Why is a Reed-Solomon kernel matrix a DFT matrix?

[ait.abd]

I am reading the following paper:
Soft-decision decoding of polar codes with Reed-Solomon kernels

On the last line of the page 319 (page 3 of the pdf) the author says "and $G$ is a Reed-Solomon kernel, which is in fact a DFT matrix".

$G$ is defined on the page 321 (page 5 of the pdf) with possible change in the order of rows as
$$
G = \left( \begin{array}{cccc}
1 & 1 & 1 & 0 \\
1 & \alpha & \alpha^2 & 0 \\
1 & \alpha^2 & \alpha & 0 \\
1 & 1 & 1 & 1\end{array} \right),
$$

where $\alpha$ is a primitive element of $\mathbf{F}_{2^2}$.

I do not understand why the author calls $G$ as a DFT matrix, because DFT matrix does not have zeros in its general form. The general form that I am considering is the following:
Wikipedia Link.

Can anyone explain the following:
1. Why $G$ is a DFT matrix?
2. If it is a DFT matrix, how can we implement it using FFT? I am looking for the butterfly structure that will implement it.

Thanks for your time.

 

[fresh_42]

I can only assume that it is sufficient to consider the DFT submatrix, because the paper is about computational complexity and zeros might not count, as they do not lead to calculation steps.