四元数体上双Hermite矩阵反问题的最小二乘解 Least-Squares Solution to the Dou-ble-Hermite Matrix Inverse Problem on the Quaternion Field

doi:10.12677/AAM.2022.118597

Advances in Applied Mathematics
Vol. 11 No. 08 ( 2022 ), Article ID: 54807 , 9 pages
10.12677/AAM.2022.118597

四元数体上双Hermite矩阵反问题的最小二乘解

王敏

●How to Cite this Article

贺州学院公共基础教学部，广西贺州

收稿日期：2022年7月11日；录用日期：2022年8月4日；发布日期：2022年8月16日

摘要

讨论四元数体上的矩阵方程组AX = Z，Y^*A = W^*的双Hermite矩阵反问题的最小二乘解及其最佳逼近解。利用双Hermite矩阵的结构特性及奇异值分解定理，将原问题转化为Hermite矩阵方程问题，得出该问题解的表达式。最后给出数值算例检验算法的正确、可行。

关键词

四元数，双Hermite矩阵，奇异值分解，反问题

Least-Squares Solution to the Double-Hermite Matrix Inverse Problem on the Quaternion Field

Min Wang

The Department of Public Basic Education, Hezhou University, Hezhou Guangxi

Received: Jul. 11^th, 2022; accepted: Aug. 4^th, 2022; published: Aug. 16^th, 2022

ABSTRACT

To discuss the least-squares solution and the best approximation solution of the double-Hermite matrix inverse problem of the matrix equation system AX = Z, Y*A = W* on the quaternion field. The original problem is transformed into an equation problem with Hermite matrix structure by using the structural properties of double-Hermite matrices and the singular value decomposition theorem. The expression for the solution to the problem is obtained. Finally, a numerical example is given to test the correctness and feasibility of the algorithm.

Keywords:Quaternion Field, Double-Hermite Matrix, Singular Value Decomposition, Inverse Problem

This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).

http://creativecommons.org/licenses/by/4.0/

1. 引言

四元数体上矩阵方程的Hermite最小二乘问题及最佳逼近问题的研究目前已经取得较多成果，例如Cao W S [1] 给出了四元数矩阵方程AXB = D的Hermite解的充要条件；袁仕芳等 [2] 得出四元数矩阵方程AXB = C的三对角Hermite极小范数最小二乘解；Zhou S等 [3] 讨论矩阵方程AX = B，XC = D的Hermitian Reflexive (Anti-Hermitian Reflexive)矩阵最小二乘解及其最佳逼近；袁仕芳等 [4] 研究了分裂四元数矩阵方程AXB + CXD = E的Hermitian解；黄敬频等 [5] 给出四元数Lyapunov方程 $A X + X A^{*} = B$ 的双自共轭解，Wang等 [6] 讨论四元数域上连续Lyapunov矩阵方程 $A^{*} X + X A = C$ 的三对角箭形矩阵约束问题；张奇梅 [7] 给出了矩阵方程组AX = Z，Y^*A = W^*的埃尔米特反自反矩阵反问题的最小二乘解的一般表达式。但在四元数体上方程组AX = Z，Y^*A = W^*的双Hermite矩阵反问题的最小二乘解问题未见研究报道。

本文的目的是讨论四元数体上的方程组

$A X = Z$ ， $Y^{*} A = W^{*}$ (1)

的双Hermitan矩阵反问题最小二乘解及最佳逼近，这里所提的反问题是指当矩阵X，Z，Y，W已知的情况下，反求矩阵A。显然当 $Z = X Λ$ ， $Λ = d i a g (λ_{1}, λ_{2}, \dots, λ_{n})$ ， $W = Σ Y$ ， $Σ = d i a g (μ_{1}, μ_{2}, \dots, μ_{m})$ ，上述问题就是双Hermitan矩阵的左右特征对反问题。四元数双Hermitan矩阵是实数域上双对称矩阵的推广。

定义1 [5] 设 $A \in Q^{n \times n}$ ，如果 $A = A^{*}$ 且 $A = S_{n} A S_{n}$ ，则称A为n阶四元数双Hermite矩阵，其中 $S_{n} = (e_{n}, e_{n - 1}, \dots, e_{1})$ 表示反对角线元素全为1的n阶方阵。全体n阶四元数双Hermite矩阵的集合表示为 $B S C_{n} (Q)$ 。

为讨论方便，四元数的全体记为Q，即

$Q = {a + b i + c j + d k | a, b, c, d \in R}$

其中 $i^{2} = j^{2} = k^{2} = - 1$ ， $i j = - j i = k$ ， $j k = - k j = i$ ， $k i = - i k = j$ 。用 $S R^{n \times n}$ ， $A S R^{n \times n}$ ， $S C_{n} (Q)$ ， $B S C_{n} (Q)$ 分别表示 $n \times n$ 对称、反对称、自共轭矩阵和双自共轭矩阵的集合； $\bar{A}$ ， $A^{Τ}$ ， $A^{*}$ ， $U^{n \times n}$ 分别表示矩阵A的共轭、转置、共轭转置和酉矩阵，本文具体研究如下问题。

问题I 给定 $X, Z \in Q^{m \times n}$ ； $Y, W \in Q^{m \times s}$ ，求 $A \in B S C_{m} (Q)$ ，使得

${‖ A X - Z ‖}^{2} + {‖ Y^{*} A - W^{*} ‖}^{2} = \min$

问题II 设 $S_{E}$ 是问题I的解集合，给定 $\tilde{M} \in Q^{n \times n}$ ，求矩阵 $\tilde{A} \in S_{E}$ 使得

$\min ‖ A - \tilde{M} ‖ = ‖ \tilde{A} - \tilde{M} ‖$

2. 问题Ⅰ的解

引理1 [8] 设 $G \in Q^{r \times r}$ ， $\sum = diag (λ_{1}, \dots, λ_{r}) > 0$ ，则对任意 $S \in S C_{r} (Q)$ ，有

