Spanning $\Bbb{R}^2$ Vector Space Does anyone have an idea about how to prove this:

Let $v_1, v_2 \in \mathbb R^2$. Prove: $\operatorname{span}\{v_1, v_2\} = \mathbb R^2$ if and only if $v_1, v_2$ are linearly independent.

I thought about showing that $v_1, v_2$ are the unit vectors of $\mathbb R^2$, but then someone told me that this assumption can be wrong.
 A: We need to break this problem into two parts.

*

*If ${v_1,v_2}$ are linearly dependent then they do not span $\mathbb{R}^2$

*If $v_1,v_2$ are linearly independent then they do span $\mathbb{R}^2$
For the first part ask yourself what does it mean for two vectors to be linearly dependent? How will the span look in this case?
For part two you need to explain why every vector in $\mathbb{R}^2$ will be in the span of $v_1$ and $v_2$. In other words, you need to explain why for every vector $v \in \mathbb{R}^2$ we can solve the equation $v = av_1 +bv_2$. Do you know how to solve an equation like this? They should have taught you how to do this in class. You need to explain why you know your method will work.
A: On the one hand, if $\mathcal{B} = \{v_{1},v_{2}\}$ is LI, we conclude $\operatorname{span}\{v_{1},v_{2}\} = \mathbb{R}^{2}$. That is because $\dim\mathbb{R}^{2} = 2$.
On the other hand, if $\operatorname{span}\{v_{1},v_{2}\} = \mathbb{R}^{2}$, then $\mathcal{B} = \{v_{1},v_{2}\}$ is LI. Otherwise we could conclude that $v_{1} = \alpha v_{2}$, that is to say, $\operatorname{span}\{v_{1},v_{2}\} = \operatorname{span}\{v_{2}\} = \mathbb{R}^{2}$, which contradicts the fact that $\dim\mathbb{R}^{2} = 2$.
