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Hi !

I have a parity check matrix $H$ for a code $C$
0 0 0 1 1 1 1 0
0 1 1 0 0 1 1 0
1 0 1 0 1 0 1 0
1 1 1 1 1 1 1 1

I am allowed to assume that
1) the dual of an $(n,k)$-code is an $[n,n-k]$-code
2) $(C^{\perp})^{\perp} = C$ (Here $\perp$ denotes the dual)

I want to prove that my code $C$ is self-dual. (ie that $C=C^{\perp}$)

Here is my logic:

I know that, since $H$ is a parity check matrix for $C$,
$H$ is a generator matrix for $C^{\perp}$.

Since $C^{\perp}$ is an $[n,n-k]$-code, the generator matrix $H$ is a matrix: $[n-k]$ x $ n$

So now looking at $H$, n=8 and k=4, so the corresponding $C$ code is a $8$x$4$ matrix.

Now let $G=[g_{i,j}]$ be the generator matrix for $C$.
$(GH^T)=0$ since every vector in the rowspace of $G$ is orthogonal to every vector in the rowspace of $H$;

Can anyone tell me what is missing to finish off my proof?

Note: i see that each row in $H$ has even number of entries and that the distance between any two rows is even. maybe this helps if I can right a definition of weight relating to duals...

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    >does it suffice? Yes it does, but see rschwieb's answer which shows one inclusion without having to look for (or look at) the generator matrix of $C$.2012-04-30

1 Answers 1

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The rows of $H$ generate $C^\perp$.

By definition of the parity check, $xH^\mathrm{T}=0$ iff $x\in C$.

What can you conclude from the fact that $HH^\mathrm{T}=[0]$?