Sylvester方程是矩阵理论研究中非常重要的一类矩阵方程, 它在特征结构配置、航天控制技术、微分方程数值解、模式识别等领域都有实际应用^[1-3].目前, 关于Sylvester方程的解与约束解人们多在实数域或复数域上讨论, 并已取得丰富的成果^[4-8], 而在四元数体上讨论该方程的约束解问题甚少.随着四元数矩阵在图像处理、飞行器姿态控制等方面的应用发展^{[9, 10]}, 讨论四元数Sylvester方程的约束解具有较大实际意义. Toeplitz矩阵是一类特殊结构矩阵, 它在信号压缩感知、超视距雷达电离层相位扰动校正等方面有重要作用^[11-13].

本文目的是把实数域上的Sylvester矩阵方程$ AX - XB = C $推广到四元数体上讨论, 给出该方程存在Toeplitz约束解的条件及解法, 同时讨论它的最佳逼近问题.

用$ {\rm {\bf R}}^{n\times n}, {\rm {\bf C}}^{n\times n}, {\rm {\bf Q}}^{n\times n} $分别表示全体$ n $阶实矩阵、复矩阵及四元数矩阵集合; $ A^T, \bar {A}, A^\ast $分别表示四元数矩阵$ A $的转置、共轭、共轭转置; $ A^ + $表示$ A $的Moore-Penrose广义逆; $ \left\| A \right\| = \sqrt {\mbox{tr}(A^\ast A)} $表示四元数矩阵$ A $的Frobenius范数; $ \mbox{vec}(A) $表示矩阵$ A $按列顺序拉直向量; $ A \otimes B $表示矩阵$ A $与$ B $的Kronecker积.下面给出相关定义和引理.

定义1.1  设$ T = (a_{ij} )_{n\times n} \in {\rm {\bf Q}}^{n\times n} $, 如果满足$ a_{ij} = a_{j - i} $, 即

$ \begin{equation} T = \left[ {{\begin{array}{*{20}c} {a_0 } \hfill & {a_1 } \hfill & \cdots \hfill & {a_{n - 2} } \hfill & {a_{n - 1} } \hfill \\ {a_{ - 1} } \hfill & {a_0 } \hfill & \cdots \hfill & {a_{n - 3} } \hfill & {a_{n - 2} } \hfill \\ \vdots \hfill & \vdots \hfill & \ddots \hfill & \vdots \hfill & \vdots \hfill \\ {a_{ - n + 2} } \hfill & {a_{ - n + 3} } \hfill & \cdots \hfill & {a_0 } \hfill & {a_1 } \hfill \\ {a_{ - n + 1} } \hfill & {a_{ - n + 2} } \hfill & \cdots \hfill & {a_{ - 1} } \hfill & {a_0 } \hfill \\ \end{array} }} \right], \end{equation} $

(1.1)

则称形如(1.1)式的矩阵为四元数Toeplitz矩阵, 全体$ n $阶四元数Toeplitz矩阵记作$ \mbox{TQ}^{n\times n} $.

显然, 一个四元数Toeplitz矩阵(1.1)由它的第$ n $行和第$ n $列共$ 2n - 1 $个元素唯一确定.记

$ \begin{eqnarray} l(T) & = & (a_{ - n + 1} , a_{ - n + 2} , \cdots , a_{ - 1} , a_0 , a_1 , \cdots , a_{n - 2} , a_{n - 1} )^T \in {\rm {\bf Q}}^{2n - 1}, \end{eqnarray} $

(1.2)

