$\mathcal{C}$ is equivalent to an ${{\mathbb F}_{q}}$-linear MRD code $\mathcal{C}^\prime$ such that $R(\mathcal{C}^\prime)=\mathcal{F}_n$.

-   The set $$U=\{f\in\mathcal{C}^\prime\colon f(1)=0\}\subseteq\mathcal{C}^\prime$$ is a $\frac{\iota rn}{\iota+1}$-dimensional $\mathcal{F}_{n,q}$-subspace of $\mathcal{C}^\prime$, and satisfies [^2] $$\label{eq:iota} \max\left\{\dim_{\mathcal{F}_{n,q}}\left(U\cap\langle f\rangle_{\mathcal{F}_n}\right)\colon f\in\mathcal{C}^\prime\right\}=\iota.$$

-   $\mathcal{C}^\prime$ is equal to $\mathcal{C}_{U,G}$, where $G:\mathcal{C}^\prime\to W$, $f\mapsto f(1)$.

Since $|\mathcal{C}|=q^{rn}$ and $t\geq n$, the Singleton-like bound of Theorem \[th:Singleton\] reads $rn\leq t(n-(n-\iota)+1)$. As $\mathcal{C}$ is MRD, this implies that $\iota+1$ divides $t$, and $t=\frac{rn}{\iota+1}$.

Since $R(\mathcal{C})\setminus\{\mathbf{0}\}$ and $\mathcal{F}_n\setminus\{\omega_0\}$ are Singer cycles of $\mathrm{GL}(n,q)$, there exists by Remark \[rem:singer\] an invertible $\mathbb{F}_q$-linear map $H\colon \mathbb{F}_{q^n}\rightarrow\mathbb{F}_{q^n}$ such that $R(\mathcal{C})= H\circ\mathcal{F}_n\circ H^{-1}$. Thus, $\mathcal{C}^\prime = \mathcal{C}\circ H$.

Clearly, $U$ is an $\mathcal{F}_{n,q}$-subspace of $\mathcal{C}^\prime$. For every $i\in\{1,\ldots,\iota\}$, we determine the size of $U_i=\{ f \in U \colon \dim_{{{\mathbb F}}_q}(\ker f)=i \}$. Let $g \in \mathcal{C}'$ be such that $\dim_{{{\mathbb F}}_q}(\ker g)=i$. As $\dim_{{{\mathbb F}}_q}(\ker g)>0$, there exists $\alpha \in {{\mathbb F}}_{q^n}^*$ such that $g(\alpha)=0$, that is $g\circ \omega_\alpha (1)=0$. As $\operatorname{\mathcal{C}}'$ is a right vector space over $\mathcal{F}_n$, it follows that $g\circ \omega_\alpha \in \operatorname{\mathcal{C}}'$ and, in particular, $g\circ \omega_\alpha \in U_i$. This implies that $$\{ f \circ \omega_\alpha \colon f \in U_i,\,\,\alpha \in {{\mathbb F}}_{q^n}^* \}$$ coincides with the set of all the elements in $\operatorname{\mathcal{C}}'$ of 