${‖ S Σ - G ‖}^{2} = {‖ S Σ - (W * (G Σ + Σ G^{*}) Σ) ‖}^{2} + {‖ G - (W * (G Σ + Σ G^{*}) Σ) ‖}^{2}$ ，

其中 $A * B$ 表示矩阵的Hadamard积， $W = (w_{i j}) \in R^{r \times r}$ ， $w_{i j} = 1 / (λ_{i}^{2} + λ_{j}^{2})$ ， $1 \leq i, j \leq r$ 。于是 ${‖ S Σ - G ‖}^{2} = \min$ 存在唯一极小F范数自共轭解： $S = W * (G Σ + Σ G^{*})$ 。

设 $S_{k} = (e_{k}, e_{k - 1}, \dots, e_{1})$ ，且

$T_{2 k} = \frac{1}{\sqrt{2}} (\begin{matrix} I_{k} & I_{k} \\ S_{k} & - S_{k} \end{matrix}), T_{2 k + 1} = \frac{1}{\sqrt{2}} (\begin{matrix} I_{k} & 0 & I_{k} \\ 0 & \sqrt{2} & 0 \\ S_{k} & 0 & - S_{k} \end{matrix})$ (2)

显然 $T_{2 k} \in U^{2 k \times 2 k}$ ， $T_{2 k + 1} \in U^{(2 k + 1) \times (2 k + 1)}$ 。于是有

引理2 [6] 设 $T_{2 k}, T_{2 k + 1}$ 如(2)所示酉阵，则双自共轭四元数矩阵的集合可表示为

$B S C_{2 k} (Q) = {T_{2 k} (\begin{matrix} M + H & 0 \\ 0 & M - H \end{matrix}) T_{2 k}^{*} | M, H \in S C_{k} (Q)}$ (3)

$B S C_{2 k + 1} (Q) = {T_{2 k + 1} (\begin{matrix} M + H & \sqrt{2} C & 0 \\ \sqrt{2} C^{*} & ρ & 0 \\ 0 & 0 & M - H \end{matrix}) T_{2 k + 1}^{*} | M, H \in S C_{k} (Q), C \in Q^{k}, ρ \in R}$ (4)

下面讨论问题I的解。当m = 2k时，令

$T_{2 k}^{*} X = (\begin{matrix} X_{1} \\ X_{2} \end{matrix})$ ， $T_{2 k}^{*} Z = (\begin{matrix} Z_{1} \\ Z_{2} \end{matrix})$ ， $T_{2 k}^{*} Y = (\begin{matrix} Y_{1} \\ Y_{2} \end{matrix})$ ， $T_{2 k}^{*} W = (\begin{matrix} W_{1} \\ W_{2} \end{matrix})$ ， (5)

其中 $X_{1}, X_{2} \in Q^{k \times n}$ ， $Z_{1}, Z_{2} \in Q^{k \times n}$ ， $Y_{1}, Y_{2} \in Q^{k \times s}$ ， $W_{1}, W_{2} \in Q^{k \times s}$ ，设分块矩阵 $(X_{1}, Y_{1})$ ， $(Z_{1}, W_{1})$ ， $(X_{2}, Y_{2})$ ， $(Z_{2}, W_{2})$ 的奇异值分解分别为

$(X_{1}, Y_{1}) = U (\begin{matrix} Σ_{1} & 0 \\ 0 & 0 \end{matrix}) V^{*}$ ， $(Z_{1}, W_{1}) = P (\begin{matrix} Σ_{2} & 0 \\ 0 & 0 \end{matrix}) Q^{*}$ ， (6)

$(X_{2}, Y_{2}) = R (\begin{matrix} Γ_{1} & 0 \\ 0 & 0 \end{matrix}) S^{*}$ ， $(Z_{2}, W_{2}) = E (\begin{matrix} Γ_{2} & 0 \\ 0 & 0 \end{matrix}) F^{*}$ ，(7)

其中

$\begin{array}{l} U = (\underset{r_{1}}{U_{1}}, \underset{k - r_{1}}{U_{2}}) \in U^{k \times k}, V = (\underset{r_{1}}{V_{1}}, \underset{n + s - r_{1}}{V_{2}}) \in U^{(n + s) \times (n + s)}, \\ R = (\underset{t_{1}}{R_{1}}, \underset{k - t_{1}}{R_{2}}) \in U^{k \times k}, S = (\underset{t_{1}}{S_{1}}, \underset{n + s - t_{1}}{S_{2}}) \in U^{(n + s) \times (n + s)}, \\ P = (\underset{r_{2}}{P_{1}}, \underset{k - r_{2}}{P_{2}}) \in U^{k \times k}, Q = (\underset{r_{2}}{Q_{1}}, \underset{n + s - r_{2}}{Q_{2}}) \in U^{(n + s) \times (n + s)}, \\ E = (\underset{t_{2}}{E_{1}}, \underset{k - t_{2}}{E_{1}}) \in U^{k \times k}, F = (\underset{t_{2}}{F_{1}}, \underset{n + s - t_{2}}{F_{2}}) \in U^{(n + s) \times (n + s)}, \end{array}$

${\begin{cases} Σ_{1} = diag (σ_{1}, \dots, σ_{r_{1}}) > 0, 令 φ_{i j} = \frac{1}{σ_{i}^{2} + σ_{j}^{2}}, 1 \leq i, j \leq r_{1}, Φ = (φ_{i j}) \in Q^{r_{1} \times r_{1}}, \\ Σ_{2} = diag (γ_{1}, \dots, γ_{r_{2}}) > 0, 令 ψ_{i j} = \frac{1}{γ_{i}^{2} + γ_{j}^{2}}, 1 \leq i, j \leq r_{2}, Ψ = (ψ_{i j}) \in Q^{r_{2} \times r_{2}}, \\ Γ_{1} = diag (η_{1}, \dots, η_{t_{1}}) > 0, Γ_{2} = diag (μ_{1}, \dots, μ_{t_{2}}) > 0, \end{cases}$ (8)

