Partial derivative of a complex conjugate vector

Question

I need to find the partial derivative of $\frac{\partial{f}}{\partial{x}}$, $f = x^* A^ * y - y^ * A x $ where $x$ and $y$ are complex-valued vectors and * represents the complex conjugate. A is a complex-valued matrix.

Answer 1

$ \def\LR#1{\left(#1\right)} \def\op#1{\operatorname{#1}} \def\trace#1{\op{Tr}\LR{#1}} \def\qiq{\quad\implies\quad} \def\p{\partial} \def\grad#1#2{\frac{\p #1}{\p #2}} $Let $\LR{A^T,A^*,A^H}$ denote the transpose, complex and hermitian conjugates of $A$, i.e. $$\eqalign{ A^* &= \LR{A^T}^H &= \LR{A^H}^T \\ A^T &= \LR{A^H}^* &= \LR{A^*}^H \\ A^H &= \LR{A^*}^T &= \LR{A^T}^* \\ A = \LR{A^*}^* &= \LR{A^H}^H &= \LR{A^T}^T \\ }$$ For typing convenience, define $\,w=\LR{A^Hy}\:$ and Frobenius product $(:)$ $$\eqalign{ A:B &= \sum_{i=1}^m\sum_{j=1}^n A_{ij}B_{ij} \;=\; \trace{A^TB} \\ A:A &= \|A\|^2_F \qquad \{ {\rm Frobenius\:norm} \}\\ }$$ Write the function using the above notation and differentiate (in the Wirtinger sense) $$\eqalign{ f &= \LR{w^Hx}^* - \LR{w^Hx} \\ &= \LR{w:x^*} - \LR{w^*:x} \\ df &= \LR{w:dx^*} - \LR{w^*:dx} \\ }$$ The gradients are therefore $$\eqalign{ \grad{f}{x^*} &= w = A^Hy, \qquad\quad \grad{f}{x} &= -w^* = -A^Ty^* \\ }$$