本文讨论长方形矩阵约束最小二乘问题, 即考虑下列约束逼近问题

$ \begin{equation}\label{shi1} \left\{ \begin{array}{ll} \min&\|AX-B\|^2, \ \ A, B\in \mathbb{R}^{m\times n}, \\ {\hbox{s.t.}}&X\in \mathcal {P}, \end{array} \right. \end{equation} $

(1.1)

其中$m>n$, $\mathcal {P}\subset \mathbb{R}^{m\times n}$是具有某种特定模式的矩阵构成的子空间. $\|A\|$表示实矩阵$A$的Frobennius范数, 定义为$ \|A\|^2=\langle A, A\rangle=\sum\limits_{i, j=1}^{n}(A_{ij})^2, $其中的内积为$\langle A, B\rangle$=trace$(A^TB)$.

对于问题(1.1) 一系列的矩阵逼近问题已经被研究.若对$X$没有任何约束限制($ (\mathcal {P}=\mathbb{R}^{n\times n})$), 文献[1]给出了标准的最小二乘研究, 得到了通解$X=A^+B$, 其中$A^+$表示矩阵$A$的Moore-penrose逆.若$ \mathcal {P}$为正交矩阵集合则生成正交Procrustes问题^{[2, 3]}.若$ \mathcal {P}$为对称矩阵集合则生成对称Procrustes问题, 此类问题起源于弹性结构的研究, 根据关系式$Xf_i=d_i$假设测量力向量函数$f_i$是与测量位移向量$d_i$有关的, 其中$X$是对称应变矩阵. Higham ^[4]通过奇异值分解分析了对称Procrustes问题并给出了稳定的算法.当$\mathcal {P}$为一些闭凸锥, 比如对称半正定锥和(对称)非负锥, 则生成相应约束的Procrustes问题, Andersson and Elfving^[5]解决了此类问题, 得到了最优矩阵, 并给出了解决此类问题的投影类型算法.

在工程计算和微分方程求数值解时, 常会碰到实对称三对角或实对称五对角矩阵.但是关于实对称五对角矩阵的最小二乘问题研究比较少, 相关结论也不多.本文我们研究此类问题, 即实对称五对角矩阵Procrustes问题:

$ \begin{equation}\label{shi2} \left\{ \begin{array}{ll} \min&\|AX-B\|^2, \ \ A, B\in \mathbb{R}^{m\times n}, \\ {\hbox{s.t.}}&X\in \mathcal {T}, \end{array} \right. \end{equation} $

(1.2)

其中$\mathcal {T}\subset \mathbb{R}^{n\times n}$为实对称五对角矩阵构成的子空间.所谓实对称五对角矩阵即具有以下对称形式的矩阵

$ \begin{equation*} \label{t} T= \left(\begin{array}{ccccccc} a_1&b_1&c_1\\ b_1&a_2&b_2&c_2&\\ c_1&b_2&a_3&b_3&\ddots\\ &c_2&b_3&a_4&\ddots&\ddots&\\ \ddots&\ddots&\ddots&\ddots&c_{n-2}\\ &\ddots&\ddots&\ddots&b_{n-1}\\ c_{n-2}&b_{n-1}&a_n \end{array}\right). \end{equation*} $

本文是对文献[48]的拓展研究, 然而是用一种完全不同并且相对简单的方法, 虽然实对称五对角是特殊的对称矩阵.本文也是文[9]的推广研究, 文[9]研究了Toeplitz矩阵Procrustes问题, 本文将此方法应用于实对称五对角矩阵, 即首先通过矩阵的奇异值分解简化问题, 然后用分析方法和结合高等矩阵分析知识解决此类问题.

在这一节中我们描述问题(1.2) 中可行集$\mathcal {T}$.

给定矩阵$A, \ B\in \mathbb{R}^{m\times n}$, 子空间$\mathcal {T}$可以表述成

$ \begin{eqnarray*} \mathcal {T}=\left\{ X\in \mathbb{R}^{n\times n}: X=\sum\limits_{g=1}^{n}a_gG_g+\sum\limits_{r=1}^{n-1}b_rR_r+\sum\limits_{s=1}^{n-2}c_sS_s \right\}, \end{eqnarray*} $

其中$a_g, b_r, c_s\in \mathbb{R}$, 矩阵$G_g, P_r, S_s\in \mathbb{R}^{n\times n}$定义如下

$ \begin{eqnarray*} \begin{array}{l} \underset{1\leq g\leq n}{(G_g)_{ij}}=\left\{\begin{array}{ll} 1,&\ i=j=g, \\ 0, &\text{其它}, \end{array} \right.\ \ \ \underset{1\leq l\leq n-1}{(R_r)_{ij}}=\left\{\begin{array}{ll} 1,&\ i=r, j=r+1, \\ 1,&\ i=r+1, j=r, \\ 0, &\text{其它}, \end{array} \right. \\[0.5ex] \underset{1\leq s\leq n-2}{(S_s)_{ij}}=\left\{\begin{array}{ll} 1,&\ i=s, j=s+2, \\ 1,&\ i=s+2, j=s, \\ 0, &\text{其它}. \end{array} \right. \end{array} \end{eqnarray*} $

显然, $G_g, P_r, S_s$构成子空间$\mathcal {T}$的一组基.

首先把问题(1.2) 转化为一个较简单的问题.假设矩阵$A$具有下列形式的奇异值分解

$ \begin{equation}\label{svd} A=P\begin{bmatrix}\Sigma\\0\end{bmatrix}Q^T, \end{equation} $

(2.1)

其中$P\in R^{m\times m}$和$Q\in R^{n\times n}$是正交矩阵, $\Sigma={\hbox{diag}}(\sigma_1, \sigma_2, \cdots, \sigma_n), \sigma_1\geq \cdots\geq\sigma_n\geq 0$.

