## 4 The Non-Convex Approximate RPCA Algorithm
### Derivation of the Overall Formula for NCA-RPCA Algorithm
\begin{array}{ll}\mathbf{L}^{(k+1)}&=\underset{L}{\text{argmin}}\ \mathcal{L}(\mathbf{L},\mathbf{S}^{(k+1)},\mathbf{K}^{(k)}, \rho^{(k)})\\ &=\underset{L}{\text{argmin}}\ \underset{L}{\sum}g(\sigma^{(k)}_{j})+ \Big{\langle}\bigtriangledown g(\sigma^{(k)}_{j}),(\sigma_{j}-\sigma^{(k)}_{j} )\Big{\rangle}\\ &\qquad\qquad\qquad\qquad+\frac{\rho^{(k)}}{2}\Bigg{\|}(\mathbf{M}-\mathbf{S}^{(k+1) }+\frac{\mathbf{K}^{(k)}}{\rho^{(k)}})-\mathbf{L}\Bigg{\|}_{F}^{2}\end{array}  \tag{23}\]  
Due to the first-order Taylor expansion of \(\underset{j}{\sum}g(\sigma^{(k)}_{j})+\Big{\langle}\bigtriangledown g(\sigma ^{(k)}_{j}),(\sigma_{j}-\sigma^{(k)}_{j})\Big{\rangle}\) being equivalent to \(g(\mathbf{L})\) and \(\underset{j}{\sum}g(\sigma^{(k)}_{j})\) being a constant in Equation (26), during the process of solving for \(L\), it is possible to omit \(\underset{j}{\sum}g(\sigma^{(k)}_{j})\), meaning that \(\underset{j}{\sum}g(\sigma^{(k)}_{j})+\Big{\langle}\bigtriangledown g(\sigma ^{(k)}_{j}),(\sigma_{j}-\sigma^{(k)}_{j})\Big{\rangle}\) is effectively equivalent to \(\Big{\langle}\bigtriangledown g(\sigma^{(k)}_{j}),(\sigma_{j}-\sigma^{(k)}_{ j})\Big{\rangle}\). The above equations are derived and simplified. \[ \begin{array}{ll}\mathbf{L}^{(k+1)}&=\underset{L}{\text{argmin}}\ \mathcal{L}(\mathbf{L},\mathbf{S}^{(k+1)},\mathbf{K}^{(k)},\rho^{(k)})\\ &=\underset{L}{\text{argmin}}\ \Big{\langle}\bigtriangledown g(\sigma^{(k)}_{j} ),(\sigma_{j}-\sigma^{(k)}_{j})\Big{\rangle}+\frac{\rho^{(k)}}{2}\Big{\|}(\bm {M}-\mathbf{S}^{(k+1)}+\frac{\mathbf{K}^{(k)}}{\rho^{(k)}})-\mathbf{L}\Big{\|}_{F}^{2}\end{array}  \tag{24}\]  
In the third step, with the updated \(L\) and \(S\) fixed, the gradient ascent method is applied to update \(K\), updating the penalty parameter \(\rho\) using the step size \(\phi\). \[ \begin{array}{ll}\mathbf{K}^{(k+1)}&=\mathbf{K}^{(k)}+\rho^{(k)}(\mathbf{M}-\mathbf{L}^{(k+ 1)}-\mathbf{S}^{(k+1)})\\ \rho^{(k+1)}&=\phi\rho^{(k)}\end{array}  \tag{25}\]  
where \(k\) denotes the iteration number. The specific solving process is as follows:  
Step one, fixing \(L\) and \(K\), update \(S\):  
\[ \begin{array}{ll}\mathbf{S}^{(k+1)}=\underset{S}{\text{argmin}}\ \lambda\sum_{ij} \bigtriangledown f_{\lambda}(s^{(k)}_{ij})\big{|}s_{ij}\big{|}+\frac{\rho^{( k)}}{2}\Big{\|}\mathbf{Z}^{(k)}-\mathbf{S}\Big{\|}_{F}^{2}\end{array}  \tag{26}\]  
\[ \begin{array}{ll}\text{Let}\ \mathbf{Z}^{(k)}=\mathbf{M}-\mathbf{L}^{(k)}+\frac{\mathbf{K}^{(k)}}{ \rho^{(k)}},\epsilon^{(k)}_{ij}=\lambda\bigtriangledown f_{\lambda}(s^{(k)}_ {ij})/\rho^{(k)}\end{array} \]  
and decompose the above equation into a subproblem regarding \(s^{(k+1)}_{ij}\), then the equation becomes:  
\[ \begin{array}{ll}s^{(k+1)}=\underset{S}{\text{argmin}}\ \varepsilon^{(k)}|s|+ \frac{1}{2}\Big{\|}s-z^{(k)}\Big{\|}_{F}^{2}\end{array}  \tag{27}\]  
where \(s^{(k+1)}\) and \(z^{(k)}\) are elements of \(\mathbf{S}^{(k+1)}\) and \(\mathbf{Z}^{(k)}\), respectively. For the sake of simplicity in notation, the subscripts i and j are omitted. Regarding the equation, the soft-thresholding operator [43] can be applied to solve for \(s^{(k+1)}\), which can be obtained from the following equation:  
\[ \begin{array}{ll}s^{(k+1)}:=\text{sgn}(z)(|z|-\varepsilon,0)_{+}\\ \text{Here, }(|z|-\varepsilon,0)_{+}\text{ represents }\max(|z|-\varepsilon,0).\end{array}