# Mixed Alternating Projections with Application to Hankel Low-Rank Approximation
## 4 Linearized Cadzow Iterations in the Signal-Estimation Problem
### Intersection of Subspaces \(\mathcal{H}\cap\mathcal{N}(\mathsf{S}_{0})\)  
For a compact representation of the further results, let us describe the intersection of the tangent subspace to the set of matrices of finite rank \(\mathcal{M}_{r}\) and the space of Hankel matrices \(\mathcal{H}\). Let us gather necessary definitions from [12]. Definition 2: We say that a series \(\mathsf{S}_{0}\in\mathbb{R}^{N}\) is governed by a generalized linear recurrence relation (GLRR) with coefficients \(\mathbf{a}\), where \(\mathbf{a}\in\mathbb{R}^{r+1}\) and \(\mathbf{a}\
eq\mathbf{0}_{r+1}\), if \(\mathbf{a}^{\mathsf{T}}\mathcal{T}_{r+1}(\mathsf{S}_{0})=\mathbf{0}_{N-r}^{ \mathsf{T}}\). We denote by \(\mathcal{Z}_{N}(\mathbf{a})=\mathcal{Z}(\mathbf{a})=\{\mathsf{S}\in\mathbb{R}^ {N}:\mathbf{a}^{\mathsf{T}}\mathcal{T}_{r+1}(\mathsf{S})=\mathbf{0}_{N-r}^{ \mathsf{T}}\}\) the subspace of time series governed by the GLRR(\(\mathbf{a}\)). Consider an equivalent form of \(\mathcal{Z}(\mathbf{a})\). Denote by \(\mathbf{Q}^{M,d}\) the operator \(\mathbb{R}^{d+1}\to\mathbb{R}^{M\times(M-d)}\) defined as follows:  
\[\mathbf{Q}^{M,d}(\mathbf{b})=\left( \begin{array}{cccc}b_{1}&0&\dots&0\\ b_{2}&b_{1}&0&\vdots\\ \vdots&b_{2}&\ddots&0\\ b_{d+1}&\vdots&\ddots&b_{1}\\ 0&b_{d+1}&\ddots&b_{2}\\ \vdots&0&\ddots&\vdots\\ 0&\dots&0&b_{d+1}\end{array} \right), \tag{15}\]  
where \(\mathbf{b}=(b_{1},\dots,b_{d+1})^{\mathsf{T}}\in\mathbb{R}^{d+1}\). Then, \(\mathcal{Z}(\mathbf{a})\) can be represented in another form as  
\[\mathcal{Z}(\mathbf{a})=\{\mathsf{S}\in\mathbb{R}^{N}:\mathbf{Q}^{\mathsf{T}} (\mathbf{a})\mathsf{S}=0_{N-r}\}, \tag{16}\]  
where \(\mathbf{Q}=\mathbf{Q}^{N,r}\)[20]. Remark 3: _Since the columns and rows of the trajectory matrix of a time series in \(\mathcal{D}_{r}\) satisfy the same GLRR(\(\mathbf{a}\)) as the original series does, we can take \(\mathbf{Q}^{L,r}(\mathbf{a})\) and \(\mathbf{Q}^{K,r}(\mathbf{a})\) as \(\mathbf{U}\) and \(\mathbf{V}\) in (12), respectively._  
Definition 3: Let \(\mathbf{a}=(a_{1},\dots,a_{k})^{\mathsf{T}}\in\mathbb{R}^{k}\). Denote by \(\mathbf{a}^{2}\in\mathbb{R}^{2k-1}=(a_{1}^{(2)},\dots,a_{2k-1}^{(2)})^{\mathsf{ T}}\) the acyclic convolution of the vector with itself:  
\[a_{i}^{(2)}=\sum_{j=\max(1,j-k+1)}^{\min(i,k)}a_{j}a_{i-j+1}. \tag{17}\]  
As shown in [12], the tangent subspace to \(\mathcal{D}_{r}\) at \(\mathsf{S}_{0}\) coincides with \(\mathcal{Z}(\mathbf{a}_{0}^{2})\), \(\dim\mathcal{Z}(\mathbf{a}_{0}^{2})=2r\). The following theorem establishes the relation between the tangent subspaces to the manifold of time series governed by GLRRs and the manifold of their trajectory matrices. **Theorem 3**. : _Let \(\mathbf{S}_{0}\in\mathcal{D}_{r}\), i.e., let \(\mathbf{S}_{0}\) be a series of rank \(r\), \(r<\min(L,K)\), and \(\mathbf{S}_{0}\in\mathcal{Z}(\mathbf{a}_{0})\)--i.e., let \(\mathbf{S}_{0}\) be governed by the \(\text{GLRR}(\mathbf{a}_{0})\) for a vector \(\mathbf{a}_{0}\in\mathbb{R}^{r+1}\). Then, \(\mathcal{T}_{L}^{-1}(\mathcal{H}\cap\mathcal{N}(\mathbf{S}_{0}))=\mathcal{Z} (\mathbf{a}_{0}^{2})\), where \(\mathbf{S}_{0}=\mathcal{T}_{L}(\mathbf{S}_{0})\)._  
Proof. : Consider a matrix \(\mathbf{X}\) in the intersection of \(\mathcal{H}\) and \(\mathcal{N}(\mathbf{S}_{0})\). In particular, there exists \(\mathbf{X}=(x_{1},\ldots,x_{N})^{\text{T}}\): \(\mathbf{X}=\mathcal{T}_{L}(\mathbf{X})\). Let us calculate the \((i,j)\)-th element of the matrix \(\mathbf{U}^{\text{T}}\mathbf{X}\mathbf{V}\) from (12) following Remark 3:  
\[\sum_{l=1}^{r+1}\sum_{p=1}^{r+1}a_{l}a_{p}x_{i+l-1+p+j-1-1}=\sum_{l=1}^{r+1} \sum_{p=1}^{r+1}a_{l}a_{p}x_{(i+j-1)+p+l-2}=\sum_{k=1}^{2r+1}a_{k}^{(2)}x_{(i+ j-1)+k-1}=0, \tag{18}\]  
where \(a_{i}^{(2)}\) are coefficients of the vector \(\mathbf{a}_{0}^{2}\); see definition (17). The values of \(i+j-1\), where \(i\) runs from \(1\) to \(L-r\) and \(j\) runs from \(1\) to \(K-r\), cover the entire interval from \(1\) to \(N-2r\), and therefore, (18) is equivalent to \(\big{(}\mathbf{Q}(\mathbf{a}^{2})\big{)}^{\text{T}}\mathbf{X}=\mathbf{0}_{N-2r}\).