因为Frobenius范数的正交不变性, 我们得到

$ \begin{eqnarray*} \|AX-B\|^2&=&\|P\begin{bmatrix}\Sigma\\0\end{bmatrix}Q^TX-B\|^2=\|\begin{bmatrix}\Sigma\\0\end{bmatrix}Q^TX-P^TB\|^2 =\|\begin{bmatrix}\Sigma\\0\end{bmatrix}Q^TX-C\|^2\\&=& \|(\Sigma Q^T)X-C_1\|^2+\|C_2\|^2=\|T-C_1\|^2+\|C_2\|^2, \end{eqnarray*} $

其中$ T=\Sigma Q^TX\in R^{n\times n}, \ \ \ C=\begin{bmatrix}C_1\\C_2\end{bmatrix}=P^TB, \ \ \ C_1\in R^{n\times n}, $则问题(1.2) 等价于

$ \begin{equation}\label{equivalent_problem} \left\{ \begin{array}{ll} \min&\|T-C_1\|^2, \\ {\hbox{s.t.}} &T\in \mathcal {T}^{'}, \end{array} \right. \end{equation} $

其中$\mathcal {T}^{'}$是下列子空间

$ \begin{eqnarray*} \mathcal {T}^{'}=\{ T\in R^{n\times n}: T=\Sigma Q^T(\sum\limits_{g=1}^{n}a_gG_g+\sum\limits_{r=1}^{n-1}b_rR_r+\sum\limits_{s=1}^{n-2}c_sS_s )\}. \end{eqnarray*} $

我们引入符号$P_\mathcal {U}(A)$表示矩阵$A$在集合$\mathcal {U}$上的投影.下列定理描述了矩阵$C_1$在子空间$\mathcal {T}^{'}$上的投影.

定理2.1  如果$C_1 \in R^{n\times n}$, 那么问题(1.2) 的通解为

$ \begin{array}{*{20}{l}} {{P_{\mathcal {T}'}}({C_1}) = \Sigma {Q^T}(\sum\limits_{g = 1}^n {a_g^*} {G_g} + \sum\limits_{r = 1}^{n - 1} {b_r^*} {R_r} + \sum\limits_{s = 1}^{n - 2} {c_s^*} {S_s}),}\\ {a_p^* = \left\{ {\begin{array}{*{20}{l}} {\frac{{\sum\limits_{i = 1}^n {{\sigma _i}} {Q_{p,i}}{{({C_1})}_{i,p}}}}{{\sum\limits_{i = 1}^n {\sigma _i^2} Q_{p,i}^2}},\sum\limits_{i = 1}^n {\sigma _i^2} Q_{p,i}^2 \ne 0,}\\ {\;\;\;\;\;\;\;\text{任意},\;\;\;\;\;\text{其它}} \end{array}} \right.(1 \le p \le n),}\\ {b_q^* = \left\{ {\begin{array}{*{20}{l}} {\frac{{\sum\limits_{i = 1}^n {{\sigma _i}} \left[ {{Q_{q + 1,i}}{{({C_1})}_{i,q}} + {Q_{q,i}}{{({C_1})}_{i,q + 1}}} \right]}}{{\sum\limits_{i = 1}^n {\sigma _i^2} (Q_{q + 1,i}^2 + Q_{q,i}^2)}},\sum\limits_{i = 1}^n {\sigma _i^2} (Q_{q + 1,i}^2 + Q_{q,i}^2) \ne 0,}\\ {\;\;\;\;\;\;\;\text{任意},\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{其它}} \end{array}} \right.(1 \le q \le n - 1),}\\ {c_t^* = \left\{ {\begin{array}{*{20}{l}} {c_t^* = \frac{{\sum\limits_{i = 1}^n {{\sigma _i}} \left[ {{Q_{t + 2,i}}{{({C_1})}_{i,t}} + {Q_{t,i}}{{({C_1})}_{i,t + 2}}} \right]}}{{\sum\limits_{i = 1}^n {\sigma _i^2} (Q_{t + 2,i}^2 + Q_{t,i}^2)}},\sum\limits_{i = 1}^n {\sigma _i^2} (Q_{t + 2,i}^2 + Q_{t,i}^2) \ne 0,}\\ {\;\;\;\;\;\;\;\text{任意},\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{其它}} \end{array}} \right.(1 \le t \le n - 2).} \end{array} $

证  目标函数$f:\mathbb{R}^{3n-3}\rightarrow \mathbb{R}, $

$ \begin{eqnarray*} f(a, b, c)&=&f\{a_1, \cdots, a_n, b_1, \cdots, b_{n-1}, c_1, \cdots, c_{n-2}\} =\dfrac{1}{2}\|T-C_1\|^2 =\dfrac{1}{2}\|\Sigma Q^TX-C_1\|^2\\&=&\dfrac{1}{2}\|\Sigma Q^T(\sum\limits_{g=1}^{n}a_gG_g+\sum\limits_{r=1}^{n-1}b_rR_r+\sum\limits_{s=1}^{n-2}c_sS_s) -C_1\|^2 \\ &=&\dfrac{1}{2}\sum\limits_{i, j=1}^{n}[(\Sigma Q^T(\sum\limits_{g=1}^{n}a_gG_g+\sum\limits_{r=1}^{n-1}b_rR_r+\sum\limits_{s=1}^{n-2}c_sS_s))_{i, j}-(C_1)_{i, j}]^2. \end{eqnarray*} $

记$\Delta=\sum\limits_{g=1}^{n}a_gG_g+\sum\limits_{r=1}^{n-1}b_rR_r+\sum\limits_{s=1}^{n-2}c_sS_s$.显然$f$是二阶连续可微的, 我们可以计算$\dfrac{\partial f}{\partial a_p}(a, b, c)$使得对所有的$1\leq p\leq n$有

$ \begin{equation}\label{a_different} \begin{array}{ll} \dfrac{\partial f}{\partial {a_p}}&=\dfrac{\partial }{\partial a_p}\left\{\dfrac{1}{2}\sum\limits_{i, j=1}^{n}\left[\left(\Sigma Q^T\Delta\right)_{i, j}-(C_1)_{i, j}\right]^2\right\}\\ &=\sum\limits_{i, j=1}^{n}\left[\left( \left(\Sigma Q^T\Delta\right)_{i, j}-(C_1)_{i, j}\right)\cdot \dfrac{\partial }{\partial {a_p}}\left(\Sigma Q^T\Delta\right)_{i, j}\right]. \end{array} \end{equation} $

