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My question is how do I find the standard generator matrix of a binary [7,6,2] code? From what I understand a generator matrix for $C$ is any $ k \times n$ matrix $ G$ with entries in $ \mathbb{F}_q$ such that the rows of $ G$ form a basis for $ C$. I understand that but where does the 2 (d) come into play? Can someone please explain this to me? Thanks!

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A "standard" generator matrix of a $[n,k]$ code usually means that we seek a $k\times n$ matrix $G$ of the form $$G = \left[I_{k\times k}\ P_{k\times (n-k)}\right]$$ where $I_{k\times k}$ denotes a $k\times k$ identity matrix. For a $[7,6,2]$ code, $P_{6\times 1}$ is just a column of $6$ bits, and I leave it to you to figure what the six bits must be in order for the code to have minimum distance $2$. Do it; write out an arbitrary 6-bit column on the right of a $6\times 6$ identity matrix. Then, step back and admire the $6\times 7$ matrix that you have written down, and ponder the fact that each row of $G$ is a codeword.

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  • $\begingroup$ \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 1 \\ \end{pmatrix} $\endgroup$ Commented Sep 18, 2014 at 1:33
  • $\begingroup$ Is that really my answer? $\endgroup$ Commented Sep 18, 2014 at 1:33
  • $\begingroup$ Yes!!! As long as you understand why the rightmost column has to be all $1$s. $\endgroup$ Commented Sep 18, 2014 at 1:50
  • $\begingroup$ Because the minimum weight has to be 2!!! Thanks so much! $\endgroup$ Commented Sep 18, 2014 at 2:03

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