the projection step. The objective is $$J\!=\!\sum_{n=1}^{N}\!\left\Vert \mtbfX_{n}^{\intercal}\mtbfh_{n}\!-\!\left(\mtbfr_{n}\!+\!\gamma\mtbfY_{n}^{\intercal}\mtbfw_{n}\right)\right\Vert _{2}^{2}+\mu_{1}\mttrace\left(\mtbfH\mtbfL\mtbfH^{\intercal}\right)\!+\!\zeta_{1}\left\Vert \mtbfH\right\Vert _{F}^{2},\label{eq:critic_obj_Graph}$$ where $\left\Vert \cdot\right\Vert _{F}^{2}$ is the Frobenius norm of a matrix, $${\displaystyle \mttrace\left(\mtbfH\mtbfL\mtbfH^{\intercal}\right)=\sum_{i=1}^{N}\sum_{j=1}^{N}c_{ij}\left\Vert \mtbfh_{i}-\mtbfh_{j}\right\Vert _{2}^{2}}\label{eq:Deviation_Graph_Constraint}$$ and $\mtbfL=\mtbfD-\mtbfC\in\mtbbR^{N\times N}$ is a graph laplacian; $\mtbfD$ is a diagonal matrix whose elements are column (or row, as $\mtbfC$ is a symmetric matrix) sums of $\mtbfC$, i.e. $d_{ii}=\sum_{i}c_{ij}$. The partial derivative of $J_{1}$, i.e. the 1st term in , with respect to $\mtbfh_{n}$ is $$\frac{\partial J_{1}}{\partial\mtbfh_{n}}=2\mtbfX_{n}\mtbfX_{n}^{\intercal}\mtbfh_{n}-2\mtbfX_{n}\mtbfr_{n}-2\gamma\mtbfX_{n}\mtbfY_{n}^{\intercal}\mtbfw_{n}.$$ Summarizing the partial derivatives with respect to all the variables in $\mtbfH=\left(\mtbfh_{1},\cdots,\mtbfh_{N}\right)\in\mtbbR^{K\times N}$, we have $$\begin{aligned}
\frac{\partial J_{1}}{\partial\mtvec\left(\mtbfH\right)}=\  & 2\left(\sum_{n=1}^{N}\mtbfE_{n}\otimes\left(\mtbfX_{n}\mtbfX_{n}^{\intercal}\right)\right)\mtvec\left(\mtbfH\right)\nonumber \\
 & -2\left(\sum_{n=1}^{N}\mtbfE_{n}\otimes\mtbfX_{n}\right)\mtvec\left(\mtbfR\right)\nonumber \\
 & -2\gamma\left(\sum_{n=1}^{N}\mtbfE_{n}\otimes\left(\mtbfX_{n}\mtbfY_{n}^{\intercal}\right)\right)\mtvec\left(\mtbfW\right)\end{aligned}$$ where $\mtvec\left(\mtbfH\right)\!=\!\left[\mtbfh_{1}^{\intercal},\cdots,\mtbfh_{N}^{\intercal}\right]^{\intercal}\!\in\!\mtbbR^{uN}$ is the vectorization process for a matrix; $\mtbfE_{n}=\mtdiag\left(0,\cdots,1,\cdots,0\right)\in\mtbbR^{N\times N}$ is a diagonal matrix with the $n$-th diagonal element equal to 1, all the other equal to zero; $\otimes$ indicates the Kronecker product between two matrices resulting in a block matrix. Let $\mtbfF_{1}=\sum_{n}\mtbfE_{n}\otimes\left(\mtbfX_{n}\mtbfX_{n}^{\intercal}\right)$, $\mtbfF_{2}=\sum_{n}\mtbfE_{n}\otimes\mtbfX_{n}$ and $\mtbfF_{3}=\sum_{n}\mtbfE_{n}\otimes\left(\mtbfX_{n}\mtbfY_{n}^{\intercal}\right)$. We have a simpler formulation for the $\partial J_{1}/\partial\mtvec\left(\mtbfH\right)$ as follows $$\frac{\partial J_{1}}{\partial\mtvec\left(\mtbfH\right)}=2\mtbfF_{1}\mtvec\left(\mtbfH\right)-2\left[\mtbfF_{2}\mtvec\left(\mtbfR\right)+\gamma\mtbfF_{3}\mtvec\left(\mtbfW\right)\right].$$ The partial derivatives of the 2nd term in , i.e. $J_{2}=\mu_{1}\mttrace\left(\mtbfH\mtbfL\mtbfH^{\intercal}\right)+\zeta_{1}\left\Vert \mtbfH\right\Vert _{F}^{2}$, with respect to $\mtbfH$ is $$\frac{\partial J_{2}}{\partial\mtbfH}=2\mu_{1}\mtbfH\mtbfL+2\zeta_{1}\mtbfH.$$ According to the Encapsulating Sum [@Petersen_2012_MatrixCookbook], we have $$\frac{\partial J_{2}}{\partial\mtvec\left(\mtbfH\right)}=2\left[\left(\mu_{1}\mtbfL^{\intercal}+\zeta_{1}\mtbfI_{N}\right)\otimes\mtbfI_{u}\right]\mtvec\left(\mtbfH\right)\label{eq:gradient_with_Graph_constraint}$$ where $\mtbfI_{N}\in\mtbbR^{N\times N}$ and $\mtbfI_{u}\in\mtbbR^{u\times u}$ are identical matrices. Setting the 