Question about eigendecomposition, signal subspace and their properties
My first swing at the answer had some very incorrect claims.


I do not have access to the article, so I am inferring some things from the portion posted in the question.


NOTA BENE: My arguments assume that the eigenvectors of $\mathbf{R}$ are arranged so that the first $n$ belong to the signal subspace and that the last $m-n$ belong to the noise subspace. That is how the formulas appear so clean.

Since $\mathbf{R}$ is a covariance matrix, it is a positive-definite Hermitian matrix. This ensures that


all of its eigenvalues are positive,
it is unitarily diagonalizable.


That last property means that there is a unitary matrix $\mathbf{U}$, whose columns are orthonormal eigenvectors of $\mathbf{R}$, such that $\mathbf{U}^{\dagger}\mathbf{R}\mathbf{U}$ is equal to a diagonal matrix that has $\mathbf{R}$'s eigenvalues on its diagonal:
\begin{eqnarray}
\mathbf{U}^{\dagger}\mathbf{R}\mathbf{U} = \textrm{diag}(\lambda_1,\ldots,\lambda_n,\underbrace{\sigma^2,\ldots,\sigma^2}_{\textrm{$m-n$ $\sigma^2$s}}).
\end{eqnarray}

Because the article mentions $\mathbf{E}_s^{\dagger}$ and $\mathbf{E}_n^{\dagger}$ and not $\mathbf{E}_s^{-1}$ and $\mathbf{E}_n^{-1}$ in the eigenvalue-eigenvector decomposition of $\mathbf{R}$, I suspect that the unitary diagonzaling matrix of $\mathbf{R}$ is $\mathbf{E}_s + \mathbf{E}_n$, where $\mathbf{E}_s^{\dagger}\mathbf{E}_n = \mathbf{O}$, an all-zero matrix. I will demonstrate using the decomposition given in the article.
\begin{eqnarray}
&amp;&amp; (\mathbf{E}_s^{\dagger} + \mathbf{E}_n^{\dagger})\color{red}{\mathbf{R}}(\mathbf{E}_s + \mathbf{E}_n)\\
&amp;=&amp; (\mathbf{E}_s^{\dagger} + \mathbf{E}_n^{\dagger})\color{red}{(\mathbf{E}_s\mathbf{\Lambda}_s\mathbf{E}_s^{\dagger} + \sigma^2\mathbf{E}_n\mathbf{E}_n^{\dagger})}(\mathbf{E}_s + \mathbf{E}_n)\\
&amp;=&amp; (\mathbf{E}_s^{\dagger}\mathbf{E}_s\mathbf{\Lambda}_s\mathbf{E}_s^{\dagger} + \sigma^2\mathbf{E}_n^{\dagger}\mathbf{E}_n\mathbf{E}_n^{\dagger})(\mathbf{E}_s + \mathbf{E}_n),
\end{eqnarray}
where I have already dropped $\mathbf{E}_s^{\dagger}\mathbf{E}_n$ and $\mathbf{E}_n^{\dagger}\mathbf{E}_s$, both of which are all-zero matrices. The next step is
\begin{equation}
\mathbf{E}_s^{\dagger}\mathbf{E}_s\mathbf{\Lambda}_s\mathbf{E}_s^{\dagger}\mathbf{E}_s + \sigma^2\mathbf{E}_n^{\dagger}\mathbf{E}_n\mathbf{E}_n^{\dagger}\mathbf{E}_n.
\end{equation}

The OP states that $\mathbf{E}_s$ is $m\times n$ and has rank $n$, where $n &lt; m$. But I suspect (hope) that $\mathbf{E}_s$ is $m\times m$ with rank $n$ and that and $\mathbf{E}_n$ is $m\times m$ with rank $m-n$, with the following structures:
\begin{eqnarray}
\mathbf{E}_s &amp;=&amp; (\mathbf{u}_1\cdots\mathbf{u}_n\underbrace{\mathbf{0}\cdots\mathbf{0}}_{m-n}),\\
\mathbf{E}_n &amp;=&amp; (\underbrace{\mathbf{0}\cdots\mathbf{0}}_{n}\mathbf{u}_{n+1}\cdots\mathbf{u}_m)
\end{eqnarray}
That is, $\mathbf{E}_s$ should have $n$ orthonormal columns followed by $m-n$ all-zero columns, while $\mathbf{E}_n$ starts with $n$ all-zero columns and ends with $m-n$ orthonormal columns.

If these conditions were true, then we would have
\begin{eqnarray}
\mathbf{E}_s^{\dagger}\mathbf{E}_s &amp;=&amp; \textrm{diag}(\underbrace{1,\ldots,1}_{n},\underbrace{0,\ldots,0}_{m-n}),\\
\mathbf{E}_n^{\dagger}\mathbf{E}_n &amp;=&amp; \textrm{diag}(\underbrace{0,\ldots,0}_{n},\underbrace{1,\ldots,1}_{m-n})
\end{eqnarray}
and
\begin{eqnarray}
&amp;&amp; (\mathbf{E}_s^{\dagger} + \mathbf{E}_n^{\dagger})\color{red}{\mathbf{R}}(\mathbf{E}_s + \mathbf{E}_n)\\
&amp;=&amp; \mathbf{E}_s^{\dagger}\mathbf{E}_s\mathbf{\Lambda}_s\mathbf{E}_s^{\dagger}\mathbf{E}_s + \sigma^2\mathbf{E}_n^{\dagger}\mathbf{E}_n\mathbf{E}_n^{\dagger}\mathbf{E}_n\\
&amp;=&amp; \textrm{diag}(\lambda_1,\ldots,\lambda_n,\underbrace{\sigma^2,\ldots,\sigma^2}_{m-n})
\end{eqnarray}




Even if $\mathbf{E}_s$ is $m\times n$ and $\mathbf{E}_n$ is $m\times(m-n)$, we still have many of the same properties as long as
\begin{eqnarray}
\mathbf{E}_s &amp;=&amp; (\mathbf{u}_1\cdots\mathbf{u}_n),\\
\mathbf{E}_n &amp;=&amp; (\mathbf{u}_{n+1}\cdots\mathbf{u}_m).
\end{eqnarray}
In particular, we have
\begin{equation}
\mathbf{E}_s^{\dagger}\mathbf{E}_n = \mathbf{O}_{n\times(m-n)}
\end{equation}
and
\begin{equation}
\mathbf{E}_s\mathbf{\Lambda}_s\mathbf{E}_s^{\dagger} + \sigma^2\mathbf{E}_n\mathbf{E}_n^{\dagger}
=
\sum_{k=1}^{n}\lambda_k\mathbf{u}_k\mathbf{u}_k^{\dagger}
+
\sum_{k=n+1}^{m}\sigma^2\mathbf{u}_k\mathbf{u}_k^{\dagger},
\end{equation}
which is a spectral decomposition of a Hermitian matrix such as $\mathbf{R}$.