I have an exercise that says, "Let $A$ be a matrix, and let $C$ be the code consisting of all solutions to $Ax = 0$. If $A$ has neither a column of zeros nor two equal columns, prove that the distance of $C$ is at least $3$. (Hint: If $v$ has weight $1$ or weight $2$, look at how $Av$ can be written in terms of the columns of $A$.)"
Finston, David R.. Abstract Algebra (Springer Undergraduate Texts in Mathematics and Technology) (p. 27). Springer International Publishing. Kindle Edition.
This is under a topic of binary code, so everything is in regards to Z2
I have no idea how to start it or what I am trying to get to. I'm also terrible at proving things. Any ideas?
The minimum distance of any code $C \subseteq \Bbb Z_2^n$ is $$\min \{d_H(u,v): u,v \in C , u \neq v\}$$ but because the code $C$ is a linear subspace we can also see it as
$$\min \{\|v\|_H: v \in C, v \neq 0\}$$
where $\|v\|= d_H(v,0)$ the number of $1$'s in the codeword: if $u$, $v$ realise the minimum, $u-v$ realises the minimal norm and is also in $C$ (of course in this field $u-v=u+v$ anyway) and the converse is true because $0 \in C$.
And a vector $(c_1,c_2,\ldots,c_n)$ is in $C$, the null space of $A$ iff $\sum_{i \in J} A_i = 0$ where $A_i$ is the $i$-th column of $A$, and $J=\{i: c_i=1\}$. This is just by how matrix multiplication over $\Bbb Z_2$ works.
So if we have a vector of minimal Hamming-norm $d$, it has exactly $d$ ones, and so there are $d$ many columns of $A$ that sum to $0$. And if we have $d$ many columns that sum to $0$, the corresponding vector that has $1$ at those indices and $0$ elsewhere lies in the null space of $A$ and so in $C$ and the minimum distance (norm) is $\le d$. These observations establish you claim: if the minimum dustance is $3$ all columns are non-zero and no pair of them can sum to $0$, e.g. be equal, and conversely.