当 $A \in S C_{2 k} (Q)$ 时，由引理2可得

$\begin{array}{l} {‖ A X - Z ‖}^{2} + {‖ Y^{*} A - W^{*} ‖}^{2} \\ = {‖ T_{2 k} (\begin{matrix} M + H & 0 \\ 0 & M - H \end{matrix}) T_{2 k}^{*} X - Z ‖}^{2} + {‖ Y^{*} T_{2 k} (\begin{matrix} M + H & 0 \\ 0 & M - H \end{matrix}) T_{2 k}^{*} - W^{*} ‖}^{2} \\ = ‖ (\begin{matrix} M + H & 0 \\ 0 & M - H \end{matrix}) (\begin{matrix} X_{1} \\ X_{2} \end{matrix}) - (\begin{matrix} Z_{1} \\ Z_{2} \end{matrix}) ‖ + {‖ (\begin{matrix} M + H & 0 \\ 0 & M - H \end{matrix}) (\begin{matrix} Y_{1} \\ Y_{2} \end{matrix}) - (\begin{matrix} W_{1} \\ W_{2} \end{matrix}) ‖}^{2} \\ = {‖ (M + H) (X_{1}, Y_{1}) - (Z_{1}, W_{1}) ‖}^{2} + {‖ (M - H) (X_{2}, Y_{2}) - (Z_{2}, W_{2}) ‖}^{2} \end{array}$

所以 ${‖ A X - Z ‖}^{2} + {‖ Y^{*} A - W^{*} ‖}^{2} = \min$ 等价于

${‖ (M + H) (X_{1}, Y_{1}) - (Z_{1}, W_{1}) ‖}^{2} + {‖ (M - H) (X_{2}, Y_{2}) - (Z_{2}, W_{2}) ‖}^{2} = \min$ (9)

利用 $(X_{1}, Y_{1}), (Z_{1}, W_{1}), (X_{2}, Y_{2}), (Z_{2}, W_{2})$ 的奇异值分解(6)和(7)代入(9)可得

$\begin{array}{l} {‖ (M + H) (X_{1}, Y_{1}) - (Z_{1}, W_{1}) ‖}^{2} + {‖ (M - H) (X_{2}, Y_{2}) - (Z_{2}, W_{2}) ‖}^{2} \\ = {‖ U^{*} (M + H) U (\begin{matrix} Σ_{1} & 0 \\ 0 & 0 \end{matrix}) - U^{*} P (\begin{matrix} Σ_{2} & 0 \\ 0 & 0 \end{matrix}) Q^{*} V ‖}^{2} + {‖ R^{*} (M - H) R (\begin{matrix} Γ_{1} & 0 \\ 0 & 0 \end{matrix}) - R^{*} E (\begin{matrix} Γ_{2} & 0 \\ 0 & 0 \end{matrix}) F^{*} S ‖}^{2} \\ = {‖ (\begin{matrix} U_{1}^{*} \\ U_{2}^{*} \end{matrix}) (M + H) (\begin{matrix} U_{1} & U_{2} \end{matrix}) (\begin{matrix} Σ_{1} & 0 \\ 0 & 0 \end{matrix}) - (\begin{matrix} U_{1}^{*} \\ U_{2}^{*} \end{matrix}) (\begin{matrix} P_{1} & P_{2} \end{matrix}) (\begin{matrix} Σ_{2} & 0 \\ 0 & 0 \end{matrix}) (\begin{matrix} Q_{1}^{*} \\ Q_{2}^{*} \end{matrix}) (\begin{matrix} V_{1} & V_{2} \end{matrix}) ‖}^{2} \\ + {‖ (\begin{matrix} R_{1}^{*} \\ R_{2}^{*} \end{matrix}) (M - H) (\begin{matrix} R_{1} & R_{2} \end{matrix}) (\begin{matrix} Γ_{1} & 0 \\ 0 & 0 \end{matrix}) - (\begin{matrix} R_{1}^{*} \\ R_{2}^{*} \end{matrix}) (\begin{matrix} E_{1} & E_{2} \end{matrix}) (\begin{matrix} Γ_{2} & 0 \\ 0 & 0 \end{matrix}) (\begin{matrix} F_{1}^{*} \\ F_{2}^{*} \end{matrix}) (\begin{matrix} S_{1} & S_{2} \end{matrix}) ‖}^{2} \end{array}$

$\begin{array}{l} = {‖ U_{1}^{*} (M + H) U_{1} Σ_{1} - U_{1}^{*} P_{1} Σ_{2} Q_{1}^{*} V_{1} ‖}^{2} + {‖ - U_{1}^{*} P_{1} Σ_{2} Q_{1}^{*} V_{2} ‖}^{2} \\ + {‖ U_{2}^{*} (M + H) U_{1} Σ_{1} - U_{2}^{*} P_{1} Σ_{2} Q_{1}^{*} V_{1} ‖}^{2} + ‖ - U_{2}^{*} P_{1} Σ_{2} Q_{1}^{*} V_{2} ‖ \\ + {‖ R_{1}^{*} (M - H) R_{1} Γ_{1} - R_{1}^{*} E_{1} Γ_{2} F_{1}^{*} S_{1} ‖}^{2} + {‖ - R_{1}^{*} E_{1} Γ_{2} F_{1}^{*} S_{2} ‖}^{2} \\ + {‖ R_{2}^{*} (M - H) R_{1} Γ_{1} - R_{2}^{*} E_{1} Γ_{2} F_{1}^{*} S_{1} ‖}^{2} + {‖ - R_{2}^{*} E_{1} Γ_{2} F_{1}^{*} S_{2} ‖}^{2} \end{array}$ (10)

