Let $C $ be a self-dual binary $[n,k,d]$ code.

I want to show that if $ c=(c_1, \dots, c_n) \in C $ then $ \sum_{i=1}^n c_i=0$ and that all the words of the code have an even weight.

We know that $ x \in C^{\perp} \Leftrightarrow Gx=0$ where $ G $ is the generator matrix.

Since $ C=C^{\perp}$ we have that $ G=H $ where $ H$ is the parity matrix of the code.

So we have that $ c=(c_1, \dots, c_n) \in C \Rightarrow c \in C^{\perp} \Rightarrow Gx=0 \Rightarrow Hx=0 $

Do we get from that, that $ \sum_{i=1}^n c_i=0$?

If so how?


By definition, the dual code $C^\perp$ consists of all words $x$ such that $\langle x,y \rangle = 0$ for all $y \in C$. If $C = C^\perp$ then in particular $\langle x,x \rangle = 0$ for all $x \in C$, which implies that $x$ has even weight.

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  • $\begingroup$ Sorry, I had a typo in my solution. Now it makes more sense. $\endgroup$ – Yuval Filmus Mar 13 '16 at 16:02
  • $\begingroup$ It doesn't necessarily hold that the all one vector is in $C$. I'll let you find an example. It's also certainly not the case that $\langle x,x \rangle = 0$ iff $x = 0$; this only holds over characteristic zero. Try to think stuff over for a few hours. $\endgroup$ – Yuval Filmus Mar 13 '16 at 16:13
  • $\begingroup$ I believe you can solve this yourself. Just try a bit longer. $\endgroup$ – Yuval Filmus Mar 13 '16 at 17:18
  • $\begingroup$ That's the basic idea, though we care about the remainder modulo 4 rather than modulo 2. At any rate, I believe you can solve it on your own, and will not provide any more help. $\endgroup$ – Yuval Filmus Mar 13 '16 at 17:29
  • $\begingroup$ Evinda, I cannot act as your TA. $\endgroup$ – Yuval Filmus Mar 13 '16 at 22:46

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