next up previous contents
Next: Basis Up: Spring 2004 Final Report Previous: Finite Fields   Contents

Vector Spaces

Vector spaces over $ \FF[2]$. The n-dimensional space, $ \FF[2]^n$, over $ \FF[2]$ is the set of vectors with n elements in $ \FF[2]$.
$ \FF[2]^n = \left\{q \left( \begin{array}{c} a_1\\
a_2\\ \vdots\\ a_n \end{array}\right) = \vec a \vert a_i \epsilon \FF[2] \right\}$


We can add vectors as follows:
$ \vec a +\vec b $ = \begin{displaymath}\left(
\begin{array}{c} a_1\\ \vdots\\ a_n \end{array} \righ...
...+b_1\\
\vdots\\ a_n+b_n \end{array} \right) \epsilon \FF[2]^n\end{displaymath}


We can scalar multiply by $ \lambda\epsilon\FF[2]$ as:
\begin{displaymath}\lambda \vec a = \lambda \left(
\begin{array}{c}
a_1\\ \vdo...
...1\\ \vdots\\ \lambda a_n \end{array} \right) \epsilon
\FF[2]^n\end{displaymath}
Make note that $ \FF[2]^n$ has $ 2^n$ elements, and is not equal to $ \FF[2^n]$ There is nothing special about $ \FF[2]$, $ \FF[8], \FF[27], \FF[5]$ etc., and the rules over those fields are no different. The different between $ \FF[2]^n$ and $ \FF[2^n]$ is described in the finite field sections. $ \FF[2]^2 = \{ {0 \choose 0}, {0 \choose 1} , {1 \choose 0} , {1
\choose 1} \}$ has four elements.


Subsections

Frederick Leitner 2004-05-12