2021-10-23, 21:49 | #1 |
Mar 2016
3^{2}×41 Posts |
2*2 Matrix with determinant 1
A peaceful night for you,
do I see it right that 2*2 matrix with determinant 1 build a subgroup in linear algebra and that calculation in the group is better for factoring than calculation in all 2*2 matrix ? Do I see right that the matrix (2 0) (0 -1) has the determinant -2 and that an exponentation with 2*p of Mp mod Mp will result in a matrix with determinant 1 Is it rigth that there are 3 pauli-matrixs https://en.wikipedia.org/wiki/Pauli_matrices with determinant 1 and that a parallel multiplication of a start matrix with determinant 1 and a pauli matrix will speed up a possible factorisation ? Would be nice to get a reply. P.S. I do not know why or how, but I am sure we will find the next Mp before next christmas. Last fiddled with by bhelmes on 2021-10-23 at 21:51 |
2021-10-24, 07:45 | #2 |
Dec 2012
The Netherlands
1758_{10} Posts |
Over any commutative ring R, the 2x2 matrices with determinant 1 form a subgroup \(SL_2(R)\)
of the unit group of the ring of all 2x2 matrices. The Pauli matrices have determinant -1, however. It is important to decide carefully which commutative ring R you are working over: the integers, the Gaussian integers, the complex numbers, etc. Some of these can be constructed using matrices: for example, matrices of the form \[ \left(\begin{array}{rr}x & -y \\ y & x\end{array}\right)\] over the real numbers form a copy of the complex numbers. |
2021-10-24, 23:21 | #3 |
Feb 2017
Nowhere
12015_{8} Posts |
Let's see. det(M^{k}) = (det(M))^{k}.
So if det(M) = -2, det(M^{2p}) = (-2)^{2p} = 2^{2p} == 1 (mod 2^{p} - 1). Check. I don't see any relation to factorization. |
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