I'm currently working on a project focused on using roots of complex numbers and linear transformations as a means of generating 2D shapes and I've mainly been using basic linear algebra. A question that came to my mind in the process is whether it's possible to generate any kind of polygon by such means? While there doesn't seem to be a way to generate every possible sort of polygon with more than $3$ vertices (due to the transformations being linear), I'm not sure if that would also be the case with triangles. Assuming this to be true for triangles, it could potentially mean that every possible 2D shape could be generated by solving for, and linearly transforming, cubic roots of unity, thanks to triangulation (and neglecting how daunting and ineffective this would be).
So as stated in the title, would it be possible to generate any kind of triangle by linearly transforming cubic roots of unity? If so, how do I prove it?
Edit: This might as well simplify to analytical geometry as I think about it.
go backwards. Given any triangle, decide on your favorite base. A linear transformation of the "shear" type takes this to an isosceles triangle. Then an expansion or contraction in the vertical direction gives an equilateral triangle. A scalar multiple is equilateral of the appropriate side length
Quadrilaterals won't wrok. For eaxmple, you cannot map a convex quadrilateral to a non-convex one