Mixing
What is the number of elements in ${A=(a_{ij}) in Gl_n(mathbb{F}_q):a_{11}=1}$
1
$begingroup$
Let $mathbb{F}_q$ be the finite field with $q$ elements. I want to calculate the number of elements of the set $S={A=(a_{ij}) in Gl_n(mathbb{F}_q):a_{11}=1}.$
We know that $|Gl_n(mathbb{F}_q)|=(q^n-1)(q^n-q)cdots(q^n-q^{n-1}).$ Trying to define some group homomorphism from $Gl_n(mathbb{F}_q)$ to $mathbb{F}_q^{times}$ whose kernel is $S,$ but failed to do so. Any help will be appreciated. Thanks.
linear-algebra abstract-algebra matrices linear-transformations finite-fields
asked Jan 15 at 8:44
user371231user371231
1511
$endgroup$
|
show 1 more comment
1
$begingroup$
Let $mathbb{F}_q$ be the finite field with $q$ elements. I want to calculate the number of elements of the set $S={A=(a_{ij}) in Gl_n(mathbb{F}_q):a_{11}=1}.$
We know that $|Gl_n(mathbb{F}_q)|=(q^n-1)(q^n-q)cdots(q^n-q^{n-1}).$ Trying to define some group homomorphism from $Gl_n(mathbb{F}_q)$ to $mathbb{F}_q^{times}$ whose kernel is $S,$ but failed to do so. Any help will be appreciated. Thanks.
linear-algebra abstract-algebra matrices linear-transformations finite-fields
asked Jan 15 at 8:44
user371231user371231
1511
$endgroup$
$begingroup$
This is probably dumb but my guess is that this contains the same amount of elements $S={A=(a_{ij}) in Gl_n(mathbb{F}_q):a_{11}=u}$ for any $uinmathbb F_q$ and these are disjoints sets that partition $Gl_n(mathbb{F}_q)$ hence they each have a fraction $frac{1}{q}$ of the elements of the later. Hence $frac{1}{q}(q^n-1)(q^n-q)dots (q^n-q^{n-1})$. This seems a bit simple though and this may not be true for $u=0$ for some reasons unknown to me.
$endgroup$
– P. Quinton
Jan 15 at 9:58
$begingroup$
Follow up question on that : does this set contains the same amount of elements as any set of the form ${A=(a_{ij}) in Gl_n(mathbb{F}_q):f(A)=0}$ where $f(A)$ is a non constant linear function of the elements of $A$ ? and what if we have several linear constraints ?
$endgroup$
– P. Quinton
Jan 15 at 10:03
$begingroup$
Recall the proof of the $left|operatorname{GL}_nleft(mathbb{F}_qright)right|$ formula, where you build your matrix column by column. Now adapt it to the $a_{11} = 1$ requirement.
$endgroup$
– darij grinberg
Jan 15 at 17:10
$begingroup$
@darij grinberg: So will it be $q^{n-1}(q^n-q)cdots (q^n-q^{n-1})$ ?
$endgroup$
– user371231
Jan 15 at 22:05
$begingroup$
Yes, that's it exactly.
$endgroup$
– darij grinberg
Jan 15 at 22:05
|
show 1 more comment
1
1
1
2
$begingroup$
Let $mathbb{F}_q$ be the finite field with $q$ elements. I want to calculate the number of elements of the set $S={A=(a_{ij}) in Gl_n(mathbb{F}_q):a_{11}=1}.$
We know that $|Gl_n(mathbb{F}_q)|=(q^n-1)(q^n-q)cdots(q^n-q^{n-1}).$ Trying to define some group homomorphism from $Gl_n(mathbb{F}_q)$ to $mathbb{F}_q^{times}$ whose kernel is $S,$ but failed to do so. Any help will be appreciated. Thanks.
linear-algebra abstract-algebra matrices linear-transformations finite-fields
asked Jan 15 at 8:44
user371231user371231
1511
$endgroup$
Let $mathbb{F}_q$ be the finite field with $q$ elements. I want to calculate the number of elements of the set $S={A=(a_{ij}) in Gl_n(mathbb{F}_q):a_{11}=1}.$
We know that $|Gl_n(mathbb{F}_q)|=(q^n-1)(q^n-q)cdots(q^n-q^{n-1}).$ Trying to define some group homomorphism from $Gl_n(mathbb{F}_q)$ to $mathbb{F}_q^{times}$ whose kernel is $S,$ but failed to do so. Any help will be appreciated. Thanks.
