A)^{-1})_{\theta^{-1}} \circ \Theta \right)= \mathscr{C}^\perp\\
&\Longleftrightarrow \mathscr{C}^\perp \star \left(  \Theta\circ
({}^t \! A)^{-1}   \right)= \mathscr{C}^\perp ,
 \end{aligned}$$ where $$({}^t \! A)^{-1}=\left( \begin{array}{ c | c c c } 0 & 1 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \\ \hline
\alpha^{-1} & 0 & \cdots & 0 \end{array}\right).$$ Thus we can conclude that $\mathscr{C}^\perp$ is a skew $\alpha^{-1}$-cyclic code.

Finally, having in mind that $(\mathscr{C}^\perp)^\perp
=\mathscr{C}$, the converse of the statement can be immediately obtained.

Let $\C$ be a skew $\alpha$-cyclic code. If $\mathscr{C} = \mathscr{C}^\perp$, then $n$ is even and $\alpha=\pm 1$.

First of all, observe that $n=\dim\mathscr{C}+\dim\mathscr{C}^{\perp}=2\dim\mathscr{C}$. Finally, let $g(x)$ be the generator polynomial of $\mathscr{C}$. By Theorem \[GMaruta:1e\] we have $$X^n-\alpha=h_1(x)g(x)\qquad \mathrm{and} \qquad X^n-\alpha^{-1}=h_2(x)g(x)\ ,$$ for some $h_1(x),h_2(x)\in R$. So we get $\deg \left[ \left( h_1(x)-h_2(x)\right) g(x)\right]=0$ and since $\deg g(x)>0$, we conclude that $h_2(x)=h_1(x)$ and this shows that $\alpha = \alpha^{-1}$, i.e. $\alpha^2=1$.

With the next result one can write directly the generator polynomial of the dual code of any skew constacyclic code.

\[asterisco\] Let $\C$ be a skew $\alpha$-cyclic code generated by the polynomial $g(x)=g_0+g_1x+\cdots+g_{n-k}x^{n-k}$. Then the dual code $\mathscr{C}^\perp$ is generated 