定理1 给定 $X, Z \in Q^{m \times n}$ ； $Y, W \in Q^{m \times s}$ ，则问题I的双自共轭反问题解可以表示为

$A = T_{2 k} (\begin{matrix} A_{11} & 0 \\ 0 & A_{22} \end{matrix}) T_{2 k}^{*} + T_{2 k} (\begin{matrix} U_{2} G_{1} U_{2}^{*} & 0 \\ 0 & P_{2} G_{2} P_{2}^{*} \end{matrix}) T_{2 k}^{*}$ (11)

其中

$A_{11} = U (\begin{matrix} Φ * (U_{1}^{*} P_{1} Σ_{2} Q_{1}^{*} V Σ_{1} + Σ_{1} {(U_{1}^{*} P_{1} Σ_{2} Q_{1}^{*} V)}^{*}) & Σ_{1}^{- 1} V_{1}^{*} Q_{1} Σ_{2} P_{1}^{*} U_{2} \\ U_{2}^{*} P_{1} Σ_{2} Q_{1}^{*} V_{1} Σ_{1}^{- 1} & 0 \end{matrix}) U^{*}$ ，(12)

$A_{22} = R (\begin{matrix} Ψ * (R_{1}^{*} E_{1} Γ_{2} F_{1}^{*} S_{1} Γ_{1} + Γ_{1} {(R_{1}^{*} E_{1} Γ_{2} F_{1}^{*} S_{1})}^{*}) & Γ_{1}^{- 1} S_{1}^{*} F_{1} Γ_{2} E_{1}^{*} R_{2} \\ R_{2}^{*} E_{1} Γ_{2} F_{1}^{*} S_{1} Γ_{1}^{- 1} & 0 \end{matrix}) R^{*}$ ，(13)

且 $G_{1} \in S C_{k - r_{1}} (Q), G_{2} \in S C_{k - r_{2}} (Q), U_{2} \in Q^{k \times (k - r_{1})}, R_{2} \in Q^{k \times (k - r_{2})}$ 为列正交矩阵。

证明由上面讨论知

${‖ A X - Z ‖}^{2} + {‖ Y^{*} A - W^{*} ‖}^{2} = \min$ 有解 $\Leftrightarrow$ (9)式有解 $\Leftrightarrow$ (10)式有解。

由引理2知 $M + H \in S C_{k} (Q)$ ，故由引理1知，(10)式可写为

$\begin{array}{l} {‖ U_{1}^{*} (M + H) U_{1} Σ_{1} - (Φ * (U_{1}^{*} P_{1} Σ_{2} Q_{1}^{*} V Σ_{1} + Σ_{1} {(U_{1}^{*} P_{1} Σ_{2} Q_{1}^{*} V)}^{*}) Σ_{1}) ‖}^{2} \\ + {‖ U_{1}^{*} P_{1} Σ_{2} Q_{1}^{*} V_{1} - (Φ * (U_{1}^{*} P_{1} Σ_{2} Q_{1}^{*} V Σ_{1} + Σ_{1} {(U_{1}^{*} P_{1} Σ_{2} Q_{1}^{*} V)}^{*}) Σ_{1}) ‖}^{2} \\ + {‖ - U_{1}^{*} P_{1} Σ_{2} Q_{1}^{*} V_{2} ‖}^{2} + {‖ U_{2}^{*} (M + H) U_{1} Σ_{1} - U_{2}^{*} P_{1} Σ_{2} Q_{1}^{*} V_{1} ‖}^{2} + ‖ - U_{2}^{*} P_{1} Σ_{2} Q_{1}^{*} V ‖ \\ + {‖ R_{1}^{*} (M - H) R_{1} Γ_{1} - (Ψ * (R_{1}^{*} E_{1} Γ_{2} F_{1}^{*} S_{1} Γ_{1} + Γ_{1} {(R_{1}^{*} E_{1} Γ_{2} F_{1}^{*} S_{1})}^{*}) Γ_{1}) ‖}^{2} \\ + {‖ R_{1}^{*} E_{1} Γ_{2} F_{1}^{*} S_{1} - (Ψ * (R_{1}^{*} E_{1} Γ_{2} F_{1}^{*} S_{1} Γ_{1} + Γ_{1} {(R_{1}^{*} E_{1} Γ_{2} F_{1}^{*} S_{1})}^{*}) Γ_{1}) ‖}^{2} \\ + {‖ - R_{1}^{*} E_{1} Γ_{2} F_{1}^{*} S_{2} ‖}^{2} + {‖ R_{2}^{*} (M - H) R_{1} Γ_{1} - R_{2}^{*} E_{1} Γ_{2} F_{1}^{*} S_{1} ‖}^{2} + {‖ - R_{2}^{*} E_{1} Γ_{2} F_{1}^{*} S_{2} ‖}^{2} \end{array}$

故 ${‖ A X - Z ‖}^{2} + {‖ Y^{*} A - W^{*} ‖}^{2} = \min$ 当且仅当

${\begin{cases} U_{1}^{*} (M + H) U_{1} = Φ * (U_{1}^{*} P_{1} Σ_{2} Q_{1}^{*} V_{1} Σ_{1} + Σ_{1} {(U_{1}^{*} P_{1} Σ_{2} Q_{1}^{*} V_{1})}^{*}) \\ U_{2}^{*} (M + H) U_{1} = U_{2}^{*} P_{1} Σ_{2} Q_{1}^{*} V_{1} Σ_{1}^{- 1} \\ R_{1}^{*} (M - H) R = Ψ * (R_{1}^{*} E_{1} Γ_{2} F_{1}^{*} S_{1} Γ_{1} + Γ_{1} {(R_{1}^{*} E_{1} Γ_{2} F_{1}^{*} S_{1})}^{*}) \\ R_{2}^{*} (M - H) R_{1} = R_{2}^{*} E_{1} Γ_{2} F_{1}^{*} S_{1} Γ_{1}^{- 1} \end{cases}$

