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My question specifically relates to the case where the vector that the matrix blocks were formed from have lower orders than the dimensions of the sub matrices.

Consider a vector (filter) $a[k]$ of length $\ell_a$. We wish to convolve this vector with another $x[k]$ of length $\ell_x > \ell_a$. The output is a vector $b[k]$ of length $\ell_b = \ell_a +\ell_x -1$. This convolution can be performed as a matrix multiplication. \begin{eqnarray} \textbf{A}\textbf{x} = \textbf{b} \end{eqnarray}

$\textbf{A}$ is a $\ell_b \times \ell_x$ where both are greater than $\ell_a$ so there is zeropadding.

Now consider a set of these matrices (not all the same but each made with different $a[k]$'s but which all have the same length $\ell_a$) are arranged into a block matrix where each block is a toeplitz convolution matrix. \begin{eqnarray} \hat{\textbf{A}} = \begin{bmatrix}&\textbf{A}_{11} & \textbf{A}_{11} & \dots & \textbf{A}_{1M} \\ & \textbf{A}_{21} & \textbf{A}_{22} & \dots & \textbf{A}_{2M}\end{bmatrix} \end{eqnarray}

I present here a $2\ell_b \times M\ell_x$ case because thats what Im interested in but It could be generalized to the $N\ell_b \times M\ell_x$ case.

I assume that the vectors/filters $a_{nm}[k]$ are not relatively prime but are not identical (maybe this could even be the case if they are identical).

My question is: Are all columns or rows in $\hat{\textbf{A}}$ linearly independent. i.e. does $\hat{\textbf{A}}$ have rank equal to one of its dimensions?

An answer would mean the world to me!

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