Coding Theory- minimum weight of dual codes
Let C be a linear code in (F^n)q
with minimum weight w(C) = 1.
Prove that the dual code C^⊥ has no codewords of weight n.
I think that I might have to do a proof by contradiction but i'm not sure where to start, apart from beginning to prove that c^⊥ has a codeword of weight n.
coding-theory
asked Dec 3 '18 at 0:53
Math_Nerd
12
add a comment |
Let C be a linear code in (F^n)q
with minimum weight w(C) = 1.
Prove that the dual code C^⊥ has no codewords of weight n.
I think that I might have to do a proof by contradiction but i'm not sure where to start, apart from beginning to prove that c^⊥ has a codeword of weight n.
coding-theory
asked Dec 3 '18 at 0:53
Math_Nerd
12
You can typeset your formulas using MathJax.
– Morgan Rodgers
Dec 3 '18 at 6:33
1
Also, you would start your proof by contradiction by assuming $C$ has $w(C) = 1$ (and so has a codeword $mathbf{c}$ with weight 1), and that $C^{perp}$ has a codeword $mathbf{c^{prime}}$ of weight $n$.
– Morgan Rodgers
Dec 3 '18 at 6:34
add a comment |
Let C be a linear code in (F^n)q
with minimum weight w(C) = 1.
Prove that the dual code C^⊥ has no codewords of weight n.
I think that I might have to do a proof by contradiction but i'm not sure where to start, apart from beginning to prove that c^⊥ has a codeword of weight n.
coding-theory
asked Dec 3 '18 at 0:53
Math_Nerd
12
Let C be a linear code in (F^n)q
with minimum weight w(C) = 1.
Prove that the dual code C^⊥ has no codewords of weight n.
I think that I might have to do a proof by contradiction but i'm not sure where to start, apart from beginning to prove that c^⊥ has a codeword of weight n.
coding-theory
coding-theory
asked Dec 3 '18 at 0:53
Math_Nerd
12
asked Dec 3 '18 at 0:53
Math_Nerd
12
asked Dec 3 '18 at 0:53
Math_Nerd
12
asked Dec 3 '18 at 0:53
Math_Nerd
12
asked Dec 3 '18 at 0:53
Math_Nerd
12
12
You can typeset your formulas using MathJax.
– Morgan Rodgers
Dec 3 '18 at 6:33
1
Also, you would start your proof by contradiction by assuming $C$ has $w(C) = 1$ (and so has a codeword $mathbf{c}$ with weight 1), and that $C^{perp}$ has a codeword $mathbf{c^{prime}}$ of weight $n$.
– Morgan Rodgers
Dec 3 '18 at 6:34
add a comment |
You can typeset your formulas using MathJax.
– Morgan Rodgers
Dec 3 '18 at 6:33
1
Also, you would start your proof by contradiction by assuming $C$ has $w(C) = 1$ (and so has a codeword $mathbf{c}$ with weight 1), and that $C^{perp}$ has a codeword $mathbf{c^{prime}}$ of weight $n$.
– Morgan Rodgers
Dec 3 '18 at 6:34
You can typeset your formulas using MathJax.
– Morgan Rodgers
Dec 3 '18 at 6:33
You can typeset your formulas using MathJax.
– Morgan Rodgers
Dec 3 '18 at 6:33
1
1
Also, you would start your proof by contradiction by assuming $C$ has $w(C) = 1$ (and so has a codeword $mathbf{c}$ with weight 1), and that $C^{perp}$ has a codeword $mathbf{c^{prime}}$ of weight $n$.
– Morgan Rodgers
Dec 3 '18 at 6:34
Also, you would start your proof by contradiction by assuming $C$ has $w(C) = 1$ (and so has a codeword $mathbf{c}$ with weight 1), and that $C^{perp}$ has a codeword $mathbf{c^{prime}}$ of weight $n$.
– Morgan Rodgers
Dec 3 '18 at 6:34
add a comment |
1 Answer
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Take a codeword $cin C$ with minimum weight 1, say $c=alpha e_i$, where $alphane0$ is a scalar and $e_i$ is the $i$th unit vector.
For each word $w=sum_j alpha_je_j$ with full weight $n$, the scalar product will give $langle c,wrangle = alphaalpha_ine 0$. So $w$ cannot lie in the dual code.
answered Dec 5 '18 at 9:43
Wuestenfux
3,6361411
add a comment |
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1 Answer
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Take a codeword $cin C$ with minimum weight 1, say $c=alpha e_i$, where $alphane0$ is a scalar and $e_i$ is the $i$th unit vector.
For each word $w=sum_j alpha_je_j$ with full weight $n$, the scalar product will give $langle c,wrangle = alphaalpha_ine 0$. So $w$ cannot lie in the dual code.
answered Dec 5 '18 at 9:43
Wuestenfux
3,6361411
add a comment |
Take a codeword $cin C$ with minimum weight 1, say $c=alpha e_i$, where $alphane0$ is a scalar and $e_i$ is the $i$th unit vector.
For each word $w=sum_j alpha_je_j$ with full weight $n$, the scalar product will give $langle c,wrangle = alphaalpha_ine 0$. So $w$ cannot lie in the dual code.
answered Dec 5 '18 at 9:43
Wuestenfux
3,6361411
add a comment |
Take a codeword $cin C$ with minimum weight 1, say $c=alpha e_i$, where $alphane0$ is a scalar and $e_i$ is the $i$th unit vector.
For each word $w=sum_j alpha_je_j$ with full weight $n$, the scalar product will give $langle c,wrangle = alphaalpha_ine 0$. So $w$ cannot lie in the dual code.
answered Dec 5 '18 at 9:43
Wuestenfux
3,6361411
Take a codeword $cin C$ with minimum weight 1, say $c=alpha e_i$, where $alphane0$ is a scalar and $e_i$ is the $i$th unit vector.
For each word $w=sum_j alpha_je_j$ with full weight $n$, the scalar product will give $langle c,wrangle = alphaalpha_ine 0$. So $w$ cannot lie in the dual code.
answered Dec 5 '18 at 9:43
Wuestenfux
3,6361411
answered Dec 5 '18 at 9:43
Wuestenfux
3,6361411
answered Dec 5 '18 at 9:43
Wuestenfux
3,6361411
answered Dec 5 '18 at 9:43
Wuestenfux
3,6361411
3,6361411
add a comment |
add a comment |
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You can typeset your formulas using MathJax.
– Morgan Rodgers
Dec 3 '18 at 6:33
1
Also, you would start your proof by contradiction by assuming $C$ has $w(C) = 1$ (and so has a codeword $mathbf{c}$ with weight 1), and that $C^{perp}$ has a codeword $mathbf{c^{prime}}$ of weight $n$.
– Morgan Rodgers
Dec 3 '18 at 6:34