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".

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Linear in/dependency has to do with determining the dimension of the vector space spanned. – DonAntonio May 28 '14 at 3:58
This is a nice question. Linear dependence and span do have something to do with each other. They're tied up together in the concept of a basis (and dimension, in a sense.) I encourage you to read more about it! I can also write a little more, let me know if you are interested. – Alex Wertheim May 28 '14 at 3:59

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.
Ok so, the dimension of $\mathbb{R}^2$ is 2, therefore its basis must have $2$ vectors. Since a basis contains the minimum number of vectors needed to span $\mathbb{R}^2$, any set of vectors that supposedly span $\mathbb{R}^2$ must have, as a minimum, 2 vectors. Then $\{(1,1),(2,2)\}$ can't be a span because it has only one significant vector, whereas it needs at least 2, right? So a set of 3 vectors can span $\mathbb{R}^2$, as long as two of its vectors are linearly independent, yes? – Zol Tun Kul May 28 '14 at 4:17
You seem to have the idea. Yes, a set of three vectors can span $\mathbb{R}^2$ as long as it has two linearly independent vectors. Just a small thing about terminology, I wouldn't say things like "one significant vector". I understand what you mean, but try to use the terms that have been defined. – Michael Albanese May 28 '14 at 4:24
I would say "$\{(1, 1), (2, 2)\}$ doesn't span $\mathbb{R}^2$ because it is linearly dependent." – Michael Albanese May 28 '14 at 4:27