linear-algebra abstract-algebra matrices linear-transformations finite-fields
linear-algebra abstract-algebra matrices linear-transformations finite-fields
asked Jan 15 at 8:44
user371231user371231
1511
asked Jan 15 at 8:44
user371231user371231
1511
asked Jan 15 at 8:44
user371231user371231
1511
asked Jan 15 at 8:44
user371231user371231
1511
asked Jan 15 at 8:44
user371231user371231
1511
1511
$begingroup$
This is probably dumb but my guess is that this contains the same amount of elements $S={A=(a_{ij}) in Gl_n(mathbb{F}_q):a_{11}=u}$ for any $uinmathbb F_q$ and these are disjoints sets that partition $Gl_n(mathbb{F}_q)$ hence they each have a fraction $frac{1}{q}$ of the elements of the later. Hence $frac{1}{q}(q^n-1)(q^n-q)dots (q^n-q^{n-1})$. This seems a bit simple though and this may not be true for $u=0$ for some reasons unknown to me.
$endgroup$
– P. Quinton
Jan 15 at 9:58
$begingroup$
Follow up question on that : does this set contains the same amount of elements as any set of the form ${A=(a_{ij}) in Gl_n(mathbb{F}_q):f(A)=0}$ where $f(A)$ is a non constant linear function of the elements of $A$ ? and what if we have several linear constraints ?
$endgroup$
– P. Quinton
Jan 15 at 10:03
$begingroup$
Recall the proof of the $left|operatorname{GL}_nleft(mathbb{F}_qright)right|$ formula, where you build your matrix column by column. Now adapt it to the $a_{11} = 1$ requirement.
$endgroup$
– darij grinberg
Jan 15 at 17:10
$begingroup$
@darij grinberg: So will it be $q^{n-1}(q^n-q)cdots (q^n-q^{n-1})$ ?
$endgroup$
– user371231
Jan 15 at 22:05
$begingroup$
Yes, that's it exactly.
$endgroup$
– darij grinberg
Jan 15 at 22:05
|
show 1 more comment
$begingroup$
This is probably dumb but my guess is that this contains the same amount of elements $S={A=(a_{ij}) in Gl_n(mathbb{F}_q):a_{11}=u}$ for any $uinmathbb F_q$ and these are disjoints sets that partition $Gl_n(mathbb{F}_q)$ hence they each have a fraction $frac{1}{q}$ of the elements of the later. Hence $frac{1}{q}(q^n-1)(q^n-q)dots (q^n-q^{n-1})$. This seems a bit simple though and this may not be true for $u=0$ for some reasons unknown to me.
$endgroup$
– P. Quinton
Jan 15 at 9:58
$begingroup$
Follow up question on that : does this set contains the same amount of elements as any set of the form ${A=(a_{ij}) in Gl_n(mathbb{F}_q):f(A)=0}$ where $f(A)$ is a non constant linear function of the elements of $A$ ? and what if we have several linear constraints ?
$endgroup$
– P. Quinton
Jan 15 at 10:03
$begingroup$
Recall the proof of the $left|operatorname{GL}_nleft(mathbb{F}_qright)right|$ formula, where you build your matrix column by column. Now adapt it to the $a_{11} = 1$ requirement.
$endgroup$
– darij grinberg
Jan 15 at 17:10
$begingroup$
@darij grinberg: So will it be $q^{n-1}(q^n-q)cdots (q^n-q^{n-1})$ ?
$endgroup$
– user371231
Jan 15 at 22:05
$begingroup$
Yes, that's it exactly.
$endgroup$
– darij grinberg
Jan 15 at 22:05
$begingroup$
This is probably dumb but my guess is that this contains the same amount of elements $S={A=(a_{ij}) in Gl_n(mathbb{F}_q):a_{11}=u}$ for any $uinmathbb F_q$ and these are disjoints sets that partition $Gl_n(mathbb{F}_q)$ hence they each have a fraction $frac{1}{q}$ of the elements of the later. Hence $frac{1}{q}(q^n-1)(q^n-q)dots (q^n-q^{n-1})$. This seems a bit simple though and this may not be true for $u=0$ for some reasons unknown to me.
