V \rightarrow \mathbb{F}_{q^n}$ be a non-degenerate reflexive sesquilinear form over $V=V(r,q^n)$ and define $\sigma' \colon V \times V \rightarrow \mathbb{F}_q, \, (\mathbf{u},\mathbf{v})\mapsto \mathrm{Tr}_{q^n/q}(\sigma(\mathbf{u},\mathbf{v}))$. Once we regard $V$ as an $rn$-dimensional ${{\mathbb F}}_q$-vector space, $\sigma^\prime$ turns out to be a non-degenerate reflexive sesquilinear form over $V=V(rn,q)$. Let $\perp$ and $\perp'$ be the orthogonal complement maps defined by $\sigma$ and $\sigma'$ on the lattices of the ${{\mathbb F}}_{q^n}$-subspaces and the ${{\mathbb F}}_q$-subspaces of $V$, respectively. The following properties hold (see [@Polverino Section 2] for the details).

-   $\dim_{{{\mathbb F}}_{q^n}}(W)+\dim_{{{\mathbb F}}_{q^n}}(W^\perp)=r$, for every ${{\mathbb F}}_{q^n}$-subspace $W$ of $V$.

-   $\dim_{{{\mathbb F}}_{q}}(U)+\dim_{{{\mathbb F}}_{q}}(U^{\perp'})=nr$, for every ${{\mathbb F}}_{q}$-subspace $U$ of $V$.

-   $W^\perp=W^{\perp'}$, for every ${{\mathbb F}}_{q^n}$-subspace $W$ of $V$.

-   Let $W$ and $U$ be an ${{\mathbb F}}_{q^n}$-subspace and an ${{\mathbb F}}_q$-subspace of $V$ of dimension $s$ and $t$, repsectively. Then $$\label{eq:dualweight} \dim_{{{\mathbb F}}_q}(U^{\perp'}\cap W^{\perp'})-\dim_{{{\mathbb F}}_q}(U\cap W)=rn-\dim_{{{\mathbb F}}_q}(U)-sn.$$

-   Let $\sigma$, $\sigma_1$ be non-degenerate reflexive sesquilinear forms over $V$ and define $\sigma^\prime$, $\sigma_1^\prime$, $\perp$, $\perp_1$, $\perp'$ and $\perp_1'$ as above. Then there exists an invertible ${{\mathbb F}}_{q^n}$-linear map $f$ such that $f(U^{\perp'})=U^{\perp_1'}$, i.e. $U^{\perp'}$ and $U^{\perp_1'}$ are $\mathrm{GL}(V)$-equivalent.

When $U$ is an ${{\mathbb F}}_q$-subspace of $V$, we denote by $U^{\perp_O}$ one of the ${{\mathbb 