verifying G*(transpose(H)) = 0 for generator matrix G and parity check matrix H.












0












$begingroup$


I've got a non systematic generator matrix G given by,
G= [1 1 0 1 0 0 0;0 1 1 0 1 0 0;0 0 1 1 0 1 0;0 0 0 1 1 0 1]
I've converted this matrix by the following operations,
(1)exchanging C3 and C6
(2)exchanging C4 and C7
(3) adding R1+R2 and changing R1.
The obtained matrix is given by Gs=[1 0 0 0 1 1 1;0 1 0 0 1 1 0;0 0 1 0 0 1 1;0 0 0 1 1 0 1].The parity check matrix of Gs is Hs= [1 0 0 1 1 0 1;0 1 0 1 1 1 0;0 0 1 1 0 1 1].Now I'm not clear how to verify G*(transpose(H)) = 0. Should we undo the column operation in the H matrix? And the row spaces of which two matrices are orthogonal?It will be of great help if someone can help.Thanks.










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$endgroup$












  • $begingroup$
    Its really hard to read. Better have the matrix form.
    $endgroup$
    – Wuestenfux
    Jan 6 at 14:59
















0












$begingroup$


I've got a non systematic generator matrix G given by,
G= [1 1 0 1 0 0 0;0 1 1 0 1 0 0;0 0 1 1 0 1 0;0 0 0 1 1 0 1]
I've converted this matrix by the following operations,
(1)exchanging C3 and C6
(2)exchanging C4 and C7
(3) adding R1+R2 and changing R1.
The obtained matrix is given by Gs=[1 0 0 0 1 1 1;0 1 0 0 1 1 0;0 0 1 0 0 1 1;0 0 0 1 1 0 1].The parity check matrix of Gs is Hs= [1 0 0 1 1 0 1;0 1 0 1 1 1 0;0 0 1 1 0 1 1].Now I'm not clear how to verify G*(transpose(H)) = 0. Should we undo the column operation in the H matrix? And the row spaces of which two matrices are orthogonal?It will be of great help if someone can help.Thanks.










share|cite|improve this question









$endgroup$












  • $begingroup$
    Its really hard to read. Better have the matrix form.
    $endgroup$
    – Wuestenfux
    Jan 6 at 14:59














0












0








0





$begingroup$


I've got a non systematic generator matrix G given by,
G= [1 1 0 1 0 0 0;0 1 1 0 1 0 0;0 0 1 1 0 1 0;0 0 0 1 1 0 1]
I've converted this matrix by the following operations,
(1)exchanging C3 and C6
(2)exchanging C4 and C7
(3) adding R1+R2 and changing R1.
The obtained matrix is given by Gs=[1 0 0 0 1 1 1;0 1 0 0 1 1 0;0 0 1 0 0 1 1;0 0 0 1 1 0 1].The parity check matrix of Gs is Hs= [1 0 0 1 1 0 1;0 1 0 1 1 1 0;0 0 1 1 0 1 1].Now I'm not clear how to verify G*(transpose(H)) = 0. Should we undo the column operation in the H matrix? And the row spaces of which two matrices are orthogonal?It will be of great help if someone can help.Thanks.










share|cite|improve this question









$endgroup$




I've got a non systematic generator matrix G given by,
G= [1 1 0 1 0 0 0;0 1 1 0 1 0 0;0 0 1 1 0 1 0;0 0 0 1 1 0 1]
I've converted this matrix by the following operations,
(1)exchanging C3 and C6
(2)exchanging C4 and C7
(3) adding R1+R2 and changing R1.
The obtained matrix is given by Gs=[1 0 0 0 1 1 1;0 1 0 0 1 1 0;0 0 1 0 0 1 1;0 0 0 1 1 0 1].The parity check matrix of Gs is Hs= [1 0 0 1 1 0 1;0 1 0 1 1 1 0;0 0 1 1 0 1 1].Now I'm not clear how to verify G*(transpose(H)) = 0. Should we undo the column operation in the H matrix? And the row spaces of which two matrices are orthogonal?It will be of great help if someone can help.Thanks.







coding-theory






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asked Jan 6 at 9:59









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  • $begingroup$
    Its really hard to read. Better have the matrix form.
    $endgroup$
    – Wuestenfux
    Jan 6 at 14:59


















  • $begingroup$
    Its really hard to read. Better have the matrix form.
    $endgroup$
    – Wuestenfux
    Jan 6 at 14:59
















$begingroup$
Its really hard to read. Better have the matrix form.
$endgroup$
– Wuestenfux
Jan 6 at 14:59




$begingroup$
Its really hard to read. Better have the matrix form.
$endgroup$
– Wuestenfux
Jan 6 at 14:59










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