I have to show that the Nordstrom-Robinson code $N$ is non-linear. I wanted to show this by finding a counter example to linearity:
Suppose for contradiction that $N$ is a linear $(16, 256, 6)$-code. Then by definition of linear code, $N \subseteq \Bbb{Z} {_2}^{16}$ is also a subspace of $\Bbb{Z} {_2}^{16}$.
By a theorem, since $N$ is a subspace of $\Bbb{Z} {_2}^{16}$, then it is closed under addition, such that if $x,y \in N$ then $x+y \in N$, and under scalar multiplication, such that if $a \in \Bbb{Z}_{3}$ and $x \in N$ then $ax \in N$.
I wanted to find a counter example for this, however I am having a hard time doing so without the generator matrix of $N$.
I know that $N$ is the code whose codewords are obtained by deleting the first 8 coordinate positions from the 256 words whose first 8 coordinates are one of $00000000, 10000001, 01000001, 00100001, 00010001, 00001001, 00000101, 00000011$ generated by the generator matrix $G'$, where $G'$ is obtained from the generator matrix of the Golay code $g_{24}$.
I think there is something simple that I am missing, any help is appreciated. Thank you!
Edit: I know there is no linear $(16, 256, 6)$ in general, but I am asked to prove that in another problem.
Consider two codewords $\mathbf{r_1}$ and $\mathbf{r_2}$ represented as the first two rows of $G'$. We may write $$\mathbf{r_1} = 10000001\mathbf{c} \qquad \text{and} \qquad \mathbf{r_2} = 01000001\mathbf{d}$$ where it's clear that $\mathbf{c}, \mathbf{d} \in \mathcal{N}$. Then the sum $\mathbf{r_1} + \mathbf{r_2} = 11000000(\mathbf{c} + \mathbf{d})$ has such a prefix that implies $(\mathbf{c} + \mathbf{d}) \not\in \mathcal{N}$. Thus $\mathcal{N}$ is not linear.