Mixing
Find the cardinality of a subset of $GL_n( mathbb F_p)$
$begingroup$
Let $m,n in mathbb N$. Let $mathbb F_p$ denote the prime field of characteristic $p$. Consider the set $$ X_m = {A in GL_n( {mathbb F_p}): A^m=1 }$$
Compute the cardinality of $X_m$.
Its clear that $vert X_m vert < infty$ since cardinality of $GL_n( mathbb F_p)$ itself is $(p^n-1)(p^n-p)...(p^n-p^{n-1})$. Moreover, suppose $A in X_m$ then $(x^m-1)$ kills $A$.
First I tried to understand the case when $m=p$. In this case if $A in X_p$ then $(x^p-1)$ kills $A$ and since $x^p-1=(x-1)^p$ hence $(x-1)^n$ also kills $A$. Also, minimal polynomial of $A$ is of the form $(x-1)^k$ for some $k leq p$. Any ideas to proceed further?
linear-algebra matrices finite-groups field-theory
edited Jan 30 at 12:21
user549397
1,6611418
asked Feb 11 '16 at 11:07
Arpit KansalArpit Kansal
7,01411135
$endgroup$
add a comment |
$begingroup$
Let $m,n in mathbb N$. Let $mathbb F_p$ denote the prime field of characteristic $p$. Consider the set $$ X_m = {A in GL_n( {mathbb F_p}): A^m=1 }$$
Compute the cardinality of $X_m$.
Its clear that $vert X_m vert < infty$ since cardinality of $GL_n( mathbb F_p)$ itself is $(p^n-1)(p^n-p)...(p^n-p^{n-1})$. Moreover, suppose $A in X_m$ then $(x^m-1)$ kills $A$.
First I tried to understand the case when $m=p$. In this case if $A in X_p$ then $(x^p-1)$ kills $A$ and since $x^p-1=(x-1)^p$ hence $(x-1)^n$ also kills $A$. Also, minimal polynomial of $A$ is of the form $(x-1)^k$ for some $k leq p$. Any ideas to proceed further?
linear-algebra matrices finite-groups field-theory
edited Jan 30 at 12:21
user549397
1,6611418
asked Feb 11 '16 at 11:07
Arpit KansalArpit Kansal
7,01411135
$endgroup$
$begingroup$
This might be of interest: mathoverflow.net/questions/221374/…
$endgroup$
– Calle
Sep 9 '17 at 16:39
$begingroup$
$GL_n$ is in bijection with the set of ordered bases of the vector space $F^n$ (think of the columns as a basis). So maybe try to enumerate/count all bases of $F^n$.
$endgroup$
– Nick
Feb 26 '18 at 20:44
1
$begingroup$
Dear @Nick, what you said is obvious but it does not really help. ( atleast to me) What does the condition that order of $A$ is divisible by $m$ Says in terms of basis ?
$endgroup$
– Arpit Kansal
Mar 9 '18 at 1:30
$begingroup$
For $m$ a large enough power of $p$, $X_m$ is in bijection with the set of nilpotent matrices (via $Nmapsto I_n + N$) , and we can compute this cardinality (it's $p^{n^2-n}$) . I don't have an answer for more general $n$, but seeing that a general answer would yield the cardinality of nilpotent matrices which is not a trivial fact, I think a general answer must be complicated
$endgroup$
– Max
Nov 3 '18 at 20:48
2
$begingroup$
@ArpitKansal For specific $m,n,p$ (also work for $mathbb{F}_q$ for $q=p^r$), list the possible rational canonical forms and compute the size of conjugacy classes. I doubt a "general formula" exist.
$endgroup$
– user10354138
Nov 3 '18 at 21:21
add a comment |
$begingroup$
Let $m,n in mathbb N$. Let $mathbb F_p$ denote the prime field of characteristic $p$. Consider the set $$ X_m = {A in GL_n( {mathbb F_p}): A^m=1 }$$
Compute the cardinality of $X_m$.
Its clear that $vert X_m vert < infty$ since cardinality of $GL_n( mathbb F_p)$ itself is $(p^n-1)(p^n-p)...(p^n-p^{n-1})$. Moreover, suppose $A in X_m$ then $(x^m-1)$ kills $A$.
