The Fixed Points of a Collineation

A point A is fixed by a collineation with matrix H exactly when $H A\sim A$, i.e. $H A = \alpha A$ for some scalar $\alpha$. In other words, A must be a right eigenvector of H. Since an $(n+1)\times(n+1)$ matrix typically has n+1 distinct eigenvalues, a collineation in ${I\!\!P}^{n}$ typically has n+1 fixed points, although some of these may be complex.

H maps the line through any two fixed points A and B onto itself: $H(\lambda A + \mu B) = \alpha\lambda A + \beta\mu B$. In addition, if A and B have the same eigenvalue ( $\alpha=\beta$), $\lambda A + \mu B$ is also an eigenvector and the entire line AB is pointwise invariant. In fact, the pointwise fixed subspaces of ${I\!\!P}^{n}$under H correspond exactly to the eigenspaces of H's repeated eigenvalues (if any).

Exercise 2.5   : Show that the matrix associated with a plane translation has a triple eigenvalue, but that the corresponding eigenspace is only two dimensional. Provide a geometric interpretation of this.

Exercise 2.6   : In 3D Euclidean space, consider a rotation by angle $\theta$ about the $\vec{z}$ axis. Find the eigenvalues and eigenspaces, prove algebraically that the rotation axis is pointwise invariant, and show that in addition the ``circular points'' with complex coordinates $(1, i, 0, 0)^{\top}$ and $(1, -i, 0, 0)^{\top}$ are fixed.