Prove that $S = {A in GL_n(K) | AJA^t = J}$ is a subgroup of $GL_n(K)$ The 2019 Stack Overflow...
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Prove that $S = {A in GL_n(K) | AJA^t = J}$ is a subgroup of $GL_n(K)$
The 2019 Stack Overflow Developer Survey Results Are InWe can't give the structure of linear space on $mathbb{Q}(sqrt{2})$ for a set of rational numbers$GL_n(mathbb F_q)$ has an element of order $q^n-1$Matrix with irreducible minimal polynomial gives rise to a fieldProve that for any prime $p$, the set $mathbb{Z_p}$ with the addition mod $p$ and multiplication mod $p$, and congruence mod $p$, is a fieldProof the set $mathbb{Z}_p$ is a fieldShow that $(S, cdot)$ is a subgroup of $(GL_3(mathbb{R}), cdot))$$GL_n(F)$ acts on the flag varietyIf $det A neq 0, AB= (det A)E_n $ then $ A in GL_n( mathbb Q [X])$Proving something is a linearly independent subsetLet $K$ be a field. Prove that the field of all polynomials over $K$ is a vector space over $K$.
$begingroup$
Given a field $K$, $n in mathbb{N}$ and $J in K^{n times n}$, show that:
$$
S = {A in GL_n(K) | AJA^t = J}
$$
is a subgroup of $GL_n(K)$.
To show that S is a subgroup, we must show:
It must include the neutral element: $e in S$
For $a,b in S$, $S$ must include: $a cdot b^{-1} in S$.
1.) Showing that $e in S$:
Since $J$ is an element over the field $K$, the multiplication has a neutral element. Therefore we choose $J = E_n$. The neutral element must also be part of the linear group over $K$, therefore if we choose $A = E_n$, we can conclude that the neutral element is in $S$.
2.)
Given an $a := AJA^t = J , b := BIB^t = I$ we will show that $a cdot b^{-1}$ will also be in $S$. The inverse of $b$ is defined as: $b^{-1} = (B^t)^{-1}I^{-1}B^{-1}$. Therefore:
$$
AJA^t(B^t)^{-1}I^{-1}B^{-1} = JI^{-1}
$$
But I am not sure how to proceed ...
linear-algebra matrices discrete-mathematics field-theory
asked Mar 21 at 6:02
ViperViper
223
$endgroup$
add a comment |
$begingroup$
Given a field $K$, $n in mathbb{N}$ and $J in K^{n times n}$, show that:
$$
S = {A in GL_n(K) | AJA^t = J}
$$
is a subgroup of $GL_n(K)$.
To show that S is a subgroup, we must show:
It must include the neutral element: $e in S$
For $a,b in S$, $S$ must include: $a cdot b^{-1} in S$.
1.) Showing that $e in S$:
Since $J$ is an element over the field $K$, the multiplication has a neutral element. Therefore we choose $J = E_n$. The neutral element must also be part of the linear group over $K$, therefore if we choose $A = E_n$, we can conclude that the neutral element is in $S$.
2.)
Given an $a := AJA^t = J , b := BIB^t = I$ we will show that $a cdot b^{-1}$ will also be in $S$. The inverse of $b$ is defined as: $b^{-1} = (B^t)^{-1}I^{-1}B^{-1}$. Therefore:
$$
AJA^t(B^t)^{-1}I^{-1}B^{-1} = JI^{-1}
$$
But I am not sure how to proceed ...
linear-algebra matrices discrete-mathematics field-theory
asked Mar 21 at 6:02
ViperViper
223
$endgroup$
1
$begingroup$
In (2) you need to show that if $AJA^t=B^JB^t=J$ then $CJC^t=J$ where $C=AB^{-1}$.
$endgroup$
– Lord Shark the Unknown
Mar 21 at 6:09
add a comment |
$begingroup$
Given a field $K$, $n in mathbb{N}$ and $J in K^{n times n}$, show that:
$$
S = {A in GL_n(K) | AJA^t = J}
$$
is a subgroup of $GL_n(K)$.
