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Reversible code if reciprocal value of root is a root

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1

$begingroup$

I know that a code C is called reversible if $(a_0, ..., a_n-1) in C$ implies that $(a_n-1, ..., a_1, a_0) in C$.

Now, how can I show that a cyclic code C = g is reversible iff with each root of g also the reciprocal value of that root is a root of g?

linear-algebra abstract-algebra

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asked Mar 22 at 10:39

JohnDJohnD

341112

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1

$begingroup$

I know that a code C is called reversible if $(a_0, ..., a_n-1) in C$ implies that $(a_n-1, ..., a_1, a_0) in C$.

Now, how can I show that a cyclic code C = g is reversible iff with each root of g also the reciprocal value of that root is a root of g?

linear-algebra abstract-algebra

share|cite|improve this question

asked Mar 22 at 10:39

JohnDJohnD

341112

$endgroup$

add a comment | 

$begingroup$

I know that a code C is called reversible if $(a_0, ..., a_n-1) in C$ implies that $(a_n-1, ..., a_1, a_0) in C$.

Now, how can I show that a cyclic code C = g is reversible iff with each root of g also the reciprocal value of that root is a root of g?

linear-algebra abstract-algebra

share|cite|improve this question

asked Mar 22 at 10:39

JohnDJohnD

341112

$endgroup$

share|cite|improve this question

asked Mar 22 at 10:39

JohnDJohnD

341112

share|cite|improve this question

asked Mar 22 at 10:39

JohnDJohnD

341112

asked Mar 22 at 10:39

JohnDJohnD

341112

asked Mar 22 at 10:39

JohnDJohnD

341112

1 Answer
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This is true only if you assume the multiplicities of the roots also match (i.e., modify the statement to: $a$ is a root of $g(x)$ with multiplicity $m$ if and only if $dfrac 1 a$ is a root of $g(x)$ with multiplicity $m$).

For $c = (a_0, ldots, a_n - 1) in C$, the associated polynomial is $p_c(x) = a_0 + a_1 x + cdots a_n - 1 x^n - 1$.

For the reverse word $d = (a_n - 1, ldots, a_0)$, the associated polynomial is $$p_d(x) = a_n - 1 + a_n - 2 x + cdots + a_0 x^n - 1 = x^n - 1left[ a_n-1left(dfrac 1 xright)^n-1 + a_n - 1 left(dfrac 1 xright)^n - 2 + cdots + a_0right] = x^n - 1p_cleft( dfrac 1 x right)$$

from which it is clear that the roots of $p_d(x)$ are exactly the reciprocals of the roots of the (non-zero) roots of $p_c(x)$.

Now if $C = langle g rangle$ is reversible, it is obvious from the above argument that $g(r) = 0 iff gleft(dfrac 1 rright) = 0$. Conversely if $g(r) = 0 iff gleft(dfrac 1 rright) = 0$, and $r$ and $dfrac 1 r$ have the same multiplicities, then again it is clear that $x^n - 1gleft(dfrac 1 xright) in C$. Thus, $C$ is reversible.

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answered Mar 22 at 11:55

M. VinayM. Vinay

7,33322136

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0

$begingroup$

This is true only if you assume the multiplicities of the roots also match (i.e., modify the statement to: $a$ is a root of $g(x)$ with multiplicity $m$ if and only if $dfrac 1 a$ is a root of $g(x)$ with multiplicity $m$).

For $c = (a_0, ldots, a_n - 1) in C$, the associated polynomial is $p_c(x) = a_0 + a_1 x + cdots a_n - 1 x^n - 1$.

For the reverse word $d = (a_n - 1, ldots, a_0)$, the associated polynomial is $$p_d(x) = a_n - 1 + a_n - 2 x + cdots + a_0 x^n - 1 = x^n - 1left[ a_n-1left(dfrac 1 xright)^n-1 + a_n - 1 left(dfrac 1 xright)^n - 2 + cdots + a_0right] = x^n - 1p_cleft( dfrac 1 x right)$$

from which it is clear that the roots of $p_d(x)$ are exactly the reciprocals of the roots of the (non-zero) roots of $p_c(x)$.

