Recent content by vr88

I'm more interested in a general solution with a linear algebra or a simultaneous equations approach in the hopes that such an approach will work for whatever initial matrices I start with.

Stephen- That's essentially what I'm looking for. I am more interested in the case of finite groups. I don't know how much permutations help, though the permutation matrices are orthogonal. For example, if G=A5 is the alternating group on 5 points, and I have two 3x3 matrices which generate this...

I mean that a matrix M is orthogonal if M-1=Mt.

Let A and B be nxn matrices which generate a group under matrix multiplication. Assume A and B are not orthogonal. How can I determine an nxn matrix X such that X-1AX and X-1BX are both orthogonal matrices? Is it possible to do this without any special knowledge of the group in question?

Perhaps you should state that you're issuing a math challenge and that you actually know what the answer is.

The first equality is wrong. Essentially what you want to show is that hkh-1k-1=e. But H∩K={e}, so just show that hkh-1k-1∈ H and hkh-1k-1∈ K.

Note that those two vectors are scalar multiples of each other, and eigenvectors are determined up to scalar multiples.

I believe that would be Artin's Constant.

I think this is right. As you said, there's a homomorphism from G to S5. The image of the G is a subgroup of S5, call it H. This subgroup can't be S5 itself, since S5 has 120 elements, and A5 has only 60. If |H| < 60 then the kernel of the homomorphism is more than just the identity, and is...

Yep. It's been done. Since any subgroup of a abelian group is normal, a simple abelian group must have only {e} and itself as subgroups. Thus the only simple abelian groups are the (cyclic) groups of order p, where p is a prime.

For 2*3*5, you would have primes greater than 5 in the form 30k±1, 30k±7, 30k±11, 30k±13