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I know Mathematica can change the coordinate system with the command ''CoordinateTransform''.For example, you can change the coordinates of an $(x,y,z)$ vector in cartesian coordinates like this.

CoordinateTransform[ "Cartesian" -> "Cylindrical", {x, y,z}]

And Mathematica will return the vector in the cylindrical coordinates system $(\rho,\phi,z)$.

However, I think it only works with some well-know basis such as the polar, the cylindrical the spherical... which the software has as default. I was wondering if this change of coordinates could be done for an arbitrary choice of the basis. For instance, say I want to plot several $\mathbb{R}^2$ vectors in this particular basis $\{(-1,-1),(1,1)\}$ for some reason. Can Mathematica do it?

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This answer is similar to @DanielHuber's answer, but using the relevant linear algebra more explicitly:

basis = {{-1, 1}, {1, 1}}; vectors = {{1, 1}, {3, 2}, {0, 1}}; Inverse[basis] . # & /@ vectors (* {{0, 1}, {-(1/2), 5/2}, {1/2, 1/2}} *)

If the vectors themselves are given in a non-standard basis, we first convert to the canonical basis:

basis0 = {{0, 1}, {1, 0}}; basis1 = {{-1, 1}, {1, 1}}; vectors = {{1, 1}, {3, 2}, {0, 1}}; Inverse[basis1] . basis0 . # & /@ vectors (* {{0, 1}, {1/2, 5/2}, {-(1/2), 1/2}} *)

This makes use of the fact that converting a vector $\vec v$ from a given basis $(\vec b_i)_i$ to the canonical basis $(\hat{\vec e}_i)_i$ is simply given by the matrix-vector product,

$$\tilde{\vec v}=B\cdot\vec v$$

where $B$ is the matrix with the basis vectors $\vec b_i$ as rows (using Mathematica's convention of row/column vectors in relation to the dot product). Likewise, converting the other was is done by the inverse of the matrix,

$$\vec v=B^{-1}\cdot\tilde{\vec v}$$

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  • $\begingroup$ Thank you both for the comments!! They've helped me a lot. Just one more thing: Do you know if it is possible to plot the vectors using the new basis (axis, gridline) just like PolarPlot does with the polar system?? $\endgroup$ Commented Dec 1, 2022 at 14:28
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To start with, {(−1,−1),(1,1)} is not a basis because the two vectors are antiparallel. I therefore take {(−1,1),(1,1)} as basis for an example.

Call the original bas bas0 and the new base, written in the old coordinates, bas1:

bas0 = {{1, 0}, {0, 1}}; bas1 = {{-1, 1}, {1, 1}};

We now need to write the old base vectors {1,0} and {0, 1} in the new base bas1.

{1, 0} == (-bas1[[1]] + bas1[[2]] )/2 {0, 1} == (bas1[[1]] + bas1[[2]] )/2

A vector: v=x{1,0}+y{0,1} in the old coordinates: {x,y} can be written in the new basis:v= xn{-1,1} + yn{1,1} with new coordinates:{xn,yn}. Therefore, to get the new from the old coordinates:

Solve[{x, y} . bas0 == {xn, yn} . bas1, {xn, yn}] (* {{xn -> 1/2 (-x + y), yn -> (x + y)/2}} *)

We may define a function for this:

old2new[p:{x_,y_}]= {{-1,1}/2,{1,1}/2}.p

We may now test if a vector, written in the new coordinates, is still the same vector by:

old2new[{x, y}] . bas1 == {x, y} . bas0 (* True *)
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