To show that S is a subgroup, we must show:
It must include the neutral element: $e in S$
For $a,b in S$, $S$ must include: $a cdot b^{-1} in S$.
1.) Showing that $e in S$:
Since $J$ is an element over the field $K$, the multiplication has a neutral element. Therefore we choose $J = E_n$. The neutral element must also be part of the linear group over $K$, therefore if we choose $A = E_n$, we can conclude that the neutral element is in $S$.
2.)
Given an $a := AJA^t = J , b := BIB^t = I$ we will show that $a cdot b^{-1}$ will also be in $S$. The inverse of $b$ is defined as: $b^{-1} = (B^t)^{-1}I^{-1}B^{-1}$. Therefore:
$$
AJA^t(B^t)^{-1}I^{-1}B^{-1} = JI^{-1}
$$
But I am not sure how to proceed ...
linear-algebra matrices discrete-mathematics field-theory
asked Mar 21 at 6:02
ViperViper
223
$endgroup$
Given a field $K$, $n in mathbb{N}$ and $J in K^{n times n}$, show that:
$$
S = {A in GL_n(K) | AJA^t = J}
$$
is a subgroup of $GL_n(K)$.
To show that S is a subgroup, we must show:
It must include the neutral element: $e in S$
For $a,b in S$, $S$ must include: $a cdot b^{-1} in S$.
1.) Showing that $e in S$:
Since $J$ is an element over the field $K$, the multiplication has a neutral element. Therefore we choose $J = E_n$. The neutral element must also be part of the linear group over $K$, therefore if we choose $A = E_n$, we can conclude that the neutral element is in $S$.
2.)
Given an $a := AJA^t = J , b := BIB^t = I$ we will show that $a cdot b^{-1}$ will also be in $S$. The inverse of $b$ is defined as: $b^{-1} = (B^t)^{-1}I^{-1}B^{-1}$. Therefore:
$$
AJA^t(B^t)^{-1}I^{-1}B^{-1} = JI^{-1}
$$
But I am not sure how to proceed ...
linear-algebra matrices discrete-mathematics field-theory
linear-algebra matrices discrete-mathematics field-theory
asked Mar 21 at 6:02
ViperViper
223
asked Mar 21 at 6:02
ViperViper
223
asked Mar 21 at 6:02
ViperViper
223
asked Mar 21 at 6:02
ViperViper
223
asked Mar 21 at 6:02
ViperViper
223
223
1
$begingroup$
In (2) you need to show that if $AJA^t=B^JB^t=J$ then $CJC^t=J$ where $C=AB^{-1}$.
$endgroup$
– Lord Shark the Unknown
Mar 21 at 6:09
add a comment |
1
$begingroup$
In (2) you need to show that if $AJA^t=B^JB^t=J$ then $CJC^t=J$ where $C=AB^{-1}$.
$endgroup$
– Lord Shark the Unknown
Mar 21 at 6:09
1
1
$begingroup$
In (2) you need to show that if $AJA^t=B^JB^t=J$ then $CJC^t=J$ where $C=AB^{-1}$.
$endgroup$
– Lord Shark the Unknown
Mar 21 at 6:09
$begingroup$
In (2) you need to show that if $AJA^t=B^JB^t=J$ then $CJC^t=J$ where $C=AB^{-1}$.
$endgroup$
– Lord Shark the Unknown
Mar 21 at 6:09
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The following method shows that $S$ is the kernel of a homomorphism defined on $GL_{n}(K)$. The kernel of a homomorphism is always a normal subgroup.
Regard $K^{n times n}$ as a vector space (of dimension $n^2$) over $K$.
For each $A in GL_{n}(K)$, we define a linear transformation $phi[A]$ on the vector space $K^{n times n}$ as follows:
$$
phi[A] J = A J A^{t}.
$$
Furthermore, $phi[A]$ is invertible, and
$$
phi[B] phi[A] = phi[BA].
$$
Therefore, $phi$ is a homomorphism from group $GL_{n}(K)$ to the group of the invertible linear transformations on $K^n$.
