JDoolin said:
Okay, that's interesting. The space of the set of functions representable by a 12th order polynomial is a 12 dimensional, non-directional vector space. Am I correct that the space of the 12th order polynomial functions is different from the 12th order polynomial functions themselves? The space of the polynomial functions is just a list of the 12 coefficients.
As IsometricPion mentioned the vector space is the set of all 12 th order polynomials, e.g. the polynomial 5x^{12}-4x^{11}+8x^{10}+... is a vector in the space, but list of numbers (5,-4,8,...) is not.
You can consider many different sets of basis vectors for this space, the most obvious being (x^{12},x^{11},x^{10},...), and in terms of that basis you can write the vector as (5,-4,8,...) for notational convenience, but the 13 numbers are no longer just an arbitrary list of numbers, but are instead coordinates in a specified basis.
You could consider different basis vectors, such as the Legendre polynomials. Then you could write the same vector as a different list of numbers corresponding to coordinates in a different basis. Although the list of numbers would be different, the vector is the same, since the vector is the polynomial and not the list of numbers.
JDoolin said:
I'm still a little confused about how an election, ...
I am still not interested in this game. It seems clear to me that they are not, and I gave several good reasons above. I am not going to critique your ideas one by one. If you think that they form a vector space then please
1) Define the bijection between different election results and different vectors
2) Demonstrate the correspondence between real-world operations on elections and operations on the vectors
3) Prove that the vectors and operations satisfy all of the axioms of vector spaces (
http://en.wikipedia.org/wiki/Vector_space)