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Say we have $n$-gons $P$ and $Q$. Is there any necessary condition for $Q = f(P)$, for some linear transformation $f : \mathbb{R}^2 \to \mathbb{R}^2$?

Sorry if this is too elementary / general.

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If you label the vertices, then an obvious necessary condition is that if you pick any three vertices and take the unique linear map that takes those to their correspondingly labelled vertices, then that map has to take all the other vertices to their corresponding vertices. –  gowers Jul 25 '10 at 17:59
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That extends easily to a deterministic test for existence, at least under the assumption that three consecutive vertices cannot be collinear. Fix three consecutive vertices $abc$ in $P,$ and for each vertex $v$ in $Q$ take a triple of consecutive vertices $uvw$ centered at $v.$ If the maps taking $abc$ to $uvw$ and $wvu$ both fail to extend to a map taking all of $P$ to all of $Q,$ then you can discard $v.$ If you check all $v$ and get failure that's it. –  Will Jagy Jul 25 '10 at 18:14

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up vote 5 down vote accepted

Jesse Douglas studied linear transformations of polygons on the complex plane in 1930s. He proved, in particular, that a transformation $z_i{}'=\sum_{i=1}^na_{ij}z_j$ (all numbers are complex) will transform a polygon $\pi=(z_1,\cdots,z_n)$ into a polygon $\pi'=(z_1{}',\cdots,z_n{}')$ if, and only if, the matrix $a_{ij}$ is cyclic, that is, if, and only if, $a_{ij}=\alpha_{j-i}$, $\alpha_{j-i}=\alpha_k$ if $k\equiv j-1\ (\text{mod}\,n)$. (See his article "On linear polygon transformations", Bull. Amer. Math. Soc. 46, (1940) pp. 551 - 560.)

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By the way, is there any result on linear transformation of polyhedra in $\mathbb{R}^n$?

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You should post this as a new question. –  Adeel Dec 30 '12 at 23:22

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