And $S$ is the kernel of that homomorphism, hence is a subgroup of $GL_{n}(K)$, and even a normal subgroup.
answered Mar 21 at 6:27
avsavs
4,012515
$endgroup$
add a comment |
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oldest
votes
$begingroup$
The following method shows that $S$ is the kernel of a homomorphism defined on $GL_{n}(K)$. The kernel of a homomorphism is always a normal subgroup.
Regard $K^{n times n}$ as a vector space (of dimension $n^2$) over $K$.
For each $A in GL_{n}(K)$, we define a linear transformation $phi[A]$ on the vector space $K^{n times n}$ as follows:
$$
phi[A] J = A J A^{t}.
$$
Furthermore, $phi[A]$ is invertible, and
$$
phi[B] phi[A] = phi[BA].
$$
Therefore, $phi$ is a homomorphism from group $GL_{n}(K)$ to the group of the invertible linear transformations on $K^n$.
And $S$ is the kernel of that homomorphism, hence is a subgroup of $GL_{n}(K)$, and even a normal subgroup.
answered Mar 21 at 6:27
avsavs
4,012515
$endgroup$
add a comment |
$begingroup$
The following method shows that $S$ is the kernel of a homomorphism defined on $GL_{n}(K)$. The kernel of a homomorphism is always a normal subgroup.
Regard $K^{n times n}$ as a vector space (of dimension $n^2$) over $K$.
For each $A in GL_{n}(K)$, we define a linear transformation $phi[A]$ on the vector space $K^{n times n}$ as follows:
$$
phi[A] J = A J A^{t}.
$$
Furthermore, $phi[A]$ is invertible, and
$$
phi[B] phi[A] = phi[BA].
$$
Therefore, $phi$ is a homomorphism from group $GL_{n}(K)$ to the group of the invertible linear transformations on $K^n$.
And $S$ is the kernel of that homomorphism, hence is a subgroup of $GL_{n}(K)$, and even a normal subgroup.
answered Mar 21 at 6:27
avsavs
4,012515
$endgroup$
add a comment |
$begingroup$
The following method shows that $S$ is the kernel of a homomorphism defined on $GL_{n}(K)$. The kernel of a homomorphism is always a normal subgroup.
Regard $K^{n times n}$ as a vector space (of dimension $n^2$) over $K$.
For each $A in GL_{n}(K)$, we define a linear transformation $phi[A]$ on the vector space $K^{n times n}$ as follows:
$$
phi[A] J = A J A^{t}.
$$
Furthermore, $phi[A]$ is invertible, and
$$
phi[B] phi[A] = phi[BA].
$$
Therefore, $phi$ is a homomorphism from group $GL_{n}(K)$ to the group of the invertible linear transformations on $K^n$.
And $S$ is the kernel of that homomorphism, hence is a subgroup of $GL_{n}(K)$, and even a normal subgroup.
answered Mar 21 at 6:27
avsavs
4,012515
$endgroup$
The following method shows that $S$ is the kernel of a homomorphism defined on $GL_{n}(K)$. The kernel of a homomorphism is always a normal subgroup.
Regard $K^{n times n}$ as a vector space (of dimension $n^2$) over $K$.
For each $A in GL_{n}(K)$, we define a linear transformation $phi[A]$ on the vector space $K^{n times n}$ as follows:
$$
phi[A] J = A J A^{t}.
$$
Furthermore, $phi[A]$ is invertible, and
$$
phi[B] phi[A] = phi[BA].
$$
Therefore, $phi$ is a homomorphism from group $GL_{n}(K)$ to the group of the invertible linear transformations on $K^n$.
And $S$ is the kernel of that homomorphism, hence is a subgroup of $GL_{n}(K)$, and even a normal subgroup.
answered Mar 21 at 6:27
avsavs
4,012515
answered Mar 21 at 6:27
avsavs
4,012515
answered Mar 21 at 6:27
avsavs
4,012515
answered Mar 21 at 6:27
avsavs
4,012515
4,012515
add a comment |
add a comment |
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1
$begingroup$
In (2) you need to show that if $AJA^t=B^JB^t=J$ then $CJC^t=J$ where $C=AB^{-1}$.
$endgroup$
– Lord Shark the Unknown
Mar 21 at 6:09