[GAP Forum] Matrices that satisfy MM^T=I or MM^T=\lambda*I
Asst. Prof. Dmitrii (Dima) Pasechnik
dima at ntu.edu.sg
Wed Dec 22 14:15:46 GMT 2010
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 <kmorris2 at 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
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--
Dmitrii Pasechnik
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