$ \begin{eqnarray} T_n & = & \left[ {{\begin{array}{*{20}c} {e_n } \hfill & {e_{n - 1} } \hfill & \cdots \hfill & {e_2 } \hfill & {e_1 } \hfill & 0 \hfill & \cdots \hfill & 0 \hfill & 0 \hfill \\ 0 \hfill & {e_n } \hfill & \cdots \hfill & {e_3 } \hfill & {e_2 } \hfill & {e_1 } \hfill & \cdots \hfill & 0 \hfill & 0 \hfill \\ \vdots \hfill & \vdots \hfill & \ddots \hfill & \vdots \hfill & \vdots \hfill & \vdots \hfill & \ddots \hfill & \vdots \hfill & \vdots \hfill \\ 0 \hfill & 0 \hfill & \cdots \hfill & {e_n } \hfill & {e_{n - 1} } \hfill & {e_{n - 2} } \hfill & \cdots \hfill & {e_1 } \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & \cdots \hfill & 0 \hfill & {e_n } \hfill & {e_{n - 1} } \hfill & \cdots \hfill & {e_2 } \hfill & {e_1 } \hfill \\ \end{array} }} \right] \in {\rm {\bf Q}}^{n^2\times (2n - 1)}, \end{eqnarray} $

(1.3)

其中$ e_i $为单位矩阵$ I_n $的第$ i $列.易知$ T_n \in {\rm {\bf R}}^{n^2\times (2n - 1)} $是列正交的.于是有

引理1.1  设$ T = (a_{ij} )_{n\times n} \in {\rm {\bf Q}}^{n\times n} $, 则

$ \begin{equation} T \in \mbox{TQ}^{n\times n} \Leftrightarrow \mbox{vec}(T) = T_n \cdot l(T), \end{equation} $

(1.4)

其中$ l(T), T_n $分别如(1.2), (1.3)式所示.

引理1.2^[14]  复数域上矩阵方程$ AX = C $有解的充要条件是$ AA^ + C = C $.此方程的通解和最小二乘解集均可表示为$ X = A^ + C + (I - A^ + A)Y, $其中$ Y \in {\rm {\bf C}}^{n\times n} $是任意矩阵, 且存在唯一极小范数最小二乘解$ X_{0} = A^ + C{.} $

本文主要讨论如下3个问题.

问题Ⅰ  给定四元数矩阵$ A, B, C \in {\rm {\bf Q}}^{n\times n} $, 求矩阵$ X \in \mbox{TQ}^{n\times n} $, 使得$ AX - XB = C $.

问题Ⅱ  对给定的$ A, B, C \in {\rm {\bf Q}}^{n\times n} $, 求矩阵$ X \in \mbox{TQ}^{n\times n} $, 使得$ \vert \vert AX - XB - C\vert \vert = \min $.

问题Ⅲ  设问题I的解集$ S_E \ne \emptyset $, $ M \in \mbox{TQ}^{n\times n} $是已知Toeplitz矩阵, 求$ \tilde {X} \in S_E $, 使得$ \mathop \min\limits_{X \in S_E } {\vert \vert }X - M {\vert \vert } = {\vert \vert }\tilde {X} - M {\vert \vert .} $

设$ X \in \mbox{TQ}^{n\times n} $, 它在实数域$ {\rm {\bf R}} $上的分解式为$ X = X_0 + X_1 i + X_2 j + X_3 k, $其中$ X_i \in {\rm {\bf R}}^{n\times n}(i = 0, 1, 2, 3) $均是实Toeplitz矩阵.又设$ A, B, C \in {\rm {\bf Q}}^{n\times n} $在实数域$ {\rm {\bf R}} $上的分解式为$ A = A_0 + A_1 i + A_2 j + A_3 k, B = B_0 + B_1 i + B_2 j + B_3 k, C = C_0 + C_1 i + C_2 j + C_3 k $, 其中$ A_i , B_i , C_i \in {\rm {\bf R}}^{n\times n}(i = 0, 1, 2, 3) $, 则四元数体上Sylvester方程

$ \begin{eqnarray} AX - XB = C \end{eqnarray} $

(2.1)

等价于

$ \begin{eqnarray} &&(A_0 + A_1 i + A_2 j + A_3 k)(X_0 + X_1 i + X_2 j + X_3 k) \\ &&- (X_0 + X_1 i + X_2 j + X_3 k)(B_0 + B_1 i + B_2 j + B_3 k) \\ & = & (C_0 + C_1 i + C_2 j + C_3 k). \end{eqnarray} $

