Using Steinitz Exchange Lemma to prove that a set of four vectors in $\mathbb{R}^2$ is linearly dependent Cheers, so I am asked to prove that in the vector space of $V = \mathbb{R}^2$, every set of 4 vectors is linearly dependent. I tried solving it using Steinintz Exchange Lemma, so:
Let $V = \mathbb{R}^2$ be the vector space of interest, and a set $\{ v_1, v_2 \}$ be a basis for our vector space (e.g. $\{ (1,0), (0,1) \}$). Now let $A \subset V, A = \{ a_1 , a_2 , a_3 , a_4 \} $. As $A \subset V$ we can say that: $$
a_i=\sum_{j=1}^2 \gamma_{ij}v_j \qquad(i=1,\dots,4)
$$ I then got a bit stuck, and saw a solution that proceeded by saying that:
If $\alpha_1a_1+\dots+\alpha_4a_4=0$, then
$$
0=\sum_{i=1}^4\alpha_ia_i=
\sum_{i=1}^4\alpha_i\biggl(\,\sum_{j=1}^2\gamma_{ij}v_j\biggr)=
\sum_{j=1}^2\biggl(\,\sum_{i=1}^4\alpha_i\gamma_{ij}\biggr)v_j
$$
so
$$
\sum_{i=1}^4\alpha_i\gamma_{ij}=0 \qquad(j=1,2)
$$ and since $4 > 2$ there are infinetely many solutions so A is not linearly independent.
Although, I understand the whole logic here, and why the result solves our question, why did we suppose that $\alpha_1a_1+\dots+\alpha_4a_4=0$, and especially why did we use the elements of A in tuples? Could it be done with any other way? Thanks for the help!
 A: You need to find scalars $\alpha_1, \alpha_2,\alpha_3,\alpha_4$ not all zero such that $\alpha_1a_1+\dots+\alpha_4a_4=0$. You found infinitely many solutions for $\alpha_1, \alpha_2,\alpha_3,\alpha_4$ so $\alpha_1=\alpha_2=\alpha_3=\alpha_4=0$ is not the only solution.
Here is another way, which doesn't use any theory regarding linear systems of equations. Let $$\begin{bmatrix} x_1 \\ y_1\end{bmatrix}, \begin{bmatrix} x_2 \\ y_2\end{bmatrix}, \begin{bmatrix} x_3 \\ y_3\end{bmatrix}, \begin{bmatrix} x_4 \\ y_4\end{bmatrix} \in \Bbb{R}^2$$
be arbitrary.

*

*If $x_1=x_2=y_1=y_2 = 0$, they are clearly linearly dependent.


*If any of $x_1,x_2,y_1,y_2$ is nonzero and $x_2y_1 = x_1y_2$ then from
$$x_2 \begin{bmatrix} x_1 \\ y_1\end{bmatrix} - x_1 \begin{bmatrix} x_2 \\ y_2\end{bmatrix} = y_2 \begin{bmatrix} x_1 \\ y_1\end{bmatrix} - y_1 \begin{bmatrix} x_2 \\ y_2\end{bmatrix} = 0$$
it follows that they are linearly dependent.


*If $x_2y_1 \ne x_1y_2$ then
$$(x_2y_4-x_4y_2)\begin{bmatrix} x_1 \\ y_1\end{bmatrix} + (x_4y_1-x_1y_4)\begin{bmatrix} x_2 \\ y_2\end{bmatrix} + \underbrace{(x_1y_2-x_2y_1)}_{\ne 0}\begin{bmatrix} x_4 \\ y_4\end{bmatrix} = 0$$
so they are linearly dependent.