又因为

$U^{*} (M + H) U = (\begin{matrix} U_{1}^{*} (M + H) U_{1} & U_{1}^{*} (M + H) U_{2} \\ U_{2}^{*} (M + H) U_{1} & U_{2}^{*} (M + H) U_{2} \end{matrix})$

$R^{*} (M - H) R = (\begin{matrix} R_{1}^{*} (M - H) R_{1} & R_{1}^{*} (M - H) R_{2} \\ R_{2}^{*} (M - H) R_{1} & R_{2}^{*} (M - H) R_{2} \end{matrix})$

令 $G_{1} = U_{2}^{*} (M + H) U_{2} \in S C_{k - r_{1}} (Q)$ ， $G_{2} = R_{2}^{*} (M + H) R_{2} \in S C_{k - r_{2}} (Q)$ ，于是有

$(M + H) = U (\begin{matrix} Φ * (U_{1}^{*} P_{1} Σ_{2} Q_{1}^{*} V_{1} Σ_{1} + Σ_{1} {(U_{1}^{*} P_{1} Σ_{2} Q_{1}^{*} V_{1})}^{*}) & Σ_{1}^{- 1} V_{1}^{*} Q_{1} Σ_{2} P_{1}^{*} U_{2} \\ U_{2}^{*} P_{1} Σ_{2} Q_{1}^{*} V_{1} Σ_{1}^{- 1} & G_{1} \end{matrix}) U^{*}$

$(M - H) = R (\begin{matrix} Ψ * (R_{1}^{*} E_{1} Γ_{2} F_{1}^{*} S_{1} Γ_{1} + Γ_{1} {(R_{1}^{*} E_{1} Γ_{2} F_{1}^{*} S_{1})}^{*}) & Γ_{1}^{- 1} S_{1}^{*} F_{1} Γ_{2} E_{1}^{*} R_{2} \\ R_{2}^{*} E_{1} Γ_{2} F_{1}^{*} S_{1} Γ_{1}^{- 1} & G_{2} \end{matrix}) R^{*}$

所以有

$A = T_{2 k} (\begin{matrix} A_{11} & 0 \\ 0 & A_{22} \end{matrix}) T_{2 k}^{*} + T_{2 k} (\begin{matrix} U_{2} G_{1} U_{2}^{*} & 0 \\ 0 & R_{2} G_{2} R_{2}^{*} \end{matrix}) T_{2 k}^{*}$

其中A₁₁，A₂₂如(12)，(13)所示。

对于 $n = 2 k + 1$ 的情形，当 $A \in B S C_{2 k + 1} (Q)$ 时，由引理2有

$A = T_{2 k + 1} (\begin{matrix} M + H & \sqrt{2} C & 0 \\ \sqrt{2} C^{*} & ρ & 0 \\ 0 & 0 & M - H \end{matrix}) T_{2 k + 1}^{*} = T_{2 k + 1} (\begin{matrix} Y & 0 \\ 0 & Z \end{matrix}) T_{2 k + 1}^{*}$

其余过程及记号完全类似 $n = 2 k$ 的情形，可证当 $n = 2 k + 1$ 时，定理1结论成立。

3. 问题Ⅱ的解

设 $\tilde{M} \in Q^{n \times n}$ 是给定的四元数矩阵，问题I的解集为 $S_{E}$ ，求矩阵 $\tilde{A} \in S_{E}$ 使得 $\min ‖ A - \tilde{M} ‖ = ‖ \tilde{A} - \tilde{M} ‖$ 。设 $T_{2 k}, T_{2 k + 1}$ 如(2)，当 $m = 2 k$ 时，记

$A_{0} = T_{2 k} (\begin{matrix} A_{11} & 0 \\ 0 & A_{22} \end{matrix}) T_{2 k}^{*}$ (14)

$\overset{⌢}{X} = T_{2 k}^{*} (A_{0} - \tilde{M}) T_{2 k} = (\begin{matrix} {\overset{⌢}{A}}_{11} & {\overset{⌢}{A}}_{12} \\ {\overset{⌢}{A}}_{21} & {\overset{⌢}{A}}_{22} \end{matrix})$ (15)

${\begin{cases} {\overset{⌢}{A}}_{11} = \frac{1}{2} (I_{k} \begin{matrix} \end{matrix} S_{k}) (A_{0} - \tilde{M}) (\begin{matrix} I_{k} \\ S_{k} \end{matrix}), {\overset{⌢}{A}}_{12} = \frac{1}{2} (I_{k} \begin{matrix} \end{matrix} S_{k}) (A_{0} - \tilde{M}) (\begin{matrix} I_{k} \\ - S_{k} \end{matrix}) \\ {\overset{⌢}{A}}_{21} = \frac{1}{2} (I_{k} \begin{matrix} \end{matrix} - S_{k}) (A_{0} - \tilde{M}) (\begin{matrix} I_{k} \\ S_{k} \end{matrix}), {\overset{⌢}{A}}_{22} = \frac{1}{2} (I_{k} \begin{matrix} \end{matrix} - S_{k}) (A_{0} - \tilde{M}) (\begin{matrix} I_{k} \\ - S_{k} \end{matrix}) \end{cases}$ (16)

当 $m = 2 k + 1$ 时，记

$\overset{⌢}{A} = T_{2 k + 1}^{*} (A_{0} - \tilde{M}) T_{2 k + 1} = (\begin{matrix} {\overset{⌢}{A}}_{11} & {\overset{⌢}{A}}_{12} \\ {\overset{⌢}{A}}_{21} & {\overset{⌢}{A}}_{22} \end{matrix})$ (17)