(2.2)

将(2.2)式左边展开, 并根据四元数矩阵实分解的唯一性, 可得

$ \begin{equation} \left\{ {\begin{array}{l} A_0 X_0 - X_0 B_0 - A_1 X_1 + X_1 B_1 - A_2 X_2 + X_2 B_2 - A_3 X_3 + X_3 B_3 = C_0 , \\ A_1 X_0 - X_0 B_1 + A_0 X_1 - X_1 B_0 - A_3 X_2 - X_2 B_3 + A_2 X_3 + X_3 B_2 = C_1 , \\ A_2 X_0 - X_0 B_2 + A_3 X_1 + X_1 B_3 + A_0 X_2 - X_2 B_0 - A_1 X_3 - X_3 B_1 = C_2 , \\ A_3 X_0 - X_0 B_3 - A_2 X_1 - X_1 B_2 + A_1 X_2 + X_2 B_1 + A_0 X_3 - X_3 B_0 = C_3 , \\ \end{array}} \right. \end{equation} $

(2.3)

由于$ X_i \in {\rm {\bf R}}^{n\times n}(i = 0, 1, 2, 3) $均是实Toeplitz矩阵, 因此由引理1.1可得

$ \begin{equation} \mbox{vec}(X_i ) = T_n \cdot l(X_i ){\begin{array}{*{20}c} \hfill \\ \end{array} }(i = 0, 1, 2, 3), \end{equation} $

(2.4)

其中$ T_n $如(1.3)式所示.记

$ \begin{eqnarray} && G = \left[ {{\begin{array}{*{20}c} {I \otimes A_0 - B_0^T \otimes I} \hfill & { - I \otimes A_1 + B_1^T \otimes I} \hfill & { - I \otimes A_2 + B_2^T \otimes I} \hfill & { - I \otimes A_3 + B_3^T \otimes I} \hfill \\ {I \otimes A_1 - B_1^T \otimes I} \hfill & {I \otimes A_0 - B_0^T \otimes I} \hfill & { - I \otimes A_3 - B_3^T \otimes I} \hfill & {I \otimes A_2 + B_2^T \otimes I} \hfill \\ {I \otimes A_2 - B_2^T \otimes I} \hfill & {I \otimes A_3 + B_3^T \otimes I} \hfill & {I \otimes A_0 - B_0^T \otimes I} \hfill & { - I \otimes A_1 - B_1^T \otimes I} \hfill \\ {I \otimes A_3 - B_3^T \otimes I} \hfill & { - I \otimes A_2 - B_2^T \otimes I} \hfill & {I \otimes A_1 + B_1^T \otimes I} \hfill & {I \otimes A_0 - B_0^T \otimes I} \hfill \\ \end{array} }} \right], \\ &&{ \tilde {G} = G\left[ {{\begin{array}{*{20}c} {T_n } \hfill & \hfill & \hfill & \hfill \\ \hfill & {T_n } \hfill & \hfill & \hfill \\ \hfill & \hfill & {T_n } \hfill & \hfill \\ \hfill & \hfill & \hfill & {T_n } \hfill \\ \end{array} }} \right], {\begin{array}{*{20}c} \hfill \\ \end{array} }L = \left[ {{\begin{array}{*{20}c} {\mbox{vec}(C_0 )} \hfill \\ {\mbox{vec}(C_1 )} \hfill \\ {\mbox{vec}(C_2 )} \hfill \\ {\mbox{vec}(C_3 )} \hfill \\ \end{array} }} \right], {\begin{array}{*{20}c} \hfill \\ \end{array} }v = \left[ {{\begin{array}{*{20}c} {l(X_0 )} \hfill \\ {l(X_1 )} \hfill \\ {l(X_2 )} \hfill \\ {l(X_3 )} \hfill \\ \end{array} }} \right].} \end{eqnarray} $

(2.5)

