A Lie matrix group is a continuous subgroup of the group of all non-singular matrices over a field , where is either R or C. ``Continuous'' really is a shorthand for saying that the Lie group is a manifold. The rough idea is that the components of a matrix in the group can vary smoothly; thus, concepts like ``differentiable function '' should make sense. I'll just say ``Lie group'' for Lie matrix group, though many mathematicians would groan at this.
Example: is the group of all orthogonal matrices, i.e. all matrices with real components such that . This is just the group of all isometries of which leave the origin fixed. (Standard calculation: let be a column vector. Then , i.e., the norm of equals the norm of .) Note also that the equations and follow from .
If , then we have immediately , i.e., . is the subgroup of all matrices in with determinant 1. Fact: is connected, and is in fact the connected component of 1in . I will focus initially on and . These are, colloquially, the groups of rotations in 2-space and 3-space. is the group of reflections and rotations.
Digression: the well-known puzzle, ``why do mirrors reverse left and right,
but not up and down?'' is resolved mathematically by pointing out that
a mirror perpendicular to the y-axis performs the reflection:
Example: , rotations in 2-space. Since , it is easy to
write down the components of . Equating these to , we see
that has the form:
Elements of have real components, but it is enlightening to
consider as a subgroup of the group of all non-singular complex
matrices. Fact: any matrix in is similar to a matrix
of the form
Example: , rotations in 3-space. Fact: any element of leaves a line through the origin fixed. This seems obvious to anyone who has rolled a ball, but it is not totally trivial to prove. I will outline two arguments, both instructive.
First, an approach Euclid might have followed. Fact: any isometry of 3-space that leaves the origin fixed is the composition of at most three reflections. For consider the initial and final positions of the x, y, and z axes. One reflection moves the x-axis to its final position, pointing in the correct direction; a second reflection takes care of the y-axis; a third may be needed to reverse the direction of the z-axis. We see that an orientation-preserving isometry is the composition of two reflections. The intersection of the planes of these two reflections gives a line that is left fixed.
Next, a linear algebra approach. We want to show that leaves a vector fixed (), or in other words that 1 is an eigenvalue. Note first that the characteristic polynomial for is a cubic with real coefficients. Hence it has at least one real root. Furthermore, implies that trivially, since all components of are real. By the spectral theorem, can be diagonalized (working over C) and its eigenvalues are all on the unit circle. Since the characteristic polynomial has real coefficients, non-real roots appear in conjugate pairs. It follows that is similar to a matrix of the form or . Since , and the determinant is the product of the roots, at least one eigenvalue must be 1.
The first argument can also be cast in linear algebra form. This leads to Householder transformations.
© 2001 Michael Weiss