(2.3)

(2.3) 式右边的乘积因子

$ \begin{equation*} \begin{array}{ll} \dfrac{\partial }{\partial {a_p}}(\Sigma Q^T\Delta)_{i, j}& =\dfrac{\partial }{\partial {a_p}}\left( \sum\limits_{l=1}^{n}\left((\Sigma Q^T)_{i, l}\cdot \Delta_{l, j}\right)\right) =\sum\limits_{l=1}^{n}\left((\Sigma Q^T)_{i, l}\cdot\dfrac{\partial }{\partial a_p}(\Delta)_{l, j}\right)\\ &=\sum\limits_{l=1}^{n}(\Sigma Q^T)_{i, l}(G_p)_{l, j}=(\Sigma Q^TG_p)_{i, j}, \end{array} \end{equation*} $

那么(2.3) 变成

$ \begin{equation}\label{a_different2} \dfrac{\partial f}{\partial {a_p}}= \sum\limits_{i, j=1}^{n} \left(\Sigma Q^T\Delta\right)_{i, j}\cdot (\Sigma Q^TG_p)_{i, j}-\sum\limits_{i, j=1}^{n}(C_1)_{i, j}\cdot (\Sigma Q^TG_p)_{i, j}. \end{equation} $

(2.4)

注意到$\left(\Sigma Q^T\Delta\right)_{i, j}$是$\Sigma Q^T$的第$i$行与$\Delta$的第$j$列的内积, 如果令$c_{-1}=0, $ $b_0=0, $ $c_0=0, $ $c_{n-1}=0, $ $b_n=0, $ $c_n=0$, 则对所有的$1\leq i, j\leq n$, 有

$ \begin{equation}\label{Sigma} \left(\Sigma Q^T\Delta\right)_{i, j}=\sigma_i(c_{j-2}Q_{j-2, i}+b_{j-1}Q_{j-1, i}+a_jQ_{j, i}+b_jQ_{j+1, i}+c_jQ_{j+2, i}). \end{equation} $

(2.5)

类似地, 元素$(\Sigma Q^TG_p)_{i, j}$是$\Sigma Q^T$的第$i$行与第$j$列的内积, 因此

$ \begin{equation*} (\Sigma Q^TG_p)_{i, j}=\left\{\begin{array}{lll} \sigma_iQ_{p, i}, j=p, \\ 0, j\neq p. \end{array} \right. \end{equation*} $

那么(2.4) 式右边的第一个和式可以写成

$ \begin{eqnarray*} \sum\limits_{i=1}^{n} \left(\Sigma Q^T\Delta\right)_{i, p}\cdot (\Sigma Q^TG_p)_{i, p}\\ &=&\sum\limits_{i=1}^{n}\sigma_i^2(c_{p-2}Q_{p-2, i}+b_{p-1}Q_{p-1, i}+a_pQ_{p, i}+b_pQ_{p+1, i}+c_pQ_{p+2, i})Q_{p, i}. \end{eqnarray*} $

定义$\sum\limits_{i=1}^{n}Q_{m, i}Q_{k, i}\sigma_i^2=\nu^T\theta$, 其中$\nu^T=(Q_{m, 1}Q_{k, 1}, Q_{m, 2}Q_{k, 2}, \cdots, Q_{m, n}Q_{k, n})$, $\theta=(\sigma_1^2, \sigma_2^2, \cdots, \sigma_n^2)$, 我们知道$\nu$的所有元素之和是正交矩阵$Q$其中两列的内积, 且当其列序号$m\neq k$时其值为0.因此对$m\neq k$, 有

$ \begin{equation*} 0=\left(\sum\limits_{i=1}^{n}\nu_i\right)\sigma_n^2\leq \nu^T\theta=\sum\limits_{i=1}^{n}\nu_i\sigma_i^2\leq \left(\sum\limits_{i=1}^{n}\nu_i\right)\sigma_1^2=0, \end{equation*} $

因此对所有的$m\neq k$都有$\nu^T\theta=0$, 且

$ \begin{equation*} \sum\limits_{i=1}^{n} \left(\Sigma Q^T\Delta\right)_{i, p}\cdot (\Sigma Q^TG_p)_{i, p}=\sum\limits_{i=1}^{n}a_p\sigma_i^2Q_{p, i}^2. \end{equation*} $

这时(2.4) 式变成

$ \frac{{\partial f}}{{\partial {a_p}}} = \sum\limits_{i = 1}^n {{a_p}} \sigma _i^2Q_{p,i}^2 - \sum\limits_{i = 1}^n {{\sigma _i}} {({C_1})_{i,p}}{Q_{p,i}} $

(2.6)

对所有的$p=1, \cdots, n$都成立.根据多元函数求极小值的一阶必要条件

$ \frac{{\partial f}}{{\partial {a_p}}} = 0,\;\;\;\;p = 1, \cdots ,n, $

得到

$ \begin{eqnarray}\label{a_p} a_p^*=\frac{\sum\limits_{i=1}^{n}\sigma_iQ_{p, i}(C_1)_{i, p}}{\sum\limits_{i=1}^{n}\sigma_i^2Q_{p, i}^2}. \end{eqnarray} $

(2.7)

注意到当rank$(A)=n$时, (2.7) 式的分母恒不为0, 因为$A$的奇异值全大于0, 则

$ \sum\limits_{i=1}^{n}\sigma_i^2Q_{p, i}^2=0 $

意味着$ \sigma_iQ_{p, i}=0 \rightarrow\ [Q_{p, 1}, Q_{p, 2}, \cdots, Q_{p, n}]=0, $这与$Q$是正交矩阵矛盾!但是当rank$(A)=k<n$时, (2.7) 式的分母有可能为0, 只需令$Q$的第$p$行的前面$k$个元素全为0, 后面的$n-k$个元素至少有一个非零元即可.因此当(2.7) 式的分母有为0时, 我们令$a_p^*$可取任意值.

