Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 1 / 10G (Coding & Cryptography)

A linear code of length $n$ and rank $k$ is a $k$-dimensional linear subspace of ${\mathbb{F}}_2^n$. The Hamming weight of a vector counts its nonzero coordinates. Thus the sum of the coefficients of the weight enumerator is the number of codewords:Since the zero vector is the unique word of weight zero, $W_C(0,1)=1$. Symmetry of the enumerator therefore implies $W_C(1,0)=1$. Conversely $W_C(1,0)=1$ means that the all-one vector $1$ belongs to $C$. Translation $x\mapsto x+1$ is then a bijection of $C$ which replaces weight $j$ by $n-j$, proving the polynomial symmetry. For length zero the same conclusion holds with the unique empty vector.

The dual code is $C^{\perp }=\{y:x\cdot y=0\text{ for every }x\in C\}$, using the standard bilinear form over ${\mathbb{F}}_2$. If $y\in C^{\perp }$, each summand of the character sum equals one. Otherwise choose $x_0\in C$ with $x_0\cdot y=1$; replacing $x$ by $x+x_0$ permutes the summands and reverses their sign. The sum equals its own negative and must be zero. ConsequentlyThis is character orthogonality for the additive group of the code.

For any vector $y\in {\mathbb{F}}_2^n$, use its Hamming weight $w(y)=\#\{j:y_j=1\}$. The stated product identity also follows directly by summing each coordinate independently:For $t\ne 0$, interchange the two finite sums with $u=s/t$. Character orthogonality evaluates the inner sum over codewords, whereas the product identity evaluates the sum over all binary vectors. ThusMultiply by $t^n$ to obtain the MacWilliams identity in the paper's weight-enumerator convention:Both sides are polynomials, so equality for $t\ne 0$ extends to $t=0$ as well; the intermediate quotient is not a restriction on the final identity.