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Let $C*$ be the length 8 binary code obtained by adding a parity check symbol to each word in $C$. (so a word $c_1, c_2, c_3, c_4, c_5, c_6, c_7$ is extended to the word $c_1, c_2, c_3, c_4, c_5, c_6, c_7, c_8$ for which $\sum_{i=0}^8 c_i =0$. Show that $C*$ is self dual (I.e show that $C*=C*^{\bot}$.

Where $C*^{\bot}$ is the orthogonal complement of $C*$

I am given the generator matrix for $C$:

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

So I think $C*$ is given by:

$$G*= \begin{bmatrix} 1 & 1 & 0 & 1 & 0 & 0 & 0 & 1\\ 0 & 1 & 1 & 0 & 1 & 0 & 0 & 1\\ 0 & 0 & 1 & 1 & 0 & 1 & 0 & 1\\ 0 & 0 & 0 & 1 & 1 & 0 & 1 &1 \end{bmatrix}$$

I can show $C*$ and $C*^{\bot}$ have the same dimensions but don't know what to do from here.

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$G^*$ is also a parity check matrix of the code $C^*$ since $G^*G^{*T}=0$ and $\dim\ker\left(G^{*}\right)=\text{rank}\left(G^*\right)=4$. Since a parity check matrix of $C^*$ is a generator matrix of $C^{*\perp}$, we get that $C^*$ is a self dual code.

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