Code is not cyclic for any q

Question

I have code $C$ over $F_p$ with generator matrix which looks like

$G = \begin{pmatrix} 0 &0& 0& 1& 0& 1& 1 &1\\ 1& 0 &0& 0 &1 &0 &1& 1\\ 1& 1& 0& 0& 0& 1& 0& 1\\ 1 &1& 1& 0 &0 &0& 1 &0\end{pmatrix}$

I need to show that this code is not cyclic for any $p$.

I constructed vector which looks like $$(c_2+c_3+c_4, c_3+c_4, c_4, c_1, c_2, c_1+c_3, c_1+c_2+c_4, c_1+c_2+c_3)$$ and its shifts but it didn't help me a lot.

Does somebody have any ideas how it can be proven?

Answer 1

Consider the word $(3,2,1,1,1,2,3,3)$ in $C$, which is the sum of the four rows. Now consider the cyclically shifted word $(3,3,3,2,1,1,1,2)$. If the code would be cyclic, this must be again of the form $(c_2+c_3+c_4, c_3+c_4, c_4, \ldots ,c_1+c_2+c_3)$. Hence we would obtain \begin{align*} c_2+c_3+c_4 & =3 \\ c_3+c_4 & = 3 \\ c_4 & = 3 \end{align*} so that $c_2=c_3=0$, $c_4=3$. Furthermore $c_1=2$. But then it follows that $(3,3,3,2,1,1,1,2)=(3,3,3,2,0,2,5,2)$, a contradiction in any field, because of $0\neq 1$.

Answer 2

A base for your code is a base for $\ker G$. So the code has dimension 4, and a base is $$ v_1=\begin{pmatrix}0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \\ 0 \\ -1\end{pmatrix} \quad v_2=\begin{pmatrix}-1 \\ 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \\ 0\end{pmatrix} \quad v_3=\begin{pmatrix}0 \\ -1 \\ 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0\end{pmatrix} \quad v_4=\begin{pmatrix}0 \\ 0 \\ -1 \\ 0 \\ 0 \\ 0 \\ 1 \\ 0\end{pmatrix} $$

The code was cyclic if and only if the base, shifted by one, is still a base of the code, but

$$ v_2=\begin{pmatrix}-1 \\ 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \\ 0\end{pmatrix} \quad v_3=\begin{pmatrix}0 \\ -1 \\ 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0\end{pmatrix} \quad v_4=\begin{pmatrix}0 \\ 0 \\ -1 \\ 0 \\ 0 \\ 0 \\ 1 \\ 0\end{pmatrix} \quad v_1=\begin{pmatrix}0 \\ 0 \\ 0 \\ -1 \\ 0 \\ 0 \\ 0 \\ 1\end{pmatrix} $$ Is a base, so it is cyclic for every p

Answer 3

$C$ must contain every cyclic shift of its codewords. Consider the cyclic shift of the last row of $G$, $x = \left( {01110001} \right)$. If $C$ is a cyclic code, $x$ must be a linear combination of the rows of $G$, i.e.

${x^T} = a\left( {\begin{array}{*{20}{c}} 0\\ 0\\ 0\\ 1\\ 0\\ 1\\ 1\\ 1 \end{array}} \right) + b\left( {\begin{array}{*{20}{c}} 1\\ 0\\ 0\\ 0\\ 1\\ 0\\ 1\\ 1 \end{array}} \right) + c\left( {\begin{array}{*{20}{c}} 1\\ 1\\ 0\\ 0\\ 0\\ 1\\ 0\\ 1 \end{array}} \right) + d\left( {\begin{array}{*{20}{c}} 1\\ 1\\ 1\\ 0\\ 0\\ 0\\ 1\\ 0 \end{array}} \right)$ for some $a,b,c,d \in {F_p}$.

Then, it is not hard to see that $b + c + d = 0\left( p \right)$ and $c + d = 1\left( p \right)$ from the sum of first and second components of the rows, respectively. From the fifth components, $b=0$. So, $c + d = 1\left( p \right)$ and $c + d = 0\left( p \right)$. This is not possible for all prime $p$.