The parity check matrix of a generator matrix $G$ is given as $$H=\left[\begin{array}{ccccccc} 0 & 1 & 1 & 1 & 1 & 0 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1 & 0\\ 1 & 1 & 0 & 1 & 0 & 0 & 1 \end{array} \right]$$
I am trying to find $d=d(C)$.Where $C$ is a linear $[n,K] $ code.
The solutions say in $H$ every two columns are different but some $3$ are dependant for example $C_1+ C_6+C_7=0$.
How does $C_1+ C_6+C_7=0$?
In general how do you find $d(C)$?
It holds that $$d(C)=\min \{ d \in \mathbb{N} | \text{ there are } d \text{ linearly dependent columns of H} \}$$
So firstly you check if you have a column only with $0$s. If not then all the $\{ C_i \}$ are linearly independent. Then $d \neq 1$.
After that you check if there is an $\alpha \in \mathbb{R}$ such that $C_i= \alpha C_j$ for some $i,j$ such that $i \neq j$ and $1 \leq i,j \leq 7$. If not then $d \neq 2$.
Then you check if there are $\alpha, \beta \neq 0$ such that $\alpha C_i+ \beta C_j=C_k$ for any $1 \leq k \leq 7, i, j \neq k, 1 \leq i,j \leq 7$.
If so, then $d=3$. Otherwise you continue in the same way by checking if there are non-zero $\alpha, \beta, \gamma$ such that $C_k=\alpha C_1+ \beta C_2+ \gamma C_3$. If so, then $d$ will be equal to $4$.
Otherwise, you continue checking the same thing for four columns.