I don't understand this proof of the theorem "If $C$ is a cyclic code, then its dual $C^{\perp}$ is cyclic too"
The proof is: " Let $c=(c_0,c_1,\dots,c_{n-1})\in C$ and $h=(h_0,h_1,\dots,h_{n-1})\in C^{\perp}$.
$\sigma(c)\cdot\sigma(h)=c\cdot h=\sum_i c_ih_i=0, \, \forall c\in C.$ ($\sigma$ is the shift)
For $C=\sigma(C)$ ($C$ is cyclic), $\sigma(h)\perp C$, so $\sigma(h)\subseteq C^{\perp}$, that is $\sigma(C^{\perp})\subset C^{\perp}$."
I really don't understand this proof. If I have to prove that $C^{\perp}$ is cyclic, how can I use $\sigma(h)=h$?
And how does conclusion $\sigma(C^{\perp})\subset C^{\perp}$ tell me that $C^{\perp}$ is cyclic?
Could someone explain please? Or does someone know another kind of proof?
Thanks
$\sigma(C^{\perp})\subset C^{\perp}$ implies $\sigma(C^{\perp})=C^{\perp}$ by a counting argument since $\sigma(c)=\sigma(c')$ implies $c=c'.$ Since $C$ and $C^{\perp}$ are subspaces (each cyclic code is also linear) $\dim C+ \dim C^{\perp}=n.$
And $\sigma(h)\neq h$ in general, unless $h$ is subperiodic which is excluded if the codelength $n$ is a prime. All the first equation says is that the inner product of cyclic shifts by 1 of the two codewords is equal to the original inner product.