Several exercises in the assignment due March 8 deal with applications of the following corollary given there:
The basic idea in applying this theorem to find a stabilized line is to observe, first of all, that the coefficient vector a of the equation is normal to the line and, therefore, determines the parallel class of the line. The lines in a parallel class are distinguished by different values of c for a given vector a. The corollary states that a must be an eigenvector of the matrix transp(U) , and unless the corresponding eigenvalue is 1, the scalar c is determined by a through the formula c = (1 - lambda)^{-1} a . v.
For example, if 1 is not an eigenvalue of U (which has the same eigenvalues as transp(U) though not the same eigenvectors), then there is one and only one line stabilized by f for each parallel class of eigenvectors a of transp(U) . When U has 1 as an eigenvalue, there will still be a unique stabilized line normal to the eigenvector of transp(U) for an eigenvalue different from 1 if there is one.
There will be a stabilized line normal to an eigenvector a of transp(U) for the eigenvalue 1 if and only if a is normal to the vector v. In this case the stabilized line, which is normal to a, must be parallel to v, and every line parallel to v is stabilized by f.
A fixed point c of f is characterized by the equation (1 - U)c = v, and, therefore, there is a unique fixed point -- hence no line that is fixed -- when 1 is not an eigenvalue of U. If, on the other hand, 1 is an eigenvalue of U but U is not the identity matrix, then 1 - U is a rank 1 matrix and there is a line of fixed points c of f if and only if v is in the column space of 1 - U.
Two of the exercises in question were:
In both cases the matrix U must be a matrix of order 2 with det U < 0. Since U^{2} = 1, (det U)^{2} = 1, and, therefore, det U = -1. Since (1 + U)(1 - U) = (1 - U)(1 + U) = 1 - U^{2} = 0, it follows that both 1 + U and 1 - U are rank 1 matrices since neither is invertible (and neither is 0 lest it be {+/-} 1 and so have determinant 1). Therefore, the eigenspace of U for the eigenvalue 1, which is the nullspace of 1 - U, is the column space of 1 + U, and that for the eigenvalue -1, which is the nullspace of the matrix 1 + U, is the column space of the matrix 1 - U. Moreover, all of the statements of this paragraph continue to hold if U is replaced by transp(U) . Finally, the column space of 1 - transp(U) is orthogonal to that of 1 + U, and the column space of 1 + transp(U) is orthogonal to that of 1 - U.
In the case of a glide reflection (No. 5) U is a symmetric orthogonal matrix, the axis of f is parallel to the eigenspace of U for the eigenvalue 1, and a coefficient vector a for the equation of the axis is normal to the axis, hence, in the eigenspace for the eigenvalue -1 of U = transp(U) , and, therefore, the axis of the glide reflection is the unique stabilized line with coefficient vector a in the eigenspace for the eigenvalue -1. If there is a stabilized line for which the coefficient vector a is an eigenvector for the eigenvalue 1 of transp(U) = U, then by the corollary above v must be perpendicular to a, hence, perpendicular to the eigenspace of U for the eigenvalue 1, and so f would be a reflection, not a glide reflection. Therefore, the axis of a glide reflection is the only line stabilized by a glide reflection.
The case of an orientation-reversing affine transformation of order 2 (No. 4) contains the case of a reflection, in which case it is obvious that any line perpendicular to the axis of the reflection is stabilized. The general situation is somewhat similar.
Since f has order 2, one has U v = -v, and, therefore, (1 + U)v = 0, from which it follows that v is in the column space of (1 - U). Therefore, as above, there is a line of fixed points c of f that is a translate of the 1-dimensional nullspace of 1 - U. What is the equation of this line? In particular, what vector may be used as its coefficient (normal) vector? A vector parallel to the nullspace of 1 - U is perpendicular to the row space of 1 - U. The row space of 1 - U is the same as the column space of 1 - transp(U) , and the relation (1 + transp(U) )(1 - transp(U) ) = 0 between two rank 1 matrices shows that the column space of 1 - transp(U) is equal to the nullspace of 1 + transp(U) . Hence, a vector normal to the fixed line of f is characterized as an eigenvector for the eigenvalue -1 of transp(U) . It was explained above that there is one and only one stabilized line with this coefficient vector, and it is, therefore, the fixed line.
Of course, 1 is also an eigenvalue of transp(U) , and, as explained above, any line of the form a . x = c with a an eigenvector of transp(U) for the eigenvalue 1 is a stabilized line provided that a is normal to v. This latter condition is seen as follows: from the fact that a is in the nullspace of 1 - transp(U) it follows that a is normal to the row space of 1 - transp(U) , which is the same as the column space of 1 - U and the nullspace of 1 + U where v is known to reside. The stabilized lines a . x = c, where transp(U) a = a, are not fixed lines since there is only one fixed line, as discussed above.