Notion of co- and contravariance of vectors dependent on the convention used for the change of basis matrix? Upon reading a bit about general relativity I came across the notion of co- and contravariance of vectors. All sources I've found agree that vectors (or rather their coordinates) transform contravariantly while covectors (linear forms) transform covariantly. However, I have a feeling that this transformation behavior is only a consequence of what one defines to be a change of basis. Let me clarify this point.
Let $V$ be a vector space and $B$ a basis for $V$. For each $v \in V$ we then have a coordinate representation $[v]_B$ of $v$ with respect to the basis $B$. If we now take another Basis of $V$, say $B'$, we can find a matrix $A$, such that $A[v]_B = [v]_{B'}$. That is $A$ converts a coordinate representation with respect to the "old" Basis $B$ to the corresponding representation of $v$ relative to $B'$. This matrix $A$ is what I consider to be the Change-of-Basis-matrix from $B$ to $B'$. Personally, I think this is the most "natural" way to define it. However, it's easy to see, that (by definition) the coordinates transform with $A$, i.e. we obtain the "new" from the "old" coordinates by multiplying with $A$. With this in mind I think it's just as good to call vectors covariant. 
Is my reasoning correct? Is the notion of co- and contravariance only dependent on the fact that (apparently) physicists define the Change-of-Basis-Matrix from $B$ to $B'$ the other way round, that is $A[v]_{B'} = [v]_B\iff [v]_{B'} = A^{-1}[v]_B$. I'm asking this question mainly because I cannot wrap my head around this convention. Not at all. Maybe anyone can clarify this a bit for me?
 A: Yes, if I understood correctly your discomfort, you are correct and I agree with you. In physics usually is created the concept of a "background" where things happen and "events" occur, as if there were a place like this in our reality. Some words like "space" and "time" are used. In general relativity, the background is called "spacetime" and is modelled as a $4$-dimensional smooth manifold. In what follows I will draw some basic facts about an $m$-dimensional smooth manifold $M$, $m \in \mathbb{N}^*$.
Let $ \ p \in M$, $x = (x^1,...,x^m) : V \to \mathbb{R}^m \ $ and $ \ y = (y^1 , ... , y^m) : W \to \mathbb{R}^m \ $ be charts (local coordinate systems) around $p$ and $ \ v \in TM|_p$. Then $V$ and $W$ are opens sets of $M$, $p \in V \cap W \ $ and $v$ is a derivation acting on germs of smooth functions. Hence we have the two basis $\displaystyle \ B_x = \left\{ \left. \frac{\partial}{\partial x^1} \right|_p , ... , \left. \frac{\partial}{\partial x^m} \right|_p \right\} \ $ and $\displaystyle \ B_y = \left\{ \left. \frac{\partial}{\partial y^1} \right|_p , ... , \left. \frac{\partial}{\partial y^m} \right|_p \right\} \ $ of $ \, TM|_p \, $ and we write $$v = \sum_{\mu = 1}^{m} v(x^{\mu}) \cdot \left. \frac{\partial}{\partial x^{\mu}} \right|_p = \sum_{\mu = 1}^{m} v(y^{\mu}) \cdot \left. \frac{\partial}{\partial y^{\mu}} \right|_p$$ Then, $\forall \nu \in \{ 1,...,m \}$, we have $$ \left. \frac{\partial}{\partial x^{\nu}} \right|_p  = \sum_{\mu = 1}^{m} \left( \left. \frac{\partial}{\partial x^{\nu}} \right|_p \right) \! (y^{\mu}) \cdot \left. \frac{\partial}{\partial y^{\mu}} \right|_p = \sum_{\mu = 1}^{m} \frac{\partial y^{\mu}}{\partial x^{\nu}} (p) \cdot \left. \frac{\partial}{\partial y^{\mu}} \right|_p$$ So, the usual change-of-basis-matrix, from $B_x$ to $B_y$, is $$S = S_{B_y}^{B_x} = \left[ \frac{\partial y^{\mu}}{\partial x^{\nu}} (p) \right]_{m \times m} = \left[ \begin{array}{cccc}
