Let $A_i$ be the number of codewords in $C$ of weight $i$, with $1 \leq i \leq n$. We defined the list of $A_i$'s as the weight distribution of the code.
Now, let $C$ be an [n, k, d] code over $F_q$. If C is a binary code containing $1 = (1, ..., 1)$, why is then the weight distribution of C symmetric?
I began as follows that, if C is symmetric, it should hold that $A_i = A_{n-i}$ for $1 \leq i \leq n$. But until here, it's just using the definitions...how sould I proceed to show the claim?
Thanks for any help!
Well, if the all-1 vector lies in the binary code, then for each codeword $c$, $$c' = (1,\ldots,1) - c$$ is also a codeword with the property that the Hamming weight of $c'$ is $n$ minus the Hamming weight of $c$.
If $C$ is binary and self-orthogonal, then $0=\langle c,c\rangle = c_1^2+\ldots+c_n^2 = c_1+\ldots+c_n$ and so $c$ must be even.