# Mixed Alternating Projections with Application to Hankel Low-Rank Approximation
## 4 Linearized Cadzow Iterations in the Signal-Estimation Problem
### Limit Property of the Composition of Projections
#### 4.4.1 Example of Bad Convergence as \(\alpha\) Tends to Zero  
Using Proposition 3 and the fact that the cosine of the \((2r+1)\)-th principal angle determines the convergence rate (Proposition 1), we show by a specific example that this cosine tends to one as \(\alpha\) tends to zero; this makes the convergence of the linearized Cadzow algorithm as slow as possible. Let \(N=2L\) and \(K=L+1\). Let the matrix weights \(\mathbf{L}=\mathbf{I}_{L}\) be the identity matrix, which corresponds to the case of white noise. For simplicity, we make the matrix \(\mathbf{R}=\mathbf{R}(\alpha)\in\mathbb{R}^{K\times K}\) equal to \(\mathbf{R}=\text{diag}(1,\alpha,\dots,\alpha,1)\) according to [13]. Numerical experiments show that similar matrix weights (up to multiplying a constant) are obtained when applying the method from [12] of finding weights for small \(\alpha\). Consider the degenerate case \(\mathbf{R}(0)\) studied in [13]. **Proposition 4**. : _For any matrix \(\mathbf{X}\in\mathbb{R}^{L\times(L+1)}\), we can find a matrix \(\mathbf{Y}\in\mathcal{H}\) such that \(\|\mathbf{X}-\mathbf{Y}\|_{\mathbf{I}_{L}\mathbf{R}(0)}=0\)._  
Proof. : The inner product (4) of matrices \(\mathbf{Z}_{1}\) and \(\mathbf{Z}_{2}\) with the matrix weights \(\mathbf{L}=\mathbf{I}_{L}\) and \(\mathbf{R}(0)=\text{diag}(1,0,\ldots,0,1)\) depends only on the first and last columns of these matrices due to the form of \(\mathbf{R}(0)\). Therefore, the required matrix \(\mathbf{Y}\) can be written in the form \(\mathbf{Y}=\Pi_{\mathcal{H},(\mathbf{I}_{L}\mathbf{R}(0))}\mathbf{X}\), where the projection is calculated uniquely using the values from the first and last columns of the matrix \(\mathbf{X}\). **Corollary 2**. : _Let \(\mathbf{X}\in\mathbb{R}^{L\times(L+1)}\) be an arbitrary matrix, \(\widehat{\mathbf{X}}=\mathbf{X}-\Pi_{\mathcal{T}_{L}(\mathcal{Z}(\mathbf{a}_ {0}^{2})),(\mathbf{I}_{L}\mathbf{R}(0))}\mathbf{X}\), \(\widehat{\mathbf{Y}}=\Pi_{\widehat{\mathcal{H}}(\mathbf{I}_{L}\mathbf{R}(0)),(\mathbf{I}_{L}\mathbf{R}(0))}\widehat{\mathbf{X}}\). Then, \(\|\widehat{\mathbf{X}}-\widehat{\mathbf{Y}}\|_{\mathbf{I}_{L}\mathbf{R}(0)}=0\)._  
Proof. : The corollary is proved by the following substitution:  
\(\Pi_{\widehat{\mathcal{H}}(\mathbf{I}_{L}\mathbf{R}(0)),(\mathbf{I}_{L} \mathbf{R}(0))}=\Pi_{\mathcal{H},(\mathbf{I}_{L}\mathbf{R}(0))}-\Pi_{\mathcal{ T}_{L}(\mathcal{Z}(\mathbf{a}_{0}^{2})),(\mathbf{I}_{L}\mathbf{R}(0))}\). **Corollary 3**. : _Let \(\mathbf{N}\in\mathcal{N}(\mathbf{S}_{0})\). Denote \(\widehat{\mathbf{N}}(\mathbf{a})=\Pi_{\widehat{\mathcal{N}}(\mathbf{I}_{L} \mathbf{R}(\mathbf{a})),(\mathbf{I}_{L}\mathbf{R}(\mathbf{a}))}\mathbf{N}\) and \(\widehat{\mathbf{H}}(\mathbf{a})=\Pi_{\widehat{\mathcal{H}}(\mathbf{I}_{L} \mathbf{R}(\mathbf{a})),(\mathbf{I}_{L}\mathbf{R}(\mathbf{a}))}\widehat{ \mathbf{N}}(\mathbf{a})\). Then, the following is fulfilled:_  
\[\|\widehat{\mathbf{N}}(\mathbf{a})\|_{\mathbf{I}_{L}\mathbf{R}( \mathbf{a})}\xrightarrow[a\to 0]{}\|\widehat{\mathbf{N}}(0)\|_{\mathbf{I}_{L}\mathbf{R}(0)},\] \[\|\widehat{\mathbf{H}}(\mathbf{a})\|_{\mathbf{I}_{L}\mathbf{R}( \mathbf{a})}\xrightarrow[a\to 0]{}\|\widehat{\mathbf{H}}(0)\|_{\mathbf{I}_{L}\mathbf{R}(0)},\] \[\|\widehat{\mathbf{N}}(0)\|_{\mathbf{I}_{L}\mathbf{R}(0)}=\| \widehat{\mathbf{H}}(0)\|_{\mathbf{I}_{L}\mathbf{R}(0)},\] \[\langle\widehat{\mathbf{N}}(\mathbf{a}),\widehat{\mathbf{H}}( \mathbf{a})\rangle_{\mathbf{I}_{L}\mathbf{R}(\mathbf{a})}\xrightarrow[a\to 0]{}\|\widehat{\mathbf{N}}(0)\|_{\mathbf{I}_{L} \mathbf{R}(0)}^{2}.\]  
Proof. : Since \(\widehat{\mathbf{N}}(\mathbf{a})=\mathbf{N}-\Pi_{\mathcal{T}_{L}(\mathcal{Z} (\mathbf{a}_{0}^{2})),(\mathbf{I}_{L}\mathbf{R}(\mathbf{a}))}\mathbf{N}\), this expression is smooth in \(\mathbf{a}\) and uniquely defined when \(\mathbf{a}=0\). The projection \(\Pi_{\widehat{\mathcal{H}}(\mathbf{I}_{L}\mathbf{R}(\mathbf{a})),(\mathbf{I}_ {L}\mathbf{R}(\mathbf{a}))}\) is also smooth in \(\mathbf{a}\) and uniquely defined at \(\mathbf{a}=0\).