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:

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