# Are all columns or rows in a block toeplitz convolution matrix linearly independent?

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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