有

${\begin{cases} {\overset{⌢}{A}}_{11} = \frac{1}{2} (\begin{matrix} I_{k} & 0 & S_{k} \\ 0 & \sqrt{2} & 0 \end{matrix}) (A_{0} - \tilde{M}) (\begin{matrix} I_{k} & 0 \\ 0 & \sqrt{2} \\ S_{k} & 0 \end{matrix}), {\overset{⌢}{A}}_{12} = \frac{1}{2} (\begin{matrix} I_{k} & 0 & S_{k} \\ 0 & \sqrt{2} & 0 \end{matrix}) (A_{0} - \tilde{M}) (\begin{matrix} I_{k} \\ 0 \\ - S_{k} \end{matrix}) \\ {\overset{⌢}{A}}_{21} = \frac{1}{2} (\begin{matrix} I_{k} & 0 & - S_{k} \end{matrix}) (A_{0} - \tilde{M}) (\begin{matrix} I_{k} & 0 \\ 0 & \sqrt{2} \\ S_{k} & 0 \end{matrix}), {\overset{⌢}{A}}_{22} = \frac{1}{2} (\begin{matrix} I_{k} & 0 & - S_{k} \end{matrix}) (A_{0} - \tilde{M}) (\begin{matrix} I_{k} \\ - S_{k} \end{matrix}) \end{cases}$ (18)

于是，关于问题II有以下结果：

定理2 设 $S_{E}$ 是问题3-I的解集合，给定 $\tilde{M} \in Q^{n \times n}$ ，则问题II存在唯一最佳逼近解，且解可表示为

$\hat{A} = A_{0} + T (\begin{matrix} \tilde{U} \frac{{\overset{⌢}{A}}_{11} + {\overset{⌢}{A}}_{11}^{*}}{2} \tilde{U} & 0 \\ 0 & \tilde{P} \frac{{\overset{⌢}{A}}_{22} + {\overset{⌢}{A}}_{22}^{*}}{2} \tilde{P} \end{matrix}) T^{*}$ (19)

当 $m = 2 k$ 时， $A_{0}$ 如(14)式， $T = T_{2 k}, {\overset{⌢}{A}}_{11}, {\overset{⌢}{A}}_{12}, {\overset{⌢}{A}}_{21}, {\overset{⌢}{A}}_{22}$ 如(16)的形式；当 $m = 2 k + 1$ 时， $T = T_{2 k + 1}, {\overset{⌢}{A}}_{11}, {\overset{⌢}{A}}_{12}, {\overset{⌢}{A}}_{21}, {\overset{⌢}{A}}_{22}$ 如(18)的形式， $\tilde{U}, \tilde{R}$ 如(6)(7)式。

证明根据F范数性质及定理1的双自共轭最小二乘解表达式可得

$\begin{matrix} {‖ A - \tilde{M} ‖}^{2} = {‖ T (\begin{matrix} A_{11} & 0 \\ 0 & A_{22} \end{matrix}) T^{*} + T (\begin{matrix} U_{2} G_{1} U_{2}^{*} & 0 \\ 0 & P_{2} G_{2} P_{2}^{*} \end{matrix}) T^{*} - \tilde{M} ‖}^{2} \\ = {‖ T^{*} (A_{0} - \tilde{M}) T - (\begin{matrix} U_{2} G_{1} U_{2}^{*} & 0 \\ 0 & P_{2} G_{2} P_{2}^{*} \end{matrix}) ‖}^{2} = {‖ (\begin{matrix} {\overset{⌢}{A}}_{11} & {\overset{⌢}{A}}_{12} \\ {\overset{⌢}{A}}_{21} & {\overset{⌢}{A}}_{22} \end{matrix}) - (\begin{matrix} U_{2} G_{1} U_{2}^{*} & 0 \\ 0 & P_{2} G_{2} P_{2}^{*} \end{matrix}) ‖}^{2} \\ = {‖ {\overset{⌢}{A}}_{11} - U_{2} G_{1} U_{2}^{*} ‖}^{2} + {‖ {\overset{⌢}{A}}_{12} ‖}^{2} + {‖ {\overset{⌢}{A}}_{21} ‖}^{2} + {‖ {\overset{⌢}{A}}_{22} - P_{2} G_{2} P_{2}^{*} ‖}^{2} \end{matrix}$

因此 $‖ A - \tilde{M} ‖ = \min$ 等价于

$‖ {\overset{⌢}{A}}_{11} - U_{2} G_{1} U_{2}^{*} ‖ = \min$ (20)

$‖ {\overset{⌢}{A}}_{22} - P_{2} G_{2} P_{2} ‖ = \min$ (21)

又因为

$\begin{matrix} {‖ {\overset{⌢}{A}}_{11} - U_{2} G_{1} U_{2}^{*} ‖}^{2} = {‖ U^{*} {\overset{⌢}{A}}_{11} U - U^{*} U_{2} G_{1} U_{2}^{*} U ‖}^{2} \\ = {‖ (\begin{matrix} U_{1}^{*} {\overset{⌢}{A}}_{11} U_{1} & U_{1}^{*} {\overset{⌢}{A}}_{11} U_{2} \\ U_{2}^{*} {\overset{⌢}{A}}_{11} U_{1} & U_{2}^{*} {\overset{⌢}{A}}_{11} U_{2} - G_{1} \end{matrix}) ‖}^{2} \\ = {‖ U_{1}^{*} {\overset{⌢}{A}}_{11} U_{1} ‖}^{2} + {‖ U_{1}^{*} {\overset{⌢}{A}}_{11} U_{2} ‖}^{2} + ‖ U_{2}^{*} {\overset{⌢}{A}}_{11} U_{1} ‖ + {‖ U_{2}^{*} {\overset{⌢}{A}}_{11} U_{2} - G_{1} ‖}^{2} \end{matrix}$

