Why does $\mathbb{R}^n$ have a "canonical" basis if vector spaces Why does $\mathbb{R}^n$ have a "canonical" basis, if vector spaces are supposed to have none?
More generally, for every field $F$ and every integer $n$, why does the "space of $n$-tuples" $F^n$ have a "canonical" basis (as an $n$-dimensional vector space over $F$), if vector spaces are supposed to have none?
(I don't know if the same thing can be asked of modules, but modules are weird. Maybe of free modules?)
Set $n$ to be $2$.
In $\mathbb{R}^2$ you have the basis $\{(1,0), (0,1)\}$. Since $\mathbb{R}^2$ is a (2D) vector space (over $\mathbb{R}$), this basis is just as arbitrary as any other basis. But it certainly doesn't look arbitrary. (The basis $\{(\pi,e), (\sqrt{2},\sqrt[3]{216156757151})\}$ is just as much of a basis, but it seems "more arbitrary" than the other one, and one ever uses it.)
Sure, I know this isn't really the "canonical" basis for $\mathbb{R}^2$ (or $F^2$), but people even go as far as calling is the standard basis! Might as well call it canonical. It even features the additive and multiplicative identities of the field, and nothing else. It's as if it were the basis with "minimal complexity", or something like that, but such notions should make no sense in vector spaces.
So what's going on with the basis $\{(1,0), (0,1)\}$, and why is it "special", if bases are supposed to be the opposite of special?
(Ie. all bases are "equivalent", all are related by the action of the automorphism group $GL(n,\mathbb{R})$ on $\mathbb{R}^n$, every $n$-dimensional vector space over $\mathbb{R}$ is isomorphic to $\mathbb{R}^n$ (although the isomorphism depends on the basis), etc.)
Nevermind that $\{(1,0), (0,1)\}$ isn't even a basis, because $(1,0)$ and $(0,1)$ aren't even vectors, they're tuples, so the best one can say is that $(1,0)$ is the coordinate tuple of infinitely many elements of the 2D $\mathbb{R}$-vector space $\mathbb{R}^2$, one for each basis. So the elements of the vector space $\mathbb{R}^2$ don't even look like $(1,0)$ or $(0,1)$, but they look like $a_0\vec b_0 + a_1\vec b_1$, where $\vec b_0,\vec b_1$ are vectors that form a basis and $a_0,a_1$ are scalars.
So then people say that vectors in $\mathbb{R}^2$ are linear combinations $a_0 \vec e_0 + a_1 \vec e_1$, where $\{\vec e_0, \vec e_1\}$ is the "standard" basis, and sometimes they'll define $\vec e_0$ as the "vector" $\begin{bmatrix}
1 \\
0 
\end{bmatrix}$, but that's not even a vector, it's a $2 \times 1$ matrix, and matrices need a basis in order to be well-defined, so the whole definition is circular.
Something fishy is going on with bases and vector space isomorphims/automorphisms...
(Consider this question a continuation of this)
 A: The Euclidean plane as a 2-dimensional real vector space (with standard operations) doesn't have a canonical basis. The real vector space of 2-tuples of real numbers (with standard operations) does. The basis you give as the canonical basis is objectively less arbitrary than your other example because the elements $0$ and $1$ are singled out as special elements in the definition of a field.
The moment you slap a Cartesian coordinate system onto your plane, this sets up an explicit bijection between points and tuples, and the canonical basis transfers from one to the other. Or is it that once you've chosen a basis for your plane, the canonical basis of the tuples gives you a bijection between the two spaces, which again gives you a Cartesian coordinate system? Maybe a bit of both?
A: Let's say you are a "purist" and develop linear algebra based on the notion of an abstract vector space and are not even introduced to $\mathbb{F}^n$ with the standard vector space structure as an example. So you covered fields, vector spaces, subspaces, linear maps and so on. You might not even seen a concrete example of a vector space or a basis.
At some point, you define the notion of a basis and prove that if $V$ has a finite (ordered) basis $\mathcal{B} = (b_1,\dots,b_n)$ then any vector $v \in V$ has a unique representation as $v = x_1 b_1 + \dots + x_n b_n$ with $x_1,\dots,x_n \in \mathbb{F}$. This sets up a bijection of sets between $V$ and the space of tuples $\mathbb{F}^n$. Let's call this bijection which depends on $\mathcal{B}$ by $\varphi_{\mathcal{B}} \colon \mathbb{F}^n \rightarrow V$. It is given by
$$ \varphi_{\mathcal{B}} \left( x_1, \dots, x_n \right) = x_1 b_1 +_{V} \dots +_{V} x_n b_n. $$
At this point, $\mathbb{F}^n$ is merely a set. Since $\varphi_{\mathcal{B}}$ is a bijection, by abstract nonsense you know that you can endow $\mathbb{F}^n$ with the structure of a vector space over $\mathbb{F}$ so that $\varphi_{\mathcal{B}}$ becomes an isomorphism. So you turn to find what is this structure and discover that it is actually the "standard" vector space structure on $\mathbb{F}^n$. While the bijection $\varphi_{\mathcal{B}}$ heavily depends on $\mathcal{B}$, the induced structure on $\mathbb{F}^n$ does not depend on the specific basis so it indeed deserves the title "standard".
So now $\varphi_{\mathcal{B}}$ and $\varphi_{\mathcal{B}}^{-1}$ (which is often denoted by $[v]_{\mathcal{B}}$ and maps a vector $v \in V$ to its coordinate vector) are isomorphisms and so they must send a basis to a basis. To which basis $\varphi_{\mathcal{B}}^{-1}$ sends the original basis $\mathcal{B}$? To the basis $\mathcal{E} = (e_1,\dots,e_n)$! Now again, no matter which basis $\mathcal{B}$ you use for $V$, the bijection $\varphi_{\mathcal{B}}^{-1}$ will map $\mathcal{B}$ to the same basis $\mathcal{E} = (e_1,\dots,e_n)$. Isn't it enough to give it the title "standard"?
Finally, here is something interesting. After you endowed $\mathbb{F}^n$ with a vector space structure, you can take $V = \mathbb{F}^n$, take some arbitrary basis $\mathcal{B}$ for $\mathbb{F}^n$ and consider the map $\varphi_{\mathcal{B}} \colon \mathbb{F}^n \rightarrow \mathbb{F}^n$. Each (ordered) basis $\mathcal{B}$ gives you a different automorphism $\varphi_{\mathcal{B}}$ and the "standard" basis (with the standard ordering) is the only one which gives you the "most standard" automorphism $\operatorname{id}_{\mathbb{F}^n}$. Phrased differently, the standard basis $\mathcal{E}$ for $\mathbb{F}^n$ is the only basis with respect to which we have $[v]_{\mathcal{E}} = v$ for all $v \in \mathbb{F}^n$ (so the vector $v$ and its coordinate representation with respect to $\mathcal{E}$ coincide!).
