Repeated projection of points onto lines Consider a point $P$ on the Euclidean plane, and lines $l_1,l_2,\ldots,l_n$. Project $P$ onto $l_1$. Then project the resulting point onto $l_2$. Then project the resulting point onto $l_3$, and so on, up to $l_n$. Then project the resulting point onto $l_1$ again, then $l_2$, ..., repeating the process.
Will the set of projected points always be bounded? That is, does there always exist $r$ (possibly depending on $P,l_1,\ldots,l_n$) such that the resulting points always stay within the disk of radius $r$ centered at $P$?
 A: Using affine geometry with homogeneous coordinates where a point in $r\in\mathbb{R}^2$ is represented by a 3-vector $(r,1)$, projection onto a line $\ell_i$ defined by a point $p_i$ and normal $\hat{n}_i$ is
$$ \begin{bmatrix}r'\\ 1\end{bmatrix} = \operatorname{proj}_{\ell_i} r = \begin{bmatrix}I - \hat{n}_i\hat{n}_i^T & \hat{n}_i (\hat{n}_i\cdot p_i) \\ 0 &1\end{bmatrix} \begin{bmatrix}r\\ 1\end{bmatrix}$$
Call the projection matrix $P_i$. Note that the eigenvalues of $P_i$ are less than or equal to 1, since the bottom right element is 1 (one eigenvalue is always 1), and the upper left block is the identity minus an orthogonal matrix. A product of such $P_i$ will therefore converge to a projection matrix with eigenvalues $1$ or $0$. The limit eigenvectors of such eigenvalues (with last component scaled to unity) will give the limit points.
The above shows that the iteration process converges to a certain point, but that point could be the point at infinity $(\mathrm{anything},0)$.
The boundedness of the limit points depends on the relative magnitude of components of the limit eigenvectors. Note that an iterated product of $P_\infty = \prod P_i$ preserves the block structure, so we have
$$ P_\infty = \begin{bmatrix} A &u\\0&1 \end{bmatrix} $$
Solving for the eigenvector corresponding to the unit eigenvalue,
$$ (P_\infty -I)\begin{bmatrix}v\\ 1\end{bmatrix} = \begin{bmatrix} A-I &u\\0&0 \end{bmatrix}\begin{bmatrix}v\\ 1\end{bmatrix}$$
Which results in $v = (A-I)^{-1} u$. We need to make sure that there is always a nonzero $v$ for any $A$ and $u$. Note that the $A$ block is formed from products of reflectors $\prod I - \hat{n}_i\hat{n}_i^T $, so unless all lines are parallel, the eigenvalues of $A$ will be bounded away from 1. Thus we have our proof that the iteration always converges to a point not at infinity.
There might be some holes in this proof. And you can probably extend this to figure out how to determine $r$ as a function of the line parameters.
