I tried to solve this problem by directly expanding it. But I need some opinions on my solution. Here is the problem definition:
$\lambda=\frac{1}{|| \mathbf{H}^H\mathbf{u}||^2}\mathbf{u}^H\mathbf{H}\mathbf{G}^H\mathbf{G}\mathbf{H}^H\mathbf{u}$ where $\mathbf{H}\in \mathcal{C}^{N_B \times N_A}$ and $\mathbf{G}\in \mathcal{C}^{N_E \times N_A}$. Elements of $\mathbf{H}$ and $\mathbf{G}$ independently follow the complex Gaussian with zero mean and unit variance, i.e., \begin{equation} h_{i,j}\sim CN(0,1), \text{ for $i=1,\dots,N_B$ and $j=1,\dots,N_A$} \end{equation} \begin{equation} g_{m,n}\sim CN(0,1), \text{ for $m=1,\dots,N_E$ and $n=1,\dots,N_A$} \end{equation}
Also, $\mathbf{H}$ and $\mathbf{G}$ are independent, and each row vector of the matrices, $\mathbf{h}_i$ for $i=1,\dots,N_B$ and $\mathbf{g}_m$ for $m=1,\dots,N_E$, are independent of each other in their matrices. $\mathbf{u}$ is an eigenvector of $\mathbf{H}\mathbf{H}^H$, so $||\mathbf{u}||^2 = 1$.
Here is my first question: Do $\mathbf{h}_i\mathbf{g}^H_m$ and $\mathbf{g}_m\mathbf{h}^H_i$ follow the complex Gaussian, i.e., $\mathbf{g}_m\mathbf{h}^H_i$ and $\mathbf{h}_i\mathbf{g}^H_m \sim CN(0,1)$ for all $i$ and $m$? And why?
Assuming they do, the next step for solution is:
It is known that $\frac{1}{|| \mathbf{H}^H\mathbf{u}||^2}=\frac{1}{N_AN_B}$. And when I define $X_{i,m} = \mathbf{h}_i\mathbf{g}^H_m=\mathbf{g}_m\mathbf{h}^H_i$, then all $X_{i,m}$ are complex Gaussian with zero mean and unit variance.
Then the diagonal part of $\mathbf{H}\mathbf{G}^H\mathbf{G}\mathbf{H}^H$ is: \begin{equation} diag\left[\sum^{N_E}_{m=1}X^2_{1,m},\sum^{N_E}_{m=1}X^2_{2,m}, \dots,\sum^{N_E}_{m=1}X^2_{N_B,m}\right] \ where\ X_{i,m} \sim CN(0,1) \end{equation}
And the $(l,k)$-th non-diagonal part $(l\neq k)$ of $\mathbf{H}\mathbf{G}^H\mathbf{G}\mathbf{H}^H$ is: \begin{equation} \sum^{N_E}_{m=1}X_{l,m}X_{k,m} \ where\ X_{i,m} \sim CN(0,1) \end{equation}
Here I have the second question: Can I assume the non-diagonal elements of $\mathbf{H}\mathbf{G}^H\mathbf{G}\mathbf{H}^H$ to be zero as $N_E$ and $N_A$?
If so, here is the next step:
$\mathbf{H}\mathbf{G}^H\mathbf{G}\mathbf{H}^H$ is approximated as a diagonal matrix, i.e., . As a result, when the diagonal elements are combined by $\mathbf{u}^H$ and $\mathbf{u}$ at each end, considering the normalizing factor $\frac{1}{|| \mathbf{H}^H\mathbf{u}||^2}=\frac{1}{N_AN_B}$, $\lambda \sim \chi^2(k=2N_E)$, i.e., $\lambda$ follows the chi-squared distribution with $2N_E$ degree of freedom.
Am I on the right track? For your convenience, I also attach my hand-written derivation.
Please note that the problem starts from a wireless communication problem.
