# Intelligent Identification of Trend Components in Singular Spectrum Analysis
## 5 Justification of EOSSA
### Justification of Algorithm 2  
**Proposition 1**. : _Let the time series have the form_  
\[s_{n}=\sum_{j=1}^{r}A_{j}\mu_{j}^{n}, \tag{3}\]  
\(\mathbf{S}\) _be its \(L\)-trajectory matrix, \(\min(L,K)>r\). Denote_  
\[\boldsymbol{\Phi}=[\Phi_{1}:\ldots:\Phi_{r}]= \begin{pmatrix}c_{1}&\ldots&1\\ c_{1}\mu_{1}&\ldots&c_{r}\mu_{r}\\ \vdots&\ddots&\vdots\\ c_{1}\mu_{1}^{L-1}&\ldots&c_{r}\mu_{r}^{L-1}\end{pmatrix} , \tag{4}\]\[\mathbf{\Psi}^{\mathrm{T}}=[\Psi_{1}:\ldots:\Psi_{r}]^{\mathrm{T}}= \begin{pmatrix}h_ {1}&h_{1}\mu_{1}&\ldots&h_{1}\mu_{1}^{K-1}\\ \vdots&\vdots&\ddots&\vdots\\ h_{r}&h_{r}\mu_{r}&\ldots&h_{r}\mu_{r}^{K-1}\end{pmatrix} . \tag{5}\]  
_Then \(\mathbf{S}=\mathbf{\Phi}\mathbf{\Psi}^{\mathrm{T}}\) and \(\mathsf{S}^{(j)}=\mathcal{T}^{-1}(\[PERSON]_{j}\Psi_{j}^{\mathrm{T}})\), where \(s_{n}^{(j)}=A_{j}\mu_{j}^{n}\), \(j=1,\ldots,r\)._  
Proof.: Note that the application of \(\mathcal{T}^{-1}\) is correct, since the matrix \(\Phi_{j}\Psi_{j}^{\mathrm{T}}\) is Hankel. The statement of the proposition is checked by straightforward calculations.  
**Remark 1**.: _Under conditions of Proposition 1, let \(G\subset\{1,\ldots,r\}\). Then for the time series \(\mathsf{S}_{(G)}\) with the terms \(s_{(G),n}=\sum_{j\in G}A_{j}\mu_{j}^{n}\) we obtain \(\mathsf{S}_{(G)}=\mathcal{T}^{-1}(\mathbf{\Phi}[G]\mathbf{\Psi}[G]^{\mathrm{ T}})\)._  
Proposition 2.: _Let the time series \(\mathsf{S}\) be given in (3). Then in Algorithm 2 with elementary clustering, \(\mathbf{\Phi}\) and \(\mathbf{\Psi}\) have the form (4) and (5) respectively._  
Proof.: It follows from [11; 16] that there exists \(\mathbf{M}\) such that \(\underline{\mathbf{P}}\mathbf{M}=\overline{\mathbf{P}}\) and its eigenvalues are \(\mu_{1},\ldots,\mu_{r}\). Therefore, the eigendecomposition of \(\mathbf{M}\) is \(\mathbf{M}=\mathbf{T}\operatorname{diag}(\mu_{1},\ldots,\mu_{r})\mathbf{T}^{-1}\). Set \(\mathbf{\Phi}=\mathbf{P}\mathbf{T}\). Then  
\[\underline{\mathbf{\Phi}}=\mathbf{P}\mathbf{T}\Rightarrow\underline{\mathbf{ \Phi}}\operatorname{diag}(\mu_{1},\ldots,\mu_{r})\mathbf{T}^{-1}=\overline{ \mathbf{P}}\Rightarrow\underline{\mathbf{\Phi}}\operatorname{diag}(\mu_{1},\ldots,\mu_{r})=\overline{\mathbf{\Phi}}.\]  
Since \(\mu_{i}\underline{\Phi}_{i}=\overline{\Phi_{i}}\), we obtain  
\[ \begin{pmatrix}\mu_{i}\phi_{1,i}\\ \mu_{i}\phi_{2,i}\\ \ldots\\ \mu_{i}\phi_{L-1,i}\end{pmatrix} = \begin{pmatrix}\phi_{2,i}\\ \phi_{3,i}\\ \ldots\\ \phi_{L,i}\end{pmatrix} \Rightarrow\phi_{1,i}=\frac{\phi_{2,i}}{\mu_{i}^{2}}= \cdots=\frac{\phi_{L,1}}{\mu_{i}^{L-1}}.\]  
Therefore, \(\Phi_{i}=c_{i}(\mu_{i}\mu_{i}^{2},\ldots,\mu_{i}^{L})^{\mathrm{T}}\) for some \(c_{i}\
eq 0\). Find the structure of the matrix \(\mathbf{\Psi}\). We have \(\mathbf{S}=\mathbf{\Phi}\mathbf{\Psi}^{\mathrm{T}}\). Therefore, the \(j\)th lagged vector of \(\mathsf{S}\) has the form \(S_{j}=\psi_{j,1}\Pi_{1}+\cdots+\psi_{j,r}\Pi_{r}\). Set \(\psi_{j,i}=\frac{A_{i}}{c_{i}}\mu_{i}^{j-1}\). Since \(\mathbf{\Psi}\) is uniquely defined by the equality \(\mathbf{S}=\mathbf{\Phi}\mathbf{\Psi}^{\mathrm{T}}\), \(\overline{\mathbf{v}}_{i}\) is proportional to \((\mu_{i}\mu_{i}^{2},\ldots,\mu_{i}^{K})^{\mathrm{T}}\). **Remark 2**. : _Note that \(\Phi_{j}\), \(j=1,\ldots,r\), constitute the basis of the column space of \(\mathbf{S}\), and \(\Phi_{i}\), \(i\in G\), gives the basis of the column space of \(\mathbf{S}_{(G)}\), where \(\mathbf{S}_{(G)}\) is the L-trajectory matrix of the time series \(\mathsf{S}_{(G)}\). The same is true for \(\Psi_{j}\) and the row spaces. Let \(\mathbf{C}\) be a nonsingular matrix \(g\times g\), where \(g=|G|\). Then multiplication by \(\mathbf{C}\) or by its inverse can be considered as a transition to another basis.