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...