TITLE: Complex ($\mathbb C$) least squares derivation
QUESTION [5 upvotes]: I know how to derive the least squares in the real domain.
If a tall matrix $A$ and a column vector $b$ are real,
then the solution of the least squares problem $Ax = b$ can be derived as:
$$\begin{align}
\{E(x)\}^2 &= ||Ax - b||^2 \\
&= (Ax-b)^T (Ax-b) \\
&= x^T A^T Ax - x^T A^T b - b^T Ax + b^T b \\
&= x^T A^T Ax - 2 x^T A^T b + b^T b \qquad (\because (Ax)^T b = b^T (Ax))
\end{align}$$
Differentiating both sides with respect to $x$,
$$\begin{align}
\frac{d \{E(x)\}^2}{dx} &= 2A^T Ax - 2 A^T b
\end{align}$$
Setting $\frac{d \{E(x)\}^2}{dx} = 0$ to find when we get the minimum $E(x)$,
$$
2A^T Ax - 2 A^T b = 0 \\
A^T Ax = A^T b \\
x = (A^T A)^{-1} A^T b
$$
Now, we turn to the complex-valued situation.
Assume $A$ and $b$ are complex,
$$\begin{align}
\{E(x)\}^2 &= ||Ax - b||^2 \\
&= (Ax-b)^H (Ax-b) \\
&= x^H A^H Ax - x^H A^H b - b^H Ax + b^H b \\
\end{align}$$
Here, I have some problems.
First, $x^H A^H b \neq b^H Ax$ unless $(Ax)^H b$ is real.
Most of all, I don't know how to differentiate the complex matrices above.
How to proceed the derivation?
There are plenty of derivations in the real domain in Google, but I couldn't find detailed explanation of the general complex case.

REPLY [8 votes]: Denote the complex conjugate, transpose, and conjugate transpose of the matrix $A$ as $(A^*, A^T, A^H)$ respectively.
Use the Frobenius (:) Inner Product to write the function and take its differential
$$\eqalign{
  f &= (Ax-b)^*:(Ax-b) \cr\cr
 df &= (Ax-b)^*:A\,dx \cr
    &= A^T(Ax-b)^*:dx \cr
}$$
Since $df=\Big(\frac{\partial f}{\partial x}:dx\Big),\,$ the gradient must be
$$\eqalign{
 \frac{\partial f}{\partial x} &= A^T(Ax-b)^* \cr
}$$
Set the gradient to zero, take the complex conjugate, and solve for $x$
$$\eqalign{
 A^T(Ax)^* &= A^Tb^* \cr
 A^HAx &= A^Hb \cr
 x &= (A^HA)^{-1}A^Hb \cr
   &= A^{+}b \cr
}$$
Notice that $x$ and $x^*$ are treated as independent variables for the purpose of differentiation.