The first 'primordial' basis of a finite vector space Let's take a vector space $V $ and set $V= \mathbb{R}^3 $ for ease of mind. 
Usually we equip $V $ with the standard basis $\{e_1,e_2,e_3\} $ and we express all our vectors in that basis:
$$v = (a,b,c) = ae_1 + be_2 + ce_3 $$
Now, we might choose a different basis $\{f_1,f_2,f_3\} $, for example:
\begin{align*}
 f_1 = (1,0,0) \\
 f_2 = (1,1,0) \\
 f_3 = (1,1,1)
\end{align*}
but here I did what seems to me a sinister trick. I gave you the new basis vectors in terms of the old one! I have $f_3 $ as $e_1 + e_2 + e_3 $. 
The same thing happens for polynomial spaces, given $\{1,x,x^2 \} $ as a basis, I might want to choose $(1,x+1,1+x+\frac{1}{2}x^2) $ as a different basis, and that's also expressed in terms of the old basis.
Now my mean old uncle Bob comes along and tells me that he has a new basis:
\begin{align*}
 g_1 = (1,0,0) \\
 g_2 = (0,1,0) \\
 g_3 = (0,0,1)
\end{align*}
but what he doesn't tell me is that he's using $f_i $ as the basis to express his new basis, so that $g_i = f_i $ for all $i $! And remember, the $f_i $'s are really combinations of $e_i $'s.
It seems to me, then, that when people tell me that $e_1 = (1,0,0) $ it is without any reference to any other basis! It's like it's self-referential somehow, or it doesn't come from anywhere! How can I make sense of this?
It seems to me that $\{e_1,e_2,e_3\}$ is then some kind of 'special primordial' basis, and everything we do is implicitely in that basis, even when people choose to work in some other basis, for example the $f_i$'s! But that just doesn't feel right!
Thanks in advance
 A: All of this is just an artifact of the fact that you chose to consider the vector space $V=\mathbb{R}^3$.  By definition, any element of $\mathbb{R}^3$ has three coordinates.  The basis $(e_1,e_2,e_3)$ has the property that the coefficients when expressing an element $v=(v_1,v_2,v_3)\in V$ with respect to this basis are just the coordinates $v_1$, $v_2$, and $v_3$.  But this doesn't mean that your definition of the $f_i$ is in terms of the $e_i$.  You have defined $f_3=(1,1,1)$, which is a perfectly explicit element of $V$ whose definition in no way involves the $e_i$.  You're just observing that it is very simple to read off from this definition how to express $f_3$ as a linear combination of the $e_i$, namely $f_3=e_1+e_2+e_3$.
What your uncle Bob is doing is abusing notation.  He really shouldn't write $g_3=(0,0,1)$, since that literally means that $g_3$ is the triple $(0,0,1)\in\mathbb{R}^3$.  Instead what he should write is $g_3=f_3$, or $g_3=A(0,0,1)$ where $A:V\to V$ is the unique linear map such that $A(e_i)=f_i$ for each $i$.  Note that the $e_i$ are again creeping into the definition, but this is only a matter of convention: by convention we say that a triple $(a,b,c)$ "represents" a vector $v$ with respect to a basis $(f_1,f_2,f_3)$ if $v=A(a,b,c)$ where $A$ is the linear map sending $e_i$ to $f_i$.  There's nothing special about our choice of $e_i$ in this convention; it's just an arbitrary (but convenient) choice.
To put it another way, the $e_i$ are not "primordial", but we have simply chosen them as our "default" basis of $\mathbb{R}^3$ because they are convenient to think about.  It's a social convention, not any deep mathematical truth.
A: @Ian's comment above is spot-on, except for one small point. You can't say $\mathbb R^3$ was given to you as a real vector space of dimension $3$, because even though all such spaces are isomorphic to $\mathbb R^3$ only one of them actually is $\mathbb R^3$, which is defined to be the set of all ordered 3-tuples of real numbers with addition defined term-wise and scalar multiplication defined by 
$$
t (a, b, c) = (ta, tb, tc).
$$
(Some folks will say that $\mathbb R^3$ is the set of all functions from $\{0, 1, 2\}$ to $\mathbb R$, but folks who say that don't need an explanation of this question about bases, I believe.)
In general, $F^n$, where $F$ is any field, is defined analogously, and on such a "coordinate space", there's an interesting map from pairs of vectors to the underlying field, namely,
$$
(x_1, \ldots, x_n) \cdot (y_1, \ldots, y_n) \mapsto x_1y_1 + \ldots + x_n y_n.
$$
Under certain conditions on the field, this "dot product" operation happens to match the criteria for being an "inner product" on the vector space, but even when it's not, it may prove useful in various contexts. 
My colleague Phil Klein has written a linear algebra book (Coding The Matrix) that discusses some of this. 
