Reduce computational cost of IPA

[tky823]

Summary

We might reduce the computational cost of IPA using the representation of block matrix.

[tky823]

$$\begin{align} \boldsymbol{\Pi}_{n} &= (\bar{\boldsymbol{E}}_{n}; \boldsymbol{e}_{n}) \\\ \boldsymbol{\Pi}_{n}\tilde{\boldsymbol{U}}_{in}\boldsymbol{\Pi}_{n}^{\mathsf{T}} &= \left( \begin{array}{cc} \boldsymbol{\Phi}_{in} & \boldsymbol{\mu}_{in} \\\ \boldsymbol{\mu}_{in}^{\mathsf{H}} & \eta_{in} \end{array} \right) \end{align}$$ $$\left\{(\boldsymbol{\Pi}_{n}\tilde{\boldsymbol{U}}_{in}\boldsymbol{\Pi}_{n}^{\mathsf{T}})^{*}\right\}^{-1} ={\displaystyle \left( \begin{array}{cc} \left(\boldsymbol{\Phi}_{in}^{*} - \dfrac{\boldsymbol{\mu}_{in}^{*}\boldsymbol{\mu}_{in}^{\mathsf{T}}}{s_{in}}\right)^{-1} & -\dfrac{(\boldsymbol{\Phi}_{in}^{*})^{-1}\boldsymbol{\mu}_{in}^{*}}{s_{in}} \\\ -\dfrac{\boldsymbol{\mu}_{in}^{\mathsf{T}}(\boldsymbol{\Phi}_{in}^{*})^{-1}}{s_{in}} & \dfrac{1}{s_{in}} \end{array} \right) },$$

where $s_{in}$ is a scalar value:

$$s_{in} := \eta_{in} - \boldsymbol{\mu}_{in}^{\mathsf{T}}(\boldsymbol{\Phi}_{in}^{*})^{-1}\boldsymbol{\mu}_{in}^{*}.$$

Since $\bar{\boldsymbol{E}}_{n}$ is equal to

$$\bar{\boldsymbol{E}}_{n}=\boldsymbol{\Pi}_{n}^{-1}\bar{\boldsymbol{E}}_{N},$$

$\boldsymbol{C}_{in}$ can be rewritten as

$$\begin{align} \boldsymbol{C}_{in} &= \bar{\boldsymbol{E}}_{n}^{\mathsf{T}}(\tilde{\boldsymbol{U}}_{in}^{-1})^{*}\bar{\boldsymbol{E}}_{n} \\\ &= (\boldsymbol{\Pi}_{n}^{-1}\bar{\boldsymbol{E}}_{N})^{\mathsf{T}}(\tilde{\boldsymbol{U}}_{in}^{-1})^{*}\boldsymbol{\Pi}_{n}^{-1}\bar{\boldsymbol{E}}_{N} \\\ &= \bar{\boldsymbol{E}}_{N}^{\mathsf{T}}\boldsymbol{\Pi}_{n}^{-\mathsf{T}}(\tilde{\boldsymbol{U}}_{in}^{-1})^{*}\boldsymbol{\Pi}_{n}^{-1}\bar{\boldsymbol{E}}_{N} \\\ &= \bar{\boldsymbol{E}}_{N}^{\mathsf{T}}\left\{(\boldsymbol{\Pi}_{n}\tilde{\boldsymbol{U}}_{in}\boldsymbol{\Pi}_{n}^{\mathsf{T}})^{*}\right\}^{-1}\bar{\boldsymbol{E}}_{N} \\\ &= \left(\boldsymbol{\Phi}_{in}^{*} - \dfrac{\boldsymbol{\mu}_{in}^{*}\boldsymbol{\mu}_{in}^{\mathsf{T}}}{s_{in}}\right)^{-1}. \end{align}$$