The partial derivative of the complex quadratic statement

Question

Assume a complex-valued function $ h\left( {a,\theta } \right) = f\left( {a,\theta } \right) + ig\left( {a,\theta } \right) $ where $ a = \left[ {\begin{array}{*{20}{c}} {{a_1}}& \cdots &{{a_n}} \end{array}} \right] $ and $ \theta = \left[ {\begin{array}{*{20}{c}} {{\theta _1}}& \cdots &{{\theta _n}} \end{array}} \right] $ has been represented by a quadratic matrix form $$ h\left( {a,\theta } \right) = {x^H}Ax = \left[ {\begin{array}{*{20}{c}} {{a_1}{e^{ - j{\theta _1}}}}& \cdots &{{a_n}{e^{ - j{\theta _n}}}} \end{array}} \right]A\left[ {\begin{array}{*{20}{c}} {{a_1}{e^{j{\theta _1}}}} \\ \vdots \\ {{a_n}{e^{j{\theta _n}}}} \end{array}} \right] $$ where $A$ is a complex n by n matrix, neither Hermitian nor skew-Hermitian so the value of $ {x^H}Ax $ is a complex number. What are the closed forms of $\frac{{\partial f\left( {a,\theta } \right)}}{{\partial \theta }} + i\frac{{\partial g\left( {a,\theta } \right)}}{{\partial \theta }} = \frac{{\partial {x^H}Ax}}{{\partial \theta }}$ and $\frac{{\partial f\left( {a,\theta } \right)}}{{\partial a}} + i\frac{{\partial g\left( {a,\theta } \right)}}{{\partial a}} = \frac{{\partial {x^H}Ax}}{{\partial a }}$ in terms of the matrix $A$ and vector $x$. Actually, I could not use the Wirtinger derivatives because I do not know how to use them in polar from.

Answer 1

Let the symbols $(\circ)/(:)$ denote the Hadamard/trace products respectively. $$\eqalign{ &c = a\circ b &\implies c_k = a_k b_k \cr &\lambda = a:b &= \sum_k a_k b_k \cr &(x\circ y):z &= x:(y\circ z) \cr }$$ For convenience, define some new variables and their differentials $$\eqalign{ e &= \exp(j\theta) &\implies de = e\circ j\,d\theta \cr x &= a\circ e &\implies dx = x\circ j\,d\theta \cr }$$ Write the function in terms of these new variables, then find its differential $$\eqalign{ h &= x^HAx = x^*:Ax \cr dh &= x^*:A\,dx + dx^*:Ax \cr &= A^Tx^*:dx + Ax:dx^*\cr &= A^Tx^*:(jx\circ d\theta) + Ax:(jx\circ d\theta)^*\cr &= jx\circ(A^Tx^*):d\theta + (jx)^*\circ(Ax):d\theta^* \cr }$$ Assuming that $\theta\in{\mathbb R}^n$, we can simplify the differential and isolate the gradient $$\eqalign{ dh &= \Big(jx\circ(A^Tx^*) - jx^*\circ(Ax)\Big):d\theta \cr \frac{\partial h}{\partial \theta} &= jx\circ(A^Hx)^* - jx^*\circ(Ax) \cr }$$ So that's the gradient wrt $\theta$, the gradient wrt $a$ can be found by a similar process. $$\eqalign{\cr a\circ\frac{\partial h}{\partial a} &= x\circ(A^Hx)^* + x^*\circ(Ax) \cr }$$