## 2 Analysis of Two-Step Method
### Convergence Conditions  
The following theorem on the convergence of an iterative method, as presented by (6), can be formulated  
**Theorem 1**. : _For \(h>0\), \(\beta_{1}\in[0,1)\) and \(\beta_{2}\geq 0\), the following inequality takes place:_  
\[h<\frac{2(1+\beta_{1})}{(1+2\beta_{2})L}. \tag{7}\]  
_Then, method (6) converges to \(x^{*}\) for any \(x^{0}\)._  
**Proof of Theorem 1. ** (1) Let the new variable \(z^{k}=(x^{k}-x^{*},x^{k-1}-x^{*})^{T}\) be introduced. Then, method (6) can be rewritten as a single-step method:  
\[z^{k+1}=Tz^{k},\]  
where matrix \(T\) is written as  
\[T= \begin{pmatrix}(1+\beta_{1})E-h(1+\beta_{2})A&h\beta_{2}A-\beta_{1}\ E\\ E&0_{d\times d}\end{pmatrix} .\]  
This method converges if, and only if, \(r(T)\) (spectral radius of matrix \(T\)) is strictly less than unity [3]. Matrix \(A\) can be represented by the spectral decomposition \(A=S\Lambda S^{T}\), where \(\Lambda\) is the diagonal matrix of eigenvalues of \(A\), \(S\) is a matrix of eigenvectors, and \(SS^{T}=S^{T}S=E\). The following transformation of \(T\) can be introduced: \(\overline{T}=\Sigma^{T}T\Sigma\), where  
\[\Sigma= \begin{pmatrix}S&0_{d\times d}\\ 0_{d\times d}&S\end{pmatrix} ,\quad\overline{T}= \begin{pmatrix}(1+\beta_{1})E- h(1+\beta_{2})\Lambda&h\beta_{2}\Lambda-\beta_{1}\ E\\ E&0_{d\times d}\end{pmatrix} .\]  
Matrix \(\overline{T}\) has the same eigenvalues, as matrix \(T\). Let us demonstrate that \(\overline{T}\) has the same spectrum as the following matrix:  
\[\bar{T}= \begin{pmatrix}T_{1}&0_{2\times 2}&\ldots&0_{2\times 2}\\ 0_{2\times 2}&T_{2}&\ldots&0_{2\times 2}\\ \ldots&\ldots&\ldots&\ldots\\ 0_{2\times 2}&0_{2\times 2}&\ldots&T_{d}\end{pmatrix} .\]  
where \(T_{i}\) are \(2\times 2\) matrices, which are presented as  
\[T_{i}= \begin{pmatrix}1+\beta_{1}-h(1+\beta_{2})\lambda_{i}&h\beta_{2}\lambda_ {i}-\beta_{1}\\ 1&0\end{pmatrix} .\]  
Matrix \(\overline{T}-\zeta E\) is presented as  
\[\overline{T}-\zeta E= \begin{pmatrix}T_{11}&T_{12}\\ T_{21}&T_{22}\end{pmatrix} ,\]where \(T_{11}=(1+\beta_{1})E-h(1+\beta_{2})\Lambda-\zeta E\), \(T_{12}=h\beta_{2}\Lambda-\beta_{1}E\), \(T_{21}=E\), \(T_{22}=-\zeta E\). The determinant of this matrix is computed by the following rule [30]:  
\[\det(\bar{T}-\zeta E)=\det(T_{11})\det(T_{22}-T_{21}T_{11}^{-1}T_{22})=\]  
\[\det(T_{11})\det \begin{pmatrix}-\zeta+\frac{\beta_{1}-h\beta_{2}\lambda_{1}}{1+ \beta_{1}-h(1+\beta_{2})\lambda_{1}-\zeta}&0&\ldots&0\\ 0&-\zeta+\frac{\beta_{1}-h\beta_{2}\lambda_{2}}{1+\beta_{1}-h(1+\beta_{2}) \lambda_{2}-\zeta}&\ldots&0\\ \ldots&\ldots&\ldots&\ldots\\ 0&0&\ldots&-\zeta+\frac{\beta_{1}-h\beta_{2}\lambda_{d}}{1+\beta_{1}-h(1+ \beta_{2})\lambda_{d}-\zeta}\end{pmatrix} =\]  
\[(\beta_{1}-h\beta_{2}\lambda_{1}-\zeta\chi_{1})\ldots(\beta_{1}-h\beta_{2} \lambda_{d}-\zeta\chi_{d}),\]  
where \(\chi_{i}=1+\beta_{1}-h(1+\beta_{2})\lambda_{i}-\zeta\), \(i=\overline{1,d}\). The determinant of the block-diagonal matrix \(\bar{T}-\zeta E\) is written as  
\[\det(\bar{T}-\zeta E)=\det(T_{1}-\zeta E_{2\times 2})\det(T_{2}-\zeta E_{2 \times 2})\ldots\det(T_{d}-\zeta E_{2\times 2}),\]  
and, as can be seen, it is equal to \(\det(\bar{T}-\zeta E)\). So, both matrices have the same eigenvalues \(\zeta_{k}\), \(k=\overline{1,2d}\) and these eigenvalues are computed as eigenvalues of matrices \(T_{i}\). 2. According to the result presented above, the analysis of eigenvalues of \(T\) leads to the analysis of roots of an algebraic equation: \[\zeta^{2}-(1+\beta_{1}-h(1+\beta_{2})\lambda)\zeta+\beta_{1}-h\beta_{2} \lambda=0.\] (8) In order to guarantee convergence, parameters should be chosen in a way which guarantees that \(|\zeta_{1,2}|<1\).