This may sound like a silly question, but:
I know that for every $n$ dimensional vector space $V$ you can get its equations by rewriting $V=S[v_1,...,v_n]$ with $\{v_1,...,v_n \}$ a basis for V and then writing $(x_1,...,x_n) \in V \iff (x_1,...,x_n) = \lambda_1v_1+...+\lambda_nv_n$ and then working that out. But now I am trying to do this for $\mathbb{R^2}$ and I get: $(x_1,x_2) \in \mathbb{R^2} \iff x_1=\lambda_1 ,x_2=\lambda_2$. This makes sense ofcourse, because every $(x_1,x_2)$ with components element of the field the vector space is made over is in $\mathbb{R^2}$, but I would like to get to one equation.
For example: $2x_0+x_1-3x_2+2x_3=0$ is a subspace of $\mathbb{R^4}$ that uses 4 variables and has $0$ on the right side of the equation. I want to express $\mathbb{R^2}$ in an equation that uses 2 variables and has 0 on the right side aswel!
How do I get there?
EDIT: Jeroen's comment cleared some things up for me, as did jorki's answer. However I still have a question regarding representing a vector space with homogenous linear equations:
If we want to find the equation for the two-dimensional vector subspace of $\mathbb{R^4}$ that is $W=S[(1,-2,0,-14),(0,0,1,5)]$ we do this as follow:
$\begin{array}{l}(x_1,...,x_4) \in W \\ \iff \exists \lambda,\mu \in \mathbb{R} : (x_1,...,x_4) = \lambda(1,-2,0,-14)+ \mu(0,0,1,5) \\ \iff \begin{cases} x_1=\lambda \\ x_2=-2\lambda \\ x_3=\mu \\ x_4=-14\lambda+5\mu \end{cases} \\ \iff \begin{cases} 2x_1+x_2=0 \\ 14x_1 -5x_3+x_4=0 \end{cases} \end{array}$
These final two equations are the result.
HOWEVER, when I try to do this with the following subspace of $\mathbb{R^4, W=S[(1,2,1,3),(2,4,3,5)]}$, I get stuck at iff number two, because I can't get the $\lambda, \mu$ out of the system. How can I find the equation for that plane? Thanks a lot!