Rather an elementary question but I couldn't find one definite answer, but while working through my script I got the following intention :
For a Subspace of $R^2$ I do need two linear independent vectors. Even if both of the vectors have more than two dimensions, like $\begin{pmatrix}1\\ 0\\ 1\end{pmatrix} , \begin{pmatrix}0\\ 1\\ 0 \end{pmatrix} $
is that correct ?
What makes me wonder here, is that any linear combination of these two is never possible to draw in a just two dimensional system, but it was a correct answer to one of our exercises.
Edit: So I have got a formal definition from a very common textbook for advanced mathematics:
" The dimension of a vector space, is the number of elements in the set of a basis" , while a basis was defined as " a set of linear independent elements".
So I think I am mixing up two different terms here: Dimension and space of R^2:
The dimensions of the two vectors above is 2, while the subspace they are in is R^3 - is it like that?
Let $X$ be some (linear) space of dimension $n \in \mathbb{N}$. Then the definition of $Y$ being a subspace of $X$ always contains the condition $$ Y \subseteq X.$$ This is why your example does not work. A basis of a subspace of $\mathbb{R}^2$ has to be a set of elements of $\mathbb{R}^2$ as well. Returning to your question we have the following:
If $Y$ is a $m$-dimensional subspace of $X$ then any basis of $Y$ consists of (exactly) $m$ linearly independant vectors (with $m \leq n$).