next up previous
Next: Constructing G (and hence, Up: A Linear Algebra Point Previous: A Linear Algebra Point

Proof

Consider the distances between the vectors of C and 0 (who is a codeword since it can be generated by G). The smallest of these distances will be the minimum weight of C since this distance is a count of nonzero positions. Thus the minimum weight of C will be greater than or equal to the minimum distance of C (since it's still possible for a smaller distance to exist between two other vectors of C).

To finish the proof, we need to show that there exists a codeword $ {\bf c_{0}}$ whose weight equals the mininum distance of C. Choose c and c' from C such that $ d({\bf C})=d({\bf c},{\bf c}')$. Set $ {\bf c_{0}}={\bf c}-{\bf c}'$ (a linear combination). Then we have $ w({\bf c_{0}})= d({\bf c}-{\bf c}',{\bf0})=d({\bf c}-{\bf c}',{\bf c}'-{\bf c}')=d({\bf c},{\bf c'})=d({\bf C})$.

We define the dual code as $ {\bf C}^{\bot}:=\{{\bf v}\epsilon {\bf V}\vert{\bf v} \cdot {\bf c}=0, \forall {\bf c} \epsilon {\bf C}\}$. Thus the dual code consists of the vectors orthogonal to all the codewords.

If C is a linear code of dimension $ k$, then $ {\bf C}^{\bot}$ is a subspace of V of dimension $ n-k$. Proof on page 187 of book.

Let C be a linear code. Then the matrix H whose rows form a basis of the dual code $ {\bf C}^{\bot}$ is called a parity check matrix of C.


next up previous
Next: Constructing G (and hence, Up: A Linear Algebra Point Previous: A Linear Algebra Point
Bill Cherowitzo 2001-12-11