I was trying to visualise Lorentz transformations and parametrized a surface by using the matrix $A$ below, with $-k$ as rapidity.
$$A = \begin{bmatrix} \cosh(k) & \sinh(k) \\ \sinh(k) & \cosh(k) \end{bmatrix}$$
I took the product $$A \times R$$ (rotation matrix) to rotate gridlines, but this is just applying the Lorentz transform to a rotated surface and thus the geometry got distorted. The product $R \times A$ just rotates the Lorentz transformed square so that didn't help.
Is there a way to preserve the shape and just rotate gridlines? Like a parametric square using the identity matrix but then rotating the gridlines in it without moving it around.
I also tried using a polar parametrisation but couldn’t find the explicit polar equation of a square and I also don’t know if this will work. Is there another method I can use, I think being limited to active linear transformations in geogebra (plz correct me if I'm wrong here) is causing this when I need a passive transform or maybe even something else entirely (like polar parametrisation I guess?)
Thanks for the help.
edit: Added the pics, here I set the rotation as pi/4 and zero and then multiplied by A (k=-1) so I get gridlines parallel to eigenvectors and the comoving frames respectively. I wanted to see a the same gridlines as in image one but with the outline as in grid 2.