由于(2.4)式等价于

$ \left[ {{\begin{array}{*{20}c} {\mbox{vec}(X_0 )} \hfill \\ {\mbox{vec}(X_1 )} \hfill \\ {\mbox{vec}(X_2 )} \hfill \\ {\mbox{vec}(X_3 )} \hfill \\ \end{array} }} \right] = \left[ {{\begin{array}{*{20}c} {T_n l(X_0 )} \hfill \\ {T_n l(X_1 )} \hfill \\ {T_n l(X_2 )} \hfill \\ {T_n l(X_3 )} \hfill \\ \end{array} }} \right] = \left[ {{\begin{array}{*{20}c} {T_n } \hfill & \hfill & \hfill & \hfill \\ \hfill & {T_n } \hfill & \hfill & \hfill \\ \hfill & \hfill & {T_n } \hfill & \hfill \\ \hfill & \hfill & \hfill & {T_n } \hfill \\ \end{array} }} \right]\left[ {{\begin{array}{*{20}c} {l(X_0 )} \hfill \\ {l(X_1 )} \hfill \\ {l(X_2 )} \hfill \\ {l(X_3 )} \hfill \\ \end{array} }} \right], $

因此方程组(2.3)等价于

$ \begin{equation} \tilde {G}v = L. \end{equation} $

(2.6)

此外, 利用矩阵Frobenius范数可得

$ \begin{eqnarray} && \vert \vert AX - XB - C\vert \vert ^2\\ & = & \vert \vert A_0 X_0 - X_0 B_0 - A_1 X_1 + X_1 B_1 - A_2 X_2 + X_2 B_2 - A_3 X_3 + X_3 B_3 - C_0 \vert \vert ^2+ \\ && \vert \vert A_1 X_0 - X_0 B_1 + A_0 X_1 - X_1 B_0 - A_3 X_2 - X_2 B_3 + A_2 X_3 + X_3 B_2 - C_1 \vert \vert ^2+ \\ && \vert \vert A_2 X_0 - X_0 B_2 + A_3 X_1 + X_1 B_3 + A_0 X_2 - X_2 B_0 - A_1 X_3 - X_3 B_1 - C_2 \vert \vert ^2+ \\ && \vert \vert A_3 X_0 - X_0 B_3 - A_2 X_1 - X_1 B_2 + A_1 X_2 + X_2 B_1 + A_0 X_3 - X_3 B_0 - C_3 \vert \vert ^2 \\ & = & \vert \vert \tilde {G}v - L\vert \vert ^2. \end{eqnarray} $

(2.7)

于是, 关于问题Ⅰ–Ⅱ的解, 有如下的结果.

定理2.1  给定四元数矩阵$ A, B, C \in {\rm {\bf Q}}^{n\times n} $, 则Sylvester方程(2.1)存在四元数Toeplitz解的充要条件是$ \tilde {G}\tilde {G}^ + L = L $.有解时, 它的一般Toeplitz解为

$ \begin{equation} X = X_0 + X_1 i + X_2 j + X_3 k, \end{equation} $

(2.8)

无解时, 它的最小二乘Toeplitz解仍为(2.8)式, 其中

$ \begin{eqnarray} &&v = {\tilde {G}}^ + L + (I - {\tilde {G}}^ + \tilde {G})Y, \forall Y \in {\rm {\bf R}}^{(8n - 4)\times 1}, \\ &&{l(X_0 ) = v(1:2n - 1), l(X_1 ) = v(2n:4n - 2)}, \\ && {l(X_2 ) = v(4n - 1:6n - 3), l(X_3 ) = v(6n - 2:8n - 4)}, \\ && \mbox{vec}(X_i ) = T_n \cdot l(X_i ), i = 0, 1, 2, 3, \end{eqnarray} $

这里$ \tilde {G} \in {\rm {\bf R}}^{4n^2\times 4(2n - 1)}, L \in {\rm {\bf R}}^{4(2n - 1)\times 1} $如(2.5)式所示, $ v(1:2n - 1) $表示由向量的第$ 1 $至$ 2n - 1 $个元素组成的$ 2n - 1 $维向量.

