Let $q$ be a power of $2$. Assume that elements of the finite field $\operatorname{GF}(q)$ are denoted by $\beta_i$ for $0\leq i \leq q-1$. We divide elements of $\operatorname{GF}(q)$ in two parts as follows; $x_i=\beta_i$, $0\leq i \leq k-1$ and $y_j=\beta_{k+j}$, for $0\leq j \leq q-k-1$.
From elements $x_i$ and $y_j$ we construct an $k \times (q-k)$ Cauchy matrix $A=\frac{1}{x_i+y_j}$ for $0\leq i \leq k-1$ and $0\leq j \leq q-k-1$. We can verify that all square sub-matrices of $A$ are non-singular Cauchy matrices over $\operatorname{GF}(q)$, since $x_i$'s and $y_j$'s are distinct elements and hence $x_i+y_j\neq 0$ for all $i$ and $j$ (proofwiki)
The matrix $A$ in coding theory is called a super-regular matrix and also in the cryptography is known as an $\operatorname{MDS}$ matrix. Consider the generator matrix $G=(I_k\mid A)$ where $I_k$ is the identity matrix of order $k$. It can be checked that $G$ generates an $\operatorname{MDS}$ code, denoted with $C$, with parameters $(n,k,d)$ page 321, Theorem 8 where $n=q$ and $d=n-k+1$. I dont know if the $\operatorname{MDS}$ code $C$ have a special name.
My question: How to extend the length of code $C$ to obtain a code $\widehat{C}$ such that the extended code $\widehat{C}$ be an $\operatorname{MDS}$ code.
My try: I think, this question is equivalent to ask how to add some columns to $A$ such that $A$ remain a super-regular matrix. The matrix $A$ consist of all elements of $\operatorname{GF}(q)$ and i dont know which column can be added to $A$ such that all square sub-matrices of the new matrix be non-singular.
I asked this question in math.stackexchange but I still haven't received an answer.
Thanks for any suggestions.