类似于$\dfrac{\partial f}{\partial {a_p}}$, 计算$\dfrac{\partial f}{\partial {b_p}}$, 对所有的$1\leq q\leq n-1$,

$ \begin{equation}\label{b_different} \begin{array}{ll} \dfrac{\partial f}{\partial {b_p}}=\sum\limits_{i, j=1}^{n}\left[\left( \left(\Sigma Q^T\Delta\right)_{i, j}-(C_1)_{i, j}\right)\cdot \dfrac{\partial }{\partial {b_p}}\left(\Sigma Q^T\Delta\right)_{i, j}\right]. \end{array} \end{equation} $

(2.8)

同样

$ \begin{equation}\label{b_different3} \begin{array}{ll} \dfrac{\partial }{\partial {b_p}}(\Sigma Q^T\Delta)_{i, j}&=\dfrac{\partial }{\partial {b_p}}\left( \sum\limits_{l=1}^{n}\left((\Sigma Q^T)_{i, l}\cdot \Delta_{l, j}\right)\right) =\sum\limits_{l=1}^{n}\left((\Sigma Q^T)_{i, l}\cdot\dfrac{\partial }{\partial {b_p}}(\Delta)_{l, j}\right)\\ &=(\Sigma Q^TR_q)_{i, j}=\left\{\begin{array}{lll} \sigma_iQ_{q, i}, j=q+1;\\ \sigma_iQ_{q+1, i}, j=q;\\ 0, {\hbox {否则}}. \end{array} \right. \end{array} \end{equation} $

(2.9)

因此结合(2.5), (2.9), (2.8) 式变成

$ \begin{array}{l} \frac{{\partial f}}{{\partial {b_p}}} = \sum\limits_{i = 1}^n {{\sigma _i}} {\left( {\Sigma {Q^T}\Delta } \right)_{i,q}}{Q_{q + 1,i}} + \sum\limits_{i = 1}^n {{\sigma _i}} {\left( {\Sigma {Q^T}\Delta } \right)_{i,q + 1}}{Q_{q,i}} - \sum\limits_{i = 1}^n {{\sigma _i}} {({C_1})_{i,q}}{Q_{q + 1,i}}\\ \;\;\;\;\;\;\;\;\; - \sum\limits_{i = 1}^n {{\sigma _i}} {({C_1})_{i,q + 1}}{Q_{q,i}}\\ \;\;\;\;\;\;\; = \sum\limits_{i = 1}^n {\sigma _i^2} ({c_{q - 2}}{Q_{q - 2,i}} + {b_{q - 1}}{Q_{q - 1,i}} + {a_q}{Q_{q,i}} + {b_q}{Q_{q + 1,i}} + {c_q}{Q_{q + 2,i}}){Q_{q + 1,i}}\\ \;\;\;\;\;\;\;\;\; + \sum\limits_{i = 1}^n {\sigma _i^2} ({c_{q - 1}}{Q_{q - 1,i}} + {b_q}{Q_{q,i}} + {a_{q + 1}}{Q_{q + 1,i}} + {b_{q + 1}}{Q_{q + 2,i}} + {c_{q + 1}}{Q_{q + 3,i}}){Q_{q,i}}\\ \;\;\;\;\;\;\;\;\; - \sum\limits_{i = 1}^n {{\sigma _i}} \left[ {{{({C_1})}_{i,q}}{Q_{q + 1,i}} + {{({C_1})}_{i,q + 1}}{Q_{q,i}}} \right]\\ \;\;\;\;\;\;\; = \sum\limits_{i = 1}^n {\sigma _i^2} {b_q}(Q_{q + 1,i}^2 + Q_{q,i}^2) - \sum\limits_{i = 1}^n {{\sigma _i}} \left[ {{{({C_1})}_{i,q}}{Q_{q + 1,i}} + {{({C_1})}_{i,q + 1}}{Q_{q,i}}} \right]. \end{array} $

(2.10)

同样根据多元函数一阶必要条件$\dfrac{\partial f}{\partial {b_p}}=0, \ \ q=1, \cdots, n-1$, 有

$ \begin{equation}\label{b_q2} b_q^*=\frac{\sum\limits_{i=1}^{n}\sigma_i\left[Q_{q+1, i}(C_1)_{i, q}+Q_{q, i}(C_1)_{i, q+1}\right]}{\sum\limits_{i=1}^{n}\sigma_i^2(Q_{q+1, i}^2+Q_{q, i}^2)}. \end{equation} $

(2.11)

同样当rank$(A)=n$时, (2.11) 式的分母不为0, 当rank$(A)=k<n$时, (2.11) 式的分母可能为0, 可能的取法为$Q$的第$q$行和第$q+1$行的前面$k$个元素全为0, 后面的$n-k$个元素每行都至少含有一个非零元.因此当(2.11) 的分母为0时, 我们令$b_q^*$取任意值.

类似于$\dfrac{\partial f}{\partial {a_p}}$, $\dfrac{\partial f}{\partial {b_p}}$, 对所有的$1\leq t\leq n-2, $

$ \begin{equation}\label{c_different} \begin{array}{ll} \dfrac{\partial f}{\partial c_t}=\sum\limits_{i, j=1}^{n}\left[\left( \left(\Sigma Q^T\Delta\right)_{i, j}-(C_1)_{i, j}\right)\cdot \dfrac{\partial }{\partial c_t}\left(\Sigma Q^T\Delta\right)_{i, j}\right], \end{array} \end{equation} $

(2.12)

$ \begin{equation}\label{c_different3} \begin{array}{ll} \dfrac{\partial }{\partial c_t}(\Sigma Q^T\Delta)_{i, j}=(\Sigma Q^TS_t)_{i, j}=\left\{\begin{array}{lll} \sigma_iQ_{t, i}, j=t+2, \\ \sigma_iQ_{t+2, i}, j=t, \\ 0, {\hbox{否则}}. \end{array}\right. \end{array} \end{equation} $

(2.13)

