Invariant planes and the classification

We have now shown how Möbius transformations are mapped to $3\times3$ matrices $A$ in $SO(2,1)$. The classification of Möbius transformations is determined by the properties of these $3\times3$ matrices.

First, we must show that each such matrix $A$ has an eigenvector $\mathbf{n}$ with eigenvalue 1. Since it is a $3\times3$ matrix, it will have three eigenvalues, counting multiplicity. Suppose $\lambda$ is an eigenvalue (possibly complex), and let $\mathbf{x}$ be an eigenvector for $\lambda$. Then $A\mathbf{x}= \lambda\mathbf{x}$. Since $A^T JA=J$, we have $J\mathbf{x}=A^TJA\mathbf{x}=\lambda\,A^TJ\mathbf{x}$. Then $J\mathbf{x}$ is an eigenvector of $A^T$ with eigenvalue $1/\lambda$. Since $A$ and $A^T$ have the same eigenvalues, we conclude that $1/\lambda$ is also an eigenvalue of $A$. Thus, if $\lambda\neq 1$, then the third eigenvalue must be 1, in order for the determinant of $A$ (which is always equal to the product of all the eigenvalues) to be 1. This proves $A$ has an eigenvalue equal to 1.

Also, if $A$ has more than one independent eigenvector with eigenvalue 1, then it must be the identity matrix. Then for non-identity matrices $A$, let $\mathbf{n}$ be a unit eigenvector with eigenvalue 1, i.e. such that $A\mathbf{n}=\mathbf{n}$. All the planes $\{\mathbf{n}^T J \mathbf{x}=c\}$ are invariant under $A$, since

$$\mathbf{n}^T J A\mathbf{x}= \mathbf{n}^T (A^T)^{-1} J \mathbf{x}= (A^{-1}\mathbf{n})^T J\mathbf{x} =\mathbf{n}^T J\mathbf{x}.$$

This is a family of parallel planes all invariant under $A$. Then we say that $A$ is elliptic, parabolic or hyperbolic if and only if the corresponding cross-sections of the standard right circular cone are ellipses, parabolas, or hyperbolas, respectively. It's really that simple. However, there is quite a bit of checking involved in the statements in this section.