Transformation matrices for basis and coordinate transformation in non-orthonormal coordinates The transformation matrices for covariant and contravariant vectors are different but in orthonormal coordinate system numerical values in matrices turn out to be same although in mathematical proof they are related by the operation of transpose of inverse of one of the matrix.


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*I wanna know how in any non-orthogonal coordinate system these two transformation matrices won't be same? 

*Also how are they same in any orthonormal coordinate system except Cartesian type coordinate system?
 A: Suppose we start with an orthonormal cartesian coordinate basis with unit vectors $\left( \hat{\mathbf{x}}, \hat{\mathbf{y}}, \hat{\mathbf{z}} \right)$.  Suppose we wish to rotate to a new coordinate system (may or may not be a coordinate basis, see the following answer which has a good discussion on the difference) with unit vectors $\left( \hat{\boldsymbol{\eta}}, \hat{\boldsymbol{\beta}}, \hat{\boldsymbol{\zeta}} \right)$.
If we define a rotation matrix, $\mathbb{A}$, from our original basis to the new coordinate system as the following:
$$
\mathbb{A} = \left[
\begin{array}{ c c c }
  \eta{\scriptstyle_{x}}     & \eta{\scriptstyle_{y}}     & \eta{\scriptstyle_{z}}      \\
  \beta{\scriptstyle_{x}} & \beta{\scriptstyle_{y}} & \beta{\scriptstyle_{z}}  \\
  \zeta{\scriptstyle_{x}} & \zeta{\scriptstyle_{y}} & \zeta{\scriptstyle_{z}}
\end{array} \right]  \tag{1}
$$
then we expect that $\mathbb{A}$ should satisfy the following if it is an orthogonal matrix:
$$
\begin{align}
  \mathbb{A} \cdot \hat{\boldsymbol{\eta}} & = \langle 1, 0, 0 \rangle  \tag{2a}  \\
  \mathbb{A} \cdot \hat{\boldsymbol{\beta}} & = \langle 0, 1, 0 \rangle  \tag{2b}  \\
  \mathbb{A} \cdot \hat{\boldsymbol{\zeta}} & = \langle 0, 0, 1 \rangle  \tag{2c}
\end{align}
$$
If Equations 2a-2c are not satisfied, that implies that $\mathbb{A}$ is not orthogonal.  One consequence of this is that the magnitude of its determinant is not unity.
In the event that $\mathbb{A}$ is not orthogonal, but is still invertible, then the proper rotation matrix from the original $\left( \hat{\mathbf{x}}, \hat{\mathbf{y}}, \hat{\mathbf{z}} \right)$ coordinate system to the new $\left( \hat{\boldsymbol{\eta}}, \hat{\boldsymbol{\beta}}, \hat{\boldsymbol{\zeta}} \right)$ coordinate system is given by the inverse of the transpose of $\mathbb{A}$, or:
$$
\mathbb{R} = \left( \mathbb{A}^{T} \right)^{-1} \tag{3}
$$
We know that any invertible matrix satisfies $\left( \mathbb{A}^{T} \right)^{-1} = \left( \mathbb{A}^{-1} \right)^{T}$ and orthogonal matrices satisfy $\mathbb{A}^{T} = \mathbb{A}^{-1}$.  Thus, if $\left( \hat{\boldsymbol{\eta}}, \hat{\boldsymbol{\beta}}, \hat{\boldsymbol{\zeta}} \right)$ were an orthonormal basis we would have:
$$
\begin{align}
  \mathbb{R} & = \left( \mathbb{A}^{-1} \right)^{T} \tag{4a} \\
  & = \left( \mathbb{A}^{T} \right)^{T} \tag{4b} \\
  & = \mathbb{A} \tag{4c}
\end{align}
$$