因此结合(2.5), (2.13), (2.12) 式变成

$ \begin{array}{l} \frac{{\partial f}}{{\partial {c_t}}}{\rm{ = }}\sum\limits_{i = 1}^n {{\sigma _i}} {\left( {\Sigma {Q^T}\Delta } \right)_{i,t}}{Q_{t + 2,i}} + \sum\limits_{i = 1}^n {{\sigma _i}} {\left( {\Sigma {Q^T}\Delta } \right)_{i,t + 2}}{Q_{t,i}} - \sum\limits_{i = 1}^n {{\sigma _i}} {({C_1})_{i,t}}{Q_{t + 2,i}}\\ \;\;\;\;\;\;\;\;\;\; - \sum\limits_{i = 1}^n {{\sigma _i}} {({C_1})_{i,t + 2}}{Q_{t,i}}\\ \;\;\;\;\; = \sum\limits_{i = 1}^n {\sigma _i^2} ({c_{t - 2}}{Q_{t - 2,i}} + {b_{t - 1}}{Q_{t - 1,i}} + {a_t}{Q_{t,i}} + {b_t}{Q_{t + 1,i}} + {c_t}{Q_{t + 2,i}}){Q_{t + 2,i}}\\ \;\;\;\;\;\;\;\;\;\; + \sum\limits_{i = 1}^n {\sigma _i^2} ({c_t}{Q_{t,i}} + {b_{t + 1}}{Q_{t + 1,i}} + {a_{t + 2}}{Q_{t + 2,i}} + {b_{t + 2}}{Q_{t + 3,i}} + {c_{t + 2}}{Q_{t + 4,i}}){Q_{t,i}}\\ \;\;\;\;\;\;\;\;\;\;\; - \sum\limits_{i = 1}^n {{\sigma _i}} \left[ {{{({C_1})}_{i,t}}{Q_{t + 2,i}} + {{({C_1})}_{i,t + 2}}{Q_{t,i}}} \right]\\ \;\;\;\;\;\;\; = \sum\limits_{i = 1}^n {\sigma _i^2} {c_t}(Q_{t,i}^2 + Q_{t + 2,i}^2) - \sum\limits_{i = 1}^n {{\sigma _i}} \left[ {{{({C_1})}_{i,t}}{Q_{t + 2,i}} + {{({C_1})}_{i,t + 2}}{Q_{t,i}}} \right]. \end{array} $

(2.14)

同样根据多元函数极小值问题一阶必要条件$\dfrac{\partial f}{\partial c_t}=0, \ \ t=1, \cdots, n-2$, 有

$ \begin{equation}\label{c_q2} c_t^*=\frac{\sum\limits_{i=1}^{n}\sigma_i\left[(C_1)_{i, t}Q_{t+2, i}+(C_1)_{i, t+2}Q_{t, i}\right]}{\sum\limits_{i=1}^{n}\sigma_i^2(Q_{t, i}^2+Q_{t+2, i}^2)}. \end{equation} $

(2.15)

类似于(2.11) 式中对分母的讨论, 当rank$(A)=n$时它恒不为0, 当rank$(A)=k<n$时, 它有可能为0.同样当(2.15) 式的分母为0时, 我们令$c_t^*$取任意值.现在我们需要证明$a_p^*(1\leq p\leq n)$, $b_q^*(1\leq q\leq n-1)$, $c_t^*(1\leq t\leq n-2)$是$f$的极小值.计算

$ \begin{equation*} \begin{array}{c} \dfrac{\partial^2 f}{\partial a_p^2}(a_p^*)=\sum\limits_{i=1}^{n}\sigma_i^2Q_{p, i}^2\geq 0, \ \ \ \ \dfrac{\partial^2 f}{\partial {b_p}^2}(b_q^*)=\sum\limits_{i=1}^{n}\sigma_i^2(Q_{q+1, i}^2+Q_{q, i}^2)\geq 0, \\[0.5ex] \dfrac{\partial^2 f}{\partial c_t^2}=\sum\limits_{i=1}^{n}\sigma_i^2(Q_{t, i}^2+Q_{t+2, i}^2)\geq 0, \end{array} \end{equation*} $

很容易证明上述不等式中等号不能同时成立, 因为$Q$是正交矩阵, 特别地, 当${\rm rank}(A)=n$, 上述不等式的严格不等号同时成立.同时有

$ \begin{equation*} \begin{array}{ll} \dfrac{\partial^2 f}{\partial {a_p}a_k}=0, \ \ {\hbox{如果}} \ \ k\neq p; \ \ \dfrac{\partial^2 f}{\partial {b_p}b_l}=0, \ \ {\hbox{如果}} \ \ l\neq q, \dfrac{\partial^2 f}{\partial c_tb_s}=0; \ \ {\hbox{如果}} \ \ t\neq s, \\ \dfrac{\partial^2 f}{\partial {a_p}b_q}=0, \ \ \dfrac{\partial^2 f}{\partial {a_p}c_t}=0, \ \ \dfrac{\partial^2 f}{\partial {b_p}a_p}=0, \ \ \dfrac{\partial^2 f}{\partial b_qc_t}=0, \ \ \dfrac{\partial^2 f}{\partial c_ta_p}=0, \ \ \dfrac{\partial^2 f}{\partial c_tb_q}=0. \end{array} \end{equation*} $

这说明关于二阶导数的海赛(Hessian)矩阵$\bigtriangledown^2f(a^*, b^*, c^*)$是半正定的.特别地, 当rank$(A)=n$时为正定, 因此, $a_p^*(1\leq p\leq n)$, $b_q^*(1\leq q\leq n-1)$, $c_t^*(1\leq t\leq n-2)$确实是$f$的极小值.

从定理2.1可以看出, $a_p^*(1\leq p\leq n)$, $b_q^*(1\leq q\leq n-1)$, $c_t^*(1\leq t\leq n-2)$只依赖与矩阵$A$的奇异值分解中的三个分解因子$P$, $Q$, $\Sigma$.但是不像对角因子$\Sigma$, 左右正交因子$P$, $Q$并不是唯一确定的即使矩阵$A$是列满秩的, 这种不唯一性的程度取决于奇异值的重数.类似于文献[10]中的定理3.11', 下列引理描述矩阵$A$的奇异值分解中正交因子的所有可能取法以及它们之间的关系.

