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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1$\begingroup$ 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. $\endgroup$– gowersCommented Jul 25, 2010 at 17:59
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1$\begingroup$ 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. $\endgroup$– Will JagyCommented Jul 25, 2010 at 18:14
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2 Answers
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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.)
answered Jul 25, 2010 at 18:31
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By the way, is there any result on linear transformation of polyhedra in $\mathbb{R}^n$?
