Covariant and Contravariant Transformation - Example of Polar Coordinates I would like to see the fact that the components of a vector transform differently (controvariant transformation) than the unit bases vectors (covariant transformation) for the specific case of cartesian to polar coordinate transformation. 
The polar unit vectors $\hat{r}$ and $\hat{\theta}$ can be expressed in terms of cartesian unit vectors, $\hat{x}$ and $\hat{y}$,  as the following
\begin{equation}
\hat{r}= \text{cos}\phi \ \hat{x} + \text{sin}\phi \ \hat{y} \\
\hat{\theta}= -\text{sin}\phi \ \hat{x} + \text{cos}\phi \ \hat{y} \tag{1}
\end{equation} 
Any vector, $\vec{V}$, can be expressed in the cartesian coordinate system as $\vec{V}=V_x \ \hat{x} + V_y \ \hat{y}$. The same vector can be expressed in polar coordinates as $\vec{V}=V_r \ \hat{r} + V_\theta \ \hat{\theta}$. We then have 
\begin{equation}
V_x \ \hat{x} + V_y \ \hat{y}=V_r \ \hat{r} + V_\theta \ \hat{\theta}. \tag{2}
\end{equation}
I then project both sides of (2) once onto $\hat{r}$, and once onto $\hat{\theta}$. Using (1) and (2) we get
\begin{equation}
V_r= \text{cos}\phi \ V_x+\text{sin}\phi \ V_y \\
V_\theta= -\text{sin}\phi \ V_x+\text{cos}\phi \ V_y \tag{3}
\end{equation}
Comparing (1) and (3), both the unit vectors and the components of a vector are transforming with the same rule, which is a contradiction! What am I missing here?
 A: As @TedShifrin pointed out, the correct transformation for the dual is $(A^{-1})^T$.
For instance, in special relativity contravariant vectors transform as
$ V^{\mu '} = \Lambda ^{\mu '} _{\,\nu} V^{\nu}$,
where $ \Lambda ^{\mu '} _{\,\nu}  $ is the Lorentz transformation taking component from the unprimed frame to the primed frame. For covariant components we have to use the inverse of the Lorentz transformation:
$ V_{\mu '} = \big( \Lambda^{-1} \big)^{\nu } _{\,\mu '} V_{\nu}$.
If we assume that $V'$ and $V$ are column vectors representing, $ V_{\mu '}$ and $V_{\nu}$, respectively. Then in matrix format we have:
$ V' = \big( \Lambda^{-1} \big)^T V$.
This can also be found here: https://en.wikipedia.org/wiki/Lorentz_transformation
Of course, one can get to the same conclusion in the context of group theory. If the vector $x$ transforms according to $x \rightarrow x'=Ax$ where $A$ is the group member. Then the dual $\tilde{x}$ transforms as $\tilde{x} \rightarrow \tilde{x} '= \big(A^{-1} \big)^T\tilde{x}$ so that $\tilde{x}^T x$ is an invariant in all frames.
A: I think that your answer is unnecessarily complicated for this question. In matrix notation equation (1) in the question is
\begin{equation}
\begin{pmatrix}
\hat{r} && \hat{\theta}
\end{pmatrix}
=\begin{pmatrix}
\hat{x} && \hat{y}
\end{pmatrix}
\begin{pmatrix}
cos\phi && -sin\phi \\
sin\phi && cos\phi
\end{pmatrix}
\triangleq
\begin{pmatrix}
\hat{x} && \hat{y}
\end{pmatrix}
M
\end{equation}
and equation (3) expresses the transformation rule for a row (covariant) vector
\begin{equation}
\begin{pmatrix}
v_r && v_{\theta}
\end{pmatrix}=
\begin{pmatrix}
v_x && v_y
\end{pmatrix}
M
\end{equation}
so the transformation for a column (thus contravariant) vector is
\begin{equation}
\begin{pmatrix}
v_r \\
v_{\theta}
\end{pmatrix}
=M^T
\begin{pmatrix}v_x\\
v_y\end{pmatrix}
\end{equation}
Which is indeed the inverse of $M$ as the latter is orthogonal.
