Matrix derivative with complex term question.
I have a question about how to proced with the matrix derivation where i have a complex matrix. Supose i have a function \begin{align} y = \mathrm{ln}|XX^H+\alpha I|, \end{align} where $X \in \mathbb{C}^{n \times m}$, the operator $H$ is the hermitian (conjugate transpose) and $\alpha$ is a small scalar. I need the derivative of $y$ with respect to $X$.
With $A = XX^H+\alpha I$ the function become \begin{align} y = \mathrm{ln}|A|. \end{align} The differential from wiki is \begin{align} dy &= d\ \mathrm{ln}|A|\cr &= \mathrm{Tr}(A^{-1} dA)\cr &= A^{-T}:dA\cr dA &= (dXX^H + XdX^H). \end{align}
Substituting $dA$ in $dy$ led to \begin{align} dy &= A^{-T}:dA\cr &= A^{-T}:(dXX^H + XdX^H)\cr &= A^{-T}:dXX^H + A^{-T}:XdX^H\cr \end{align}
From here i got stuck. I don't know how to deal with the hermitian term $dX^H$ and also don't know if i can apply some cyclic properties (because A is not square) like $A^{-T}:dXX^H = X^HA^{-T}:dX$.
Any help would be apreciated.
Let $A^C$ denote the complex conjugate.
Using the cyclic property of the trace (aka Frobenius product), you can rearrange the differential expression to easily spot the desired gradients. $$\eqalign{ dy &= A^{-T}X^C:dX \;+\; X^TA^{-T}:dX^H \\ \frac{\partial y}{\partial X} &= A^{-T}X^C, \qquad\qquad X^TA^{-T} = \frac{\partial y}{\partial X^H} \\ }$$ Since $A$ is hermitian
$$\eqalign{ A^H &= A,\quad &A^{-H}=A^{-1} \\ A^T&=A^C,\quad &A^{-T}=A^{-C} }$$ Furthermore, $y$ is real so you really only need to know one of the gradients and you can calculate any other gradient using the above table of $A$-equivalencies $$\eqalign{ \frac{\partial y}{\partial X^H} &= \left(\frac{\partial y}{\partial X}\right)^H = (A^{-T}X^C)^H = X^TA^{-T} \\ \frac{\partial y}{\partial X^C} &= \left(\frac{\partial y}{\partial X}\right)^C = (A^{-T}X^C)^C = A^{-1}X \\ }$$