Permutations and terminology












1












$begingroup$


Say I have the following permutation



$$sigma ={begin{pmatrix}1&2&3&4&5&6&7&8\1&2&3&8&4&5&6&7end{pmatrix}}$$



which consists to let unchanged some first elements from $1$ to $k$ and to apply a circular shift on the elements ${k+1, ldots, n}$.



What is the name for such transformation?



Thank you






permutations permutation-cycles







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    The name is "a $5$-cycle in $S_8$", namely $sigma=(48765)$.
    $endgroup$
    – Dietrich Burde
    Dec 2 '18 at 19:10












  • $begingroup$
    Thank you @DietrichBurde Does this (n-k)-cycle let unchanged the first $k$ elements?
    $endgroup$
    – Adam54
    Dec 2 '18 at 19:19












  • $begingroup$
    Yes, it keeps $1$,$2$, $3$ fixed (unchanged). Have a look at the cycle notation for permutations.
    $endgroup$
    – Dietrich Burde
    Dec 2 '18 at 19:20








  • 1




    $begingroup$
    Have a look at the cycle notation: $(45)$ means that $4$ goes to $5$ and $5$ goes to $4$. Then $(48765)$ means that $4mapsto 8mapsto 7mapsto 6mapsto 5mapsto 4mapstocdots $
    $endgroup$
    – Dietrich Burde
    Dec 2 '18 at 19:23








  • 1




    $begingroup$
    You should read en.wikipedia.org/wiki/Cyclic_permutation and en.wikipedia.org/wiki/Permutation#Cycle_notation
    $endgroup$
    – Jean-Claude Arbaut
    Dec 2 '18 at 19:31
















1












$begingroup$


Say I have the following permutation



$$sigma ={begin{pmatrix}1&2&3&4&5&6&7&8\1&2&3&8&4&5&6&7end{pmatrix}}$$



which consists to let unchanged some first elements from $1$ to $k$ and to apply a circular shift on the elements ${k+1, ldots, n}$.



What is the name for such transformation?



Thank you






permutations permutation-cycles







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    The name is "a $5$-cycle in $S_8$", namely $sigma=(48765)$.
    $endgroup$
    – Dietrich Burde
    Dec 2 '18 at 19:10












  • $begingroup$
    Thank you @DietrichBurde Does this (n-k)-cycle let unchanged the first $k$ elements?
    $endgroup$
    – Adam54
    Dec 2 '18 at 19:19












  • $begingroup$
    Yes, it keeps $1$,$2$, $3$ fixed (unchanged). Have a look at the cycle notation for permutations.
    $endgroup$
    – Dietrich Burde
    Dec 2 '18 at 19:20








  • 1




    $begingroup$
    Have a look at the cycle notation: $(45)$ means that $4$ goes to $5$ and $5$ goes to $4$. Then $(48765)$ means that $4mapsto 8mapsto 7mapsto 6mapsto 5mapsto 4mapstocdots $
    $endgroup$
    – Dietrich Burde
    Dec 2 '18 at 19:23








  • 1




    $begingroup$
    You should read en.wikipedia.org/wiki/Cyclic_permutation and en.wikipedia.org/wiki/Permutation#Cycle_notation
    $endgroup$
    – Jean-Claude Arbaut
    Dec 2 '18 at 19:31














1












1








1





$begingroup$


Say I have the following permutation



$$sigma ={begin{pmatrix}1&2&3&4&5&6&7&8\1&2&3&8&4&5&6&7end{pmatrix}}$$



which consists to let unchanged some first elements from $1$ to $k$ and to apply a circular shift on the elements ${k+1, ldots, n}$.



What is the name for such transformation?



Thank you






permutations permutation-cycles







share|cite|improve this question











$endgroup$




Say I have the following permutation



$$sigma ={begin{pmatrix}1&2&3&4&5&6&7&8\1&2&3&8&4&5&6&7end{pmatrix}}$$



which consists to let unchanged some first elements from $1$ to $k$ and to apply a circular shift on the elements ${k+1, ldots, n}$.



What is the name for such transformation?