\frac{\partial y^{1}}{\partial x^{1}} (p) & \frac{\partial y^{1}}{\partial x^{2}} (p) & \dots & \frac{\partial y^{1}}{\partial x^{m}} (p) \\
\frac{\partial y^{2}}{\partial x^{1}} (p) & \frac{\partial y^{2}}{\partial x^{2}} (p) & \dots & \frac{\partial y^{2}}{\partial x^{m}} (p) \\ \vdots & \vdots & \ddots & \vdots \\
\frac{\partial y^{m}}{\partial x^{1}} (p) & \frac{\partial y^{m}}{\partial x^{2}} (p) & \dots & \frac{\partial y^{m}}{\partial x^{m}} (p) \end{array} \right]$$
Then, by definition, changing the roles played for $x$ and $y$, the change-of-basis-matrix, from $B_y$ to $B_x$, is $$S_{B_x}^{B_y} = \big( S_{B_y}^{B_x} \big)^{-1} = S^{-1} = \left[ \frac{\partial x^{\mu}}{\partial y^{\nu}} (p) \right]_{m \times m}$$
In fact, $\forall \alpha , \beta \in \{ 1,...,m \}$, we can compute directly
\begin{align*}
\big( S_{B_y}^{B_x} \cdot S_{B_x}^{B_y} \big)^{\alpha}_{\beta} & = \sum_{\mu = 1}^{m} \big( S_{B_y}^{B_x} \big)^{\alpha}_{\mu} \big( S_{B_x}^{B_y} \big)^{\mu}_{\beta} = \\ & = \sum_{\mu = 1}^{m} \frac{\partial y^{\alpha}}{\partial x^{\mu}} (p) \cdot \frac{\partial x^{\mu}}{\partial y^{\beta}} (p) = \\ & = \sum_{\mu = 1}^{m} \left( \left. \frac{\partial}{\partial y^{\beta}} \right|_p \right) \! (x^{\mu}) \cdot \left( \left. \frac{\partial}{\partial x^{\mu}} \right|_p \right) \! (y^{\alpha}) = \\ & = \left[ \sum_{\mu = 1}^{m} \left( \left. \frac{\partial}{\partial y^{\beta}} \right|_p \right) \! (x^{\mu}) \cdot \left. \frac{\partial}{\partial x^{\mu}} \right|_p \right] \! (y^{\alpha}) = \\ & = \left( \left. \frac{\partial}{\partial y^{\beta}} \right|_p \right) \! (y^{\alpha}) = \\ & = [\partial_{\beta} (y^{\alpha} \circ y^{-1})] \big( y(p) \big) = \\ & = \big[ (y^{\alpha} \circ y^{-1})' \big( y(p) \big) \big] (e_{\beta}) = \\ & = \big[ (\pi^{\alpha} \circ y \circ y^{-1})' \big( y(p) \big) \big] (e_{\beta}) = \\ & = \big[ (\pi^{\alpha})' \big( y(p) \big) \big] (e_{\beta}) = \\ & = \pi^{\alpha} (e_{\beta}) = \\ & = \delta^{\alpha}_{\beta}
\end{align*}
Where $ \ \pi^{\alpha} : \mathbb{R}^m \to \mathbb{R} \ $ is the (linear) projection onto the $\alpha$-th coordinate, $\pi^{\alpha} (t^1 , ... , t^m) = t^{\alpha}$, $\forall t^1 , ... , t^m \in \mathbb{R}$, $\forall \alpha \in \{ 1,...,m \}$, and $ \ e_{\beta} = (\delta^{1}_{\beta} , ... , \delta^{m}_{\beta}) \in \mathbb{R}^m \ $ is the usual $\beta$-th vector of the standard basis of $\mathbb{R}^m$. Hence, $\big( S_{B_y}^{B_x} \cdot S_{B_x}^{B_y} \big)^{\alpha}_{\beta} = (Id)^{\alpha}_{\beta}$, $\forall \alpha , \beta \in \{ 1,...,m \}$. For this reason, $S_{B_y}^{B_x} \cdot S_{B_x}^{B_y} = Id_{m \times m}$. Similarly, one gets $ \ S_{B_x}^{B_y} \cdot S_{B_y}^{B_x} = Id_{m \times m}$.
I will write $\displaystyle \ S^{\mu}_{\nu} = \frac{\partial y^{\mu}}{\partial x^{\nu}} (p) = \big( S_{B_y}^{B_x} \big)^{\mu}_{\nu}$, $\forall \mu , \nu \in \{ 1,...,m \}$.
Since we have $ \ [v]^{B_x} = \big[ v(x^1) \ \dots \ v(x^m) \big]^T \ $ and $ \ [v]^{B_y} = \big[ v(y^1) \ \dots \ v(y^m) \big]^T \ $, we get the known formula $ \ S^{B_y}_{B_x} \cdot [v]^{B_x} = [v]^{B_y} \ $.
