Is this subset a basis of the vector space $V$? 
Decide if the following subset is a basis of this vector space $V$:
$$\left\{v_{1},v_{2}\right\}= \left\{\begin{pmatrix} 1 \\  2 
\end{pmatrix},\begin{pmatrix} \alpha \\  1  \end{pmatrix}\right\}
\text{ in V = }\mathbb{R}^{2}$$ for an arbitrary fixed $\alpha \in
\mathbb{R}, \alpha \neq \frac{1}{2}$

I'm not sure how to do it but I think if $v_{1}$ and $v_{2}$ are linear independent then the subset will be a basis of the vector space $V$?
So I tried to do that:
$$x\begin{pmatrix} 1 \\  2 
\end{pmatrix}+y\begin{pmatrix} \alpha \\  1  \end{pmatrix}=\begin{pmatrix} 0 \\  0  \end{pmatrix}$$
$$I: x+y\alpha=0$$
$$II: 2x+y=0$$
$II: y=-2x$
Put that in $I:x-2x\alpha=0$
We need to divide by $x$ now and we must say $x \neq 0$ here else we cannot divide, so we see already this cannot be linear independent. But let's continue:
$1-2\alpha = 0$
$2\alpha = 1$
$\alpha = \frac{1}{2}$ task excluded this solution but anyway, conclusion stays this is not linear independent and thus the subset is not a basis of $V$.
I hope this is alright or is it done completely different?
 A: Your logic is flawed, and you make several statements that have no real basis.
For example, you say 

we must say $x \neq 0$ here else we cannot divide

But this is not true. You don't have to say that. Nobody is forcing you, and in fact, saying $x\neq 0$ is only one part of the solution.
remember: what you want to prove is that $x=0,y=0$ is the only solution to the equation $x v_1 + y v_2 = 0$.

What you have to do is seperate the cases. Indeed, if $xv_1+yv_2=0$, then $x-2x\alpha = 0$.
This leaves you with two options.


*

*$x=0$. In this case, you can go on to show that $y=0$ as well.

*$x\neq 0$. In this case, you have already proven that $\alpha=\frac12$ which is impossible.


So, you can conclude that the two vectors are independent.

Alternatively, you can transform the equation $x-2x\alpha=0$ into $$x(1-2\alpha)=0$$
and then divide the equation by $(1-2\alpha)$ (why can you do that?) to get $x=0$.
A: 
Put that in $I:x-2x\alpha=0$
We need to divide by $x$ now and we must say $x \neq 0$ here else we cannot divide, so we see already this cannot be linear independent. But let's continue:

You don't have to divide; continue like this:
$$x-2x\alpha=0 \iff x(1-2\alpha)=0$$
So either $\alpha = \tfrac{1}{2}$, but this value is excluded, or $x=0$. So we have $x=0$ and from $y=-2x$, also $y=0$. But that means the vectors are in fact linearly independent!
Note that for a basis, you also need that the two vectors span $V$, but since the dimension is $2$, you may know that any 2 linearly independent vectors will indeed span the space.
A: To your proof: From  $x-2x\alpha=0$ we get $x(1- 2 \alpha)=0$. Hence $ \alpha=1/2$ or $x=0$. Since $ \alpha \ne 1/2$, we get $x=0$. From $2x+y=0$ we finally derive $y=0$. 
Thus :  $v_1$ and $v_2$ are linear independent .
