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I am asked to prove the fact that if $C$ is an $[N,k]$ code, and $C^{\perp} = x \in \mathbb{F}_2^N$ $|$ $(x,c) = 0$  $\forall c \in C$, then $\dim C + \dim C^{\perp} = N$. I am regrettably far behind in my studies and still do not know too much about codes, but isn't this basically a straight application of the rank-nullity theorem?

In particular, can I just think of it like this: $C$ has $k$ free values in its vectors, so $C^{\perp}$ has to have $k$ $0$'s in it's vector, so $C^{\perp}$ can only have $N - k$ free values? Or am I oversimplifying things? This seems like a trivial question, but maybe I am not thinking of it right...

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    Yes. You are on the right track. This is an application of rank-nullity. Consider the linear mapping from $\mathbb{F}_2^N$ to $\mathbb{F}_2^k$ defined by calculating the `inner products' with the $k$ basis vectors. Another way of looking at is to observe that the information positions (leading ones of the rows of a generator matrix in a reduced row-echelon form) become check positions of the dual code (you can solve that particular coordinate from the single linear equation involving it).2012-11-06

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