I don't really think that representation theory is required here, it's easier than that. (Although representation theory is really cool!!)
Every finite dimensional vector space over the real numbers is isomorphic to R^n, for some n. But there is not always a canonical or "unique" isomorphism. I think the real difference -- and it is a subtle one -- is that R^n always comes with an "obvious" choice of basis, and many vector spaces don't.
I think the following may be the easiest interesting example. Consider the subspace of R^3, consisting of all (x, y, z) for which
x + y + z = 0.
This is a 2-dimensional real vector space. It is isomorphic to R^2, although in some sense it does not "present as such": you have to choose an isomorphism to R^2 if you want to "treat it as R^2", and there is no single choice that stands out.
In practice, one would not necessarily construct an isomorphism to R^2, or (more or less equivalently) exhibit a basis, before working with this vector space directly.
Every finite dimensional vector space over the real numbers is isomorphic to R^n, for some n. But there is not always a canonical or "unique" isomorphism. I think the real difference -- and it is a subtle one -- is that R^n always comes with an "obvious" choice of basis, and many vector spaces don't.
I think the following may be the easiest interesting example. Consider the subspace of R^3, consisting of all (x, y, z) for which
x + y + z = 0.
This is a 2-dimensional real vector space. It is isomorphic to R^2, although in some sense it does not "present as such": you have to choose an isomorphism to R^2 if you want to "treat it as R^2", and there is no single choice that stands out.
In practice, one would not necessarily construct an isomorphism to R^2, or (more or less equivalently) exhibit a basis, before working with this vector space directly.