What are the rules for taking derivatives in linear algebra? I was reading through a paper on beamforming and came across an equation whose derivative I don't fully understand.
A cost function is given as:
$$ J(\mathbf{w}) = \mathbf{w}^HR\mathbf{w} +\lambda^*[\mathbf{w}^H\mathbf{s}(\theta_0-1] + \lambda[\mathbf{s}(\theta_0)^H\mathbf{w}-1] $$
and the derivative w.r.t. $\mathbf{w}^*$ is found and set to equal 0:
$$\frac{\partial}{\partial\mathbf{w}^*}J(\mathbf{w}) = R\mathbf{w} + \lambda^*\mathbf{s}(\theta_0) = 0$$
It seems to me that $\mathbf{w}$ and $\mathbf{w}^*$ (or $\mathbf{w}^H$) are treated as essentially independent variables, although in my (perhaps naïve) mind they should be related somehow.
Is there a fundamental reason they're separate? Perhaps some linear algebra rules for derivatives would be wonderful for me, I just don't know where to look.
Thanks!
Edit: To clarify some, $\mathbf{w}$ are the beamforming weights, $\mathbf{s}(\theta_0)$ is the steering vector towards the angle of interest, $R$ is the covariance matrix of the input signal, and $\lambda$ is a tuning parameter.
 A: The complex partial derivative is defined in such a way that $\frac{\partial z}{\partial \bar{z}} = \frac{\partial z}{\partial z^*} = 0$.  If $J(\bf w)$ is considered to be a function of $\bf w$ and not of its complex conjugate, ${\bf w}^*$, the partial derivative of $J$ wrt ${\bf w}^*$ is zero.  When one takes the partial derivative wrt a variable, the complex conjugate of the variable is treated as a constant, and vice versa.  
In your author's equation for the cost function, $\bf w$ is a column that contains the (complex) variables and ${\bf w}^H$ is a row of the complex conjugates of those variables.  When we take the derivative of the expression for the cost function, we note that the first two terms are linear in ${\bf w}^H$, so the partial derivative is just the coefficient of ${\bf w}^H$.  The last third term in the expression for $J$ is considered to be a constant, so its partial derivative is zero.  
Please note that this has not been a rigorous explanation.  
A: Let me answer this question.
In fact, this is the well known Wiener solution.
\begin{align}
J(\mathbf{w}) &= 
\mathbf{w}^HR\mathbf{w} +\lambda^*[\mathbf{w}^H\mathbf{s}-1] + \lambda[\mathbf{s}^H\mathbf{w}-1]
\end{align}
where we write $\mathbf{s}=\mathbf{s}(\theta_0)$ for simply. Then
\begin{align}
\frac{\partial}{\partial\mathbf{w}^*}J(\mathbf{w}) = 2R\mathbf{w} + 2\lambda^*\mathbf{s}= 0.
\end{align}
It is worth noting that we use
\begin{align}
\frac{\partial}{\partial\mathbf{w}^*} (\mathbf{w}^HR\mathbf{w} ) &= 2R\mathbf{w} \\
\frac{\partial}{\partial\mathbf{w}^*} (\lambda^*[\mathbf{w}^H\mathbf{s}-1]) &=\lambda^*\mathbf{s}\\
\frac{\partial}{\partial\mathbf{w}^*} (\lambda[\mathbf{s}^H\mathbf{w}-1]) 
&= (\lambda \mathbf{s}^H)^H = \lambda^*\mathbf{s}.
\end{align}
For more reference, please see the derivative of complex vector.