引理2.1  给定矩阵$A\in \mathbb{R}^{m\times n}$, 假定$A$的所有不同的奇异值为$\sigma_1> \cdots >\sigma_k>0$, 它们的重数分别为$\mu_1, \cdots, \mu_k\geq 1$, $\mu_1+\cdots+\mu_k=r$.令$A$的奇异值分解为$A=P{\hbox{diag}}(\Sigma, 0_{m-r, n-r}) Q$, 其中$\Sigma={\hbox{diag}}(\sigma_1I_{\mu_1}, \cdots, \sigma_kI_{\mu_k})\in \mathbb{R}^{r\times r}$.令$\hat{P}\in \mathbb{R}^{m\times m}$, $\hat{Q}\in \mathbb{R}^{n\times n}$为给定的正交矩阵, 那么$A=\hat{P}{\hbox{diag}}(\Sigma, 0_{m-r, n-r})\hat{W}^T$当且仅当存在正交矩阵$U_i\in \mathbb{R}^{\mu_i\times \mu_i}$, $i=1, \cdots, k$, $\tilde{V}\in \mathbb{R}^{(m-r)\times (m-r)}$, $\tilde{W}\in \mathbb{R}^{(n-r)\times (n-r)}$使得

$ \begin{equation} \label{lemma1_shizi} \hat{P}=P[U_1\oplus\cdots\oplus U_k\oplus \tilde{V}] \ \ {\hbox{和}}\ \ \hat{Q}=Q[U_1\oplus\cdots\oplus U_k\oplus \tilde{W}]. \end{equation} $

(2.16)

下面结合引理2.1我们证明当矩阵$A$列满秩时问题(1.2) 只有唯一解, 当矩阵$A$不是列满秩时, 问题(1.2) 有无穷多解.首先下列众所周知的引理保证问题(1.2) 解的存在性.

引理2.2 ^[5]  设$\mathcal{H}$为实数域上的线性子空间, $\mathcal {C}$是$\mathcal {H}$上的非空闭凸锥.假定$f$是凸函数并且强制性的(在线性子空间$\mathcal{H}$上定义内积$\langle X, Y\rangle={\hbox{trace}}(XY^T), \ \ X, Y\in \mathcal {H}$, 称函数$f$在$\mathcal {C}$上具有强制性如果$\min\limits_{\|X\|\rightarrow \infty, X\in \mathcal{C}}f(X)=\infty.$), 那么极小值问题$\min\limits_{X\in \mathcal{C}}f(X)$有解, 并且如果$f$是严格凸的, 解是唯一的.

假定$A$是列满秩的, 且$A$的另一奇异值分解为$A=\hat{P}\begin{bmatrix}\Sigma\\0\end{bmatrix}\hat{Q}$, 其中

$ \begin{equation*} \hat{P}=P\begin{pmatrix} U_1&\\ &\ddots \\ U_k&\\ &\tilde{V} \end{pmatrix} \ \ \ \ {\hbox{和}} \ \ \ \ \hat{Q}=Q\begin{pmatrix} U_1\\ &\ddots &\\ U_k\\ \end{pmatrix}, \end{equation*} $

其中矩阵$P, Q$在式2.1中给出, 分块矩阵$P=[P_1, P_2]$, $\hat{P}=[\hat{P}_1, \hat{P}_2]$, 其中$P_1\in \mathbb{R}^{m\times n}, \hat{P}_1\in \mathbb{R}^{m\times n}$则有

$ \begin{equation*} \hat{P}_1=P_1\begin{pmatrix} U_1\\ &\ddots &\\ U_k\\ \end{pmatrix}. \end{equation*} $

根据定理2.1, 我们可得到另外一组依赖于$\hat{Q}$和$\hat{C}_1(=\hat{P}_1^TB)$的解$\hat{a}_p^*(1\leq p\leq n)$, $\hat{b}_q^*(1\leq q\leq n-1)$, $\hat{c}_t^*(1\leq t\leq n-2)$, 具体表达式类似于(2.7), (2.11), (2.15) 式.但是

$ \begin{array}{l} \begin{array}{*{20}{l}} {\sum\limits_{i = 1}^n {{\sigma _i}} {{\hat Q}_{f,i}}{{({{\hat C}_1})}_{i,h}}}&{ = [{\sigma _1}{Q_{f,1}}, \cdots ,{\sigma _n}{Q_{f,n}}]\left( {\begin{array}{*{20}{c}} {{U_1}}&{}&{}\\ {}& \ddots &{}\\ {{U_k}}&{}&{} \end{array}} \right)\left( {\begin{array}{*{20}{c}} {U_1^T}&{}&{}\\ {}& \ddots &{}\\ {U_k^T}&{}&{} \end{array}} \right)P_1^T\left( {\begin{array}{*{20}{c}} {{B_{1,h}}}\\ \vdots \\ {{B_{n,h}}} \end{array}} \right)}\\ {}&{ = \sum\limits_{i = 1}^n {{\sigma _i}} {Q_{f,i}}{{({C_1})}_{i,h}},} \end{array}\\ \;\;\;\;\;\;\;\;\;\;\begin{array}{*{20}{l}} {\sum\limits_{i = 1}^n {\sigma _i^2} \hat Q_{f,i}^2}&{ = [{\sigma _1}{Q_{f,1}}, \cdots ,{\sigma _n}{Q_{f,n}}]\left( {\begin{array}{*{20}{c}} {{U_1}}&{}&{}\\ {}& \ddots &{}\\ {{U_k}}&{}&{} \end{array}} \right)\left( {\begin{array}{*{20}{c}} {U_1^T}&{}&{}\\ {}& \ddots &{}\\ {U_k^T}&{}&{} \end{array}} \right)\left( {\begin{array}{*{20}{c}} {{\sigma _1}{Q_{f,1}}}\\ \vdots \\ {{\sigma _n}{Q_{f,n}}} \end{array}} \right)}\\ {}&{ = \sum\limits_{i = 1}^n {\sigma _i^2} Q_{f,i}^2.} \end{array} \end{array} $

因此$\hat{a}_p^*=a_p^*(1\leq p\leq n)$, $\hat{b}_q^*=b_q^*(1\leq q\leq n-1)$, $\hat{c}_t^*=c_t^*(1\leq t\leq n-2)$.这说明当矩阵$A$是列满秩时, 问题(1.2) 只有唯一解, 这与引理2.2中的结论相符.从上亦可以看出, 当rank$(A)<n$, 问题(1.2) 的解是与引理2.1中的矩阵$\tilde{V}, \tilde{W}$是有关的, 因此也就说明解并不是唯一的.

