I got stuck at understanding the solution of an exercise from my coding theory course;
Question
Show that the minimum weight of a linear code C is equal to the minimum distance of C.
Solution
Let w be the minimum weight of C and d be the minimum distance
of C. If u $\in$ C, then the weight of u is equal to d(u; 0) by definition, hence w $\geq$ d.
Conversely, let u; v $\in$ C. Then d(u; v) = d(u – u; v – u). Hence d(u; v) = d(0; v – u);
which is the weight of the codeword v – u. This implies d $\leq$ w, and so we conclude d = w.
In the last sentence, How can we deduce that d $\leq$ w??
Best Answer
First of all, there's a typo in the question. You already have $w \ge d$. For the converse, you need $w \le d$, not $d \le w$.
And, this ($w \le d$) can be seen as follows. Suppose $d(u, v) = d$, i.e. $u$ and $v$ have the minimum distance. But $d(u, v) = d(0, v-u)$, which is the weight of the $v-u$ codeword, which cannot be less than $w$. Hence $w \le d$.