Appendix A: Factorization of the data matrix $\mathbf{M}$

Because equation 7 is a singular value decomposition (SVD) of the signal matrix $\mathbf{S}$ , the left matrix in equation 7 is a unitary matrix:

$\displaystyle \mathbf{I}=\mathbf{U}^S(\mathbf{U}^S)^H=[\mathbf{U}_1^S\quad \mat... ...[\begin{array}{c} (\mathbf{U}_1^S)^H \\ (\mathbf{U}_2^S)^H \end{array}\right].$

(21)

Combining equations 4, 8, and 21, we can derive:

$\begin{displaymath}\begin{split} \mathbf{M}&=\mathbf{S}+\mathbf{N} \\ &=\mathbf... ...}(\mathbf{N}^H\mathbf{U}_2^S)^H \end{array}\right], \end{split}\end{displaymath}$

(22)

where $\Sigma_1$ and $\Sigma_2$ are introduced matrices and are diagonal and positive definite.

In order to make the right matrix orthonormal, we make two assumptions:

The noise is close to white noise in the sense that $\mathbf{N}\mathbf{N}^H=\lambda\mathbf{I}$ .
The signal is orthogonal to the noise in the sense that $\mathbf{S}\mathbf{N}^H=\mathbf{0}$ .

We let $\mathbf{P}^H$ denote the right matrix of the last equation in 22, then

$\begin{displaymath}\begin{split} &\mathbf{P}^H\mathbf{P} \\ &=\left[\begin{arra... ...{11} & p_{12}\\ p_{21} & p_{22} \end{array}\right] \end{split}\end{displaymath}$

(23)

where

$\begin{displaymath}\begin{split} p_{11}&=(\Sigma_1)^{-1}((\mathbf{U}_1^S)^H\math... ...{I}+(\Sigma_1^S)^2)(\Sigma_1)^{-1} \\ &=\mathbf{I} \end{split}\end{displaymath}$

(24)

when $\Sigma_1=\sqrt{\lambda\mathbf{I}+(\Sigma_1^S)^2}$ .

$\begin{displaymath}\begin{split} p_{12}&=(\Sigma_1)^{-1}((\mathbf{U}_1^S)^H\math... ...}_1^S)^H\mathbf{N}^H\mathbf{U}_2^S )(\Sigma_2)^{-1} \end{split}\end{displaymath}$

(25)

Since $\mathbf{U}^S$ is an orthogonal matrix, then $(\mathbf{U}_1^S)^H\mathbf{U}_2^S=\mathbf{0}$ . Since $\mathbf{S}\mathbf{N}^H=\mathbf{0}$ , then $\mathbf{U}_1^S\Sigma_1^S(\mathbf{V}_1^S)^H\mathbf{N}^H=\mathbf{0}$ , thus $(\mathbf{V}_1^S)^H\mathbf{N}^H=\mathbf{0}$ . In the same way, since $\mathbf{NS}^H=\mathbf{0}$ , thus $\mathbf{NV}_1^S=\mathbf{0}$ . Then,

$\begin{displaymath}\begin{split} p_{12}&=\mathbf{0}. \end{split}\end{displaymath}$

(26)

$\begin{displaymath}\begin{split} p_{21}&=(\Sigma_2)^{-1}((\mathbf{U}_2^S)^H\math... ...{V}_1^S\Sigma_1^S)(\Sigma_1)^{-1} \\ &=\mathbf{0}. \end{split}\end{displaymath}$

(27)

$\begin{displaymath}\begin{split} p_{22}&=(\Sigma_2)^{-1}((\mathbf{U}_2^S)^H\math... ...(\lambda\mathbf{I})(\Sigma_2)^{-1}\\ &=\mathbf{I}, \end{split}\end{displaymath}$

(28)

when $\Sigma_2=\sqrt{\lambda\mathbf{I}}$ . Thus, we prove that $\mathbf{P}^H\mathbf{P}=\mathbf{I}$ when $\Sigma_1$ and $\Sigma_2$ are appropriately chosen, and $\mathbf{P}$ is orthonormal.

2020-02-21