证  由方程组(2.6)及引理1.2可得, 方程组(2.1)存在四元数Toeplitz解$ \Leftrightarrow $方程组(2.6)有解$ \Leftrightarrow \tilde {G}\tilde {G}^ + L = L $.有解时, (2.1)的Toeplitz解显然由(2.8)式给出.无解时, 由(2.7)式可得$ \vert \vert AX - XB - C\vert \vert = \min \Leftrightarrow \vert \vert \tilde {G}v - L\vert \vert = \min $.因此(2.1)式的最小二乘Toeplitz解仍为(2.8)式.证毕.

设问题I的解集$ S_E \ne \emptyset $, $ M \in \mbox{TQ}^{n\times n} $是已知Toeplitz矩阵, 现将$ M $作实分解$ M = M_0 + M_1 i + M_2 j + M_3 k, $其中$ M_i \in {\rm {\bf R}}^{n\times n}(i = 0, 1, 2, 3) $均是实Toeplitz矩阵.记

$ \begin{eqnarray} v_M = \left[ {{\begin{array}{*{20}c} {l(M_0 )} \hfill \\ {l(M_1 )} \hfill \\ {l(M_2 )} \hfill \\ {l(M_3 )} \hfill \\ \end{array} }} \right] \in {\rm {\bf R}}^{4(2n - 1)\times 1}, {\begin{array}{*{20}c} \hfill \\ \end{array} }\hat {T} = \left[ {{\begin{array}{*{20}c} {T_n } \hfill & \hfill & \hfill & \hfill \\ \hfill & {T_n } \hfill & \hfill & \hfill \\ \hfill & \hfill & {T_n } \hfill & \hfill \\ \hfill & \hfill & \hfill & {T_n } \hfill \\ \end{array} }} \right]. \end{eqnarray} $

(3.1)

则有

$ \begin{eqnarray} \vert \vert X - M\vert \vert ^2 & = &\sum\limits_{i = 0}^3 {\vert \vert X_i - M_i \vert \vert ^2} = \sum\limits_{i = 0}^3 {\vert \vert \mbox{vec}(X_i ) - \mbox{vec}(M_i )\vert \vert ^2} \\ & = &\sum\limits_{i = 0}^3 {\vert \vert T_n \cdot l(X_i ) - T_n \cdot l(M_i )\vert \vert ^2} = \vert \vert \hat {T}v - \hat {T}v_M \vert \vert ^2 \\ & = &{\vert \vert }(\hat {T}^\ast \hat {T})^{ {1 \over 2}}\mbox{(}v - v_M ){\vert \vert }^2, \end{eqnarray} $

(3.2)

其中

$ {\hat {T}^\ast \hat {T} = \left[ {{\begin{array}{*{20}c} D \hfill & \hfill & \hfill & \hfill \\ \hfill & D \hfill & \hfill & \hfill \\ \hfill & \hfill & D \hfill & \hfill \\ \hfill & \hfill & \hfill & D \hfill \\ \end{array} }} \right]}, D = \mbox{diag}(1, \cdots , n - 1, n, n - 1, \cdots , 1) \in {\rm {\bf R}}^{(2n - 1)\times (2n - 1)}. $

于是关于问题Ⅲ, 有如下结果.

