Differentiate hadamard product of square matrix $S \odot VV^T \in R^{n \times n} $ over rectangular matrix $V \in R^{n \times r}$

Question

I want to differentiate $f = \log\det(L)$ over $V$ where $L = S \odot VV^T$.

The thing that I know is $df = L^{-T} : dL = (S \odot VV^T) : (dS \odot VV^T+ S \odot (VdV^T+dVV^T ))$ where $:$is a Frobenius product and $\odot$ is a Hadamard product.

Maybe $dS$ is zero over $dV$ and $df =(S \odot VV^T) : S \odot (VdV^T+dVV^T ) $ . But I don't know how to treat the Hadarmard product in trace.

I wanna get a solution in a matrix form, it's okay if it is derived in element wise form. But I don't want the solution including the $E$ and $B$ matrices.

How can i solve the problem?

Thanks in advance.

Answer 1

The Hadamard and Frobenius products commute with each other, i.e. $$A\odot B:C = A:B\odot C$$ This property can be used to complete your derivation (assuming $dS$ is zero) $$\eqalign{ df &= L^{-T}:dL \cr &= L^{-T}:S\odot(dV\,V^T+V\,dV^T) \cr &= S\odot L^{-T}:(dV\,V^T+V\,dV^T) \cr &= (S\odot L^{-T}+S^T\odot L^{-1}):dV\,V^T \cr &= (S\odot L^{-T}+S^T\odot L^{-1})V:dV \cr\cr \frac{\partial f}{\partial V} &= (S\odot L^{-T}+S^T\odot L^{-1})V \cr\cr }$$