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Here is a problem with solution from Gadea's book. I have a little question about the solution. enter image description here My problem: I do not know why $A\left(\begin{array}{c} f_{1}\\ f_{2}\\ f_{3} \end{array}\right)=\left(\begin{array}{c} 2\\ -1\\ 3 \end{array}\right) $

It is probably something quite trivial or fundamental, but I do not get it.

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  • $\begingroup$ Are you confused about the $(2,-1,3)$ part? If so, those are the components of the original vector $X$, i.e. the coefficients in the first equation. $\endgroup$
    – Brick
    Mar 27, 2017 at 0:07
  • $\begingroup$ @Brick Ok, I probably should have been myself clearer: my question is exactly why is $A$ (the jacobian of the change of variables map) the change of basis matrix? I thought the change of basis matrix was the matrix whose column vectors are the elements of one basis written in the other basis. $\endgroup$
    – soap
    Mar 27, 2017 at 9:15

1 Answer 1

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The matrix $A$ is the jacobian of the map,

$$\Phi(\rho,\theta,z) = (\rho \cos \theta, \rho \sin \theta,z) = (x,y,z)$$ $$ A: T_{(\rho, \theta,z)}\mathbb{R}^3 \to T_{(x,y,z)}\mathbb{R}^3,\ \begin{pmatrix} \rho \\ \theta \\ z \end{pmatrix} \mapsto \begin{pmatrix} x \\ y \\ z \end{pmatrix}$$

In the above definition, we choose to not distinguish between a vector in the tangent space (which is expressed in terms of $\partial^j$) and the coefficients i.e to the vector $X$ we have the associated vector,

$$\begin{pmatrix} \ \ 2 \\ -1 \\ \ \ 3 \end{pmatrix}$$

In terms of the tangent space basis generated by $\rho, \theta,z$ we have that,

$$X = f_1 \frac{\partial}{\partial \rho} +f_2 \frac{\partial}{\partial \theta} +f_3 \frac{\partial}{\partial z} $$

Hence, if we apply $A$ to the vector $\begin{pmatrix} f_1 & f_2 & f_3 \end{pmatrix}$ then its image should be the vector $X$ i.e its coordinate representation. This is true since $\Phi$ is a diffeomorphism and so $A$ is an isomorphism i.e $(f_1, f_2, _3)^t$ is the unique preimage.

$$ A\begin{pmatrix} f_1 \\ f_2 \\ f_3 \end{pmatrix} = \begin{pmatrix} \ \ 2 \\ -1 \\ \ \ 3 \end{pmatrix}$$

$\textbf{Big Picture}$: All you need to think about is that the represenation of $X$ in cylindrical is the velocity vector of a curve $\gamma$ in $\mathbb{R}^3$ where this $\mathbb{R}^3$ is the spherical coordinate one. Next, we take $(\Phi \circ \gamma)'(0)$ which is now the representation of $X$ in cartesian coordinates i.e the flat or standard $\mathbb{R}^3$. Hence we can say $(\Phi \circ \gamma)'(0) = (2,-1,3)$ i.e,

$$ (f_1,f_2,f_3) = A^{-1}(2,-1,3)$$

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  • $\begingroup$ Ok, I probably should have been myself clearer: my question is exactly why is $A$ (the jacobian of the change of variables map) the change of basis matrix? I thought the change of basis matrix was the matrix whose column vectors are the elements of one basis written in the other basis. $\endgroup$
    – soap
    Mar 27, 2017 at 8:59
  • $\begingroup$ I think I got it: a well-known result is: $$\frac{\partial}{{\partial x_{i}}}=\frac{\partial\tilde{x_{j}}}{{\partial x_{i}}}\frac{\partial}{{\partial\tilde{x}_{j}}}$$ Where $\tilde{x}$ are the new coordinates (in this case the cylindrical coordinates). In particular, this means that... $\endgroup$
    – soap
    Mar 27, 2017 at 9:31
  • $\begingroup$ ...$\frac{\partial}{{\partial r}}=\frac{\partial x}{{\partial r}}\frac{\partial}{{\partial x}}+\frac{\partial y}{{\partial r}}\frac{\partial}{{\partial y}}+\frac{\partial z}{{\partial r}}\frac{\partial}{{\partial z}}$ And we can see that this corresponds to the first column of the Jacobian matrix. Likewise for the other cylindrical coordinates. Note: I will accept your answer anyway, because it was nonetheless a very good answer (my question was faulty) and I greatly appreciate your effort. $\endgroup$
    – soap
    Mar 27, 2017 at 9:33
  • $\begingroup$ I'm sorry, I was away. The matrix $A$ is the jacobian of the map, $$\Phi(\rho,\theta,z) = (\rho \cos \theta, \rho \sin \theta,z) = (x,y,z)$$ $\endgroup$ Mar 27, 2017 at 12:07
  • $\begingroup$ I made the appropriate edit. $\endgroup$ Mar 27, 2017 at 12:20

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