$G=(I_{k}|A)$ is a generator matrix of $\mathcal{C}$ iff $H=(-A^{T}|I_{n-k})$ is a control matrix of $\mathcal{C}$

Solution 1:

You're almost there. What you wish to show is that the codes defined by $G$ and $H$ are the same, i.e.,

$$ \{ c \in \mathbb{F}_q^n: c=uG, u \in \mathbb{F}_q^k \} = \{ c \in \mathbb{F}_q^n: cH^\mathrm{T}=0\}. $$

Denote the left set by $\mathcal{G}$ and the right one by $\mathcal{H}$.

You have already proven that $\mathcal{G} \subseteq \mathcal{H}$. Since $c\in\mathcal{G}$, it follows that $c = uG$ for some $u\in \mathbb{F}_q^k$ and thus $cH^\mathrm{T}= uGH^\mathrm{T}= u(A-A)= 0$ by definition of the matrices $G$ and $H$.

For the other direction $\mathcal{H} \subseteq \mathcal{G}$, we split the codeword into two parts, $c=(d,e)$, $d \in \mathbb{F}_q^k$, $e \in \mathbb{F}_q^{n-k}$. Observe that with this notation, $cH^\mathrm{T}=0 \Leftrightarrow -dA+e=0 \Leftrightarrow dA = e$. Thus, if we set $u=d$, we obtain $$ uG = (d,dA) = (d,e), $$ which proves that we can find a vector $u$ that generates $c$ and thus $\mathcal{H} \subseteq \mathcal{G}$.