TITLE: An Extension of an $\operatorname{MDS}$ Code over $\operatorname{GF}(2^q)$
QUESTION [1 upvotes]: 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.

REPLY [1 votes]: Your notation is nonstandard. Code length is $n,$ dimension is $k,$ minimum distance is $d=n-k+1$ when a code is MDS. The paper by Alderson available here
proves the following:
Theorem 2. A $q-$ary $(q+k−2,k)-$ MDS code can be extended to a (unique) maximum length MDS code if and only if $q$ is even.
The authors state "It is important to point out that we can not possibly improve Theorem 2 to include (q + k − 3)-MDS codes."
{"set_name": "stack_exchange", "score": 1, "question_id": 296596}