The code words of C are in the form :
$a_1a_2a_3a_4a_5a_6a_7$
Where $a_1a_2a_3a_4a_5a_6$ are any numbers in the arithmetic field: $\mathbb{Z}_{11}$
And:
$a_7=-\left( 7a_{1}+6a_{2}+5a_{3}+4a_{4}+3a_{5}+2a_{6}\right)$
Show that the code has minimum distance $d=2$
How would we do this ?
On a side note, I reading through the solution apparently $a_{7}=0$. Was just wondering how this was so ? Is it because $a_{7}$ it is not apart of $\mathbb{Z}_{11}$
On
The parity check matrix of the code is $$ H = \begin{bmatrix} 7 & 6 & 5 & 4 & 3 & 2 & 1\\ \end{bmatrix}. $$ Since all the columns are different from zero, but any pair is linearly dependent, the minimum distance is $2$.
In fact, quoting from the Wikipedia article,
the minimum distance of a code is the minimum number $d$ such that every $d - 1$ columns of a parity-check matrix $H$ are linearly independent while there exist $d$ columns of $H$ that are linearly dependent.
The parity equation is $$ 7a_{1}+6a_{2}+5a_{3}+4a_{4}+3a_{5}+2a_{6} + a_7 = \sum_{i=1}^7 (8-i)a_i= 0 \pmod{11} \tag{1}$$
Question: can there be two tuples ${\bf b}$, ${\bf c}$ that verify $(1)$ and which differ only in one position? Answer: no. If this were true, suppose in position $i\in \{1,2 \cdots 7\}$ ($b_i\ne c_i$); then, susbtracting both parity equations we'd obtain
$$(8-i)b_i = (8-i)c_i \pmod{11} \tag{2}$$ But, because $11$ is prime, we can cancel the factor, then we'd get $b_i=c_i$ - contradiction.
Then two different tuples ${\bf b}$, ${\bf c}$ must differ in at least 2 positions, hence $d_{\min} \ge 2$. That this bound is attained is easy to see. Take for example two tuples that coincide in the first 5 positions and differ in $a_6$ (they are part of the code because it includes all six-length tuples). Hence $d_{\min} = 2$.