Hello Dmitrii and Katie,
Something about Dmitrii's argument didn't seem accurate to me.
In a finite field of char. not 2, every element is a sum of two squares.
E.g. If q=5 then 2 is not a square yet 2=1^2+1^2, also 3 is not a
square yet
2^2+2^2=8=3. Etc.
So let s be and element in GF(q) and let a and b be elements such that
s= a^2 + b^2.
Then use
M=
[ a b ]
[ b -a]
So
M M^t =
[
a^2+b^2 0
0 a^2+b^2
]
=
(a^2+b^2) I_2
=
s I_2
So you can get any scalar. These are usually called "similitudes".
Look in Artin's ``Geometric Algebra'' is you want a reference Katie.
Hope that helps,
James
On 12/22/10 9:15 AM, Asst. Prof. Dmitrii (Dima) Pasechnik wrote:
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
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