Can anyone tell me what is the derivative of hadamard product

Question

I want to know the derivative of ${\bf{(}}{{\bf{w}}^{\mathop{\rm H}\nolimits} }{\bf{A}} \odot {\bf{(}}{{\bf{w}}^{\mathop{\rm H}\nolimits} }{\bf{A)^{\bf{*}} - }}{{\bf{g}}^H}{\bf{)(}}{{\bf{A}}^H}{\bf{w}} \odot {{\bf{(}}{{\bf{A}}^{\mathop{\rm H}\nolimits} }{\bf{w)}}^{\bf{*}}}{\bf{ - g)}}$ with respect to ${\bf{w}}$, where $ \odot $ is the hadamard product, $^{\mathop{\rm H}\nolimits} $ is conjugate transpose and $^{\bf*}$ is elementwise conjugate.

Answer 1

For convenience, define 2 new vector variables $$\eqalign{ x &= A^Hw &\implies x^* = A^Tw^*\cr y &= x^*\odot x - g &\implies y^* = y-g^*+g &\implies dy^* = dy \cr }$$ and denote the trace/Frobenius product by a colon, i.e. $$\eqalign{A:B={\rm tr}(A^TB)\cr}$$ Now write the function in terms of these new variables and find its differential and gradient(s) $$\eqalign{ f &= y^*:y \cr\cr df &= y^*:dy + y:dy^* \cr &= (y+y^*):dy \cr &= (y+y^*):x^*\odot dx \cr &= (y+y^*)\odot x^*:dx \cr &= (y+y^*)\odot x^*:A^H\,dw \cr &= A^*\Big((y+y^*)\odot x^*\Big):dw \cr\cr \frac{\partial f}{\partial w} &= A^*\Big((y+y^*)\odot x^*\Big), &\frac{\partial f}{\partial w^*} &= A\Big((y+y^*)\odot x\Big) \cr\cr }$$