# Show $[6,3]$ quaternary code $\mathcal{G}_6$ is self-dual.

The $[6,3]$ quaternary code $\mathcal{G}_6$ has generator matrix $G_6$ in standard form given by $G_6=\left[\begin{array}{ccc|ccc} 1 & 0 & 0 & 1 & \omega & \omega\\ 0 & 1 & 0 & \omega & 1 & \omega\\ 0 & 0 & 1 & \omega & \omega & 1 \end{array}\right]$. Show $\mathcal{G}_6$ is self-dual.

As $G_6$ is generator matrix of $\mathcal{G}_6$, $G$ is parity check matrix for $\mathcal{G}_6^{\bot}$. Also $\mathcal{G}_6,\mathcal{G}_6^{\bot}$ both are $3\times 6$ matrix. So far, I just gather all information from definitions and propositions from the book. I stuck at this step as I don't quite see any information that I can use to show $\mathcal{G}_6$ is self dual.

I found a similar question Self-Dual Code; generator matrix and parity check matrix, but I don't quite understand the answer. Can someone explain the answer in more detial or give me a hint to keep going to work on the question? Thanks.

• What is $\omega$? Commented Dec 20, 2015 at 23:21
• @GitGud I don't know, the question doesn't say anything about it. Commented Dec 20, 2015 at 23:24
• The code is quaternary. That means that the alphabet has size four. Because self-duality is discussed, the alphabet is at least a ring, most likely a field. I'm willing to wager that the alphabet here is $GF(4)$ aka $\Bbb{F}_4=\{0,1,\omega,\omega+1=\omega^2\}$. Commented Dec 21, 2015 at 7:55
• Anyway. It is obvious that $G_6$ has the correct size and rank. All you need to is to show that its rows are orthogonal to themselves and each other. In other words: calculate the 3x3 matrix $G_6G_6^T$, and check that you get all zeros. Commented Dec 21, 2015 at 7:57
• And, if you don't know what $\omega$ is and how you do arithmetic in $GF(4)$, you need to work on that first. Luckily that is not difficult at all. It is characteristic two, meaning that $\omega+\omega=0$. Commented Dec 21, 2015 at 7:59

Assuming that my educated guess about the alphabet being $GF(4)=\{0,1,\omega,\omega+1=\omega^2\}$ is correct, then the claim is false as stated.
The inner product of the 1st and 2nd rows is $$0+0+0+\omega+\omega+\omega^2=\omega^2\neq0.$$ Therefore not all words of $C$ are orthogonal to all the others.
The same holds for any pair of distinct rows of $G$.
So please check the source of the question, and give us more of context. The same calculation (more or less) shows that the resulting code is self-dual, if the alphabet is the ring of binary dual numbers $\Bbb{Z}_2[\omega]$, with $\omega+\omega=0=\omega^2$. I would need to check the source to see whether that is a live possibility. Such rings have been proposed and studied.