Minimum weight codeword in a MDS code, with $0$ in the $i$th position.

Question

Suppose there is a linear $q$-ary MDS code $C$ of length $n$ and dimension $k$ , where $1<k < n$ , show that there is at least one minimum weight codeword of $C$ with $0$ in the $i$th position.

Answer 1

So the minimum distance of $C$ is equal to $d=n-k+1$. Consider the subspace $C_i$ of words of $C$ that have a zero at the $i$th position. By basic linear algebra $\dim C_i=k-1>0$. Because every word of $C_i$ is also a word of $C$, we can say right away that the minimum distance $d_i$ of $C_i$ is at least $d$. In other words, $d_i\ge d$.

On the other hand we can view $C_i$ as a linear code $C'$ of length $n'=n-1$, dimension $k'$ and minimum distance $d'$ by throwing out that $i$th component from each word. That component is always zero, so throwing it out won't change the minimum distance. In other words, $k'=\dim C_i=k-1$ and $d'=d_i\ge d$.

But the Singleton bound applied to $C'$ says that $$ d'\le n'-k'+1=n-k+1=d. $$ Therefore $d'=d$, and the claim follows.