20220922, 03:01  #12  
Mar 2016
2·5·41 Posts 
Is it possible to calculate one belonging rot. matrix from a vector ?
Quote:
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. 

20220925, 00:50  #13 
Feb 2017
Nowhere
1759_{16} 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 PariGP 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 20220925 at 16:35 
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