How to determine whether $3$ vectors of $\mathbb R^2$ are linearly independent or dependent Say I have $3$ vectors of $\mathbb R^2$: $(-2,1), (1,3), (2,4)$. What do I have to do in order to show that these $3$ vectors of $\mathbb R^2$ are linearly independent or dependent? Thanks
 A: For linear independence you have to show that if $a(-2,1)+b(1,3)+c(2,4)=(0,0)$ then $a=b=c=0$. But, in this particular case we have $\dim\mathbb{R}^2=2$, and if these three vectors were linearly independent then you could extend them to form a basis, let's say $B$, so $\mathrm{card}(B)\geq 3$, which contradicts $\dim\mathbb{R}^2=2$.
A: You do not need to do anything. If you have $R^n$ space and have $k > n$ vectors, then at least one of them would be dependent from others
A: Let $v_1,v_2,v_3\in\mathbb{R}^2$ be any three arbitrary vectors. 
Define a matrix $A\in\mathbb{R}^{2\times 3}$ as follows.
$$A:=\begin{bmatrix} \biggl |& \biggl|&\biggl|\\v_1&v_2 &v_3\\\biggl|&\biggl|&\biggl|\end{bmatrix}$$
Let $\Psi(A)$ denote the rank of the matrix $A$. We have $\Psi(A)\le\min\{2,3\}=2$. But we know that $\dim \mathscr{N}(A)+\dim \mathscr{R}(A)=3$, where $\mathscr{N}(A)$ and $\mathscr{R}(A)$ denote the null space and row space of $A$ respectively. Now using the fact  that $\dim \mathscr{R}(A)=\Psi(A)$, we may write $ \dim \mathscr{N}(A)+\Psi(A)=3$, and hence $\dim \mathscr{N}(A)= 3-\Psi(A)$. Now because $\Psi(A)\leq 2$, we have $\dim \mathscr{N}(A)\ge 3-2=1$. In other words, the dimension of the null space of matrix $A$ is at least $1$. Therefore $x\in\mathscr{N}(A)$ for some $x\neq\mathbf{0}$, and hence $Ax=0$ for some $x\neq\mathbf{0}$. But $Ax$ is the linear combination of the column vectors $v_1,v_2,v_3$, and therefore there exists a non-zero combination of $v_1,v_2,v_3$ which gives $\mathbf{0}$. In other words, there exists $k_1,k_2,k_3$, not all of which are $0$, such that $\sum_{j} k_j v_j=0$. Thus by the definition of linear dependence, the vectors $v_1,v_2,v_3$ are linearly dependent.