所以 $‖ {\overset{⌢}{A}}_{11} - U_{2} G_{1} U_{2}^{*} ‖ = \min$ 等价于

$‖ U_{2}^{*} {\overset{⌢}{A}}_{11} U_{2} - G_{1} ‖ = \min$ (22)

上式成立当且仅当

$G_{1} = U_{2}^{*} \frac{{\overset{⌢}{A}}_{11} + {\overset{⌢}{A}}_{11}^{*}}{2} U_{2}$ 。

类似的(20)，(21)成立当且仅当 $G_{2} = R_{2}^{*} \frac{{\overset{⌢}{A}}_{22} + {\overset{⌢}{A}}_{22}^{*}}{2} R_{2}$ 。代入(11)，并且由奇异值分解式(6)有

$U_{2} U_{2}^{*} = I - (X_{1} \begin{matrix} \end{matrix} Y_{1}) {(X_{1} \begin{matrix} \end{matrix} Y_{1})}^{+} = \tilde{U}$ ， $P_{2} P_{2}^{*} = I - (X_{2} \begin{matrix} \end{matrix} Y_{2}) {(X_{2} \begin{matrix} \end{matrix} Y_{2})}^{+} = \tilde{P}$ ， (23)

因此得到

$\hat{A} = A_{0} + T (\begin{matrix} \tilde{U} \frac{{\overset{⌢}{X}}_{11} + {\overset{⌢}{X}}_{11}^{*}}{2} \tilde{U} & 0 \\ 0 & \tilde{P} \frac{{\overset{⌢}{X}}_{22} + {\overset{⌢}{X}}_{22}^{*}}{2} \tilde{P} \end{matrix}) T^{*}$ ，

其中A₀为(14)或(3.2.14)式中的形式。证毕。

4. 数值算例

算法步骤求问题I的解的步骤

1) 计算 $X_{1}, X_{2}, Y_{1}, Y_{2}, Z_{1}, Z_{2}, W_{1}, W_{2}$ ；

2) 对矩阵 $(X_{1}, Y_{1}), (Z_{1}, W_{1})$ 作奇异值分解；

3) 按(8)式构造 $Φ$ 和 $Ψ$ ；

4) 利用定理1求得问题I的解A。

算例当m = 4时，给定四元数矩阵

$X = (\begin{matrix} 1 & i + j \\ k & 1 + k \\ 2 i & 2 \\ - 1 + k & 1 + i - k \end{matrix})$

$Z = (\begin{matrix} - 1.2 + 0.5 i - 3 j + 2 k & 1 + 2.3 i - 1.7 j - 3 k \\ - 1 + 1.25 i - 1.25 j & 4.75 + 0.75 i + 0.75 j + 3.5 k \\ - 0.5 + 2 i - 0.5 j + 1.25 k & 2.75 + 2.5 j + 0.75 k \\ 4 - 0.75 j & 1 - 1.25 i + 1.25 j - 2 k \end{matrix})$

$Y = (\begin{matrix} i + j & - 1 - i + k & - 1 - k & - 2 i \\ 2 - i & 1 - 3 i + 2 k & - 1 + 2 i - k & - 2 j \end{matrix})$

$W = (\begin{matrix} 1.5 - 0.5 i - 0.2 j + 0.3 k & - 3.25 - 2 i - j - 0.25 k & - 2.75 - 1.25 i - j - 0.75 k & - 0.25 - 1.25 i + 1.25 j - 0.5 k \\ 1.6 + 2.7 i - 7 j + 2.5 k & 1.75 + i + j + 1.25 k & - 0.75 + 1.75 i - 2 j - 3.5 k & - 0.75 - 1.75 i + 1.5 j + 0.5 k \end{matrix})$

按步骤(1)计算 $X_{1}, X_{2}, Y_{1}, Y_{2}, Z_{1}, Z_{2}, W_{1}, W_{2}$ 可得

$X_{1} = (\begin{matrix} 0.7071 k & 0.7071 + 1.414 i + 0.7071 j - 0.7071 k \\ 1.414 i + 0.7071 k & 2.121 + 0.7071 k \end{matrix})$

$X_{2} = (\begin{matrix} 1.414 - 0.7071 k & - 0.7071 + 0.7071 j + 0.7071 k \\ - 1.414 i + 0.7071 k & - 0.7071 + 0.7071 k \end{matrix})$

$Y_{1} = (\begin{matrix} 1.414 i - 0.7071 j - 0.7071 k & 1.414 + 0.7071 i + 1.414 j \\ - 1.414 + 0.7071 i & 0.7071 i - 0.7071 j \end{matrix})$

$Y_{2} = (\begin{matrix} - 1.414 i - 0.7071 j - 0.7071 k & 1.414 + 0.7071 i - 1.414 j \\ 0.7071 i - 1.414 k & 1.414 + 3.535 i - 2.121 k \end{matrix})$

$Z_{1} = (\begin{matrix} 1.980 + 0.354 i - 2.652 j + 1.414 k & 1.414 + 0.743 i - 0.318 j - 3.536 k \\ - 1.061 + 2.298 i - 1.237 j + 0.884 k & 5.303 + 0.530 i + 2.298 j + 3.005 k \end{matrix})$

$W_{1} = (\begin{matrix} 0.884 + 1.237 i - 0.742 j + 0.141 k & 0.601 - 0.672 i + 3.889 j - 2.121 k \\ - 4.243 + 2.298 i + 1.414 j + 0.707 k & 0.707 - 1.944 j + 0.707 j + 1.591 k \end{matrix})$

$W_{2} = (\begin{matrix} 1.237 - 0.530 i + 1.025 j - 0.566 k & 1.662 - 3.146 i + 6.01 j - 1.414 k \\ - 0.354 + 0.53 i - 0.354 k & 1.768 + 0.53 i - 2.121 j - 3.359 k \end{matrix})$

