Transformation matrix of parabolic coordinates I am working on the exercises in "Spacetime and Geometry" from Sean Carroll. In exercise 4a) of section 3 we should find the transformation Matrix $A$ and its inverse, that transforms the parabolodidal coordinates $(u,v,\phi)$ to the Cartesian coordinates. It is given, that
\begin{align}
x = uv\cos{\phi}, \qquad y = uv\sin{\phi}, \qquad z = \frac{1}{2}(u^2-v^2)
\end{align}
First of all, I am not very experienced in transformation matrices. My first observation was, that this coordinate transformation is not linear, so why does the transformation matrix even exist? I found solutions online, where the matrix is calculated by
\begin{align}
A = \left( \begin{array}{rrr}
\partial_ux & \partial_vx & \partial_\phi x \\ 
\partial_uy & \partial_vy & \partial_\phi y \\
\partial_uz & \partial_vz & \partial_\phi z
\end{array}\right) = 
\left( \begin{array}{rrr}
vcos(\phi) & ucos(\phi) & -uvsin(\phi) \\ 
vsin(\phi) & usin(\phi) & uvcos(\phi) \\
u & -v & 0
\end{array}\right)
\end{align}
It makes sense to me to calculate the matrix this way, since $x'^\mu=\frac{\partial x^\mu}{\partial x'^\nu}x^\nu$. My question is, shouldn't this matrix $A$ satisfy
\begin{align}
x'=Ax
\end{align}
where $x'=(u,v,\phi)^T,\quad x = (x,y,z)^T$
 A: The correct version of $x'^\mu=\frac{\partial x^\mu}{\partial x'^\nu}x^\nu$ should be $$dx'^\mu=\frac{\partial x'^\mu}{\partial x^\nu}dx^\nu$$
Notice that the prime is in the numerator. In 1D this becomes the chain rule:$dy=\frac{dy}{dx}dx$, which is a way to check this equation. This explains why this transformation is possible: only for infinitesimal transformations around a point can we approximate the transformation as a linear map. You could check this equation numerically by calculating $dx'^\mu$ using two ways. First by using the Jacobian you gave (the transformation matrix A) and secondly by calculating $dx'^\mu\approx x'^\mu(x^\nu+dx^\nu)-x'^\mu(x^\nu)$, both using small values for $dx^\nu$.
A: \begin{align*}
&\text{with }\\
&\mathbf{R}=
\begin{bmatrix}
  x \\
  y \\
  z \\
\end{bmatrix}=
\begin{bmatrix}
    u\,v\,\cos(\phi) \\
    u\,v\,\sin(\phi) \\
    \frac{1}{2}(u^2-v^2)  
    \\
  \end{bmatrix}\quad,
\mathbf{q}=\begin{bmatrix}
    u \\
    v \\
    \phi
    \\
  \end{bmatrix}  \Rightarrow\\\\
&A_{ij}=\frac{\partial R_i}{\partial q_j}=
 \left[ \begin {array}{ccc} v\cos \left( \phi \right) &u\cos \left(
\phi \right) &-uv\sin \left( \phi \right) \\  v\sin
 \left( \phi \right) &u\sin \left( \phi \right) &uv\cos \left( \phi
 \right) \\  u&-v&0\end {array} \right]\\
\end{align*}
from here the metric $~\mathbf{G}$
\begin{align*}
 &\mathbf{G}=\mathbf{A}^T\,\mathbf{A}=
 \left[ \begin {array}{ccc} {u}^{2}+{v}^{2}&0&0\\  0&
{u}^{2}+{v}^{2}&0\\  0&0&{u}^{2}{v}^{2}\end {array}
 \right]
\quad\Rightarrow\quad\text{line element}\\
&ds^2=du^2(u^2+v^2)+dv^2(u^2+v^2)+d\phi^2\,u^2\,v^2
\end{align*}
the rotation  matrix (transformation matrix)  $~\mathbf{S}~$  between $~[\,u~,v~,\phi]~$
and $~[\,x~,y~,z]~$ system is:
\begin{align*}
 &\mathbf{S}=\begin{bmatrix}
              \hat{e}_u & \hat{e}_v & \hat{e}_\phi \\
            \end{bmatrix}\quad\text{where }\quad,
 \hat{e}_u=\frac{\partial \mathbf{R}}{\partial u}\quad,        
 \hat{e}_v=\frac{\partial \mathbf{R}}{\partial v}\quad,
 \hat{e}_\phi=\frac{\partial \mathbf{R}}{\partial \phi}  \\\\
 &\mathbf{S}= \left[ \begin {array}{ccc} {\frac {v\cos \left( \phi \right) }{\sqrt
{{u}^{2}+{v}^{2}}}}&{\frac {u\cos \left( \phi \right) }{\sqrt {{u}^{2}
+{v}^{2}}}}&-\sin \left( \phi \right) \\  {\frac {v
\sin \left( \phi \right) }{\sqrt {{u}^{2}+{v}^{2}}}}&{\frac {u\sin
 \left( \phi \right) }{\sqrt {{u}^{2}+{v}^{2}}}}&\cos \left( \phi
 \right) \\  {\frac {u}{\sqrt {{u}^{2}+{v}^{2}}}}&-{
\frac {v}{\sqrt {{u}^{2}+{v}^{2}}}}&0\end {array} \right]\quad,\mathbf{S}^T\,\mathbf{S}=\mathbf{I}_3\\
\end{align*}