Thank you






permutations permutation-cycles




permutations permutation-cycles






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 2 '18 at 19:25









Bernard

119k639112




119k639112










asked Dec 2 '18 at 19:09









Adam54Adam54

435




435








  • 2




    $begingroup$
    The name is "a $5$-cycle in $S_8$", namely $sigma=(48765)$.
    $endgroup$
    – Dietrich Burde
    Dec 2 '18 at 19:10












  • $begingroup$
    Thank you @DietrichBurde Does this (n-k)-cycle let unchanged the first $k$ elements?
    $endgroup$
    – Adam54
    Dec 2 '18 at 19:19












  • $begingroup$
    Yes, it keeps $1$,$2$, $3$ fixed (unchanged). Have a look at the cycle notation for permutations.
    $endgroup$
    – Dietrich Burde
    Dec 2 '18 at 19:20








  • 1




    $begingroup$
    Have a look at the cycle notation: $(45)$ means that $4$ goes to $5$ and $5$ goes to $4$. Then $(48765)$ means that $4mapsto 8mapsto 7mapsto 6mapsto 5mapsto 4mapstocdots $
    $endgroup$
    – Dietrich Burde
    Dec 2 '18 at 19:23








  • 1




    $begingroup$
    You should read en.wikipedia.org/wiki/Cyclic_permutation and en.wikipedia.org/wiki/Permutation#Cycle_notation
    $endgroup$
    – Jean-Claude Arbaut
    Dec 2 '18 at 19:31














  • 2




    $begingroup$
    The name is "a $5$-cycle in $S_8$", namely $sigma=(48765)$.
    $endgroup$
    – Dietrich Burde
    Dec 2 '18 at 19:10












  • $begingroup$
    Thank you @DietrichBurde Does this (n-k)-cycle let unchanged the first $k$ elements?
    $endgroup$
    – Adam54
    Dec 2 '18 at 19:19












  • $begingroup$
    Yes, it keeps $1$,$2$, $3$ fixed (unchanged). Have a look at the cycle notation for permutations.
    $endgroup$
    – Dietrich Burde
    Dec 2 '18 at 19:20








  • 1




    $begingroup$
    Have a look at the cycle notation: $(45)$ means that $4$ goes to $5$ and $5$ goes to $4$. Then $(48765)$ means that $4mapsto 8mapsto 7mapsto 6mapsto 5mapsto 4mapstocdots $
    $endgroup$
    – Dietrich Burde
    Dec 2 '18 at 19:23








  • 1




    $begingroup$
    You should read en.wikipedia.org/wiki/Cyclic_permutation and en.wikipedia.org/wiki/Permutation#Cycle_notation
    $endgroup$
    – Jean-Claude Arbaut
    Dec 2 '18 at 19:31








2




2




$begingroup$
The name is "a $5$-cycle in $S_8$", namely $sigma=(48765)$.
$endgroup$
– Dietrich Burde
Dec 2 '18 at 19:10






$begingroup$
The name is "a $5$-cycle in $S_8$", namely $sigma=(48765)$.
$endgroup$
– Dietrich Burde
Dec 2 '18 at 19:10














$begingroup$
Thank you @DietrichBurde Does this (n-k)-cycle let unchanged the first $k$ elements?
$endgroup$
– Adam54
Dec 2 '18 at 19:19






$begingroup$
Thank you @DietrichBurde Does this (n-k)-cycle let unchanged the first $k$ elements?
$endgroup$
– Adam54
Dec 2 '18 at 19:19














$begingroup$
Yes, it keeps $1$,$2$, $3$ fixed (unchanged). Have a look at the cycle notation for permutations.
$endgroup$
– Dietrich Burde
Dec 2 '18 at 19:20






$begingroup$
Yes, it keeps $1$,$2$, $3$ fixed (unchanged). Have a look at the cycle notation for permutations.
$endgroup$
– Dietrich Burde
Dec 2 '18 at 19:20






1




1




$begingroup$
Have a look at the cycle notation: $(45)$ means that $4$ goes to $5$ and $5$ goes to $4$. Then $(48765)$ means that $4mapsto 8mapsto 7mapsto 6mapsto 5mapsto 4mapstocdots $
$endgroup$
– Dietrich Burde
Dec 2 '18 at 19:23






$begingroup$
Have a look at the cycle notation: $(45)$ means that $4$ goes to $5$ and $5$ goes to $4$. Then $(48765)$ means that $4mapsto 8mapsto 7mapsto 6mapsto 5mapsto 4mapstocdots $
$endgroup$
– Dietrich Burde
Dec 2 '18 at 19:23






