I'm trying to proof a fact about MDS-codes:
Linear $[n,k,d]_q$-code $K$ is MDS-code if and only if it has generating matrix of the form
$$ G_{k\times n}=
\begin{pmatrix}
a_1&a_2&a_3&\dots& a_d&0&0&0&\dots&0\\
0&b_1&b_2&\dots& b_{d-1}&b_d&0&0&\dots&0\\
0&0&c_1&\dots& c_{d-2}&c_{d-1}&c_d&0&\dots&0\\
\dots&\dots&\dots&\dots&\dots&\dots&\dots&\dots&\dots&\dots\\
0&0&0&\dots& 0&0&w_1&w_2&\dots&w_d\\
\end{pmatrix}_{k\times n},
a_i,b_i,c_i,w_i \not=0.
$$
I think it's clearly that if linear $[n,k,d]_q$-code $K$ has generating matrix $G_{k\times n}$ then it's MDS. But I have no idea how to proof the converse statement.