# Aiding Dictionary Learning Through Multi-Parametric Sparse Representation
## 3 Analysis of the Multi-Parametric Formulation Induced by the DL Problem
### Multi-Parametric Interpretation
\begin{bmatrix} \left(\mathbf{\Sigma}_{12}^{-1}\right)^{\top}\mathbf{U}^{\top}\mathbf{y}+\frac{1}{2}\left( \mathbf{\Sigma}_{12}^{-2}\right)^{\top}\left(\tilde{\mathbf{V}}_{1}^{\top}\tilde{\mathbf{ \lambda}}-\mathbf{V}_{1}^{\top}\mathbf{1}\mu\right)\end{bmatrix}  =0,\] (19b) \[\tilde{\mathbf{V}}_{2}^{\top}\tilde{\mathbf{\lambda}}-\mathbf{V}_{2}^{\top} \mathbf{1}\mu =0. \tag{19c}\]  
Rearranging (19) to highlight the unknown terms (\(\hat{\mathbf{\alpha}}\), \(\tilde{\mathbf{\lambda}}\) and \(\mu\)) leads to the following compact form:  
\[ \begin{bmatrix}\tilde{\mathbf{V}}_{11}^{\top}\\ \mathbf{1}^{\top}\end{bmatrix} \hat{\mathbf{\alpha}}= \begin{bmatrix}\left(\mathbf{\Sigma} _{11}^{-1}\right)^{\top}\mathbf{U}^{\top}\\ 0\end{bmatrix} \mathbf{y}+ \begin{bmatrix}0\\ \delta\end{bmatrix} + \begin{bmatrix}\left(\frac{1}{2}\mathbf{\Sigma}_{11}^{-2}\right) ^{\top}\\ 0\end{bmatrix}\begin{bmatrix}\tilde{\mathbf{V}}_{1}^{\top}&-\mathbf{V}_{1}^{\top}\mathbf{1} \end{bmatrix}\begin{bmatrix}\tilde{\mathbf{\lambda}}\\ \mu\end{bmatrix} , \tag{20a}\] \[ \begin{bmatrix}\frac{1}{2}\left(\mathbf{\Sigma}_{12}^{-2}\right)^{ \top}\mathbf{V}_{1}^{\top}&-\frac{1}{2}\left(\mathbf{\Sigma}_{12}^{-2}\right)^{\top} \mathbf{V}_{1}^{\top}\mathbf{1}\\ \tilde{\mathbf{V}}_{2}^{\top}&-\mathbf{V}_{2}^{\top}\mathbf{1}\end{bmatrix}\begin{bmatrix} \tilde{\mathbf{\lambda}}\\ \mu\end{bmatrix} = \begin{bmatrix}\tilde{\mathbf{V}}_{12}^{\top}\\ 0\end{bmatrix} \hat{\mathbf{\alpha}}- \begin{bmatrix}\left(\mathbf{\Sigma}_{12}^{-1} \right)^{\top}\mathbf{U}^{\top}\\ 0\end{bmatrix} \mathbf{y}. \tag{20b}\]Note that the matrices appearing to the left of the equal sign in both (20a) and (20b) are full-row rank: the first is in \(\mathbb{R}^{\hat{n}\times\hat{n}}\) and the second is in \(\mathbb{R}^{(n-\hat{n}+1)\times(2n-\hat{n}+1)}\). Thus, we can proceed as follows:  
1. we obtain \(\left[\tilde{\lambda}^{\top}\quad\mu\right]^{\top}\) from (20b) as a function of \(\hat{\mathbf{\mathrm{\hat{\alpha}}}}\) and \(\mathbf{y}\);
2. we introduce it in (20a) and obtain \(\hat{\mathbf{\mathrm{\hat{\alpha}}}}\) as an affine form of \(\mathbf{y}\);
3. we go back in (20b), replace the now-known \(\hat{\mathbf{\mathrm{\hat{\alpha}}}}\) and obtain \(\left[\tilde{\lambda}^{\top}\quad\mu\right]^{\top}\) as an affine form of \(\mathbf{y}\). For ease of notation, we assume that the steps mentioned above lead to  
\[\hat{\mathbf{\mathrm{\hat{\alpha}}}}(\mathbf{y})=\tilde{M}_{\alpha}\mathbf{y}+\tilde{N}_{ \alpha},\quad\tilde{\lambda}(\mathbf{y})=\tilde{M}_{\lambda}\mathbf{y}+\tilde{N}_{ \lambda},\quad\mu(\mathbf{y})=\mathbf{M}_{\mu}\mathbf{y}+\mathbf{N}_{\mu}. \tag{21}\]  
Combining with the fact that \(\hat{\mathbf{\mathrm{\hat{\alpha}}}}=\mathbf{0},\hat{\mathbf{\mathrm{\hat{\lambda}}}}=\bm {0}\), as required in (18c)-(18d), allows for writing (by interlacing rows of zeroes in matrices \(\tilde{M}_{\alpha}\), \(\tilde{N}_{\alpha}\), \(\tilde{M}_{\lambda}\), \(\tilde{N}_{\lambda}\)) the explicit solutions for \(\mathbf{\mathrm{\alpha}},\mathbf{\mathrm{\lambda}},\mu\):  
\[\mathbf{\mathrm{\alpha}}(\mathbf{y})=\mathbf{M}_{\alpha}\mathbf{y}+\mathbf{N}_{\alpha},\quad\mathbf{ \mathrm{\lambda}}(\mathbf{y})=\mathbf{M}_{\lambda}\mathbf{y}+\mathbf{N}_{\lambda},\quad\mu(\bm {y})=\mathbf{M}_{\mu}\mathbf{y}+\mathbf{N}_{\mu}. \tag{22}\]  
Recalling that \(\tilde{\mathbf{\mathrm{\hat{\lambda}}}}>0\) and \(\hat{\mathbf{\mathrm{\hat{\alpha}}}}>0\), as required in (18c)-(18d), and using the notation from (21) gives the feasible domain over which the affine laws (21) describe the optimal solution:  
\[\mathcal{R}=\left\{\mathbf{y}:\,\tilde{M}_{\alpha}\mathbf{y}+\tilde{N}_{\alpha}>0,\, \tilde{M}_{\lambda}\mathbf{y}+\tilde{N}_{\lambda}>0\right\}=\left\{\mathbf{y}:\, \begin{bmatrix}\tilde{\mathbf{M}}_{\alpha}\\ \tilde{M}_{\lambda}\end{bmatrix} \mathbf{y}+ \begin{bmatrix}\tilde{N}_{\alpha}\\ \tilde{N}_{\lambda}\end{bmatrix} >0\right\}.