$endgroup$
– P. Quinton
Jan 15 at 9:58
$begingroup$
This is probably dumb but my guess is that this contains the same amount of elements $S={A=(a_{ij}) in Gl_n(mathbb{F}_q):a_{11}=u}$ for any $uinmathbb F_q$ and these are disjoints sets that partition $Gl_n(mathbb{F}_q)$ hence they each have a fraction $frac{1}{q}$ of the elements of the later. Hence $frac{1}{q}(q^n-1)(q^n-q)dots (q^n-q^{n-1})$. This seems a bit simple though and this may not be true for $u=0$ for some reasons unknown to me.
$endgroup$
– P. Quinton
Jan 15 at 9:58
$begingroup$
Follow up question on that : does this set contains the same amount of elements as any set of the form ${A=(a_{ij}) in Gl_n(mathbb{F}_q):f(A)=0}$ where $f(A)$ is a non constant linear function of the elements of $A$ ? and what if we have several linear constraints ?
$endgroup$
– P. Quinton
Jan 15 at 10:03
$begingroup$
Follow up question on that : does this set contains the same amount of elements as any set of the form ${A=(a_{ij}) in Gl_n(mathbb{F}_q):f(A)=0}$ where $f(A)$ is a non constant linear function of the elements of $A$ ? and what if we have several linear constraints ?
$endgroup$
– P. Quinton
Jan 15 at 10:03
$begingroup$
Recall the proof of the $left|operatorname{GL}_nleft(mathbb{F}_qright)right|$ formula, where you build your matrix column by column. Now adapt it to the $a_{11} = 1$ requirement.
$endgroup$
– darij grinberg
Jan 15 at 17:10
$begingroup$
Recall the proof of the $left|operatorname{GL}_nleft(mathbb{F}_qright)right|$ formula, where you build your matrix column by column. Now adapt it to the $a_{11} = 1$ requirement.
$endgroup$
– darij grinberg
Jan 15 at 17:10
$begingroup$
@darij grinberg: So will it be $q^{n-1}(q^n-q)cdots (q^n-q^{n-1})$ ?
$endgroup$
– user371231
Jan 15 at 22:05
$begingroup$
@darij grinberg: So will it be $q^{n-1}(q^n-q)cdots (q^n-q^{n-1})$ ?
$endgroup$
– user371231
Jan 15 at 22:05
$begingroup$
Yes, that's it exactly.
$endgroup$
– darij grinberg
Jan 15 at 22:05
$begingroup$
Yes, that's it exactly.
$endgroup$
– darij grinberg
Jan 15 at 22:05
|
show 1 more comment
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$begingroup$
This is probably dumb but my guess is that this contains the same amount of elements $S={A=(a_{ij}) in Gl_n(mathbb{F}_q):a_{11}=u}$ for any $uinmathbb F_q$ and these are disjoints sets that partition $Gl_n(mathbb{F}_q)$ hence they each have a fraction $frac{1}{q}$ of the elements of the later. Hence $frac{1}{q}(q^n-1)(q^n-q)dots (q^n-q^{n-1})$. This seems a bit simple though and this may not be true for $u=0$ for some reasons unknown to me.
$endgroup$
– P. Quinton
Jan 15 at 9:58
$begingroup$
Follow up question on that : does this set contains the same amount of elements as any set of the form ${A=(a_{ij}) in Gl_n(mathbb{F}_q):f(A)=0}$ where $f(A)$ is a non constant linear function of the elements of $A$ ? and what if we have several linear constraints ?
$endgroup$
– P. Quinton
Jan 15 at 10:03
$begingroup$
Recall the proof of the $left|operatorname{GL}_nleft(mathbb{F}_qright)right|$ formula, where you build your matrix column by column. Now adapt it to the $a_{11} = 1$ requirement.
$endgroup$
– darij grinberg
Jan 15 at 17:10
$begingroup$
@darij grinberg: So will it be $q^{n-1}(q^n-q)cdots (q^n-q^{n-1})$ ?
$endgroup$
– user371231
Jan 15 at 22:05
$begingroup$
Yes, that's it exactly.
$endgroup$
– darij grinberg
Jan 15 at 22:05