When the generator of a code satisfies $Gcdot G^T={bf 0}$
Consider $A$ is an $ktimes n-k$ matrix over $mathbb{F}_q$. Consider an $k times n$ matrix $G=(I_kmid A)$ as a generator of a code $C$.
My question: If $Gcdot G^T={bf 0}$ over $mathbb{F}_q$, then is it correct to say $G$ is not only generator of $C$, but also is parity check matrix of $C$?
Is there a special name for the code $C$ in coding theory?
Thanks
coding-theory
asked Dec 2 '18 at 13:15
user0410
1899
add a comment |
Consider $A$ is an $ktimes n-k$ matrix over $mathbb{F}_q$. Consider an $k times n$ matrix $G=(I_kmid A)$ as a generator of a code $C$.
My question: If $Gcdot G^T={bf 0}$ over $mathbb{F}_q$, then is it correct to say $G$ is not only generator of $C$, but also is parity check matrix of $C$?
Is there a special name for the code $C$ in coding theory?
Thanks
coding-theory
asked Dec 2 '18 at 13:15
user0410
1899
add a comment |
Consider $A$ is an $ktimes n-k$ matrix over $mathbb{F}_q$. Consider an $k times n$ matrix $G=(I_kmid A)$ as a generator of a code $C$.
My question: If $Gcdot G^T={bf 0}$ over $mathbb{F}_q$, then is it correct to say $G$ is not only generator of $C$, but also is parity check matrix of $C$?
Is there a special name for the code $C$ in coding theory?
Thanks
coding-theory
asked Dec 2 '18 at 13:15
user0410
1899
Consider $A$ is an $ktimes n-k$ matrix over $mathbb{F}_q$. Consider an $k times n$ matrix $G=(I_kmid A)$ as a generator of a code $C$.
My question: If $Gcdot G^T={bf 0}$ over $mathbb{F}_q$, then is it correct to say $G$ is not only generator of $C$, but also is parity check matrix of $C$?
Is there a special name for the code $C$ in coding theory?
Thanks
coding-theory
coding-theory
asked Dec 2 '18 at 13:15
user0410
1899
asked Dec 2 '18 at 13:15
user0410
1899
asked Dec 2 '18 at 13:15
user0410
1899
asked Dec 2 '18 at 13:15
user0410
1899
asked Dec 2 '18 at 13:15
user0410
1899
1899
add a comment |
add a comment |
1 Answer
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Each codeword $cin C$ has the form $c = u G$ for some vector $uin {Bbb F}_q^k$.
Since $G G^t=0$, $cG^t = (uG)G^t = u(GG^t)=u0 = 0$ and so a dimension argument shows that $G^t$ is a a parity check matrix of $G$. If so, the code is called self-dual.
answered Dec 2 '18 at 14:36
Wuestenfux
3,5791411
1
Yes, but: to be a true parity check matrix for (and for the code to self-dual) we also require that $H$ has $n-k$ rows, hence we need $k=n/2$.
– leonbloy
Dec 2 '18 at 14:49
@leonbloy You right. In self-dual codes the length of code shroud be even.
– user0410
Dec 2 '18 at 16:18
add a comment |
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1 Answer
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1 Answer
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Each codeword $cin C$ has the form $c = u G$ for some vector $uin {Bbb F}_q^k$.
Since $G G^t=0$, $cG^t = (uG)G^t = u(GG^t)=u0 = 0$ and so a dimension argument shows that $G^t$ is a a parity check matrix of $G$. If so, the code is called self-dual.
answered Dec 2 '18 at 14:36
Wuestenfux
3,5791411
1
Yes, but: to be a true parity check matrix for (and for the code to self-dual) we also require that $H$ has $n-k$ rows, hence we need $k=n/2$.
– leonbloy
Dec 2 '18 at 14:49
@leonbloy You right. In self-dual codes the length of code shroud be even.
– user0410
Dec 2 '18 at 16:18
add a comment |
Each codeword $cin C$ has the form $c = u G$ for some vector $uin {Bbb F}_q^k$.
Since $G G^t=0$, $cG^t = (uG)G^t = u(GG^t)=u0 = 0$ and so a dimension argument shows that $G^t$ is a a parity check matrix of $G$. If so, the code is called self-dual.
answered Dec 2 '18 at 14:36
Wuestenfux
3,5791411
1
Yes, but: to be a true parity check matrix for (and for the code to self-dual) we also require that $H$ has $n-k$ rows, hence we need $k=n/2$.
– leonbloy
Dec 2 '18 at 14:49
@leonbloy You right. In self-dual codes the length of code shroud be even.
– user0410
Dec 2 '18 at 16:18
add a comment |
Each codeword $cin C$ has the form $c = u G$ for some vector $uin {Bbb F}_q^k$.
Since $G G^t=0$, $cG^t = (uG)G^t = u(GG^t)=u0 = 0$ and so a dimension argument shows that $G^t$ is a a parity check matrix of $G$. If so, the code is called self-dual.
answered Dec 2 '18 at 14:36
Wuestenfux
3,5791411
Each codeword $cin C$ has the form $c = u G$ for some vector $uin {Bbb F}_q^k$.
Since $G G^t=0$, $cG^t = (uG)G^t = u(GG^t)=u0 = 0$ and so a dimension argument shows that $G^t$ is a a parity check matrix of $G$. If so, the code is called self-dual.
answered Dec 2 '18 at 14:36
Wuestenfux
3,5791411
answered Dec 2 '18 at 14:36
Wuestenfux
3,5791411
answered Dec 2 '18 at 14:36
Wuestenfux
3,5791411
answered Dec 2 '18 at 14:36
Wuestenfux
3,5791411
3,5791411
1
Yes, but: to be a true parity check matrix for (and for the code to self-dual) we also require that $H$ has $n-k$ rows, hence we need $k=n/2$.
– leonbloy
Dec 2 '18 at 14:49
@leonbloy You right. In self-dual codes the length of code shroud be even.
– user0410
Dec 2 '18 at 16:18
add a comment |
1
Yes, but: to be a true parity check matrix for (and for the code to self-dual) we also require that $H$ has $n-k$ rows, hence we need $k=n/2$.
– leonbloy
Dec 2 '18 at 14:49
@leonbloy You right. In self-dual codes the length of code shroud be even.
– user0410
Dec 2 '18 at 16:18
1
1
Yes, but: to be a true parity check matrix for (and for the code to self-dual) we also require that $H$ has $n-k$ rows, hence we need $k=n/2$.
– leonbloy
Dec 2 '18 at 14:49
Yes, but: to be a true parity check matrix for (and for the code to self-dual) we also require that $H$ has $n-k$ rows, hence we need $k=n/2$.
– leonbloy
Dec 2 '18 at 14:49
@leonbloy You right. In self-dual codes the length of code shroud be even.
– user0410
Dec 2 '18 at 16:18
@leonbloy You right. In self-dual codes the length of code shroud be even.
– user0410
Dec 2 '18 at 16:18
add a comment |
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