Derivate of determinant for square forms

Question

I'm interesting in derivate of matrix. I have proven that the derivative of a non-symmetric matrix is $d/dA|A| = |A|(A^{-1})^T$, when $A^{-1}$ exists and if the matrix is symmetric $d/dS|S| = |S|(2S^{-1}-diag(S^{-1})$. Now, I need to prove that the $d/dA |A^TSA| = 2|A^TSA|SA(A^TSA)^{-1}$ where $S$ is simetric matrix, any suggestion to prove this?.

Answer 1

Define the matrix $$\eqalign{ X &= A^TSA \\ }$$ You already know the gradient of $\det(X)$, so let's use that to calculate the differential, then perform a change of variables from $X\to A$. $$\eqalign{ \phi &= \det(X) \\ d\phi &= \phi X^{-T}:dX \\ &= \phi X^{-T}:\big(A^TS\,dA+dA^TSA\big) \\ &= \phi SAX^{-T}:dA \,+\, \phi SAX^{-1}:dA \\ &= \phi SA\big(X^{-T}+X^{-1}\big):dA \\ &= 2\phi SAX^{-1}:dA \\ \frac{\partial\phi}{\partial A} &= 2\phi SAX^{-1} \\ }$$ where a colon represents the trace/Frobenius product, i.e. $\;A:B={\rm Tr}(A^TB)$.

The properties of the trace allow the terms in a such a product to be rearranged in various ways. $$\eqalign{ A:B &= B:A \\ A:B &= A^T:B^T \\ A:BC &= B^TA:C = AC^T:B \\ }$$