定理3.1  设问题Ⅰ的解集$ S_E \ne \emptyset $, $ M \in \mbox{TQ}^{n\times n} $是已知四元数Toeplitz矩阵, 则在$ S_E $中使得$ \vert \vert X - M\vert \vert = \min $的解$ \tilde {X} $存在, 且有如下表达式

$ \begin{equation} \tilde {X} = X_0 + X_1 i + X_2 j + X_3 k, \end{equation} $

(3.3)

其中

$ \begin{eqnarray} && \tilde {v} = \tilde {G}^ + L + (I - \tilde {G}^ + \tilde {G})[(\hat {T}^\ast \hat {T})^{ {1 \over 2}}(I - \tilde {G}^ + \tilde {G})]^ + (\hat {T}^\ast \hat {T})^{ {1 \over 2}}(v_M - \tilde {G}^ + L), \\ && l(X_0 ) = \tilde {v}(1:2n - 1), {\begin{array}{*{20}c} \hfill \\ \end{array} }l(X_1 ) = \tilde {v}(2n:4n - 2), \\ && l(X_2 ) = \tilde {v}(4n - 1:6n - 3), {\begin{array}{*{20}c} \hfill \\ \end{array} }l(X_3 ) = \tilde {v}(6n - 2:8n - 4), \\ && \mbox{vec}(X_i ) = T_n \cdot l(X_i ), i = 0, 1, 2, 3, \end{eqnarray} $

这里的符号意义与前面所示相同.

证  当$ X \in S_E $时, 根据定理2.1及(3.2)式可知$ {\vert \vert }X - M{\vert \vert }^2 = \min \Leftrightarrow {\vert \vert }(\hat {T}^\ast \hat {T})^{ {1 \over 2}}[\tilde {G}^ + L + (I - \tilde {G}^ + \tilde {G})Y - v_M {]\vert \vert }^2 = \min , $其中$ \hat {T}^\ast \hat {T} $是一个正对角矩阵.当$ \tilde {G}^ + \tilde {G} \ne I $时, 由引理1.2, 上式关于$ Y $的最小二乘解为$ \tilde {Y} = [(\hat {T}^\ast \hat {T})^{ {1 \over 2}}(I - \tilde {G}^ + \tilde {G})]^ + (\hat {T}^\ast \hat {T})^{ {1 \over 2}}(v_M - \tilde {G}^ + L), $当$ \tilde {G}^ + \tilde {G} = I $时, (2.1)存在唯一解$ \tilde {v} = \tilde {G}^ + L $, 因此不论哪种情况均有

$ \begin{eqnarray} \tilde {v} & = &\tilde {G}^ + L + (I - \tilde {G}^ + \tilde {G})\tilde {Y} \\ & = & \tilde {G}^ + L + (I - \tilde {G}^ + \tilde {G})[(\hat {T}^\ast \hat {T})^{ {1 \over 2}}(I - \tilde {G}^ + \tilde {G})]^ + (\hat {T}^\ast \hat {T})^{ {1 \over 2}}(v_M - \tilde {G}^ + L), \end{eqnarray} $

于是存在$ \tilde {X} \in S_E $使得$ \vert \vert X - M\vert \vert = \min $成立, 并且$ \tilde {X} $由(3.3)式给出.证毕.

根据定理2.1和定理3.1, 我们给出问题Ⅰ–Ⅲ的求解步骤:

步1  写出四元数矩阵$ A, B, C $的实分解式, 即$ A = A_0 + A_1 i + A_2 j + A_3 k, B = B_0 + B_1 i + B_2 j + B_3 k, C = C_0 + C_1 i + C_2 j + C_3 k. $

步2  按(2.5)式写出实矩阵$ \tilde {G} $和实向量$ L $.

步3  检验条件$ \tilde {G}\tilde {G}^ + L = L $是否成立.

ⅰ)若条件成立, 说明问题Ⅰ有解, 并按(2.8)式写出其Toeplitz解集$ S_E $;

ⅱ)若条件不成立, 说明问题Ⅰ无解, 此时问题Ⅱ的最小二乘Toeplitz解集仍为$ S_E $.

