# Question about eigendecomposition, signal subspace and their properties

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

In it, it is mentioned:

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),

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!

• "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. – Marcus Müller May 27 '20 at 14:16
• OK! I will edit it, thanks! – tyrela May 28 '20 at 1:24
• 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. – Joe Mack May 28 '20 at 20:02
• $\mathbf{E}_s$ 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. – Joe Mack May 28 '20 at 20:28
• 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? – tyrela May 29 '20 at 2:59

## 1 Answer

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: $$\begin{eqnarray} \mathbf{U}^{\dagger}\mathbf{R}\mathbf{U} = \textrm{diag}(\lambda_1,\ldots,\lambda_n,\underbrace{\sigma^2,\ldots,\sigma^2}_{\textrm{m-n \sigma^2s}}). \end{eqnarray}$$

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 $$$$\mathbf{E}_s^{\dagger}\mathbf{E}_s\mathbf{\Lambda}_s\mathbf{E}_s^{\dagger}\mathbf{E}_s + \sigma^2\mathbf{E}_n^{\dagger}\mathbf{E}_n\mathbf{E}_n^{\dagger}\mathbf{E}_n.$$$$

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 $$$$\mathbf{E}_s^{\dagger}\mathbf{E}_n = \mathbf{O}_{n\times(m-n)}$$$$ and $$$$\mathbf{E}_s\mathbf{\Lambda}_s\mathbf{E}_s^{\dagger} + \sigma^2\mathbf{E}_n\mathbf{E}_n^{\dagger} = \sum_{k=1}^{n}\lambda_k\mathbf{u}_k\mathbf{u}_k^{\dagger} + \sum_{k=n+1}^{m}\sigma^2\mathbf{u}_k\mathbf{u}_k^{\dagger},$$$$ which is a spectral decomposition of a Hermitian matrix such as $$\mathbf{R}$$.

• really thanks！ I think I get it. – tyrela May 30 '20 at 11:27