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

I am trying to prove that, ''given a C [n, k, d]-linear code, $G=({I}_{k}|A)$ is a generator matrix of C iff $H=(-{A}^{T}|{I}_{n-k})$ is a control matrix of C''.

Firstly, I have supposed that $G=({I}_{k}|A)$ is a generator matrix of C; we name the columns of A as ${a}_{1},{a}_{2},..,{a}_{n-k}$; if $x=({x}_{1},...,{x}_{k})\in {\mathbb{F}}_{q}^{k}$, I can codify this into a word of the code by taking the generator matrix and multiplying it this way: $xG=({x}_{1},...,{x}_{k},x\cdot {a}_{1},...,x\cdot {a}_{n-k})\in \mathcal{C}$, where $\cdot $ is the dot product of two vectors. I also know the following result:

H is a control matrix of C if the following holds: $x\in \mathcal{C}$.

So, If I show $H({x}_{1},...,{x}_{k},x\cdot {a}_{1},...,x\cdot {a}_{n-k}{)}^{T}=(0,...,0{)}^{T}$, I would have show that $x\in \mathcal{C}$; this is very easy to show taking on account how H is constructed.

For the other direction, I would have to show that $H{x}^{T}=(0,..,0{)}^{T}$ has, as a system of equations, as solutions all the words of the code... Here I get stucked.

Secondly, I would have to prove that, assuming $H=(-{A}^{T}|{I}_{n-k})$ is a control matrix of C, $G=({I}_{k}|A)$ is a generator matrix of C. For this I suppose I would have to solve the system $H{x}^{T}=(0,...,0{)}^{T}$ (which, by hipothesis I know has as solutions the words of the code), and then show that each of that words can be generated by G, i.e., that, given a code word, I can find a vector in ${\mathbb{F}}_{q}^{k}$ such that the product of that vector by G is the code word firstly given. Nevertheless I am not sure this is the best approach...

Any help, guidance, or anything will be very helpful.