Linear independance of (binary) vectors here is what i want to do.
I have 4 vectors (in $\mathbb{Z}^{12}_{2}$), lets say
$v_1 = (1, 0, 0, 1, 0, 1, 0, 0, 1, 1, 1, 0)$
$v_2 = (1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 1, 0)$
$v_3 = (1, 1, 0, 0, 0, 1, 0, 1, 0, 1, 1, 0)$
$v_4 = (1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1)$
These vectors are supposed to build a 4-dimensional vector subspace of $\mathbb{Z}^{12}_{2}$.
For that to be the case they have to be linearly independent. 
Now i want to check if they are. I could do that with a system of linear equations but accordung to my script i should: "[...] compute all 16 linear combinations in $\mathbb{Z}_{2}$ and check if any of them occur twice. If so, $v_1,v_2,v_3,v_4$" are not linearly independent."
I thought about that for over an hour now and i dont really get what i am supposed to do and whyit is correct. Would really appreciate some help.
 A: You can immediately see that $v_4$ is independent from the others since its 12th element is $1$, whereas the others have $0$ in that position; there is no way that a combination of $0$s can create a $1$.
Similarly, $v_2$ and $v_3$ are independent of the others based on the 3rd and 2nd elements respectively.
You can then see that $v_1$ is not a combination of the others. Therefore they are all independent, and no need to compute any of the 16 combinations.
A: The subspace they span is $$\{av_1+bv_2+cv_3+dv_4:a,b,c,d\in \mathbb{Z}_2\}$$
However, since $|\mathbb{Z}_2|=2$, there are only 16 elements (at most), corresponding to choosing $0$ or $1$ for each of $a,b,c,d$.  Compute these 16, and see if there are any duplicates.  Done.
A: Compute all the $16$ possibilities, if two of them happen to be equal, then you can set them equal, subtract one from the other one and find a new linear combination that has a non-zero coefficient.
In other words, write all linear combinations of the form
$$f(\alpha_1,\alpha_2,\alpha_3,\alpha_4)= \alpha_1 v_1 + \alpha_2 v_2 + \alpha_3 v_3 + \alpha_4 v_4$$
You have $2^4$ inputs.
If $v_1,v_2,v_3,v_4$are linearly independent, then each of the vectors $f(\alpha_1,\alpha_2,\alpha_3,\alpha_4)$ must be different, because if not, we find two different representations for the same vector in the space which violates linear independence. (Remember: Linear independence gives you the ability to write vectors in its span "uniquely"). 
A: You can show that a collection of vectors is linearly independent iff there is a unique way to write each element of its span as a linear combination of those vectors. (The trick is that the difference of two duplicates is zero...) This is why searching for duplicates will tell you if the vectors are independent. 
A: @Thèophile 's answer is the best, I think. There's no way zeros can produce a $1$. So, $v_4$ is independent of the other vectors because of the last component.
Then $v_3$ is independent of $v_1$ and $v_2$ because of the sixth component. And finally, $v_2$ is independent of $v_1$ because of the third component.
