I'm given $A_i$ the number of codewords in C of weight $i$, C an [n, k, d] code over $F_q$.
In this book (https://books.google.ch/books?id=RAsjychHkLYC&pg=PA252&lpg=PA252&dq=sum+of+weight+enumerator+A0+%2B+A1+%2B+...+%2B+An+%3D+q%5Ek&source=bl&ots=GxByVqye2h&sig=ACfU3U1N9N-FZBWuvaMX5_OjQ4SfqBtI_g&hl=de&sa=X&ved=2ahUKEwiJvPePwbHiAhUuzoUKHTgIACwQ6AEwD3oECAQQAQ#v=onepage&q=sum%20of%20weight%20enumerator%20A0%20%2B%20A1%20%2B%20...%20%2B%20An%20%3D%20q%5Ek&f=false) I read exercice 373 which says:
Let $M$ be a $q^k \times n$ matrix whose rows are the codewords of an C. Let $(A_i^{\perp})$ be the weight distribution of $C^{\perp}$. Why
- does a column of $M$ either consit entirely of zeros or contain every element of $F_q$ an equal number of times?
- does $M$ has $A_i^{\perp} / (q-1)$ zero columns?
These two things are mentioned in the book, but not proved. Could anybody tell me why the statement is true? (Don't have to be a full proof!)
Thanks in advance!
Well, consider the mapping $\phi:C\rightarrow {\Bbb F}_q:c\mapsto c_i$ which maps a codeword $c$ onto its $i$th coordinate $c_i$. This mapping is a group homomorphism, $\phi(c+c') =c_i+c'_i = \phi_i(c)+\phi_i(c')$.
It is well-known that for a group homomorphism, each image value is taken on an equal number of times. More concretely, let $I=\ker \phi_i$ be the kernel of the mapping. Then the mapping $C/I\rightarrow{\Bbb F}_q: c+I\mapsto \phi(c)= c_i$ is an isomorphism provided that the mapping $\phi$ is not the zero map (i.e., all codewords have $i$th coordinate 0). Thus the coordinate $c_i\in{\Bbb F}_q$ is taken on $|c+I|=|I|$ times with $|I|=q^k/q = q^{k-1}$.
This holds for the matrix $M$ and now you have the solution for problem a).