On picking a basis in a vector space Given a finite dim. arbitrary vector space $V$, is picking a basis the same as picking an isomorphism into the set of $n$-tuples?
This really confuses me, everyone seems to pass to vectors without caring that the actual sets are different.
 A: Let $n = \dim V$. Picking an isomorphism $V \to \mathbb{R}^n$ is the same as picking an ordered basis of $V$.
If you have an isomorphism $f\colon V \to \mathbb{R}^n$, define $v_i \in V$ as the unique vector of $V$ such that $f(v_i) = e_i$, for all $i = 1, \dotsc, n$. Since $f$ is an isomorphism, $\{v_1, \dotsc, v_n\}$ is a basis of $V$. Then you can endow it with the natural order, i.e. $(v_1, \dotsc, v_n)$ is the ordered basis.
If you have an ordered basis $(v_1, \dotsc, v_n)$ of $V$, then there is a unique linear map $f\colon V \to \mathbb{R}^n$ such that $f(v_i) = e_i$ for all $i = 1, \dotsc, n$. Since $\{ e_1, \dotsc, e_n \}$ is a basis of $\mathbb{R}^n$, $f$ is an isomorphism. Notice here that we need the basis of $V$ to be ordered, otherwise we could choose many different isomorphisms.
Once you have fixed either an ordered basis of $V$ or the corresponding isomorphism $f\colon V \to \mathbb{R}^n$, you can identify each vector $v$ with $f(v)$. This is convenient because it allows you to treat any finite-dimensional vector space as a space of $n$-tuples (which are easy to handle on paper). But of course $v$ and $f(v)$ are actually different objects.