1




1




$begingroup$
You should read en.wikipedia.org/wiki/Cyclic_permutation and en.wikipedia.org/wiki/Permutation#Cycle_notation
$endgroup$
– Jean-Claude Arbaut
Dec 2 '18 at 19:31




$begingroup$
You should read en.wikipedia.org/wiki/Cyclic_permutation and en.wikipedia.org/wiki/Permutation#Cycle_notation
$endgroup$
– Jean-Claude Arbaut
Dec 2 '18 at 19:31










1 Answer
1






active

oldest

votes


















0












$begingroup$

$1, 2, 3$ each map to themselves. $1to 1$. Period. $2to 2.$ Period. $3to 3.$ Period. We can represent this as $(1)(2)(3)$, but typically this is not customary.



So we can start with $4$:



$4 to 8$, then



$8 to 7$, then



$7to 6$,



then $6to 5$, and finally,



$5 to 4$, completing the five cycle (bringing us back to the starting number for the cycle, $4$:



That gives us the permutation $sigma = (48765)$. This can also be written as $$sigma = (1)(2)(3)(48765) = (48765)$$




To get the cycle you are looking for, you'll need the following permutation:



$alpha ={begin{pmatrix}1&2&3&4&5&6&7&8\1&2&3&5&6&7&8&4end{pmatrix}} = (45678)$



The name for this transformation is merely "permutation" $alpha$ on the set $S$ of $8$ numbers, specifically, both $alpha, sigma in mathbb S_8$, the group of permutations on $S$.



If we were talking about vertices of an octagon, numbered 1-8, then $(12345678)$ could be described as a rotation of the octagon.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks @amWhy As I have written to Dietrich, I think I am reffering to another kind of cycle.
    $endgroup$
    – Adam54
    Dec 2 '18 at 19:29










  • $begingroup$
    Top number goes to bottom number in each column.
    $endgroup$
    – amWhy
    Dec 2 '18 at 19:37










  • $begingroup$
    Thank you. So writing left shift is straightforward, but for writing a right shift we need to read from top to bottom?
    $endgroup$
    – Adam54
    Dec 2 '18 at 19:42










  • $begingroup$
    Yes, the numbers in the top row represent the 8 possible numbers to permute. Where each of those numbers goes is written immediately below the number, in the second row.
    $endgroup$
    – amWhy
    Dec 2 '18 at 19:46










  • $begingroup$
    Thanks again. So, if I have a sequence of 8 unknown numbers, but I know that when I will have these numbers, I will right shift the 5 last numbers. How can I write this function, terminologically?
    $endgroup$
    – Adam54
    Dec 2 '18 at 19:49











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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

$1, 2, 3$ each map to themselves. $1to 1$. Period. $2to 2.$ Period. $3to 3.$ Period. We can represent this as $(1)(2)(3)$, but typically this is not customary.



So we can start with $4$:



$4 to 8$, then



$8 to 7$, then



$7to 6$,



then $6to 5$, and finally,



$5 to 4$, completing the five cycle (bringing us back to the starting number for the cycle, $4$:



That gives us the permutation $sigma = (48765)$. This can also be written as $$sigma = (1)(2)(3)(48765) = (48765)$$




To get the cycle you are looking for, you'll need the following permutation:



$alpha ={begin{pmatrix}1&2&3&4&5&6&7&8\1&2&3&5&6&7&8&4end{pmatrix}} = (45678)$



The name for this transformation is merely "permutation" $alpha$ on the set $S$ of $8$ numbers, specifically, both $alpha, sigma in mathbb S_8$, the group of permutations on $S$.



If we were talking about vertices of an octagon, numbered 1-8, then $(12345678)$ could be described as a rotation of the octagon.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks @amWhy As I have written to Dietrich, I think I am reffering to another kind of cycle.
    $endgroup$
    – Adam54
    Dec 2 '18 at 19:29










  • $begingroup$
    Top number goes to bottom number in each column.
    $endgroup$
    – amWhy
    Dec 2 '18 at 19:37










  • $begingroup$
    Thank you. So writing left shift is straightforward, but for writing a right shift we need to read from top to bottom?
    $endgroup$
    – Adam54
    Dec 2 '18 at 19:42










  • $begingroup$
    Yes, the numbers in the top row represent the 8 possible numbers to permute. Where each of those numbers goes is written immediately below the number, in the second row.
    $endgroup$
    – amWhy
    Dec 2 '18 at 19:46










  • $begingroup$
    Thanks again. So, if I have a sequence of 8 unknown numbers, but I know that when I will have these numbers, I will right shift the 5 last numbers. How can I write this function, terminologically?
    $endgroup$
    – Adam54
    Dec 2 '18 at 19:49
















0












$begingroup$

$1, 2, 3$ each map to themselves. $1to 1$. Period. $2to 2.$ Period. $3to 3.$ Period. We can represent this as $(1)(2)(3)$, but typically this is not customary.



