 Solution

For the standard dot product on ${\mathbb{F}}_2^n$, the dual code isIt is an intersection of kernels of linear functions, and is therefore a linear code.

Let $\sigma$ denote cyclic right shift. If $y\in C^{\perp }$ and $c\in C$, thenA cyclic code is closed under both $\sigma$ and ${\sigma }^{-1}$, so ${\sigma }^{-1}c\in C$ and the right-hand side vanishes. Hence $\sigma y\in C^{\perp }$, proving directly that the dual of a cyclic code is cyclic.

Identify words with polynomials in ${\mathbb{F}}_2[X]/(X^n-1)$. If the generator polynomial of a cyclic code is the monic divisor $g(X)$ andthen the generator polynomial of $C^{\perp }$ is the monic reciprocalEquivalently, the parity-check polynomial of one code becomes, after reversal, the generator polynomial of its dual.