步4  在问题Ⅰ有解时, 对给定的四元数Toeplitz矩阵$ M $, 按(3.1)式写出对应的实向量$ v_M . $

步5  按(3.3)式写出问题Ⅲ的最佳逼近解$ \tilde {X} $, 即由$ \mbox{vec}(X_i ) = T_n \cdot l(X_i ), i = 0, 1, 2, 3 $, 可得出4个实Toeplitz矩阵$ X_i , i = 0, 1, 2, 3 $, 从而获得$ \tilde {X} = X_0 + X_1 i + X_2 j + X_3 k. $

算例  给定下列四元数矩阵

$ \begin{eqnarray} &&{ A = \left[ {{\begin{array}{*{20}c} 1 \hfill & 0 \hfill & 0 \hfill & 1 \hfill \\ 0 \hfill & 1 \hfill & i \hfill & 0 \hfill \\ 0 \hfill & i \hfill & j \hfill & 0 \hfill \\ 1 \hfill & 0 \hfill & 0 \hfill & k \hfill \\ \end{array} }} \right], {\begin{array}{*{20}c} \hfill \\ \end{array} }B = \left[ {{\begin{array}{*{20}c} 1 \hfill & 0 \hfill & 0 \hfill & k \hfill \\ 0 \hfill & 1 \hfill & 0 \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & i \hfill & 0 \hfill \\ k \hfill & 0 \hfill & 0 \hfill & j \hfill \\ \end{array} }} \right], }\\ &&{ C = \left[ {{\begin{array}{*{20}c} { - 1 + i + j} \hfill & {0.5 + 0.5j} \hfill & {1 - i} \hfill & 1 \hfill \\ {0.5i - 0.5j} \hfill & i \hfill & { - 0.75 + 0.25i} \hfill & {0.75 - 0.5i} \hfill \\ { - 1.5 + 0.5i - 0.25j} \hfill & { - 1.5 + 0.5i} \hfill & {0.75 + 0.5i} \hfill & {0.5 - i} \hfill \\ {2 - i} \hfill & { - 0.5 - 0.75i} \hfill & {0.5 - 1.5i + k} \hfill & { - 1} \hfill \\ \end{array} }} \right].} \end{eqnarray} $

试讨论Sylvester方程(2.1)的Toeplitz解的存在性.

解  四元数矩阵$ A, B, C $的实分解矩阵分别为

$ A_0 = \left[ {{\begin{array}{*{20}c} 1 \hfill & 0 \hfill & 0 \hfill & 1 \hfill \\ 0 \hfill & 1 \hfill & 0 \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & 0 \hfill & 0 \hfill \\ 1 \hfill & 0 \hfill & 0 \hfill & 0 \hfill \\ \end{array} }} \right], A_1 = \left[ {{\begin{array}{*{20}c} 0 \hfill & 0 \hfill & 0 \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & 1 \hfill & 0 \hfill \\ 0 \hfill & 1 \hfill & 0 \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & 0 \hfill & 0 \hfill \\ \end{array} }} \right], A_2 = \left[ {{\begin{array}{*{20}c} 0 \hfill & 0 \hfill & 0 \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & 0 \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & 1 \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & 0 \hfill & 0 \hfill \\ \end{array} }} \right], $

