I've been reading about residual codes, and have come across several statements that I am having trouble showing.
(1.) By using the residual code, one can show that a $[16, 5, 8]$ binary code contains the all-one codeword.
(2.) All $[16, 5, 8]$ binary codes are equivalent.
For Part 1, I'm not sure how to use the residual codes. I'm having trouble with Part 2 as well. Any help would be greatly appreciated. Thanks in advance!
Edit: Below is the definition of a residual code.
Let $\mathcal{C}$ be an $[n,k]$ code and let $c \in \mathcal{C}$ have weight $w$. Let $\mathcal{I}$ be the set of coordinates on which $c$ is nonzero. Then, the residual code of $\mathcal{C}$ with respect to $c$ is the code of length $n-w$ punctured on all the coordinates of $\mathcal{I}$.
Partial Solution: Below is my thought process for solving Part $1$.
Let $\mathcal{C}$ be a $[16, 5, 8]$ binary code. Pick a codeword $c \in \mathcal{C}$ that has a weight of $8$. Then, the residual code $\mathcal{C}_1$ of $\mathcal{C}$ with respect to $c$ is an $[16 - 8, 5-1, d_1] = [8, 4, d_1]$ code, where $d_1 \geq \lceil{\frac{8}{2}}\rceil = 4$. By the Singleton Bound, $d_1 \leq 8 - 4 + 1 = 5$. Thus, $4 \leq d_1 \leq 5$. However, if $d_1 = 5$, then we have equality in the Singleton Bound, which implies $\mathcal{C}_1$ is MDS, which implies $\mathcal{C}_1$ is trivial, a contradiction. Therefore, $d_1 = 4$.
By permuting coordinates, we may assume $\mathcal{C}$ has a generator matrix $ G = \begin{bmatrix}1\ldots 1 & 0 \ldots 0\\G_0 & G_1\end{bmatrix} $ where the first row is $c$ and $G_1$ is a generator matrix of $C_1$.
Now, every $[8,4,4]$ binary code is equivalent to the extended Hamming code $\widehat{\mathcal{H}_3}$. Thus, $G_1$ is a generator matrix for $\widehat{\mathcal{H}_3}$. Therefore, $G_1$ can take on the following form:
$ G_1 = \begin{matrix} 1&1 &1 &1 &1 &1 &1 &1 \\\ 1 &1 &0 &1 &0 &0 &0 &1 \\\ 1& 0& 0& 0& 0& 1& 1& 1 \\\ 1& 1& 0& 0& 1& 1& 0& 0 \end{matrix} $
This is where I'm stuck. Can I say the code $\mathcal{C}_0$ corresponding to $G_0$ is also a $[8,4,4]$ code? Clearly $G_0$ is a $4 \times 8$ matrix, and by the Singleton Bound with a similar argument as above, its minimum distance $d_0 \leq 4$. If so, we can make it have the same generator matrix as $G_1$, which will give is the all-one codeword in $\mathcal{C}$.
Then, I think Part $2$ would follow since $G_1$ and $G_0$ are both equivalent to the generator matrix of $ {\mathcal{H}_3}$.