First I tried to understand the case when $m=p$. In this case if $A in X_p$ then $(x^p-1)$ kills $A$ and since $x^p-1=(x-1)^p$ hence $(x-1)^n$ also kills $A$. Also, minimal polynomial of $A$ is of the form $(x-1)^k$ for some $k leq p$. Any ideas to proceed further?
linear-algebra matrices finite-groups field-theory
edited Jan 30 at 12:21
user549397
1,6611418
asked Feb 11 '16 at 11:07
Arpit KansalArpit Kansal
7,01411135
$endgroup$
Let $m,n in mathbb N$. Let $mathbb F_p$ denote the prime field of characteristic $p$. Consider the set $$ X_m = {A in GL_n( {mathbb F_p}): A^m=1 }$$
Compute the cardinality of $X_m$.
Its clear that $vert X_m vert < infty$ since cardinality of $GL_n( mathbb F_p)$ itself is $(p^n-1)(p^n-p)...(p^n-p^{n-1})$. Moreover, suppose $A in X_m$ then $(x^m-1)$ kills $A$.
First I tried to understand the case when $m=p$. In this case if $A in X_p$ then $(x^p-1)$ kills $A$ and since $x^p-1=(x-1)^p$ hence $(x-1)^n$ also kills $A$. Also, minimal polynomial of $A$ is of the form $(x-1)^k$ for some $k leq p$. Any ideas to proceed further?
linear-algebra matrices finite-groups field-theory
linear-algebra matrices finite-groups field-theory
edited Jan 30 at 12:21
user549397
1,6611418
asked Feb 11 '16 at 11:07
Arpit KansalArpit Kansal
7,01411135
edited Jan 30 at 12:21
user549397
1,6611418
asked Feb 11 '16 at 11:07
Arpit KansalArpit Kansal
7,01411135
edited Jan 30 at 12:21
user549397
1,6611418
edited Jan 30 at 12:21
user549397
1,6611418
edited Jan 30 at 12:21
user549397
1,6611418
1,6611418
asked Feb 11 '16 at 11:07
Arpit KansalArpit Kansal
7,01411135
asked Feb 11 '16 at 11:07
Arpit KansalArpit Kansal
7,01411135
asked Feb 11 '16 at 11:07
Arpit KansalArpit Kansal
7,01411135
7,01411135
$begingroup$
This might be of interest: mathoverflow.net/questions/221374/…
$endgroup$
– Calle
Sep 9 '17 at 16:39
$begingroup$
$GL_n$ is in bijection with the set of ordered bases of the vector space $F^n$ (think of the columns as a basis). So maybe try to enumerate/count all bases of $F^n$.
$endgroup$
– Nick
Feb 26 '18 at 20:44
1
$begingroup$
Dear @Nick, what you said is obvious but it does not really help. ( atleast to me) What does the condition that order of $A$ is divisible by $m$ Says in terms of basis ?
$endgroup$
– Arpit Kansal
Mar 9 '18 at 1:30
$begingroup$
For $m$ a large enough power of $p$, $X_m$ is in bijection with the set of nilpotent matrices (via $Nmapsto I_n + N$) , and we can compute this cardinality (it's $p^{n^2-n}$) . I don't have an answer for more general $n$, but seeing that a general answer would yield the cardinality of nilpotent matrices which is not a trivial fact, I think a general answer must be complicated
$endgroup$
– Max
Nov 3 '18 at 20:48
2
$begingroup$
@ArpitKansal For specific $m,n,p$ (also work for $mathbb{F}_q$ for $q=p^r$), list the possible rational canonical forms and compute the size of conjugacy classes. I doubt a "general formula" exist.
$endgroup$
– user10354138
Nov 3 '18 at 21:21
add a comment |
$begingroup$
This might be of interest: mathoverflow.net/questions/221374/…
$endgroup$
– Calle
Sep 9 '17 at 16:39
$begingroup$
$GL_n$ is in bijection with the set of ordered bases of the vector space $F^n$ (think of the columns as a basis). So maybe try to enumerate/count all bases of $F^n$.
$endgroup$
– Nick
Feb 26 '18 at 20:44
1
$begingroup$
Dear @Nick, what you said is obvious but it does not really help. ( atleast to me) What does the condition that order of $A$ is divisible by $m$ Says in terms of basis ?
$endgroup$
– Arpit Kansal
Mar 9 '18 at 1:30
$begingroup$
For $m$ a large enough power of $p$, $X_m$ is in bijection with the set of nilpotent matrices (via $Nmapsto I_n + N$) , and we can compute this cardinality (it's $p^{n^2-n}$) . I don't have an answer for more general $n$, but seeing that a general answer would yield the cardinality of nilpotent matrices which is not a trivial fact, I think a general answer must be complicated
$endgroup$
– Max
Nov 3 '18 at 20:48
2
$begingroup$
@ArpitKansal For specific $m,n,p$ (also work for $mathbb{F}_q$ for $q=p^r$), list the possible rational canonical forms and compute the size of conjugacy classes. I doubt a "general formula" exist.
