Span of vectors with more entries in each vector than the amount of vectors I can easily visualize why $\begin{bmatrix}
    1\\ 0
\end{bmatrix}$ and $\begin{bmatrix}
    0\\ 1
\end{bmatrix}$ spans $R^2$. I have learned that we need at least two linearly independent vectors to span $R^2$, $3$ to span $R^3$ and so on... This is easy to visualize when the entries in the vectors are the same as the amount of total vectors.
But what happens if we for instance have the two vectors $\begin{bmatrix}
    1\\ 0 \\ 1 \\ 0
\end{bmatrix}$ and $\begin{bmatrix}
    0\\1\\0\\1 
\end{bmatrix}$.
Clearly,  the two vectors are linearly independent, but do they still only span $R^2$? If we add the two vectors we produce $\begin{bmatrix}
    1\\ 1\\1\\1
\end{bmatrix}$ which is a vector in $R^4$, right?
So my question boils down to what is the span of $\begin{bmatrix}
    1\\ 0 \\ 1 \\ 0
\end{bmatrix}$ and  $\begin{bmatrix}
    0\\1\\0\\1 
\end{bmatrix}$, and in general what is the span of a set of vectors with more entries in each vector than vectors in total?
 A: Your vectors live in $\mathbb{R}^4$, not in $\mathbb{R}^2$, and therefore it is out of the question that they span $\mathbb{R}^2$. Since they live in $\mathbb{R}^4$, they span a subspace of $\mathbb{R}^4$, whish happens to be$$\left\{\begin{bmatrix}x&y&x&y\end{bmatrix}^T\,\middle|\,x,y\in\mathbb{R}\right\}.$$
A: The span of the vectors you mention is a subspace of the vector space to which they belong. So $(1,0,1,0)$ and $(0,1,0,1)$ span a two dimensional subspace (since as you note the vectors are linearly independent) of $\mathbb{R}^4$. It is not $\mathbb{R}^2$ exactly as this subspace contains $4$-tuples, but it is `equivalen' (in some sense) to $\mathbb{R}^2$ if this is of interest to you.
So in general a span of $m<n$ linearly independent vectors in $\mathbb{R}^n$  is a $m$-dimensional subspace of $\mathbb{R}^n$.
A: You are right the two vectors are linearly independent and therefore they span a $2$ dimensional subspace in $\mathbb{R^4}$ that is in parametric form
$$\begin{bmatrix}
    x\\ y \\ z \\ w
\end{bmatrix}=s\cdot \begin{bmatrix}
    1\\ 0 \\ 1 \\ 0
\end{bmatrix}+t\cdot \begin{bmatrix}
    0\\ 1 \\ 0 \\ 1
\end{bmatrix}=\begin{bmatrix}
    s\\ t \\ s \\ t
\end{bmatrix}$$
with $s,t \in \mathbb{R}$, or in cartesian form


*

*$x=z$

*$y=w$