Now, $\forall \nu \in \{ 1,...,m \}$, we have
\begin{align*}
v^{\nu} & = v(y^{\nu}) = \\ & = \left[ \sum_{\mu = 1}^{m} v(x^{\mu}) \cdot \left. \frac{\partial}{\partial x^{\mu}} \right|_p \right] \! (y^{\nu}) = \\ & = \sum_{\mu = 1}^{m} v(x^{\mu}) \cdot \left( \left. \frac{\partial}{\partial x^{\mu}} \right|_p \right) \! (y^{\nu}) = \\ & = \sum_{\mu = 1}^{m} \frac{\partial y^{\nu}}{\partial x^{\mu}} (p) \cdot v(x^{\mu}) = \\ & = \sum_{\mu = 1}^{m} S^{\nu}_{\mu} \, v(x^{\mu}) = \\ & = \sum_{\mu = 1}^{m} S^{\nu}_{\mu} \, v^{\mu}
\end{align*}
Let $ \ \omega \in (TM|_p)^* = \bigwedge^1 (TM|_p)^* \ $ be a covector. Then we can write $$\omega = \sum_{\mu = 1}^{m} \omega \! \left( \left. \frac{\partial}{\partial x^{\mu}} \right|_p \right) \cdot dx^{\mu}|_p = \sum_{\mu = 1}^{m} \omega \! \left( \left. \frac{\partial}{\partial y^{\mu}} \right|_p \right) \cdot dy^{\mu}|_p$$ where $ \ B_x^* = \{ dx^1|_p , ... , dx^m|_p \} \ $ and $ \ B_y^* = \{ dy^1|_p , ... , dy^m|_p \} \ $ are basis of $ \, (TM|_p)^*$, $B_x^*$ is dual to $B_x$ and $B_y^*$ is dual to $B_y$. So we have
\begin{align*}
\omega_{\nu} & = \omega \! \left( \left. \frac{\partial}{\partial y^{\nu}} \right|_p \right) = \\ & = \left[ \sum_{\mu = 1}^{m} \omega \! \left( \left. \frac{\partial}{\partial x^{\mu}} \right|_p \right) \cdot dx^{\mu}|_p \right] \! \left( \left. \frac{\partial}{\partial y^{\nu}} \right|_p \right) = \\ & = \sum_{\mu = 1}^{m} \omega \! \left( \left. \frac{\partial}{\partial x^{\mu}} \right|_p \right) \cdot \big( dx^{\mu}|_p \big) \! \left( \left. \frac{\partial}{\partial y^{\nu}} \right|_p \right) = \\ & = \sum_{\mu = 1}^{m} \omega \! \left( \left. \frac{\partial}{\partial x^{\mu}} \right|_p \right) \cdot \left( \left. \frac{\partial}{\partial y^{\nu}} \right|_p \right) \! (x^{\mu}) = \\ & = \sum_{\mu = 1}^{m} \frac{\partial x^{\mu}}{\partial y^{\nu}} (p) \cdot \omega \! \left( \left. \frac{\partial}{\partial x^{\mu}} \right|_p \right) = \\ & = \sum_{\mu = 1}^{m} (S^{-1})^{\mu}_{\nu} \ \omega \! \left( \left. \frac{\partial}{\partial x^{\mu}} \right|_p \right) = \\ & = \sum_{\mu = 1}^{m} (S^{-1})^{\mu}_{\nu} \ \omega_{\mu}
\end{align*}
We conclude that the laws of changing coordinates, from $(x^{\mu})$ to $(y^{\nu})$, for a vector $v$ and a covector $\omega$ are, respectively, of the form $$v^{\nu} = \sum_{\mu = 1}^{m} S^{\nu}_{\mu} \, v^{\mu} \qquad \text{ and } \qquad \omega_{\nu} = \sum_{\mu = 1}^{m} (S^{-1})^{\mu}_{\nu} \ \omega_{\mu}$$ I always end confusing myself when I have to work the rules of raising and lowering indices. Some books define $ \ (S^{-1})^{\mu}_{\nu} = S_{\mu}^{\nu} \ $ and I don't like it because I keep an eye on a same index repeated upper in a term and lower in another because usually they are summed up, sometimes using the Einstein convention and omitting the big sigma sign for sums. So, I avoid changing the place of indices superscript and subscript.
A: If your question is whether the nomenclature regarding co- and contravariant vectors is based on an arbitrary choice, then the answer is a resounding "Yes!". In fact, the conventional choice of which kind of vectors to call "covariant" turns out, with hindsight, to be exactly wrong. The functor that sends a pointed smooth manifold to its tangent space is a covariant functor, while the one that sends it to its cotangent space is contravariant. Oh well. 
