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Old 2021-10-23, 21:49   #1
bhelmes's Avatar
Mar 2016

32×41 Posts
Default 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
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
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Old 2021-10-24, 07:45   #2
Nick's Avatar
Dec 2012
The Netherlands

33368 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
\[ \left(\begin{array}{rr}x & -y \\ y & x\end{array}\right)\]
over the real numbers form a copy of the complex numbers.
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Old 2021-10-24, 23:21   #3
Dr Sardonicus
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Feb 2017

22·5·257 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.
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