[GAP Forum] Matrices that satisfy MM^T=I or MM^T=\lambda*I
James Wilson
wilson at math.ohio-state.edu
Wed Dec 22 14:59:49 GMT 2010
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<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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>
>
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