$ \begin{eqnarray} &&{ A_3 = \left[ {{\begin{array}{*{20}c} 0 \hfill & 0 \hfill & 0 \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & 0 \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & 0 \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & 0 \hfill & 1 \hfill \\ \end{array} }} \right], B_0 = \left[ {{\begin{array}{*{20}c} 1 \hfill & 0 \hfill & 0 \hfill & 0 \hfill \\ 0 \hfill & 1 \hfill & 0 \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & 0 \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & 0 \hfill & 0 \hfill \\ \end{array} }} \right], B_1 = \left[ {{\begin{array}{*{20}c} 0 \hfill & 0 \hfill & 0 \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & 0 \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & 1 \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & 0 \hfill & 0 \hfill \\ \end{array} }} \right]}, \\ &&{B_2 = \left[ {{\begin{array}{*{20}c} 0 \hfill & 0 \hfill & 0 \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & 0 \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & 0 \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & 0 \hfill & 1 \hfill \\ \end{array} }} \right], B_3 = \left[ {{\begin{array}{*{20}c} 0 \hfill & 0 \hfill & 0 \hfill & 1 \hfill \\ 0 \hfill & 0 \hfill & 0 \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & 0 \hfill & 0 \hfill \\ 1 \hfill & 0 \hfill & 0 \hfill & 0 \hfill \\ \end{array} }} \right]}, \\ &&{ C_0 = \left[ {{\begin{array}{*{20}c} { - 1} \hfill & {0.5} \hfill & 1 \hfill & 1 \hfill \\ 0 \hfill & 0 \hfill & { - 0.75} \hfill & {0.75} \hfill \\ { - 1.5} \hfill & { - 1.5} \hfill & {0.75} \hfill & {0.5} \hfill \\ 2 \hfill & { - 0.5} \hfill & {0.5} \hfill & { - 1} \hfill \\ \end{array} }} \right], C_1 = \left[ {{\begin{array}{*{20}c} 1 \hfill & 0 \hfill & { - 1} \hfill & 0 \hfill \\ {0.5} \hfill & 1 \hfill & {0.25} \hfill & { - 0.5} \hfill \\ {0.5} \hfill & {0.5} \hfill & {0.5} \hfill & { - 1} \hfill \\ { - 1} \hfill & { - 0.75} \hfill & { - 1.5} \hfill & 0 \hfill \\ \end{array} }} \right]}, \\ &&{C_2 = \left[ {{\begin{array}{*{20}c} 1 \hfill & {0.5} \hfill & 0 \hfill & 0 \hfill \\ { - 0.5} \hfill & 0 \hfill & 0 \hfill & 0 \hfill \\ { - 0.25} \hfill & 0 \hfill & 0 \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & 0 \hfill & 0 \hfill \\ \end{array} }} \right], C_3 = \left[ {{\begin{array}{*{20}c} 0 \hfill & 0 \hfill & 0 \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & 0 \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & 0 \hfill & 0 \hfill \\ 0 \hfill & 0 \hfill & 1 \hfill & 0 \hfill \\ \end{array} }} \right]}. \end{eqnarray} $

按(2.5)式写出实矩阵$ \tilde {G} $和实向量$ L $并直接计算可知$ \tilde {G}\tilde {G}^ + L = L $且$ \tilde {G}^ + \tilde {G} = I $, 因此根据定理2.1, 所给的Sylvester方程(2.1)存在唯一Toeplitz解$ X $, 且由公式(2.8)可得

$ X = \left[ {{\begin{array}{*{20}c} {0.5 + 0.5i + k} \hfill & { - 0.25i + 0.5j - 0.5k} \hfill & {0.5 - 0.5i} \hfill & { - 0.5j - 0.5k} \hfill \\ 1 \hfill & {0.5 + 0.5i + k} \hfill & { - 0.25i + 0.5j - 0.5k} \hfill & {0.5 - 0.5i} \hfill \\ {0.5 + 0.5j} \hfill & 1 \hfill & {0.5 + 0.5i + k} \hfill & { - 0.25i + 0.5j - 0.5k} \hfill \\ { - 0.5 + 0.5i + j} \hfill & {0.5 + 0.5j} \hfill & 1 \hfill & {0.5 + 0.5i + k} \hfill \\ \end{array} }} \right]. $

本文提出一种判断四元数Sylvester方程是否具有Toeplitz约束解的方法.我们根据Toeplitz矩阵的结构特点, 给出四元数Toeplitz矩阵的新刻划.利用四元数矩阵的实分解和矩阵的Kronecker积, 把约束方程问题转化为无约束方程问题, 解决了四元数乘法非交换的限制, 得到了四元数Sylvester方程具有Toeplitz解的充要条件, 以及它的Toeplitz解集和最小二乘解集$ S_E $.此外, 利用矩阵Frobenius范数性质, 在Toeplitz解集$ S_E \ne \emptyset $的条件下, 获得$ S_E $与预先给定的四元数Toeplitz矩阵$ M $有极小Frobenius范数的最佳逼近解.本文结果可为解决相关约束四元数矩阵方程问题提供有益参考.