I know that we can define two vectors to be orthogonal only if they are elements of a vector space with an inner product.
So, if $\vec x$ and $\vec y$ are elements of $\mathbb{R}^n$ (as a real vector space), we can say that they are orthogonal iff $\langle \vec x,\vec y\rangle=0$, where $\langle \vec x,\vec y\rangle $ is an inner product.
Usually the inner product is defined with respect to the standard basis $E=\{\hat e_1,\hat e_2 \}$ (for $n=2$ to simplify notations), the standard definition is:
$$
\langle \vec x,\vec y\rangle_E=x_1y_1+x_2y_2
$$
Where
$$
\begin{bmatrix}
x_1\\x_2
\end{bmatrix}
=[\vec x]_E
\qquad
\begin{bmatrix}
y_1\\y_2
\end{bmatrix}
=[\vec y]_E
$$
are the components of the two vectors in the standard basis and, by definition of the inner product, $\hat e_1$ and $\hat e_2$ are orho-normal.
Now, if $\vec v_1$ and $\vec v_2$ are linearly independent the set $V=\{\vec v_1,\vec v_2\}$ is a basis and we can express any vector in this basis with a couple of components:
$$
\begin{bmatrix}
x'_1\\x'_2
\end{bmatrix}
=[\vec x]_V
\qquad
\begin{bmatrix}
y'_1\\y'_2
\end{bmatrix}
=[\vec y]_V
$$
from which we can define an inner product:
$$
\langle \vec x,\vec y\rangle_V=x'_1y'_1+x'_2y'_2
$$
Obviously we have:
$$
[\vec v_1]_V=
\begin{bmatrix}
1\\0
\end{bmatrix}
\qquad
[\vec v_2]_V=
\begin{bmatrix}
0\\1
\end{bmatrix}
$$
and $\{\vec v_1,\vec v_2\}$ are orthogonal (and normal) for the inner product $\langle \cdot,\cdot\rangle_V$.
This means that any two linearly independent vectors are orthogonal with respect to a suitable inner product defined by a suitable basis. So orthogonality seems a ''coordinate dependent'' concept.
The question is: is my reasoning correct? And, if yes, what make the usual standard basis so special that we chose such basis for the usual definition of orthogonality?
I add something to better illustrate my question.
If my reasoning is correct than, for any basis in a vector space there is an inner product such that the vectors of the basis are orthogonal. If we think at vectors as oriented segments (in pure geometrical sense) this seems contradicts our intuition of what ''orthogonal'' means and also a geometric definition of orthogonality. So, why what we call a ''standard basis'' seems to be in accord with intuition and other basis are not?
Best Answer
To expand a bit on Daniel Fischer’s comment, coming at this from a different direction might be fruitful. There are, as you’ve seen, many possible inner products. Each one determines a different notion of length and angle—and so orthogonality—via the formulas with which you’re familiar. There’s nothing inherently coordinate-dependent here. Indeed, it’s often possible to define inner products in a coordinate-free way. For example, for vector spaces of functions on the reals, $\int_0^1 f(t)g(t)\,dt$ and $\int_{-1}^1 f(t)g(t)\,dt$ are commonly-used inner products. The fact that there are many different inner products is quite useful. There is, for instance, a method of solving a large class of interesting problems that involves orthogonal projection relative to one of these “non-standard” inner products.
Now, when you try to express an inner product in terms of vector coordinates the resulting formula is clearly going to depend on the choice of basis. It turns out that for any inner product one can find a basis for which the formula looks just like the familiar dot product.
You might also want to ask yourself what makes the standard basis so “standard?” If your vector space consists of ordered tuples of reals, then there’s a natural choice of basis, but what about other vector spaces? Even in the Euclidean plane, there’s no particular choice of basis that stands out a priori. Indeed, one often chooses an origin and coordinate axes so that a problem takes on a particularly simple form. Once you’ve made that choice, then you can speak of a “standard” basis for that space.