Array-processing: eigenstructure of the spatial covariance matrix

Question

I've been staring at the following underlined statement from this book for hours, and I cannot for the life of me figure out how it can be right: Since the vector $AR_{ss}Aq_s$ is a linear combination of the columns of $A$, the eigenvector $q_s$ lies in the column space of $A$

My Understanding

For some definitions:

Spatial covariance matrix: $R_{xx} = AR_{ss}A^H + \sigma^2I_{N \times N}$

The book states that $R_{xx}$ is Hermitian, which makes sense as it is a vector product of the form $\vec{x} (\vec{x}^H)$.

$A$ is an $N \times r$ matrix defined as the following matrix of linearly independent column vectors, where each vector is $N$ elements long:

$$A = \begin{bmatrix} \vec{a}(\theta_0) & \vec{a}(\theta_1) & \ldots & \vec{a}(\theta_{r-1}) \\ \end{bmatrix}$$

$R_{ss}$ is an $r \times r$ matrix (we may ignore its contents for this purpose). $q_s$ is an eigenvector of $A R_{ss} A^H$. First of all I'm convinced that when they write:

$$A R_{ss} A q_s$$

they really mean:

$$A R_{ss} A^H q_s$$

otherwise it simply would not make sense from the start. My confusion is when they say $A R_{ss} A^H q_s$ is a linear combination of the column vectors of $A$. At first I thought maybe it just wasn't obvious to me, so I started doing the calculations as an exercise.

My calculation (please don't make me type all this into a text equation editor).

My Question

I think my calculation is correct, but there was a lot to keep track of, so I did not do the multiplication with $q_s$ because it's trivial, and it doesn't solve the problem. How can the products of different elements (complex conjugated no less) of the vectors in $A$ end up as a linear combination of the columns of $A$?

Am I forgetting something fundamental here? Maybe something to do with the fact that $q_s$ is an eigenvector of $A R_{ss} A^H$?

Answer 1

Consider the first part of the statement (with the book typo corrected)

the vector $AR_{ss}A^Hq_s$ is a linear combination of the columns of $A$

You can confirm this by simply denoting the product $R_{ss}A^Hq_s$ as (vector) $z=[z_0,z_1,\ldots, z_{r-1}]^T$. It then follows that $AR_{ss}A^Hq_s=Az=\sum_{i=0}^{r-1}z_i \vec{a}(\theta_{i})$, i.e., a linear combination of the columns of $A$.

Now, since $q_s$ is an eigenvector of $AR_{ss}A^H$, it follows that $$ \begin{align} q_s &=\frac{1}{\sigma_s^2}AR_{ss}A^Hq_s \\ &=\frac{1}{\sigma_s^2}\sum_{i=0}^{r-1}z_i \vec{a}(\theta_{i}), \end{align} $$ where $\sigma_s^2$ is the corresponding eigenvalue. The above equation means that $q_s$ is also a linear combination of the columns of $A$, which confirms the second part of the statement

the eigenvector $q_s$ lies in the column space of $A$