Does linear dependency have anything to do when determining a span? 
Q: Does $\{(1,1) , (2,2)\}$ span $\mathbb{R}^2$? A: No, because they are linearly dependent.

I agree that it doesn't span $\mathbb{R}^2$, but from my understanding, linear dependency has nothing to do with that:
All that matters is whether you are capable of producing any vector in $\mathbb{R}^2$ by some sort of linear combination of the elements there, that is:
$$c_1\cdot(1,1) + c_2\cdot(2,2) = (x,y)$$
For an arbitrary $x,y \in \mathbb{R}$ and some $c_1,c_2\in\mathbb{R}$. Since those vector are linearly dependent, we can remove the second one and are left with
$$c_1\cdot(1,1) = (x,y)$$
Which clearly doesn't span $\mathbb{R}^2$, because no matter what $c_1$, you can't end up with, dunno, $(2,3)$.
So my question is: does linear dependency have anything to do when determining a span?

I saw another example, but this time about $2x2$ matrices. Given four matrices, they asked whether they spanned all the $2x2$ matrices - then they answered "no because the fourth one is a multiple of the first".
A: Suppose $V$ is an $n$-dimensional vector space and $v_1, \dots, v_n \in V$. If $\{v_1, \dots, v_n\}$ spans $V$, then $\{v_1, \dots, v_n\}$ is linearly independent. Why? Well, if $\{v_1, \dots, v_n\}$ were linearly dependent, we could find a subcollection $\{v_{i_1}, \dots, v_{i_k}\}$ with $k < n$ which is linearly independent and still spans $V$. As $\{v_{i_1}, \dots, v_{i_k}\}$ is linearly independent and spans $V$, it is a basis for $V$. But $V$ is $n$-dimensional, so any basis for $V$ has $n$ vectors, so we have a contradiction. Therefore, $\{v_1, \dots, v_n\}$ is linearly independent.
We can also use similar reason in the other direction. That is, given $n$ vectors, we can tell whether they are linearly independent or not by considering their span. In summary, we have the following:

Theorem: Let $V$ be an $n$-dimensional vector space and let $v_1, \dots, v_n \in V$ (distinct). Then $\{v_1, \dots, v_n\}$ spans $V$ if and only if $\{v_1, \dots, v_n\}$ is linearly independent.

