# Mixed Alternating Projections with Application to Hankel Low-Rank Approximation
## 5 Mixed Cadzow Iterations
### Numerical Experiments  
In this section, we demonstrate the result of Theorem 6 through a numerical experiment. Consider the problem of estimating a finite-rank signal when it is corrupted by noise. As in Section 4.4.2, the signal has the form \(\mathsf{S}_{0}=(s_{1},\ldots,s_{N})^{\mathsf{T}}\), where \(s_{k}=\sin\!\left(\frac{2\pi\tilde{k}}{6}\right)\) and \(N=40\). The observed time series is \(\mathsf{X}=\mathsf{S}_{0}+\varepsilon\), where \(\varepsilon\) is Gaussian white noise with zero mean and standard deviation \(\sigma=0.1\). Let us apply iterations given in (6) for estimating the signal \(\mathsf{S}_{0}\). Set \(L=20\) and \(r=2\). Let us take \(T_{1}\) iterations with one of three values of \(\alpha\), \(0.5\), \(0.1\) and \(0.01\), and then \(T_{2}\) iterations with \(\alpha=1\). For the second iterations, we consider the identity matrices \(\mathbf{L}_{2}\) and \(\mathbf{R}_{2}\) (this choice corresponds to \(\alpha=1\)). For the choice of \((\mathbf{L}_{1},\mathbf{R}_{1})\), we set \(\mathbf{L}_{1}\) as the identity matrix, and for a given \(\alpha\), we calculated the matrix of weights \(\mathbf{R}_{1}\) using the method described in [14]. We denote the results of the iterations \(\mathbf{Y}^{\star}\) and the signal estimate \(\widehat{\mathsf{S}}=\mathcal{T}_{L}^{-1}(\mathbf{Y}^{\star})\). For the linearized version (20), we can use the theoretical considerations. For the Cadzow mixed iterations (6), we have no theoretical results and estimated the error using \(I=1000\) simulations of time series with different realizations of noise. The accuracy of the estimate was measured by RMSE (root mean square error) and estimated as the square root of the sample mean of \(I\) realizations of \(\sum_{i=1}^{N}(\hat{s}_{i}-s_{i})^{2}/N\). The estimate of the signal \(\mathsf{S}\) using the linearized Cadzow algorithm has the form \(\widehat{\mathsf{S}}=\mathsf{S}_{0}+\Pi_{\mathcal{Z}(\mathbf{a}_{0}^{2}), \mathbf{Q}}\,\varepsilon\), where \(\Pi_{\mathcal{Z}(\mathbf{a}_{0}^{2}),\mathbf{Q}}\) was taken from the second statement of Theorem 6 and \(\operatorname{GLRR}(\mathbf{a}_{0})\) governs the signal \(\mathsf{S}_{0}\). To calculate RMSE of \(\widehat{\mathsf{S}}\), we consider its autocovariance matrix \(\boldsymbol{\Sigma}_{\widehat{\mathsf{S}}}=\boldsymbol{\Pi}_{\mathcal{Z}( \mathbf{a}_{0}^{2}),\mathbf{Q}}\,\boldsymbol{\Sigma}\,\Pi_{\mathcal{Z}( \mathbf{a}_{0}^{2}),\mathbf{Q}}^{T}\), where \(\boldsymbol{\Sigma}=\mathbf{I}_{N}\) is the autocovariance matrix of \(\varepsilon\). Then, RMSE can be calculated by taking the square root of the average value of the diagonal elements of \(\boldsymbol{\Sigma}_{\mathsf{S}}\), since the \(i\)-th element of the diagonal is the variance of \(\tilde{s}_{i}\). The results are shown in Figure 2. The legend denotes the values of \(\alpha\) for obtaining the matrix weights \(\mathbf{R}_{1}\), and which method was used (\"Linear\" corresponds to linearization (20), and \"True\" corresponds to the original Cadzow method (6)). To demonstrate the statement 2 of Theorem 6, we took a large \(T_{2}\), which numerically corresponds to the case \(T_{2}=\infty\). We see that the results predicted for the linearized Cadzow iterations and estimated by simulation for the original Cadzow  
Figure 2: RMSE (root mean square error) depending on \(T_{1}\) and \(\alpha\). The signal rank 2 and the target rank 2. To demonstrate that the proposed approach allows one to improve the computational costs while keeping the accuracy high, let us consider the dependence of RMSE on \(T_{1}+T_{2}\), where the total number of iterations \(T_{1}+T_{2}\) changes from 1 to 10 and \(\alpha=0.02\). In Figure 3, the case \(T_{1}=0\) (marked as \"\(0.02\times 0+1\times 10\)\") corresponds to the case of fast convergence but low accuracy for standard Cadzow iterations with \(\alpha=1\). The case \(T_{1}=10\) (\"\(0.02\times 10+1\times 0\)\") corresponds to the case of slow convergence but has good accuracy for Cadzow iterations with \(\alpha=0.02\).