Dear Katie, MM^T=\lambda*I holds iff \lambda is a square in GF(q), i.e. \lamda=\mu^2, as can be see by taking the determinant of both sides of your equation. Then each M can be obtained as \mu M', for M' in GO(n,GF(q)). Best, Dmitrii
On 22 December 2010 21:40, Katie Morrison <kmorr...@gmail.com> wrote: > I understand that the general orthogonal group that GAP computes is not the > group of matrices that satisfy MM^T=I because the GO group they compute > actually leaves a different bilinear form fixed than the dot product. But > is there an easy way to find the group of matrices that satisfy MM^T=I and > or to find the generalized version of this that satisfy MM^T=\lambda*I for > some nonzero \lambda \in GF(q)? A brute force search becomes completely > unwieldy for matrices larger than 3 by 3, so if I can find some easy way to > map from the general orthogonal group that GAP uses or find an efficient > algorithm for computing this other group, that would be great. > > Thanks, > Katie Morrison > _______________________________________________ > Forum mailing list > Forum@mail.gap-system.org > http://mail.gap-system.org/mailman/listinfo/forum > -- Dmitrii Pasechnik ----- DISCLAIMER: Any text following this sentence does not constitute a part of this message, and was added automatically during transmission. CONFIDENTIALITY: This email is intended solely for the person(s) named and may be confidential and/or privileged. If you are not the intended recipient, please delete it, notify us and do not copy, use, or disclose its content. Thank you. Towards A Sustainable Earth: Print Only When Necessary _______________________________________________ Forum mailing list Forum@mail.gap-system.org http://mail.gap-system.org/mailman/listinfo/forum
