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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.

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    I have a lot of questions here. Why do you say that "there doesn't seem to be a way to generate every possible sort of polygon with more than 3 vertices"? It would make more sense, perhaps, if we knew exactly how you were "using linear algebra". Also, a "cubic root of unity" is a complex number. What do you mean when you talk about "linearly transforming cubic roots of unity"? – 2017-01-08
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    My knowledge of linear algebra is very limited, most of it I got in touch with by reading on my own, and so my remark with the polygons with more than 3 vertices is a purely intuitive one. With transforming roots of unity I naturally meant doing so after representing them in matrix form. – 2017-01-08
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    Also, it might be useful to you to know that affine transformations can be computed using an augmented matrix, as explained in [this wiki article](https://en.m.wikipedia.org/wiki/Affine_transformation). – 2017-01-08
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    What does it mean to represent a root of unity in matrix form? – 2017-01-08
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    Give me a moment to type it up. – 2017-01-08
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    If a complex number $z=a+bi$ may be represented as $$\begin{bmatrix} a&-b \\ b&a \end{bmatrix}$$ then the roots of unity, given by $cos\frac{\theta+2k\pi}{n} +isin\frac{\theta+2k\pi}{n}$, may be represented as $$\begin{bmatrix} cos\frac{\theta+2k\pi}{n}&-sin\frac{\theta+2k\pi}{n} \\ sin\frac{\theta+2k\pi}{n}&cos\frac{\theta+2k\pi}{n} \end{bmatrix}$$ Is this correct? – 2017-01-08

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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

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    The implication here seems to be that the process for generating polygons is taking a regular polygon and applying a linear transformation. That's a reasonable interpretation – 2017-01-08
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    @Omnomnomnom, yes, I describe one direction, all invertible transformations. Reverse. – 2017-01-08
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    Thank you, got it now. Really nice way to think about it. – 2017-01-08