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](https://proofwiki.org/wiki/Value_of_Cauchy_Determinant))


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](https://rads.stackoverflow.com/amzn/click/0444851933) 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](https://math.stackexchange.com/questions/2702532/an-extension-of-an-mds-code) but I still haven't received an answer. 


Thanks for any suggestions. 



An Extension of an $\operatorname{MDS}$ Code over $\operatorname{GF}(2^q)$
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](https://www.unb.ca/saintjohn/sase/dept/math/_resources/alderson/mds-codes.pdf)
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."