根据上面的讨论, 给出求解问题(1.2) 的数值算法如下:

算法  给定实矩阵$A, B\in \mathbb{R}^{m\times n} (m\geq n)$,

1.得出矩阵$A$的奇异值分解分解$A=P\begin{bmatrix}\Sigma\\0\end{bmatrix}Q^T$, 分块矩阵$P=[P_1, P_2], \ P_1\in \mathbb{R}^{m\times n}$, 计算$C_1=P_1^TB$.

2.按定理(2.1) 计算$a_p^*(1\leq p\leq n)$, $b_q^*(1\leq q\leq n-1)$, $c_t^*(1\leq t\leq n-2)$.

3.得到问题(1.2) 的解, 若rank$(A)=n$, 则该解是问题的唯一解.

例  给定矩阵

$ A = \left( {\begin{array}{*{20}{l}} {{\rm{ - 4}}.{\rm{1}}\;\;\;\;\;{\rm{4}}.{\rm{9}}\;\;\;\;\;{\rm{4}}.{\rm{3}}\;\;\;\;\;{\rm{6}}.{\rm{2}}\;\;\;\;\;{\rm{6}}.{\rm{8}}\;\;\;\;\;{\rm{0}}.{\rm{5}}\;\;\;\;\;{\rm{1}}.{\rm{5}}\;\;\;\;\;{\rm{8}}.{\rm{9}}}\\ {\;{\rm{3}}.{\rm{0}}\;\;\;\;\;{\rm{2}}.{\rm{1}}\;\;\;\;\;{\rm{9}}.{\rm{3}}\;\;\;\;\;{\rm{3}}.{\rm{7}}\;\;\;\;\;{\rm{0}}.{\rm{9}}\;\;\;\;\;{\rm{ - 3}}.{\rm{6}}\;\;\;\;\;{\rm{ - 6}}.{\rm{7}}\;\;\;{\rm{2}}.{\rm{7}}}\\ {{\rm{8}}.{\rm{7}}\;\;\;\;\;\;{\rm{6}}.{\rm{4}}\;\;\;\;\;{\rm{6}}.{\rm{8}}\;\;\;\;\;{\rm{5}}.{\rm{7}}\;\;\;\;\;{\rm{0}}.{\rm{3}}\;\;\;\;\;{\rm{ - 6}}.{\rm{3}}\;\;\;\;\;{\rm{ - 6}}.{\rm{9}}\;\;\;{\rm{2}}.{\rm{5}}}\\ {\;{\rm{0}}.{\rm{1}}\;\;\;\;\;{\rm{3}}.{\rm{2}}\;\;\;\;\;\;{\rm{2}}.{\rm{1}}\;\;\;\;\;{\rm{4}}.{\rm{5}}\;\;\;\;\;{\rm{6}}.{\rm{1}}\;\;\;\;\;{\rm{ - 7}}.{\rm{1}}\;\;\;\;\;{\rm{7}}.{\rm{2}}\;\;\;{\rm{ - 8}}.{\rm{6}}}\\ {{\rm{ - 7}}.{\rm{6}}\;\;\;\;\;{\rm{ - 9}}.{\rm{6}}\;\;\;\;\;{\rm{8}}.{\rm{3}}\;\;\;\;\;{\rm{0}}.{\rm{4}}\;\;\;\;\;{\rm{ - 6}}.{\rm{0}}\;\;\;{\rm{6}}.{\rm{9}}\;\;\;\;\;{\rm{4}}.{\rm{7}}\;\;\;\;{\rm{ - 2}}.{\rm{3}}}\\ {{\rm{ - 9}}.{\rm{7}}\;\;\;\;\;{\rm{7}}.{\rm{2}}\;\;\;\;\;{\rm{6}}.{\rm{2}}\;\;\;\;\;\;{\rm{0}}.{\rm{2}}\;\;\;\;\;{\rm{0}}.{\rm{1}}\;\;\;\;\;{\rm{0}}.{\rm{8}}\;\;\;\;\;{\rm{ - 5}}.{\rm{5}}\;\;{\rm{ - 8}}.{\rm{0}}}\\ {\;{\rm{9}}.{\rm{9}}\;\;\;\;\;{\rm{ - 4}}.{\rm{1}}\;\;\;\;\;{\rm{ - 1}}.{\rm{3}}\;\;\;\;\;{\rm{ - 3}}.{\rm{1}}\;\;\;\;{\rm{0}}.{\rm{1}}\;\;\;\;\;{\rm{4}}.{\rm{5}}\;\;\;\;\;{\rm{1}}.{\rm{2}}\;\;\;\;{\rm{9}}.{\rm{0}}}\\ {\;{\rm{7}}.{\rm{8}}\;\;\;\;\;{\rm{ - 7}}.{\rm{4}}\;\;\;\;\;{\rm{ - 2}}.{\rm{1}}\;\;\;\;\;\;\;{\rm{0}}\;\;\;\;\;{\rm{1}}.{\rm{9}}\;\;\;\;\;{\rm{4}}.{\rm{4}}\;\;\;\;\;{\rm{ - 4}}.{\rm{5}}\;\;\;{\rm{2}}.{\rm{3}}}\\ {{\rm{ - 4}}.{\rm{3}}\;\;\;\;\;{\rm{ - 2}}.{\rm{6}}\;\;\;\;\;{\rm{6}}.{\rm{2}}\;\;\;\;\;\;\;{\rm{3}}.{\rm{8}}\;\;\;\;{\rm{5}}.{\rm{8}}\;\;\;\;\;{\rm{3}}.{\rm{5}}\;\;\;\;\;{\rm{7}}.{\rm{1}}\;\;\;\;\;{\rm{2}}.{\rm{3}}} \end{array}} \right) $

