I was wondering about, if we have two vectors $x=\begin{pmatrix} 2 \\ 1 \end{pmatrix}$ and $y=\begin{pmatrix} 3 \\ 2 \end{pmatrix} \in \Bbb R^2$, then $y-x=\begin{pmatrix} 1 \\ 1 \end{pmatrix}$.
If we think about the subspace spanned by this $y-x$ vector, then we know it could be represented by the line along the $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$ diagonal.
However, I’m wondering how this generalizes to higher dimensions. Imagine the $3 \times 2$ case, where $x$ and $y$ are both $3 \times 2$ matrices and would represent planes in three dimensional space. How do you represent the difference between the basis vectors? I imagine you can’t just look at $y-x$ because then you’re only looking at the difference between the first column of $y$ and the first column of $x$ and so on for the second column. I hope this idea makes sense, thanks if possible.
I think I see your confusion. It may be helpful to go back to the definitions.
A set $V$ over the field $F$ equipped with the binary operations $+:(V \times V) \rightarrow V$ and $\cdot: (F\times V) \rightarrow V$ is said to be a vector space provided (the list of 8 criteria that you're probably familiar with). The elements of $V$ are called vectors and the elements of $F$ are called scalars.
So in the hypothetical you provided, where $V$ is the set of $3 \times 2$ matrices, the matrices are vectors. It doesn't really makes sense to say something like "the matrix represents planes in two dimensions", because the matrices are just vectors in the space.
Where you may be getting confused is that linear transformations can be represented by matrices, e.g. if $A$ is $3 \times 2$, it might act on vectors in $\mathbb{R}^2$ and brings them to $\mathbb{R}^3$. In general, we can't say the geometric interpretation of this transformation is bringing two-space to two-space, since $A$ might very well take the plane to a line through the origin or to the origin itself.
Now to your question: "How do you represent the difference between the basis vectors?"
First you need to specifiy a basis for $\mathbb{R}^{3\times 2}$. A natural one might be:
$\Bigg\{\begin{gather*}\begin{bmatrix}1 & 0 \\ 0 & 0 \\ 0 & 0\end{bmatrix}, \begin{bmatrix}0 & 1 \\ 0 & 0 \\ 0 & 0\end{bmatrix}, \begin{bmatrix}0 & 0 \\ 1 & 0 \\ 0 & 0\end{bmatrix}, \begin{bmatrix}0 & 0 \\ 0 & 1 \\ 0 & 0\end{bmatrix}, \begin{bmatrix}0 & 0 \\ 0 & 0 \\ 1 & 0\end{bmatrix}, \begin{bmatrix}0 & 0 \\ 0 & 0 \\ 0 & 1\end{bmatrix}\end{gather*}\Bigg\}$
Then the difference between some basis vectors, say $e_1-e_2$, is just $\begin{bmatrix}1 & -1 \\ 0 & 0 \\ 0&0\end{bmatrix}$ as expected, and the subspace spanned by this vector is just $\begin{bmatrix}c & -c \\ 0 & 0 \\ 0&0\end{bmatrix}$ where $c\in F$ (presumably $F=\mathbb{R}$ here, though you could of course choose others).