How to calculate the gradient of matrix equation Short question: How do I calculate the gradient of the $MSE(a, b)$ equation below?

Longer explanation: This problem arises, while I'm following a derivation of a term for an optimal beamvector $a$ in a data transmission. The mean square error (MSE) of this data transmission is calculated as follows:
$$MSE(a, b) = a^H(Hbb^HH^H+R_n)a + 1 - a^HHb - b^HH^Ha$$
where:


*

*$a$, $b$: vectors, which can be chosen

*$H$, $R_n$: matrices, which are fixed

*$a^H$: denotes the Hermitian adjoint of $a$


The vector $a$ can be optimized (in dependece of $b$) by setting the gradient of the MSE to zero.
The problem is that I don't know how to calculate the gradient when the equation has the above form. The $a^H$ at the beginning and the $a$ at the end of the first summand irritates me...
The answer shall be:
$$ a^* = (Hbb^HH^H+R_n)^{-1}Hb = R_n^{-1}Hb\frac{1}{1+b^HH^HR_n^{-1}Hb}$$
But how to calculate this?

Update:
Using equations from The Matrix Cookbook I got this far:
$$\frac{\partial MSE(a, b)}{\partial a} = \frac{\partial}{\partial a} \left[ a^H\left(Hbb^HH^H+R_n\right)a\right] + \frac{\partial}{\partial a} 1 - \frac{\partial}{\partial a} \left[a^HHb\right] - \frac{\partial}{\partial a} \left[b^HH^Ha\right]$$
With


*

*$\frac{\partial}{\partial a} 1 = 0$

*$\frac{\partial b^TX^TDXx}{\partial X} = D^TXbc^T + DXcb^T$ (Cookbook (74))


I get:
$$\frac{\partial MSE(a, b)}{\partial a} = (Hbb^HH^H+R_n)^Ha + (Hbb^HH^H+R_n)a - \frac{\partial}{\partial a} \left[a^HHb\right] - \frac{\partial}{\partial a} \left[b^HH^Ha\right]$$
And that's it. I don't even know if I used equation (74) from the cookbook right, but it was the closed equation for the first summand. I'm sorry, I just don't get it...
 A: I'm not sure whether the following results hold for complex cases.
Let all the vectors and matrices be real valued. Then
$$A=a^TBa+1-a^THb-b^TH^Ta$$
where $B=Hbb^TH^T+R_n$. $B$ is symmetric if $R_n$ is symmetric. Then
$$dA=da^TBa+a^TBda-da^THb-b^TH^Tda$$
Let the gradient be zero.
$$a^T(B^T+B)-2b^TH^T=0$$
If $B$ is symmetric, we have $2Ba=2Hb$ which implies
$$a^*=B^{-1}Hb=(Hbb^TH^T+R_n)^{-1}Hb$$
But for the rest of your expected answer, I'm not sure. Because $$(cc^T+R_n)^{-1}c=\frac{R_n^{-1}c}{c^TR_n^{-1}c+1}, c=Hb$$
implies
$$cc^TR_n^{-1}=c^TR_n^{-1}cI$$
Take trace on both sides of the above equation. The equation holds only when dimension is one.
A: Define two new variables
$$\eqalign{
c = Hb,\quad M = cc^H+R_n \\
}$$
Write the cost function in terms of these new variables.
Then calculate the gradient with respect to $a$
while treating $a^H$ as a constant
(i.e. the Wirtinger derivative).
$$\eqalign{
 \phi &= a^HMa + 1 - c^Ha - a^Hc \\
d\phi
 &= a^HM\,da - c^Hda \\
 &= (M^Ha - c)^Hda \\
\frac{\partial\phi}{\partial a} &= (M^Ha - c)^H \\
}$$
Set the gradient to zero and solve
$$\eqalign{
M^Ha &= c \\
a &= (M^H)^{-1}c \\
  &= (cc^H+R_n^H)^{-1}c \\
  &= (Hbb^HH^H+R_n^H)^{-1}Hb \\
}$$
This equals the prescribed answer only if $\,R_n^H=R_n$
