## 3 On a Modification of the Splitting Iterative Scheme and Determining \(\eta_{k}\)  
Inspired by Equation (3), the iterative scheme in Equation (13) is orthogonal if its descent and residual vectors satisfy  
\[\mathbf{A}\mathbf{u}_{k}\cdot\mathbf{r}_{k+1}=0.\]Usually this is not true. For the original splitting iterative scheme in Equation (13), we have  
\[\mathbf{r}_{k+1}=\mathbf{r}_{k}-\mathbf{A}\mathbf{u}_{k}, \tag{25}\]  
which results to  
\[\mathbf{r}_{k+1}^{\mathrm{T}}\mathbf{A}\mathbf{u}_{k}=\mathbf{r}_{k}^{\mathrm{ T}}\mathbf{A}\mathbf{u}_{k}-\|\mathbf{A}\mathbf{u}_{k}\|^{2}\
eq 0. \tag{26}\]  
We define a quantity \(\eta_{k}\) to measure the degree of the non-orthogonality of the iterative scheme:  
\[\eta_{k}=\frac{\mathbf{r}_{k}^{\mathrm{T}}\mathbf{A}\mathbf{u}_{k}}{\|\mathbf{ A}\mathbf{u}_{k}\|^{2}}. \tag{27}\]  
Now the iterative scheme in Equation (13) is orthogonal if and only if \(\eta_{k}=1\). Rather than the constant value of \(\alpha\) in Equation (14), in this section we derive a more powerful version of the splitting iterative scheme (5) with a stepwise varying factor \(\eta_{k}\) in Equation (27). Theorem 2: _If a factor \(\eta_{k}\) is inserted preceding \(\mathbf{u}_{k}\) in the first one of Equation (13), then a new splitting iteration method for solving Equation (1) reads as_  
\[\mathbf{x}_{k+1}=\mathbf{x}_{k}+\eta_{k}\mathbf{u}_{k},\ \ \mathbf{M}\mathbf{u}_{k}= \mathbf{r}_{k}. \tag{28}\]  
_The optimal value for the orthogonalization factor \(\eta_{k}\) at each step is given by Equation (27)._  
Proof: Multiplying the first one in Equation (28) by \(\mathbf{A}\) and using Equation (11) yields  
\[\mathbf{r}_{k+1}=\mathbf{r}_{k}-\eta_{k}\mathbf{A}\mathbf{u}_{k};\]  
the squared norm is  
\[\|\mathbf{r}_{k+1}\|^{2}=\|\mathbf{r}_{k}\|^{2}-2\eta_{k}\mathbf{r}_{k}^{ \mathrm{T}}\mathbf{A}\mathbf{u}_{k}+\eta_{k}^{2}\|\mathbf{A}\mathbf{u}_{k}\|^ {2}.\]  
Let  
\[g=2\eta_{k}\mathbf{r}_{k}^{\mathrm{T}}\mathbf{A}\mathbf{u}_{k}-\eta_{k}^{2}\| \mathbf{A}\mathbf{u}_{k}\|^{2}; \tag{29}\]  
we have  
\[\|\mathbf{r}_{k+1}\|^{2}=\|\mathbf{r}_{k}\|^{2}-g. \tag{30}\]  
The following maximization problem is established to maximally decrease the residual vector norm at each step by  
\[\max_{\eta_{k}}\{g=2\eta_{k}\mathbf{r}_{k}^{\mathrm{T}}\mathbf{A}\mathbf{u}_{ k}-\eta_{k}^{2}\|\mathbf{A}\mathbf{u}_{k}\|^{2}\}. \tag{31}\]  
Using \(dg/d\eta_{k}=0\) for the maximality of \(g\) in Equation (31), we can derive Equation (27). Remark 2: _In contrast to the \(\alpha\)-extrapolation in Equation (14), Equation (28) is an \(\eta_{k}\)-extrapolation of Equation (13). When \(\alpha\) is a constant in Equation (14), \(\eta_{k}\) is a varying factor per step._  
Theorem 3: _The splitting iterative scheme in Equation (28) is absolute convergence, i.e.,_  
\[\mathbf{r}_{k+1}=\mathbf{r}_{k}-\mathbf{A}\mathbf{u}_{k}, \tag{25}\]0Proof. : Inserting Equation (27) for the optimal value of \(\eta_{k}\) into Equation (29) yields  
\[\mathbf{r}_{k+1}=\mathbf{r}_{k}-\mathbf{A}\mathbf{u}_{k}, \tag{25}\]1  
It follows from Equation (30) that  
\[\mathbf{r}_{k+1}=\mathbf{r}_{k}-\mathbf{A}\mathbf{u}_{k}, \tag{25}\]2  
due to \((\mathbf{r}_{k}^{\mathrm{T}}\mathbf{A}\mathbf{u}_{k})^{2}/\|\mathbf{A}\mathbf{ u}_{k}\|^{2}>0\). Hence, we proved Equation (32). In the splitting iterative scheme in Equations (28) and (27), we have  
\[\mathbf{r}_{k+1}=\mathbf{r}_{k}-\mathbf{A}\mathbf{u}_{k}, \tag{25}\]3  
which means that the consecutive residual vector \(\mathbf{r}_{k+1}\) is perpendicular to \(\mathbf{A}\mathbf{u}_{k}\). Proof. : By means of Equations (28) and (27), we have  
\[\mathbf{r}_{k+1}=\mathbf{r}_{k}-\mathbf{A}\mathbf{u}_{k}, \tag{25}\]4  
Multiplying Equation (34) by \(\mathbf{A}\) and using Equation (11) yields  
\[\mathbf{r}_{k+1}=\mathbf{r}_{k}-\mathbf{A}\mathbf{u}_{k}, \tag{25}\]5  
which being taken the inner product to \(\mathbf{A}\mathbf{u}_{k}\) renders  
\[\mathbf{r}_{k+1}=\mathbf{r}_{k}-\mathbf{A}\mathbf{u}_{k}, \tag{25}\]6  
The proof of Equation (33) for Theorem 4 is complete.