$endgroup$
– user10354138
Nov 3 '18 at 21:21
$begingroup$
This might be of interest: mathoverflow.net/questions/221374/…
$endgroup$
– Calle
Sep 9 '17 at 16:39
$begingroup$
This might be of interest: mathoverflow.net/questions/221374/…
$endgroup$
– Calle
Sep 9 '17 at 16:39
$begingroup$
$GL_n$ is in bijection with the set of ordered bases of the vector space $F^n$ (think of the columns as a basis). So maybe try to enumerate/count all bases of $F^n$.
$endgroup$
– Nick
Feb 26 '18 at 20:44
$begingroup$
$GL_n$ is in bijection with the set of ordered bases of the vector space $F^n$ (think of the columns as a basis). So maybe try to enumerate/count all bases of $F^n$.
$endgroup$
– Nick
Feb 26 '18 at 20:44
1
1
$begingroup$
Dear @Nick, what you said is obvious but it does not really help. ( atleast to me) What does the condition that order of $A$ is divisible by $m$ Says in terms of basis ?
$endgroup$
– Arpit Kansal
Mar 9 '18 at 1:30
$begingroup$
Dear @Nick, what you said is obvious but it does not really help. ( atleast to me) What does the condition that order of $A$ is divisible by $m$ Says in terms of basis ?
$endgroup$
– Arpit Kansal
Mar 9 '18 at 1:30
$begingroup$
For $m$ a large enough power of $p$, $X_m$ is in bijection with the set of nilpotent matrices (via $Nmapsto I_n + N$) , and we can compute this cardinality (it's $p^{n^2-n}$) . I don't have an answer for more general $n$, but seeing that a general answer would yield the cardinality of nilpotent matrices which is not a trivial fact, I think a general answer must be complicated
$endgroup$
– Max
Nov 3 '18 at 20:48
$begingroup$
For $m$ a large enough power of $p$, $X_m$ is in bijection with the set of nilpotent matrices (via $Nmapsto I_n + N$) , and we can compute this cardinality (it's $p^{n^2-n}$) . I don't have an answer for more general $n$, but seeing that a general answer would yield the cardinality of nilpotent matrices which is not a trivial fact, I think a general answer must be complicated
$endgroup$
– Max
Nov 3 '18 at 20:48
2
2
$begingroup$
@ArpitKansal For specific $m,n,p$ (also work for $mathbb{F}_q$ for $q=p^r$), list the possible rational canonical forms and compute the size of conjugacy classes. I doubt a "general formula" exist.
$endgroup$
– user10354138
Nov 3 '18 at 21:21
$begingroup$
@ArpitKansal For specific $m,n,p$ (also work for $mathbb{F}_q$ for $q=p^r$), list the possible rational canonical forms and compute the size of conjugacy classes. I doubt a "general formula" exist.
$endgroup$
– user10354138
Nov 3 '18 at 21:21
add a comment |
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$begingroup$
This might be of interest: mathoverflow.net/questions/221374/…
$endgroup$
– Calle
Sep 9 '17 at 16:39
$begingroup$
$GL_n$ is in bijection with the set of ordered bases of the vector space $F^n$ (think of the columns as a basis). So maybe try to enumerate/count all bases of $F^n$.
$endgroup$
– Nick
Feb 26 '18 at 20:44
1
$begingroup$
Dear @Nick, what you said is obvious but it does not really help. ( atleast to me) What does the condition that order of $A$ is divisible by $m$ Says in terms of basis ?
$endgroup$
– Arpit Kansal
Mar 9 '18 at 1:30
$begingroup$
For $m$ a large enough power of $p$, $X_m$ is in bijection with the set of nilpotent matrices (via $Nmapsto I_n + N$) , and we can compute this cardinality (it's $p^{n^2-n}$) . I don't have an answer for more general $n$, but seeing that a general answer would yield the cardinality of nilpotent matrices which is not a trivial fact, I think a general answer must be complicated
$endgroup$
– Max
Nov 3 '18 at 20:48
2
$begingroup$
@ArpitKansal For specific $m,n,p$ (also work for $mathbb{F}_q$ for $q=p^r$), list the possible rational canonical forms and compute the size of conjugacy classes. I doubt a "general formula" exist.
$endgroup$
– user10354138
Nov 3 '18 at 21:21