Now if $C = langle g rangle$ is reversible, it is obvious from the above argument that $g(r) = 0 iff gleft(dfrac 1 rright) = 0$. Conversely if $g(r) = 0 iff gleft(dfrac 1 rright) = 0$, and $r$ and $dfrac 1 r$ have the same multiplicities, then again it is clear that $x^n - 1gleft(dfrac 1 xright) in C$. Thus, $C$ is reversible.

share|cite|improve this answer

answered Mar 22 at 11:55

M. VinayM. Vinay

7,33322136

$endgroup$

add a comment | 

0

$begingroup$

This is true only if you assume the multiplicities of the roots also match (i.e., modify the statement to: $a$ is a root of $g(x)$ with multiplicity $m$ if and only if $dfrac 1 a$ is a root of $g(x)$ with multiplicity $m$).

For $c = (a_0, ldots, a_n - 1) in C$, the associated polynomial is $p_c(x) = a_0 + a_1 x + cdots a_n - 1 x^n - 1$.

For the reverse word $d = (a_n - 1, ldots, a_0)$, the associated polynomial is $$p_d(x) = a_n - 1 + a_n - 2 x + cdots + a_0 x^n - 1 = x^n - 1left[ a_n-1left(dfrac 1 xright)^n-1 + a_n - 1 left(dfrac 1 xright)^n - 2 + cdots + a_0right] = x^n - 1p_cleft( dfrac 1 x right)$$

from which it is clear that the roots of $p_d(x)$ are exactly the reciprocals of the roots of the (non-zero) roots of $p_c(x)$.

Now if $C = langle g rangle$ is reversible, it is obvious from the above argument that $g(r) = 0 iff gleft(dfrac 1 rright) = 0$. Conversely if $g(r) = 0 iff gleft(dfrac 1 rright) = 0$, and $r$ and $dfrac 1 r$ have the same multiplicities, then again it is clear that $x^n - 1gleft(dfrac 1 xright) in C$. Thus, $C$ is reversible.

share|cite|improve this answer

answered Mar 22 at 11:55

M. VinayM. Vinay

7,33322136

$endgroup$

add a comment | 

$begingroup$

This is true only if you assume the multiplicities of the roots also match (i.e., modify the statement to: $a$ is a root of $g(x)$ with multiplicity $m$ if and only if $dfrac 1 a$ is a root of $g(x)$ with multiplicity $m$).

For $c = (a_0, ldots, a_n - 1) in C$, the associated polynomial is $p_c(x) = a_0 + a_1 x + cdots a_n - 1 x^n - 1$.

For the reverse word $d = (a_n - 1, ldots, a_0)$, the associated polynomial is $$p_d(x) = a_n - 1 + a_n - 2 x + cdots + a_0 x^n - 1 = x^n - 1left[ a_n-1left(dfrac 1 xright)^n-1 + a_n - 1 left(dfrac 1 xright)^n - 2 + cdots + a_0right] = x^n - 1p_cleft( dfrac 1 x right)$$

from which it is clear that the roots of $p_d(x)$ are exactly the reciprocals of the roots of the (non-zero) roots of $p_c(x)$.

Now if $C = langle g rangle$ is reversible, it is obvious from the above argument that $g(r) = 0 iff gleft(dfrac 1 rright) = 0$. Conversely if $g(r) = 0 iff gleft(dfrac 1 rright) = 0$, and $r$ and $dfrac 1 r$ have the same multiplicities, then again it is clear that $x^n - 1gleft(dfrac 1 xright) in C$. Thus, $C$ is reversible.

share|cite|improve this answer

answered Mar 22 at 11:55

M. VinayM. Vinay

7,33322136

$endgroup$

share|cite|improve this answer

answered Mar 22 at 11:55

M. VinayM. Vinay

7,33322136

answered Mar 22 at 11:55

M. VinayM. Vinay

7,33322136

answered Mar 22 at 11:55

M. VinayM. Vinay

7,33322136