## Appendix A
\begin{split}&\mathbf{x}^{T}[(1-\omega)\mathbf{D}\mathbf{U}+(1+\lambda \omega-2\omega+\omega^{2})\mathbf{D}\mathbf{L}+(\omega-\omega^{2})\mathbf{U} \mathbf{L}+(\lambda\omega)\mathbf{L}\mathbf{D}+(\omega)\mathbf{U}^{2}\\ &\quad+(-\lambda)\mathbf{D}^{2}+(-\lambda\omega^{2})\mathbf{L}^{ 2}]\mathbf{x}=0\\ &\Rightarrow(\mathbf{x}^{T}[(1-\omega)\mathbf{D}\mathbf{U}+(1+ \lambda\omega-2\omega+\omega^{2})\mathbf{D}\mathbf{L}+(\omega-\omega^{2}) \mathbf{U}\mathbf{L}\ +(\lambda\omega)\mathbf{L}\mathbf{D}+(\omega)\mathbf{U}^{2}\\ &\quad+(-\lambda)\mathbf{D}^{2}+(-\lambda\omega^{2})\mathbf{L}^{ 2}]\mathbf{x}^{T}=0\\ &\Rightarrow\mathbf{x}^{T}[(1-\omega)\mathbf{D}\mathbf{U}+(1+ \lambda\omega-2\omega+\omega^{2})\mathbf{D}\mathbf{L}+(\omega-\omega^{2}) \mathbf{U}\mathbf{L}\ +(\lambda\omega)\mathbf{L}\mathbf{D}+(\omega)\mathbf{U}^{2}\\ &\quad+(-\lambda)\mathbf{D}^{2}+(-\lambda\omega^{2})\mathbf{L}^{ 2}]^{T}(\mathbf{x}^{T})^{T}=0\\ &\Rightarrow\mathbf{x}^{T}\big{\{}(1-\omega)\mathbf{D}\mathbf{U} \big{\}}^{T}+[(1+\lambda\omega-2\omega+\omega^{2})\mathbf{D}\mathbf{L}]^{T}+[( \omega-\omega^{2})\mathbf{U}\mathbf{L}]^{T}+[(\lambda\omega)\mathbf{L}\mathbf{ D}]^{T}+[(\omega)\mathbf{U}^{2}]^{T}\\ &\quad+[(-\lambda)\mathbf{D}^{2}]^{T}+[(-\lambda\omega^{2}) \mathbf{L}^{2}]^{T}\big{\}}\mathbf{x}=0.\end{split}  \tag{10}\]  
Therefore, we can obtain  
\[ \begin{split}&\mathbf{x}^{T}[(1-\omega)\mathbf{D}\mathbf{L}+(1+ \lambda\omega-2\omega+\omega^{2})\mathbf{D}\mathbf{U}+(\omega-\omega^{2}) \mathbf{L}\mathbf{U}+(\lambda\omega)\mathbf{U}\mathbf{D}+(\omega)\mathbf{L}^ {2}\\ &\quad+(-\lambda)\mathbf{D}^{2}+(-\lambda\omega^{2})\mathbf{U}^{2} ]\mathbf{x}=0.\end{split}  \tag{11}\]  
Next, we combine Equations (10) and (11) to produce  
\[ \begin{split}&\mathbf{x}^{T}[(1-\omega)\mathbf{D}(\mathbf{L}+ \mathbf{U})+(1+\lambda\omega-2\omega+\omega^{2})\mathbf{D}(\mathbf{L}+ \mathbf{U})+(\omega-\omega^{2})(\mathbf{U}\mathbf{L}+\mathbf{L}\mathbf{U})\\ &+(\lambda\omega)(\mathbf{L}+\mathbf{U})\mathbf{D}+(\omega)( \mathbf{U}^{2}+\mathbf{L}^{2})+(-\lambda)(\mathbf{D}^{2}+\mathbf{D}^{2})+(- \lambda\omega^{2})(\mathbf{L}^{2}+\mathbf{U}^{2})]\mathbf{x}=0.\end{split}  \tag{12}\]  
After reducing Equation (12), we can obtain Equation (11) as follows:  
\[ \begin{split}&\mathbf{x}^{T}[(2+\lambda\omega-3\omega+\omega^{2}) \mathbf{D}(\mathbf{L}+\mathbf{U})+(\omega-\omega^{2})(\mathbf{U}\mathbf{L}+ \mathbf{L}\mathbf{U})+(\lambda\omega)(\mathbf{L}+\mathbf{U})\mathbf{D}\\ &+(\omega-\lambda\omega^{2})(\mathbf{U}^{2}+\mathbf{L}^{2})+(-2 \lambda)\mathbf{D}^{2}]\mathbf{x}=0.\end{split}  \tag{13}\]  
Moreover, because \(\mathbf{W}=\mathbf{D}-\mathbf{L}-\mathbf{U}\), we can rewrite this to \(\mathbf{L}+\mathbf{U}=\mathbf{D}-\mathbf{W}\) and substitute it into Equation (13):  
\[ \begin{split}&\mathbf{x}^{T}[(2+\lambda\omega-3\omega+\omega^{2}) \mathbf{D}(\mathbf{D}-\mathbf{W})+(\omega-\omega^{2})(\mathbf{U}\mathbf{L}+ \mathbf{L}\mathbf{U})+(\lambda\omega)(\mathbf{D}-\mathbf{W})\mathbf{D}\\ &+(\omega-\lambda\omega^{2})(\mathbf{U}^{2}+\mathbf{L}^{2})+(-2 \lambda)\mathbf{D}^{2}]\mathbf{x}=0.\end{split}  \tag{14}\]  
For a clearer analysis, we can expand Equation (14) and rearrange it as follows:  
\[ \begin{split}&\mathbf{x}^{T}[(2+2\lambda\omega-3\omega+\omega^{2}- 2\lambda)\mathbf{D}^{2}+(-2-2\lambda\omega+3\omega-\omega^{2})\mathbf{D} \mathbf{W}\\ &+(\omega-\omega^{2})(\mathbf{U}\mathbf{L}+\mathbf{L}\mathbf{U}) +(\omega-\lambda\omega^{2})(\mathbf{U}^{2}+\mathbf{L}^{2})]\mathbf{x}=0. \end{split}  \tag{15}\]  
Because both \(\mathbf{D}\) and \(\mathbf{W}\) are symmetric positive definite matrices, \(\mathbf{x}^{T}\mathbf{D}^{2}\mathbf{x}\) and \(\mathbf{x}^{T}\mathbf{W}\mathbf{D}\mathbf{x}\) should be greater than zero for all non-zero \(\mathbf{x}\). Moreover, to ensure that \(\mathbf{x}^{T}\mathbf{W}\mathbf{D}\mathbf{x}>0\) always holds, \(\mathbf{W}\mathbf{D}\) must also be symmetric. Therefore, we can infer Equation (15) to the following inequalities:  
\[ \begin{split}& 2+2\lambda\omega-3\omega+\omega^{2}-2\lambda>0\\ &\lambda<\frac{\omega^{2}-3\omega+2}{2-2\omega}\end{split}