Non-orthonormal basis vectors impossible? When we speak of vectors in their component representation:
$x = \{ \xi_1, \xi_2, ..., \xi_n \}$
with respect to some basis $\{ e_1, e_2, ..., e_n \}$,
this means the linear combination
$\xi_1 e_1 + \xi_2 e_2 + ... + \xi_n e_n$.
But, this implies that if we let $y = \{ 1, 0, ..., 0 \}$, then $y = e_1$
So, 
$e_1$ must have component representation $\{1, 0, ..., 0 \}$, $e_2$ has  $\{0, 1, 0, ..., 0 \}$, and so forth.
Given this, what sense is there to speak of an orthonormal vs an "arbitrary" basis?  Isn't the axis aligned orthonormal basis the only basis possible?
 A: The coordinates  may be orthonormal but the basis vectors do not have to be orthonormal. 
Let $v_1=(1,3)$ and $v_2= (3,2)$ then the coordinates of $v_1$ with respect to the basis $\{ v_1, v_2 \}$ is $(1,0)$.
Also the coordinates of   $v_2$ with respect to the basis $\{ v_1, v_2 \}$ is $(0,1)$
The coordinates are orthonormal but the vectors are not.  
A: A basis being orthonormal is dependent on the inner product used. Have a think: why are the coordinate vectors $(1, 0, 0, \ldots, 0)$ and $(0, 1, 0 ,\ldots, 0)$ orthogonal? Traditionally, if they were just considered vectors in $\mathbb{R}^n$, then under the dot product, they are orthogonal because their dot product is $0$. But, does taking the dot product mean anything here? Dot products on coordinate vectors won't necessarily correspond to the inner product on the space.
However, and this is what I think you might be getting at, it is true that any basis can be made orthonormal with an appropriate inner product! If you take a basis $(e_1, \ldots, e_n)$ on a real vector space (without an inner product), and define an inner product by
$$\langle \lbrace \xi_1, \ldots, \xi_n \rbrace,  \lbrace \mu_1, \ldots, \mu_n \rbrace\rangle = \xi_1\mu_1 + \ldots + \xi_n\mu_n,$$
i.e. doing the dot product to the coordinate vectors, then we form an inner product under which $(e_1, \ldots, e_n)$ is orthonormal. (It's a nice little exercise to show that such a bilinear map is indeed an inner product, and that $(e_1, \ldots, e_n)$ is orthonormal.)
So basically, I wouldn't say that non-orthonormal bases are impossible, as given any inner product, most of our bases are not orthonormal. However, I would say that every basis can be made orthonormal by equipping the space with the right inner product.
A: The point is the distinction between vectors and their representation as linear combination with respect to a basis. Orthogonality  as a concept is a condition on the given two vectors. Of course this can be translated into a condition on their representation. But that translated condition will vary depending on the basis under discussion.
As an analogy think of positive integers as opposed to their representation in base 2, or base 10 or base 5 or base 7. In usual base 10 number system a number $\ne2$ being prime means its rightmost digit has to be 1, 3, 7 or 9 necessarily (but not sufficient). In base 7, a prime number can have 1, 2, 3,4 or 5 or 6  as the examples below show:
The following two-digit numbers: 41, 32, 23, 14, 25,16 interpreted as  base 7 numbers  are  the prime numbers  29, 23, 17, 11, 19, 13 (base 10).
ASIDE: If you choose a prime number $p$ as base (as we did with 7 above), every number from 1 to $p-1$ can occur as the right most digit of a suitable prime number. (This is simply Dirichlet's theorem on Prime Numbers on Arithmetic Progressions, from the year 1837).
A: If $X$ is a real (complex) n-dimensional inner product space, then you choose an orthonormal basis $\{ e_1,e_2,\cdots,e_n \}$ of $X$ and establish a map
$$
             U_e : X \rightarrow \mathbb{C}^n
$$
given by $U_ef = (\langle f,e_1\rangle,\langle f,e_2\rangle,\cdots)$. This will map $e_1$ to $(1,0,0,\cdots)$, $e_2$ to $(0,1,0,\cdots)$, etc.. But you can also do this with another orthonormal basis $\{ e_1',e_2',\cdots \}$ of $X$, and you'll have a different correspondence $U_{e'} : X\rightarrow \mathbb{C}^n$. These maps will preserve the inner product, but the basis mapped to the standard basis in $\mathbb{C}^n$ may vary from $U_e$ to $U_{e'}$. The map $U_{e'}^{-1}U_e : X\rightarrow X$ maps the orthnormal basis $e$ to $e'$; so that map is not the identity. Constructing a map $U_e : X\rightarrow \mathbb{C}^n$ requires choosing an orthonormal basis of $X$; so these maps are not all the same. The map $U_{e'}^{-1}U_{e}$ is not the identity; it is an orthogonal map (or unitary map if the space is complex.) The map $U_{e'}U_{e}^{-1} : \mathbb{C}^n\rightarrow\mathcal{C}^n$ is an orthogonal matrix if the spaces are real, and is a unitary matrix if the spaces are complex.
