0
$\begingroup$

I have a parity check matrix for a binary linear code V below:

$$ H = \begin{bmatrix} 1 & 0 & 1 & 1 & 1 & 1 \\ 1 & 1 & 0 & 1 & 1 & 1 \\ 1 & 0 & 0 & 1 & 0 & 1 \end{bmatrix} $$

I want to find a generator matrix for V. Is anything different than just converting the parity matrix to the generator since I'm trying to find one for 'V'? As in, put $H$ in standard form to get $[I\mid A]$, then $G = [-A^T \mid I]$?

Steps I did: R3 = R1 + R3 $$ H = \begin{bmatrix} 1 & 0 & 1 & 1 & 1 & 1 \\ 1 & 1 & 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 0 & 1 & 0 \end{bmatrix} $$ R1 = R1 + R3 $$ H = \begin{bmatrix} 1 & 0 & 0 & 1 & 0 & 1 \\ 1 & 1 & 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 0 & 1 & 0 \end{bmatrix} $$

R2 = R2 + R1 $$ H = \begin{bmatrix} 1 & 0 & 0 & 1 & 0 & 1 \\ 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 1 & 0 \end{bmatrix} $$

So the parity matrix is now in the form [I3 | A ] where A = $$ A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 1 & 0 \end{bmatrix} $$ $$G = [-A^T \mid I] = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 & 1 & 0 \\ 1 & 1 & 0 & 0 & 0 & 1 \end{bmatrix} $$

Is this correct?

Edit: $$G = [-A^T \mid I] = \begin{bmatrix} 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 & 0 & 1 \end{bmatrix} $$

$\endgroup$
  • 1
    $\begingroup$ You can post mathematical notation using MathJax and $\LaTeX$. I'll format the matrix $H$ for you (unless someone beats me to it!) to give you the idea. $\endgroup$ – hardmath Oct 21 '15 at 12:32
  • $\begingroup$ Oh thank you! Much easier to read. $\endgroup$ – pfinferno Oct 21 '15 at 13:19
  • 1
    $\begingroup$ The final matrix $G$ is wrong. $-A^T$ is wrong and you don't have the identity in the right block. $\endgroup$ – Sfarla Oct 21 '15 at 13:54
  • $\begingroup$ Check the last edit, is that correct? I was following an example I found which seemed to have $-A^T$ wrong $\endgroup$ – pfinferno Oct 21 '15 at 14:11
1
$\begingroup$

I think the last edit is correct. But for the row operation part I would do

R1 = R1 + R3, R2 = R2 + R3

and then swap R1 and R3 to get an [I3|A]

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

The formulas being used are incorrect.

Generator Matrix

G = [I | A]

Parity Matrix H = [ -A Transpose | I]

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.