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Old 2022-09-22, 03:01   #12
bhelmes
 
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Mar 2016

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Default Is it possible to calculate one belonging rot. matrix from a vector ?

Quote:
Originally Posted by Dr Sardonicus View Post
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.

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Old 2022-09-25, 00:50   #13
Dr Sardonicus
 
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Feb 2017
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Default

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
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