20200221, 20:08  #1 
Mar 2016
5×7^{2} Posts 
pollard rho with bilinear form and eigenvektor ?
A peaceful evening,
if i make a pollard rho algorithm with a suitable bilinear form instead of the function f(x)=x²+1, is it then better to use the eigenvektor of the matrix or not. I only know the concept of eigenvektors with a matrix and one singular vektor, but i thought that the "radius" of the bilinearform with the eigenvektor is smaller. Any ideas, suggestion or mathematical clarification ? Greetings from the eliptic curves Bernhard 
20200226, 19:46  #2  
Mar 2016
5·7^{2} Posts 
Quote:
if A is the matrix of the bilinearform and v the eigenvektor, than Av=rv, multiplication with the vektor v on the left side gives the result vAv=v*r*v = r *v *v where the bilinearform is disappearing Nevertheless you can reduce C on the unit circle, is it possible to find a subgroup of a bilinearform such as A= (1, 1) (1, 1) ? Thanks for your patience Bernhard Last fiddled with by bhelmes on 20200226 at 20:42 

20200304, 20:52  #3  
Mar 2016
365_{8} Posts 
Quote:
It is always possible to reduce the 2x2 bilinearform to a quadratic polynom by setting y=x, this is trivial. If I choose a linear substituion for y depending on x (for example y=2x) can i make any mathematical consequnences by it ? Greetings Bernhard 

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