## 3 The Proposed KBDSP Method
### Optimization of KBDSP  
To facilitate the solution of the problem (7), we transform it to the following equivalent problem by importing three new auxiliary variables:  
\[\underset{Z}{\min}\frac{1}{2}\text{Tr}(K-2 KJ+Z^{T}KJ)+\alpha\left\|K-G^{T}KH \right\|_{F}^{2}+\frac{\beta}{2}\|J-B\|_{F}^{2}+\gamma\|B\|\underset{ \framebox{$k$}}{\underbrace{\left\|\overline{k}\right\|}} \tag{8}\] \[s,t\quad B\geq 0,diag(B)=0,B^{T}=B,J=Z,\ G=Z,\ H=Z\]  
The problem (8) is solved by using ADMM. The corresponding augmented Lagrangian function is  
\[ \begin{array}{ll}&L(J,G,H,B,\lambda_{1},\lambda_{2},\lambda_{3})\\ =&\frac{1}{2}\text{Tr}(K-2 KJ+Z^{T}KJ)+\alpha\left\|K-G^{T}KH\right\|_{F}^{2}+ \frac{\beta}{2}\left\|J-B\right\|_{F}^{2}+\gamma\left\|B\right\|\right\|\\ &+\frac{\mu}{2}\left[\left\|J-Z+\frac{\lambda_{1}}{\mu}\right\|_{F}^{2}+ \left\|G-Z+\frac{\lambda_{2}}{\mu}\right\|_{F}^{2}+\left\|H-Z+\frac{\lambda_{ 3}}{\mu}\right\|_{F}^{2}\right]\end{array}  \tag{9}\]  
where \(\lambda_{1},\lambda_{2},\lambda_{3}\) are Lagrangian multipliers and \(\mu>0\) is a penalty parameter. These variables can be updated alternatively. (1) Updating \(J\)  
After deleting the irrelevant terms, the problem (9) is written as:  
\[ \begin{array}{ll}\min_{\int}\frac{1}{2}\text{Tr}(-2 KJ+J^{T}KJ)+\frac{\mu}{2} \left\|J-Z+\frac{\gamma_{1}}{\mu}\right\|_{F}^{2}+\frac{\beta}{2}\left\|J-B \right\|_{F}^{2}\\ &s,t\ \ \ B\geq 0,diag(B)=0,B^{T}=B\end{array}  \tag{10}\]  
Let the first derivative be equal to zero, we obtain:  
\[J=(K+B+\mu I)^{-1}(K+\mu Z-\lambda_{1}+\beta B) \tag{11}\]  
(2) Updating \(G\)  
\[\min_{G}\alpha\left\|K-G^{T}KH\right\|_{F}^{2}+\frac{\mu}{2}\left\|G-Z+\frac{ \lambda_{2}}{\mu}\right\|_{F}^{2} \tag{12}\]  
Let the first derivative be equal to zero, we have:  
\[G=(2\alpha KHH^{T}K^{T}+\mu I)^{-1}(2\alpha KHK^{T}+\mu Z-\lambda_{2}) \tag{13}\]  
(3) Updating \(H\)  
\[\min_{H}\alpha\left\|K-G^{T}KH\right\|_{F}^{2}+\frac{\mu}{2}\left\|H-Z+\frac{ \lambda_{3}}{\mu}\right\|_{F}^{2} \tag{14}\]  
Let the first derivative be equal to zero, we have:  
\[H=(2\alpha K^{T}GG^{T}K+\mu I)^{-1}(2\alpha K^{T}GK+\mu Z-\lambda_{3}) \tag{15}\]  
(4) Updating \(Z\)  
\[\min_{Z}\frac{3\mu}{2}\left\|Z-\frac{J+G+H+(\lambda_{1}+\lambda_{2}+\lambda_{ 3})/\mu}{3}\right\|_{F}^{2} \tag{16}\]  
Let the first derivative equal to zero, we have:  
\[Z=\frac{J+G+H+(Y_{1}+Y_{2}+Y_{3})/\mu}{3} \tag{17}\]  
(5) Updating \(B\)  
For \(B\), the subproblem is:  
\[\min_{B}\frac{\beta}{2}\left\|J-B\right\|_{F}^{2}+\gamma\left\|B\right\|_{ \left\|\frac{\gamma}{k}\right\|}\ \ \ s,t\ \ \ B\geq 0,diag(B)=0,B^{T}=B \tag{18}\]According to the Ky Fan Theorem [42], problem (18) can be rewritten as:  
\[ \begin{array}{ll}&\min\limits_{B}\frac{\beta}{2}\|J-B\|_{F}^{2}+\gamma<\text{ Diag}(B\mathbf{1})-B,\,A)>\\ &s,t\ \ \ B\geq 0,diag(B)=0,B^{T}=B,\,0\leq A\leq I,\,\,\text{Tr}(A)=k\end{array}  \tag{19}\]  
where \(A=UU^{T}\), \(U\) consists of \(k\) eigenvectors associated with the \(k\) smallest eigenvalues of Diag \((B\mathbf{1})-B\). \(I\) denotes the identity matrix and \(\mathbf{1}\) denotes a column vector with all elements being \(1\). For Equation (19), it is equivalent to:  
\[\min\limits_{B}\frac{1}{2}\bigg{\|}B-J+\frac{\gamma}{\beta}(diag(A)\mathbf{1}^ {T}-A)\bigg{\|}_{F}^{2} \tag{20}\]  
Let \(Q=J-\frac{\gamma}{\beta}(diag(A)\mathbf{1}^{T}-A),\widetilde{Q}=Q-\text{ Diag}(diag(Q))\), then \(B=\max\left(0,(Q+\widetilde{Q})/2\right)\). Once we obtain the matrix \(B\), we can calculate the similarity matrix by \((|B|+|B|^{T})/2\), and then, we can obtain the clustering results by applying the spectral clustering algorithm [20]. The updating process is terminated when the convergence condition is reached, i.e., \(\max\left(\big{\|}Z^{t+1}-Z^{t}\big{\|}_{F},\big{\|}B^{t+1}-B^{t}\big{\|}_{F} \big{\|}A^{t+1}-A^{t}\big{\|}_{F}\right)<\epsilon\), where \(\epsilon\) is the stop threshold. To make the process clear, the complete procedures to solve problem (7) is outlined as follows (Algorithm 1).