算法3.1 ：BiCR算法求解矩阵方程组(1.1)的perhermitian解

1. 设 $A_{i j} \in ℂ^{m \times n}, B_{i j} \in ℂ^{n \times l}, C_{i j} \in ℂ^{m \times l}$ ，适当维度的自反矩阵 $S \in ℂ^{n \times n}$ ， ${\tilde{U}}_{j} (0) = W_{j} (0) = 0 \in ℙ ℂ^{n \times n}, j = 1, 2, \dots, q$ ，

${\tilde{V}}_{i} (0) = T_{i} (0) = 0 \in ℂ^{m \times l}, i = 1, 2, \dots, p$ ，

$σ (0) = τ (0) = 1$ .

任意给定初始值 $X_{j} (1) \in ℙ ℂ^{n \times n}$ ， $Z_{j} (1) \in ℙ ℂ^{n \times n}$ ， $ε > 0$ 。

计算以下式子：

$R_{i} (1) = C_{i} - \sum_{j = 1}^{q} A_{i j} X_{j} (1) B_{i j}, i = 1, 2, \dots, p,$

$U_{j} (1) = Z_{j} (1), V_{i} (1) = R_{i} (1),$

${\tilde{U}}_{j} (1) = \frac{U_{j} (1)}{\sum_{j = 1}^{q} ‖ U_{j} (1) ‖}, {\tilde{V}}_{i} (1) = \frac{V_{i} (1)}{\sum_{i = 1}^{p} ‖ V_{i} (1) ‖},$

$T_{i} (1) = \sum_{j = 1}^{q} A_{i j} {\tilde{U}}_{j} (1) B_{i j}, W_{j} (1) = \frac{1}{2} \sum_{i = 1}^{p} (A_{i j}^{H} {\tilde{V}}_{i} (1) B_{i j}^{H} + S {(A_{i j}^{H} {\tilde{V}}_{i} (1) B_{i j}^{H})}^{H} S),$

$σ (1) = \sum_{i = 1}^{p} {‖ T_{i} (1) ‖}^{2}, τ (1) = \sum_{j = 1}^{q} {‖ W_{j} (1) ‖}^{2}, r (1) = \sqrt{\sum_{i = 1}^{p} {‖ R_{i} (1) ‖}^{2}},$

$k = 1.$

2. 如果 $r (k) < ε$ ，则停止；否则，转步骤3。

3. 计算以下式子：

$α (k) = \frac{\sum_{i = 1}^{p} Re (t r (T_{i}^{H} (k) R_{i} (k)))}{σ (k)},$

$X_{j} (k + 1) = X_{j} (k) - α (k) {\tilde{U}}_{j} (k),$

$R_{i} (k + 1) = R_{i} (k) - α (k) T_{i} (k),$

$M_{i} (k) = \sum_{j = 1}^{q} A_{i j} W_{j} (k) B_{i j},$

$\begin{matrix} U_{j} (k + 1) = W_{j} (k) - \frac{\sum_{i = 1}^{p} Re (t r (T_{i}^{H} (k) M_{i} (k)))}{σ (k)} {\tilde{U}}_{j} (k) \\ - \frac{\sum_{i = 1}^{p} Re (t r (T_{i}^{H} (k - 1) M_{i} (k)))}{σ (k - 1)} {\tilde{U}}_{j} (k - 1), \end{matrix}$

$N_{j} (k) = \sum_{i = 1}^{p} A_{i j}^{H} T_{i} (k) B_{i j}^{H},$

$\begin{matrix} V_{i} (k + 1) = T_{i} (k) - \frac{\sum_{i = 1}^{p} Re (t r (T_{i}^{H} (k) M_{i} (k)))}{τ (k)} {\tilde{V}}_{i} (k) \\ - \frac{\sum_{j = 1}^{q} Re (t r (W_{j}^{H} (k - 1) N_{j} (k)))}{τ (k - 1)} {\tilde{V}}_{i} (k - 1), \end{matrix}$

${\tilde{U}}_{j} (k + 1) = \frac{U_{j} (k + 1)}{\sum_{j = 1}^{q} ‖ U_{j} (k + 1) ‖}, {\tilde{V}}_{i} (k + 1) = \frac{V_{i} (k + 1)}{\sum_{i = 1}^{p} ‖ V_{i} (k + 1) ‖},$

$T_{i} (k + 1) = \sum_{j = 1}^{q} A_{i j} {\tilde{U}}_{j} (k + 1) B_{i j},$

$W_{j} (k + 1) = \frac{1}{2} \sum_{i = 1}^{p} (A_{i j}^{H} {\tilde{V}}_{i} (k + 1) B_{i j}^{H} + S {(A_{i j}^{H} {\tilde{V}}_{i} (k + 1) B_{i j}^{H})}^{H} S),$

$σ (k + 1) = \sum_{i = 1}^{p} {‖ T_{i} (k + 1) ‖}^{2},$

$τ (k + 1) = \sum_{j = 1}^{q} {‖ W_{j} (k + 1) ‖}^{2},$

$r (k + 1) = \sqrt{\sum_{i = 1}^{p} {‖ R_{i} (k + 1) ‖}^{2}} .$

4. $k = k + 1$ ，返回步骤2。