# DFT matrix on GF(4)

1. May 7, 2013

### 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:
1. Why $G$ is a DFT matrix?