$ B = \left( {\begin{array}{*{20}{l}} {{\rm{4}}.{\rm{9}}\;\;\;\;{\rm{43}}.{\rm{40}}\;\;\;\;\;{\rm{9}}.{\rm{34}}\;\;\;\;{\rm{0}}.{\rm{99}}\;\;\;\;{\rm{1}}.{\rm{55}}\;\;\;\;{\rm{4}}.{\rm{12}}\;\;\;\;\;{\rm{4}}\;\;\;\;\;\;{\rm{9}}.{\rm{2}}}\\ {{\rm{0}}.{\rm{78}}\;\;\;\;{\rm{3}}.{\rm{14}}\;\;\;\;{\rm{ - 2}}.{\rm{64}}\;\;\;\;{\rm{1}}.{\rm{37}}\;\;\;\;{\rm{1}}.{\rm{91}}\;\;\;\;\;{\rm{9}}.{\rm{01}}\;\;\;\;{\rm{1}}.{\rm{98}}\;\;\;\;{\rm{8}}.{\rm{44}}}\\ {{\rm{6}}.{\rm{40}}\;\;\;\;{\rm{3}}.{\rm{65}}\;\;\;\;\;\;\;{\rm{0}}\;\;\;\;\;\;\;{\rm{8}}.{\rm{18}}\;\;\;\;{\rm{4}}.{\rm{22}}\;\;\;\;{\rm{0}}.{\rm{05}}\;\;\;{\rm{6}}.{\rm{25}}\;\;\;\;{\rm{3}}.{\rm{68}}}\\ {{\rm{1}}.{\rm{90}}\;\;\;\;{\rm{ - 3}}.{\rm{93}}\;\;\;\;{\rm{8}}.{\rm{72}}\;\;\;\;\;\;\;{\rm{0}}\;\;\;\;\;\;{\rm{ - 8}}.{\rm{56}}\;\;\;{\rm{2}}.{\rm{97}}\;\;\;{\rm{7}}.{\rm{33}}\;\;\;\;{\rm{6}}.{\rm{20}}}\\ {{\rm{8}}.{\rm{43}}\;\;\;\;\;\;\;{\rm{0}}\;\;\;\;\;\;{\rm{2}}.{\rm{37}}\;\;\;\;{\rm{8}}.{\rm{90}}\;\;\;\;{\rm{4}}.{\rm{90}}\;\;\;\;{\rm{0}}.{\rm{49}}\;\;\;\;{\rm{3}}.{\rm{75}}\;\;\;\;{\rm{7}}.{\rm{31}}}\\ {{\rm{1}}.{\rm{73}}\;\;\;\;{\rm{1}}.{\rm{19}}\;\;\;\;\;\;{\rm{6}}.{\rm{45}}\;\;\;\;{\rm{7}}.{\rm{34}}\;\;\;\;{\rm{8}}.{\rm{15}}\;\;\;\;{\rm{6}}.{\rm{93}}\;\;\;\;\;\;{\rm{0}}\;\;\;\;\;\;{\rm{1}}.{\rm{93}}}\\ {{\rm{1}}.{\rm{70}}\;\;\;\;{\rm{0}}.{\rm{38}}\;\;\;\;\;{\rm{9}}.{\rm{66}}\;\;\;\;\;\;\;{\rm{0}}\;\;\;\;\;\;\;\;{\rm{0}}\;\;\;\;\;\;{\rm{6}}.{\rm{50}}\;\;\;\;{\rm{4}}.{\rm{19}}\;\;\;\;{\rm{9}}.{\rm{04}}}\\ {{\rm{9}}.{\rm{94}}\;\;\;\;{\rm{4}}.{\rm{58}}\;\;\;\;{\rm{ - 6}}.{\rm{64}}\;\;\;\;{\rm{3}}.{\rm{41}}\;\;\;{\rm{ - 4}}.{\rm{57}}\;\;\;{\rm{9}}.{\rm{83}}\;\;\;\;\;\;{\rm{0}}\;\;\;\;\;\;{\rm{5}}.{\rm{69}}}\\ {{\rm{4}}.{\rm{39}}\;\;\;\;{\rm{ - 8}}.{\rm{69}}\;\;\;\;{\rm{8}}.{\rm{70}}\;\;\;\;{\rm{1}}.{\rm{66}}\;\;\;\;\;\;\;{\rm{0}}\;\;\;\;\;{\rm{5}}.{\rm{52}}\;\;\;\;\;{\rm{7}}.{\rm{93}}\;\;\;\;{\rm{6}}.{\rm{38}}} \end{array}} \right) $

其中rank($A)=8$, 计算矩阵$A$的奇异值分解, 得到$A$的奇异值为

$ {\rm{(}}\;\;{\rm{25}}{\rm{.6126, }}\;\;\;{\rm{24}}{\rm{.2777, }}\;\;\;{\rm{19}}{\rm{.2189, }}\;\;\;{\rm{16}}{\rm{.4819, }}\;\;\;{\rm{12}}{\rm{.6274, }}\;\;\;{\rm{7}}{\rm{.5266, }}\;\;\;{\rm{4}}{\rm{.5672, }}\;\;\;{\rm{7 2}}{\rm{.3510), }} $

依算法得问题(1.2) 的唯一解为

$ X=\begin{pmatrix} 0.0756&0.0063 &0.0560&&&&& \\ 0.0063&0.0851&0.0649&0.0352&&&&\\ 0.0560&0.0649&0.2031&0.2108&0.1937 &&&\\ & 0.0352& 0.2108&0.2398 &0.1717 &0.2154&&\\ &&0.1937&0.1717&0.1541& 0.2087&0.1756 &\\ &&&0.2154&0.2087&0.1656& 0.1434&0.1847\\ &&&&0.1756&0.1434&0.1858&0.1393\\ &&&&&0.1847&0.1393&0.2218 \end{pmatrix}. $