}^{3\times 3}({{P}_{1}}, {{Q}_{1}}).$

利用上述算法, 令$\varepsilon =\text1.0e-11$, 可得

${{Z}_{5140501}}=\left( \begin{matrix} -2.0000-5.0000i&1.0000+4.0000i&-8.0000+7.0000i \\ 1.0000+4.0000i&2.0000+5.0000i&7.0000+8.0000i \\ 6.0000+3.0000i&-3.0000+6.0000i&0 \\ \end{matrix} \right) \\ \in C_{r}^{3\times 3}({{P}_{1}}, {{Q}_{1}}), $

相应的余项

${{R}_{5140501}}=\left\| {{A}_{1}}Z+Z{{B}_{1}}-{{C}_{1}} \right\|=\text{2}\text{.8055e-4}, $

其中下标是迭代步数.

例2  仍然考虑矩阵方程${A}_{1}Z+Z{{B}_{1}}={C}_{1}$, 其中${A}_{1}, {{B}_{1}}, {{P}_{1}}, {{Q}_{1}}, {Z}_{0}$和${Z}^{*}$同例1中的一样.在本例中, 令

${{C}_{1}}=\left( \begin{matrix} 8-i -52+3i 35+24i \\ -21+75i 28+31i -55+33i \\ 80+76i 12+29i -7-i \\ \end{matrix} \right).$

可以证明上述矩阵方程是不相容的.令$\varepsilon =\text1.0e-10$, 由上面算法得

${{Z}_{5140501}}=\left( \begin{matrix} -1.0064-4.6660i 1.8552+4.7525i -7.4849+5.1686i \\ 1.8552+4.7525i 1.0064+4.6660i 5.1686+7.4849i \\ 5.6201+2.7592i -2.7592+5.6201i 0 \\ \end{matrix} \right) \\ \in C_{r}^{3\times 3}({{P}_{1}}, {{Q}_{1}}).$

相应的余项

${{R}_{5140501}}=\left\| {{A}_{1}}Z+Z{{B}_{1}}-{{C}_{1}} \right\|=\text{2}\text{.8055e-4}.$

本文利用HSDM, 给出求解矩阵方程${A}_{1}Z+Z{{B}_{1}}={C}_{1}$广义自反最佳逼近解的迭代算法.无论方程${A}_{1}Z+Z{{B}_{1}}={C}_{1}$是否相容, 所给的算法都可以计算其广义自反的最佳逼近解.数值例子也说明了算法的可行性, 但该算法的缺点是收敛速度较慢.文中广义自反矩阵的集合是凸集, 若其它矩阵方程的未知约束矩阵是凸集, 则可以应用HSDM解决其最佳逼近问题.