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2022-09-22, 03:01   #12
bhelmes

Mar 2016

22·3·5·7 Posts
Is it possible to calculate one belonging rot. matrix from a vector ?

Quote:
 Originally Posted by Dr Sardonicus No, sir. If M is a 2x2 matrix, scalar multiplication of M by k produces a matrix with determinant k^2*det(M). Thus, if M is a nonsingular 2x2 matrix, det((1/det(M))M) is 1/det(M). If M is 2x2 and det(M) is not a square, no scalar multiple of M will have determinant 1.
A peaceful, early morning, especially for Dr Sardonicus,

let: p=31, u1=2, v1=3; so that the norm (u1,v1)=u1²+v1²=13=12⁻¹ mod 31 and 13²=20⁻¹ mod 31

Is it possible from linear algebra to calculate one belonging rotation matrix from this vector ?

The calculated target is:
13*
(27,2)* (2)=(5)
(-2,27) (3)=(9)

13*
(14,17)* (2)=(4)
(-17,14) (3)=(11)

13*
(24,8)* (2)=(6)
(-8,24) (3)=(15)

http://localhost/devalco/unit_circle/system_tangens.php

(the red coloured boxes right, all calculations checked and it seems to be all right.)

My first try:
Let p=31

Let M1=13*M1*=
13*
(a*, b*)
(-b*, a*)

1. with det (M1)=1=det (13² * det (M1*)) so that det (M1*)=20 mod 31, therefore a*²+b*²=20
2. with M1*(u1,v1)=(u2,v2) with norm (u1,v1)=(u2,v2)=u1²+v1²=u2²+v2²=13 mod p

so that

13*
(a*, b*) = (u2)
(-b*, a*) (v2)

This is more a fragment and should point in one direction, and as it is too late for me,
I hope that some one could finish the calculation.

 2022-09-25, 00:50 #13 Dr Sardonicus     Feb 2017 Nowhere 2×11×283 Posts It appears you are trying to solve R[a;b] = [c;d] where R = [x,y;-y,x]. Writing as a system of linear equations, x*a + y*b = c b*x - a*y = d which may be rewritten [a,b;b,-a][x;y]=[c;d] which is solvable if a^2 + b^2 ≠ 0. EDIT: Feeding the formula to Pari-GP produces the same values you got: Code: ? matsolve([Mod(2,31),Mod(3,31);Mod(3,31),Mod(-2,31)],Mod(1/13,31)*[Mod(5,31);Mod(9,31)]) %1 = [Mod(27, 31)] [Mod(2, 31)] ? matsolve([Mod(2,31),Mod(3,31);Mod(3,31),Mod(-2,31)],Mod(1/13,31)*[Mod(4,31);Mod(11,31)]) %2 = [Mod(14, 31)] [Mod(17, 31)] ? matsolve([Mod(2,31),Mod(3,31);Mod(3,31),Mod(-2,31)],Mod(1/13,31)*[Mod(6,31);Mod(15,31)]) %3 = [Mod(24, 31)] [Mod(8, 31)] Last fiddled with by Dr Sardonicus on 2022-09-25 at 16:35
 2022-10-02, 21:22 #14 bhelmes     Mar 2016 22·3·5·7 Posts A prediction of a rotation matrix for Mp A peaceful and pleasant night for you, If Mp is a Mersenne number with exponent p I can predict the following rotation matrix M=(Mp-2^[(p-1)/2];-2^[(p-1)/2) (2^[(p-1)/2]; Mp-2^[(p-1)/2]) with det (M)=1 Example: Mp=31 p=5 M=(27, -4) (4, 27) with det (M)=16+16=1 mod 31. This rotation matrix is not a result of two vectors (x1,y1), (x2, y2) with the same norm x1²+y²=x2²+y2²=n mod Mp, which needed to be found in advance, but it is completely predictive. I did not understand, why this matrix occurs one time for non quadratic residues and one time for quadratic residues: (27,4)* (2)=(4) (-4,27) (3)=(11) with 22+32 = 13 and jac (13, 31)=-1 (27,4)* (5)=(8) (-4,27) (7)=(14) 52+72 = 12 and jac (12, 31)=-1 (27,27)* (8)=(1) (-27,27) (15)=(3) 82+152 = 10 and jac (10, 31)=1 Is there a mathematical explication for this phenomenon ? More examples and as usual a web interface under http://devalco.de/unit_circle/system_tangens.php Enjoy the predictable rotation matrix,
 2022-10-04, 14:12 #15 Dr Sardonicus     Feb 2017 Nowhere 2×11×283 Posts If 2*x^2 == 1 (mod N) and M = x[-1,-1;1,-1] we have det(M) = 2*x^2 == 1 (mod N) and M^2 = x^2[0,2;-2,0] == [0,1;-1,0] (mod N), so that M^8 == [1,0;0,1] (mod N). One can take N = 2^n - 1 with n > 1 odd, and x = 2^((n-1)/2). However, suitable x will exist for any N composed entirely of primes congruent to 1 or 7 (mod 8).
 2022-11-24, 22:52 #16 bhelmes     Mar 2016 22·3·5·7 Posts A prediction of the rotation matrices There should be more peace in the world ! The pythagorean triples can be used to make a surjective mapping to the rotation matrices concerning p if x²+y²=z² | : z² (x/z)²+(y/z)²=1 which is the determinant of the rotation matrix a=x/z mod p; b=y/z mod p M= (a, b) (-b, a) Question: if M*(n, m)=(u, v), then n²+m²=u²+v² mod p and if jacobi (n²+m², p)=-1 resp. n²+m² a non quadratic residue concerning p and p = 3 mod 4 is there a chance that [(n+mi)/(u,vi)]^r=1 mod p [r=(p+1)/2] and r maximal and not a right divisor of (p+1)/2 ? Or in other words: Is it possible to say anything about the group order of the resulting vector which is the divisor of the two vectors connected by the rotation matrix ? A peaceful 1. advent Of course a wonderful php scripting image with more infos you will need: http://devalco.de/unit_circle/system_tangens.php

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