$$rn\leq(rn-k)(n-(n-\iota)+1).$$ Therefore, $\mathcal{C}_{U,G}$ is an MRD code if and only if $\iota+1$ divides $rn$ and $k=\frac{\iota rn}{\iota+1}$.

In this case, the parameters of $\mathcal{C}_{U,G}$ are provided by Theorem \[th:construction\] and the weight distribution of $\mathcal{C}_{U,G}$ follows from Theorem \[th:weightdistribution\]. Also, the right idealiser of $\mathcal{C}_{U,G}$ contains $\mathcal{F}_n$ by Theorem \[th:construction\], and hence is equal to $\mathcal{F}_n$ by Theorem \[th:propertiesideal\].

Different choices of the map $G$ yield equivalent codes, i.e. $\mathcal{C}_{U,G}$ is uniquely determined by $U$, up to equivalence.

Let $V=V(r,q^n)$ and $W=V(rn-k,q)$. Let $U=V(k,q)$ be an ${{\mathbb F}_{q}}$-subspace of $V$, $G:V\to W$ and $\overline{G}:V\to W$ be two ${{\mathbb F}_{q}}$-linear maps with $\ker(G)=\ker(\overline{G})=U$. Then the codes $\mathcal{C}_{U,G}$ and $\mathcal{C}_{U,\overline{G}}$ are equivalent.

Let $B_U \cup\{\mathbf{w}_1,\ldots,\mathbf{w}_{rn-k}\}$ be an ${{\mathbb F}_{q}}$-basis of $V$ such that $B_U$ is an ${{\mathbb F}_{q}}$-basis of $U$. Clearly, $G(\mathbf{w}_1),\ldots,G(\mathbf{w}_{rn-k})$ are ${{\mathbb F}_{q}}$-linearly independent, as well as $\overline{G}(\mathbf{w}_1),\ldots,\overline{G}(\mathbf{w}_{rn-k})$. Then there exists an invertible ${{\mathbb F}_{q}}$-linear map $L:V\to V$ such that $L(U)=U$ and $L(G({{\mathbf w}}_i))=\overline{G}({{\mathbf w}}_i)$ for every $i=1,\ldots,rn-k$, i.e. $L\circ G=\overline{G}$. Therefore, by choosing $R={\rm Id}_{{\mathbb{F}_{q^n}}}$ and $\sigma={\rm Id}_{{\rm Aut}({{\mathbb F}_{q}})}$, we have $$L\circ\mathcal{C}_{U,G}^{\sigma}\circ R = \left\{L\circ(G\circ\tau_{\mathbf{v}})\colon \mathbf{v} \in V\right\} =\mathcal{C}_{U,\overline{G}}.$$ The claim is proved.

We recall the following conjugacy property of Singer cycles of ${\rm GL}(n,q)$.

\[rem:singer\] The cyclic 