I'm looking for a simplification of the following expression with a "norm-writing" : $$\begin{pmatrix} (\bar{z}-\bar{\mu})^{T}& (z-\mu)^{T} \end{pmatrix} \begin{pmatrix} \Gamma &C \\ \bar{C}& \bar{\Gamma} \end{pmatrix}^{-1} \begin{pmatrix} z-\mu\\ \bar{z}-\bar{\mu} \end{pmatrix}$$ with $z\in \mathbb{C}^{K}, \mu\in \mathbb{C}^{K},\Gamma\in \mathbb{C}^{K\times K},C\in \mathbb{C}^{K\times K}$$
For example, with an easier case :
$$(x-x_{0})^{T}\Sigma^{T}\Sigma(x-x_{0})=\left \| \Sigma(x-x_{0}) \right \|_{2}^{2}$$
with $x\in \mathbb{C}^{N}, x_{0}\in \mathbb{C}^{N},\Sigma\in \mathbb{C}^{N\times N}$
Notes :
- $\bar{x}$ denotes the complex conjugated
- $x^{T}$ denotes the transpose
- I imagine a expression like this : $\left \| U(z-\mu) \right \|_{2}^{2}$ with U a complex matrix and a demonstration with the complex conjugate transpose $x^{*}$ but I can't see where to go
- If you want to know where this expression come from, it's a part of the probability density fonction for complex normal distribution (https://en.wikipedia.org/wiki/Complex_normal_distribution section "density function")
Thanks.
HINT (hopefully it serves your purpose):
According to block-based matrix inversion, say you define $$\begin{bmatrix} \Gamma & C \\ \overline{C} & \overline{\Gamma} \end{bmatrix}^{-1} = \begin{bmatrix} \Lambda_{aa} & \Lambda_{ab} \\ \Lambda_{ba} & \Lambda_{bb} \end{bmatrix} \ ,$$ then $$\left[\left(\overline{z} - \overline{u}\right)^T \ \ \left({z} - {u}\right)^T\right] \begin{bmatrix} \Lambda_{aa} & \Lambda_{ab} \\ \Lambda_{ba} & \Lambda_{bb} \end{bmatrix} \begin{bmatrix}\left(\overline{z} - \overline{u}\right) \\ \left({z} - {u}\right)\end{bmatrix} \ .$$ Now you can perform "completing the square"...
But there are some cross-coupling terms, which will remain in quadratic form.