So we can start with $4$:



$4 to 8$, then



$8 to 7$, then



$7to 6$,



then $6to 5$, and finally,



$5 to 4$, completing the five cycle (bringing us back to the starting number for the cycle, $4$:



That gives us the permutation $sigma = (48765)$. This can also be written as $$sigma = (1)(2)(3)(48765) = (48765)$$




To get the cycle you are looking for, you'll need the following permutation:



$alpha ={begin{pmatrix}1&2&3&4&5&6&7&8\1&2&3&5&6&7&8&4end{pmatrix}} = (45678)$



The name for this transformation is merely "permutation" $alpha$ on the set $S$ of $8$ numbers, specifically, both $alpha, sigma in mathbb S_8$, the group of permutations on $S$.



If we were talking about vertices of an octagon, numbered 1-8, then $(12345678)$ could be described as a rotation of the octagon.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks @amWhy As I have written to Dietrich, I think I am reffering to another kind of cycle.
    $endgroup$
    – Adam54
    Dec 2 '18 at 19:29










  • $begingroup$
    Top number goes to bottom number in each column.
    $endgroup$
    – amWhy
    Dec 2 '18 at 19:37










  • $begingroup$
    Thank you. So writing left shift is straightforward, but for writing a right shift we need to read from top to bottom?
    $endgroup$
    – Adam54
    Dec 2 '18 at 19:42










  • $begingroup$
    Yes, the numbers in the top row represent the 8 possible numbers to permute. Where each of those numbers goes is written immediately below the number, in the second row.
    $endgroup$
    – amWhy
    Dec 2 '18 at 19:46










  • $begingroup$
    Thanks again. So, if I have a sequence of 8 unknown numbers, but I know that when I will have these numbers, I will right shift the 5 last numbers. How can I write this function, terminologically?
    $endgroup$
    – Adam54
    Dec 2 '18 at 19:49














0












0








0





$begingroup$

$1, 2, 3$ each map to themselves. $1to 1$. Period. $2to 2.$ Period. $3to 3.$ Period. We can represent this as $(1)(2)(3)$, but typically this is not customary.



So we can start with $4$:



$4 to 8$, then



$8 to 7$, then



$7to 6$,



then $6to 5$, and finally,



$5 to 4$, completing the five cycle (bringing us back to the starting number for the cycle, $4$:



That gives us the permutation $sigma = (48765)$. This can also be written as $$sigma = (1)(2)(3)(48765) = (48765)$$




To get the cycle you are looking for, you'll need the following permutation:



$alpha ={begin{pmatrix}1&2&3&4&5&6&7&8\1&2&3&5&6&7&8&4end{pmatrix}} = (45678)$



The name for this transformation is merely "permutation" $alpha$ on the set $S$ of $8$ numbers, specifically, both $alpha, sigma in mathbb S_8$, the group of permutations on $S$.



If we were talking about vertices of an octagon, numbered 1-8, then $(12345678)$ could be described as a rotation of the octagon.






share|cite|improve this answer











$endgroup$



$1, 2, 3$ each map to themselves. $1to 1$. Period. $2to 2.$ Period. $3to 3.$ Period. We can represent this as $(1)(2)(3)$, but typically this is not customary.



So we can start with $4$:



$4 to 8$, then



$8 to 7$, then



$7to 6$,



then $6to 5$, and finally,



$5 to 4$, completing the five cycle (bringing us back to the starting number for the cycle, $4$:



That gives us the permutation $sigma = (48765)$. This can also be written as $$sigma = (1)(2)(3)(48765) = (48765)$$




To get the cycle you are looking for, you'll need the following permutation:



$alpha ={begin{pmatrix}1&2&3&4&5&6&7&8\1&2&3&5&6&7&8&4end{pmatrix}} = (45678)$



The name for this transformation is merely "permutation" $alpha$ on the set $S$ of $8$ numbers, specifically, both $alpha, sigma in mathbb S_8$, the group of permutations on $S$.



