mersenneforum.org > Math 2*2 Matrix with determinant 1
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 2021-10-23, 21:49 #1 bhelmes     Mar 2016 22·3·5·7 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 Nick     Dec 2012 The Netherlands 71416 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 Dr Sardonicus     Feb 2017 Nowhere 623710 Posts Let's see. det(Mk) = (det(M))k. So if det(M) = -2, det(M2p) = (-2)2p = 22p == 1 (mod 2p - 1). Check. I don't see any relation to factorization.
2022-02-14, 05:10   #4
bhelmes

Mar 2016

1101001002 Posts

Quote:
 Originally Posted by Nick 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.
Considering the matrix M of the form
(n²+1; n)
(n; 1)

=(a; b)
(c; d)

The determinant is one.

For n=0 it includes the neutral element concerning the multiplication,

is closed under exponentation,
means for M^n the elements of b and c are equal and

Is closed under multiplication for the element a=1 mod 4.
It has obviously an inverse matrix
and the multiplication of M*N seems to be kommutativ.

It is not clear for me how you calculate the group order of this sub-sub-group
if you limit all elements modulo f and limit the multiplication to the exponentation of one M

Perhaps someone could correct me and add a little bit more knowledge.

P.S. I managed to speed up my computer ; 7 days instead 8 days for an exponent at the wawefront.
(137 Watt instead 120 Watt) (you could easily speed up the calculation in the bios (Asus mainboard) )

2022-02-14, 15:04   #5
Dr Sardonicus

Feb 2017
Nowhere

34·7·11 Posts

Quote:
 Originally Posted by bhelmes It is not clear for me how you calculate the group order of this sub-sub-group if you limit all elements modulo f and limit the multiplication to the exponentation of one M
The characteristic polynomial of M = [n^2 + 1, n; n, 1] is F(x) = x^2 - (n^2 + 2)*x + 1. The discriminant is D = n^2*(n^2 + 4).

Since M is a real symmetric matrix, it is diagonalizable, with diagonal elements the roots of F(x) = 0.

If f = p, a prime number, and gcd(p,D) = 1, then the multiplicative order of M (mod p) divides p - 1 if (D/p) = +1 (quadratic character).

Because F(x) is a quadratic with constant term 1, we can say that the multiplicative order divides p + 1 if (D/p) = -1.

So we can say that if p does not divide n or n^2 + 4, the multiplicative order (mod p) divides p - (D/p).

EDIT: Assuming p does not divide D, we can drop the square factor n^2 from the quadratic residue symbol, and use $$$\frac{n^2+4}{p}$$$ instead of (D/p).

Last fiddled with by Dr Sardonicus on 2022-02-15 at 15:14 Reason: formatting; and as indicated

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