From the matrix cookbook, I got:
$\frac{\partial}{\partial X} ||X||_F^2 = \frac{\partial}{\partial X} Tr(XX^H) = 2X$
Now I want to compute $\frac{\partial}{\partial X} ||X-Y||_F^2 = \frac{\partial}{\partial X} Tr\left((X-Y)(X-Y)^H\right)$, but I am stuck here, any idea?
For convenience, define a new variable $$\eqalign{ W &= X-Y \cr }$$ Write the function in terms of this new variable and the Frobenius (:) Inner Product and find its differential $$\eqalign{ f &= W:W \cr df &= 2W:dW \cr &= 2W:dX\cr }$$Since $df=\big(\frac{\partial f}{\partial X}:dX\big),\,$ we can identify $$\eqalign{ \frac{\partial f}{\partial X} &= 2\,W \cr &= 2\,(X-Y) \cr }$$ Note that setting $Y=0,\,$ recovers the Cookbook result.