If we were talking about vertices of an octagon, numbered 1-8, then $(12345678)$ could be described as a rotation of the octagon.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 31 '18 at 21:32

























answered Dec 2 '18 at 19:26









amWhyamWhy

192k28225439




192k28225439












  • $begingroup$
    Thanks @amWhy As I have written to Dietrich, I think I am reffering to another kind of cycle.
    $endgroup$
    – Adam54
    Dec 2 '18 at 19:29










  • $begingroup$
    Top number goes to bottom number in each column.
    $endgroup$
    – amWhy
    Dec 2 '18 at 19:37










  • $begingroup$
    Thank you. So writing left shift is straightforward, but for writing a right shift we need to read from top to bottom?
    $endgroup$
    – Adam54
    Dec 2 '18 at 19:42










  • $begingroup$
    Yes, the numbers in the top row represent the 8 possible numbers to permute. Where each of those numbers goes is written immediately below the number, in the second row.
    $endgroup$
    – amWhy
    Dec 2 '18 at 19:46










  • $begingroup$
    Thanks again. So, if I have a sequence of 8 unknown numbers, but I know that when I will have these numbers, I will right shift the 5 last numbers. How can I write this function, terminologically?
    $endgroup$
    – Adam54
    Dec 2 '18 at 19:49


















  • $begingroup$
    Thanks @amWhy As I have written to Dietrich, I think I am reffering to another kind of cycle.
    $endgroup$
    – Adam54
    Dec 2 '18 at 19:29










  • $begingroup$
    Top number goes to bottom number in each column.
    $endgroup$
    – amWhy
    Dec 2 '18 at 19:37










  • $begingroup$
    Thank you. So writing left shift is straightforward, but for writing a right shift we need to read from top to bottom?
    $endgroup$
    – Adam54
    Dec 2 '18 at 19:42










  • $begingroup$
    Yes, the numbers in the top row represent the 8 possible numbers to permute. Where each of those numbers goes is written immediately below the number, in the second row.
    $endgroup$
    – amWhy
    Dec 2 '18 at 19:46










  • $begingroup$
    Thanks again. So, if I have a sequence of 8 unknown numbers, but I know that when I will have these numbers, I will right shift the 5 last numbers. How can I write this function, terminologically?
    $endgroup$
    – Adam54
    Dec 2 '18 at 19:49
















$begingroup$
Thanks @amWhy As I have written to Dietrich, I think I am reffering to another kind of cycle.
$endgroup$
– Adam54
Dec 2 '18 at 19:29




$begingroup$
Thanks @amWhy As I have written to Dietrich, I think I am reffering to another kind of cycle.
$endgroup$
– Adam54
Dec 2 '18 at 19:29












$begingroup$
Top number goes to bottom number in each column.
$endgroup$
– amWhy
Dec 2 '18 at 19:37




$begingroup$
Top number goes to bottom number in each column.
$endgroup$
– amWhy
Dec 2 '18 at 19:37












$begingroup$
Thank you. So writing left shift is straightforward, but for writing a right shift we need to read from top to bottom?
$endgroup$
– Adam54
Dec 2 '18 at 19:42




$begingroup$
Thank you. So writing left shift is straightforward, but for writing a right shift we need to read from top to bottom?
$endgroup$
– Adam54
Dec 2 '18 at 19:42












$begingroup$
Yes, the numbers in the top row represent the 8 possible numbers to permute. Where each of those numbers goes is written immediately below the number, in the second row.
$endgroup$
– amWhy
Dec 2 '18 at 19:46




$begingroup$
Yes, the numbers in the top row represent the 8 possible numbers to permute. Where each of those numbers goes is written immediately below the number, in the second row.
$endgroup$
– amWhy
Dec 2 '18 at 19:46












$begingroup$
Thanks again. So, if I have a sequence of 8 unknown numbers, but I know that when I will have these numbers, I will right shift the 5 last numbers. How can I write this function, terminologically?
$endgroup$
– Adam54
Dec 2 '18 at 19:49




$begingroup$
Thanks again. So, if I have a sequence of 8 unknown numbers, but I know that when I will have these numbers, I will right shift the 5 last numbers. How can I write this function, terminologically?
$endgroup$
– Adam54
Dec 2 '18 at 19:49


















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Aardman Animations

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Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
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Are they similar matrix

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0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
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