$(X_{1}, Y_{1}), (Z_{1}, W_{1})$ 的奇异值分解分别为

$U = (\begin{matrix} 0.7290 & - 0.6845 \\ 0.5174 - 0.0862 i - 0.4312 j - 0.0862 k & 0.5511 - 0.0918 i - 0.4592 j - 0.0918 k \end{matrix})$

$V = (\begin{matrix} - 0.0469 - 0.2657 i - 0.0156 j - 0.0696 k & - 0.0722 - 0.4094 i - 0.0241 j + 0.2758 k & - 0.2520 + 0.3951 i - 0.1446 j - 0.0186 k & - 0.1247 + 0.3716 i + 0.1069 j - 0.5165 k \\ 0.3978 - 0.3892 i - 0.3509 j - 0.0085 k & 0.2299 + 0.1663 i - 0.1577 j - 0.3962 k & 0.1763 + 0.3154 i - 0.0028 j + 0.2892 k & - 0.1536 - 0.1104 i - 0.1951 j - 0.1294 k \\ - 0.2032 - 0.3267 i + 0.2728 j + 0.2415 k & - 0.3130 + 0.2626 i + 0.0373 j - 0.0109 k & - 0.5201 + 0.1459 i + 0.2101 j + 0.3436 k & - 0.0234 - 0.3050 i + 0.0095 j + 0.0562 k \\ 0.2642 - 0.1477 i - 0.2955 j + 0.1719 k & - 0.3589 + 0.1554 i + 0.3108 j + 0.2649 k & 0.1239 + 0.1689 i + 0.0985 j - 0.1934 k & - 0.3249 + 0.2138 i + 0.2307 j + 0.4199 k \end{matrix})$

$Σ_{1} = (\begin{matrix} 3.9021 & 0 & 0 & 0 \\ 0 & 2.6969 & 0 & 0 \end{matrix}), Σ_{2} = (\begin{matrix} 11.1053 & 0 & 0 & 0 \\ 0 & 3.5713 & 0 & 0 \end{matrix})$

$P = (\begin{matrix} 0.5923 & - 0.8057 \\ - 0.0148 + 0.3338 i - 0.0164 j + 0.7330 k & - 0.0109 + 0.2454 i - 0.0121 j + 0.5389 k \end{matrix})$

$Q = (\begin{matrix} 0.2363 + 0.0327 i + 0.2665 j - 0.1781 k & - 0.1480 + 0.1976 i - 0.3124 j + 0.0844 k & 0.1126 - 0.3084 i - 0.0197 j - 0.0957 k & 0.0776 - 0.0521 i + 0.4757 j - 0.5678 k \\ 0.2793 - 0.0356 i - 0.0431 j + 0.6125 k & 0.1469 + 0.1766 i - 0.2092 j + 0.1714 k & - 0.5211 + 0.2146 i + 0.1661 j - 0.1652 k & - 0.1355 - 0.0487 i + 0.1561 j - 0.0155 k \\ 0.1665 - 0.2848 i + 0.1782 j - 0.2407 k & 0.0733 - 0.2211 i + 0.1493 j - 0.5012 k & - 0.3224 - 0.1659 i + 0.2677 j - 0.5018 k & 0.1087 - 0.0474 i - 0.0149 j + 0.0674 k \\ 0.0766 + 0.0055 i - 0.3837 j + 0.1803 k & - 0.0337 - 0.2210 i + 0.4744 j - 0.3250 k & 0.0763 + 0.1620 i - 0.1434 j + 0.0519 k & - 0.2824 - 0.0678 i + 0.2507 j - 0.4809 k \end{matrix})$

按(8)式构造 $Φ$ 和 $Ψ$ 得

$Φ = (\begin{matrix} 0.0328 & 0.0444 \\ 0.0444 & 0.0687 \end{matrix}), Ψ = (\begin{matrix} 0.0041 & 0.0073 \\ 0.0073 & 0.0392 \end{matrix})$

问题I的解

$A = (\begin{matrix} - 0.2 & i + 0.5 j & - 0.5 j - k & 1 \\ - i - 0.5 j & 1.5 & 0.75 & 0.25 i + k \\ 0.5 j + k & 0.75 & 1.5 & i + 0.5 k \\ 1 & - 0.25 i - k & - i - 0.5 k & 0 \end{matrix})$

5. 全文内容和创新点的总结

本文利用双Hermite矩阵的结构特性及奇异值分解定理，将原问题转化为Hermite矩阵方程问题，得出该问题解的表达式。本文对相关定理的证明过程比较完整和合理，文章创新性较好和技术含量较高，同时通过数值分析进一步证实了本文的研究结论。

基金项目

贺州学院校级科研项目(2020ZC12)。

文章引用

王敏. 四元数体上双Hermite矩阵反问题的最小二乘解
Least-Squares Solution to the Dou-ble-Hermite Matrix Inverse Problem on the Quaternion Field[J]. 应用数学进展, 2022, 11(08): 5660-5668. https://doi.org/10.12677/AAM.2022.118597

参考文献

1. Cao, W.S. (2002) Hermite Solutions to Quaternion Matrix Equation AXB = D. Mathematical Theory and Applications, 65-67.

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3. Zhou, S., Yang, S.T. and Wang, W. (2011) Least-Squares Solutions of Matrix Equations (AX = B, XC = D) for Hermitian Reflexive (Anti-Hermitian Reflexive) Matrices and Its Approximation. Journal of Mathematics Research and Exposition, 31, 1108-1116.

4. Yuang, S.F., Wang, Q.W. and Yu, Y.B. (2017) On Hermitian Solutions of the Split Quaternion Matrix Equation AXB + CXD = E. Advances in Applied Clifford Alge-bras, 27, 3235-3252. https://doi.org/10.1007/s00006-017-0806-y

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