Lately, I am readig a paper titled A Subspace Method for Estimating Sensor Gains and Phases.

In it, it is mentioned:

enter image description here

There are $m$ sensors in the array, $n$ known narrowband far-field signals($m \gt n$, equal power). $R$ is covariance matrix of received signal, $\Gamma \in \mathbb{C}^{m \times m}$ is the diagonal matrix of unknown sensor gains and phases, which this paper tried to solve. $A\in \mathbb{C}^{m \times n}$ is the steering matrix. $E_s$ are eigenvectors associated with principal eigenvalues. $^\dagger$ represents conjugate transpose.

It's easy to know $E_s \in \mathbb{C}^{m \times n}$ and $rank(E_s)=n.$

My first question: I can't figure out why the eq(6) holds. Did I miss some properties of the signal subspace $E_sE_s^H?$

Then the equation was transformed into eq(9),

enter image description here

where $D_i$ is each column of $A$(in diagonal form) and $v$ is column vector formed form diagonal elements in $\Gamma$.

My second question: Why $W_i=D_i^HE_sE_s^HD_i$ has n unity eigenvalues and m-n zero eigenvalues?(underlined in blue).

Thank for your patience to read my questions!

  • $\begingroup$ "I'm reading a paper": Well then, give us the full title and authors of that paper, so that we have context! Otherwise, you'd have to define what $E_S^\dagger$ means. $\endgroup$ May 27, 2020 at 14:16
  • $\begingroup$ OK! I will edit it, thanks! $\endgroup$
    – tyrela
    May 28, 2020 at 1:24
  • $\begingroup$ I lack access to the article, so I will address what I can glean. Since $\mathbf{R}$ is a covariance matrix, it is Hermitian positive-definite. This ensures that (1) it has all positive eigenvalues and (2) it is unitarily diagonalizable. There is some unitary matrix $\mathbf{U}$ such that $\mathbf{R} = \mathbf{U}\textrm{diag}(\lambda_1,\ldots,\lambda_k,\sigma^2,\ldots,\sigma^2)\mathbf{U}^{\dagger}$, where I am assuming that the signal part and the noise part are orthogonal. $\endgroup$
    – Joe Mack
    May 28, 2020 at 20:02
  • $\begingroup$ $\mathbf{E}_s$ is $m\times n$, and $n < m$. It consists of $n$ orthonormal columns $\mathbf{u}_1,\ldots,\mathbf{u}_n\in\mathbb{C}^m$. $\mathbf{E}_s$ is not unitary; it is not even square. But it has a nice property kind of like unitarity. The product $\mathbf{E}_s\mathbf{E}_s^{\dagger}$ is an $m\times n$ matrix times an $n\times m$ matrix, so it is $m\times m$. The upper-left $n\times n$ sub-matrix of this product is the $n\times n$ identity matrix $\mathbf{I}_n$. The rest of the product is all zeros. $\endgroup$
    – Joe Mack
    May 28, 2020 at 20:28
  • $\begingroup$ Really thanks! It also explained my second question. But I don't know why $E_s$ and $E_sE_s^{\dagger}$ have these properties. Is there any proof of it? $\endgroup$
    – tyrela
    May 29, 2020 at 2:59

1 Answer 1


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} && (\mathbf{E}_s^{\dagger} + \mathbf{E}_n^{\dagger})\color{red}{\mathbf{R}}(\mathbf{E}_s + \mathbf{E}_n)\\ &=& (\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)\\ &=& (\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 $n < m$. 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 &=& (\mathbf{u}_1\cdots\mathbf{u}_n\underbrace{\mathbf{0}\cdots\mathbf{0}}_{m-n}),\\ \mathbf{E}_n &=& (\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 &=& \textrm{diag}(\underbrace{1,\ldots,1}_{n},\underbrace{0,\ldots,0}_{m-n}),\\ \mathbf{E}_n^{\dagger}\mathbf{E}_n &=& \textrm{diag}(\underbrace{0,\ldots,0}_{n},\underbrace{1,\ldots,1}_{m-n}) \end{eqnarray} and \begin{eqnarray} && (\mathbf{E}_s^{\dagger} + \mathbf{E}_n^{\dagger})\color{red}{\mathbf{R}}(\mathbf{E}_s + \mathbf{E}_n)\\ &=& \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\\ &=& \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 &=& (\mathbf{u}_1\cdots\mathbf{u}_n),\\ \mathbf{E}_n &=& (\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 $\mathbf{R}$.

  • $\begingroup$ really thanks! I think I get it. $\endgroup$
    – tyrela
    May 30, 2020 at 11:27

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