Definition $\,$ Given an $[n, m, d]_q$ code , and a set $R$ of $z$ indices ${i_1,...,i_z}$, the shortened code $R$ is the set of words from which are zero at positions $i_1,...,i_z$ and whose zero coordinates are deleted, thus effectively shortening these words by $z$ positions. The shortened code $R$ has length $n−z$, dimension $≥ m − z$ and minimum distance $≥d$.
My question: Why the dimension of new code is $≥ m − z$.
Thanks
Edition: I want to answer to this question by coding theory terminology.
First Case: Consider $G$ is an generator of $C$. Let $z=1$. If the entries of $i$th column of $G$ be zero then the dimension of the new code is $m$ since we can delete the $i$th column of $G$. Consider there are some nonzero elements on $i$th column of $G$. By elementary row operations the elements $i$th column of $G$ can be zero except one entry. For simplicity, consider the first element of $i$th column of $G$ be nonzero.
Every code-word in $C$ can be obtained by this form
$$ (v_1,v_2,\cdots ,v_k)\times G_{k\times n}=(c_1,c_2,\cdots ,c_n) $$ Now the desired code-words are in the form of $(c_1,c_2,\cdots ,c_{i-1},0,c_{i+1}, \cdots ,c_n)$ that just can be obtained by words in the form of $(0,v_1,v_2, \cdots ,v_k)$. Therefore, the desired code-words can be obtained by removing the $i$th column and first row of $G$ which means the dimension of the new code is $m-1$.
Second Case: Consider we want to get code-words that are zero in the two positions $i$ and $j$. Let $C_1$ and $C_2$ are sub-codes that consist of code-words that are zero in $i$th and $j$th position which are zero. Then the dimension of the new sub-code are $$ dim(C_1\cap C_2)=dim(C_1)+dim(C_2)-dim(C_1\cup C_2)\geq m-1+m-1-m=m-2 $$ Now by induction on $z$ we can prove other cases.