Proving group action on $GL_{n}(mathbb{F})$
$begingroup$
I got asked to prove by conjugation $Pcdot A=PAP^{-1}$ that $GL_{n}(mathbb{F})curvearrowright mathbb{F}^{ntimes n}$ where $GL_{n}(mathbb{F})$ is the general liner group.
I'm not sure that I understand what does it mean to prove by conjugation. I know the theorm behind the $curvearrowright$ operator, but I'm not sure how to connect it with $Pcdot A=PAP^{-1}$.
The theorm I understand: let $G$ be a group and $X$ is a set. Then a (left) group action $phi$ of $G$ on $X$ is a function:
$$Gtimes X to X:(g,x) to phi (g,x)$$
In order to show the theorm we should prove identity and compatibility. I just don't understand how.
group-theory group-actions
asked Dec 22 '18 at 15:20
TTaJTa4TTaJTa4
1666
$endgroup$
|
show 4 more comments
$begingroup$
I got asked to prove by conjugation $Pcdot A=PAP^{-1}$ that $GL_{n}(mathbb{F})curvearrowright mathbb{F}^{ntimes n}$ where $GL_{n}(mathbb{F})$ is the general liner group.
I'm not sure that I understand what does it mean to prove by conjugation. I know the theorm behind the $curvearrowright$ operator, but I'm not sure how to connect it with $Pcdot A=PAP^{-1}$.
The theorm I understand: let $G$ be a group and $X$ is a set. Then a (left) group action $phi$ of $G$ on $X$ is a function:
$$Gtimes X to X:(g,x) to phi (g,x)$$
In order to show the theorm we should prove identity and compatibility. I just don't understand how.
group-theory group-actions
asked Dec 22 '18 at 15:20
TTaJTa4TTaJTa4
1666
$endgroup$
2
$begingroup$
$PA$ is the definition of the action, not a matrix product.
$endgroup$
– Randall
Dec 22 '18 at 15:22
$begingroup$
What do you mean with the "theorem behind the $curvearrowright$ operator"?
$endgroup$
– Hagen von Eitzen
Dec 22 '18 at 15:22
$begingroup$
At present, this question is unclear, so, regrettably, I've voted to close it.
$endgroup$
– Shaun
Dec 22 '18 at 15:25
$begingroup$
Please check out my edit
$endgroup$
– TTaJTa4
Dec 22 '18 at 15:27
1
$begingroup$
You mean you want to prove that the conjugation formula you gave acutally defines an action of the genral linear group on the groups of square matrices, is that it?
$endgroup$
– A.Rod
Dec 22 '18 at 15:30
|
show 4 more comments
$begingroup$
I got asked to prove by conjugation $Pcdot A=PAP^{-1}$ that $GL_{n}(mathbb{F})curvearrowright mathbb{F}^{ntimes n}$ where $GL_{n}(mathbb{F})$ is the general liner group.
I'm not sure that I understand what does it mean to prove by conjugation. I know the theorm behind the $curvearrowright$ operator, but I'm not sure how to connect it with $Pcdot A=PAP^{-1}$.
The theorm I understand: let $G$ be a group and $X$ is a set. Then a (left) group action $phi$ of $G$ on $X$ is a function:
$$Gtimes X to X:(g,x) to phi (g,x)$$
In order to show the theorm we should prove identity and compatibility. I just don't understand how.
group-theory group-actions
asked Dec 22 '18 at 15:20
TTaJTa4TTaJTa4
1666
$endgroup$
I got asked to prove by conjugation $Pcdot A=PAP^{-1}$ that $GL_{n}(mathbb{F})curvearrowright mathbb{F}^{ntimes n}$ where $GL_{n}(mathbb{F})$ is the general liner group.
I'm not sure that I understand what does it mean to prove by conjugation. I know the theorm behind the $curvearrowright$ operator, but I'm not sure how to connect it with $Pcdot A=PAP^{-1}$.
The theorm I understand: let $G$ be a group and $X$ is a set. Then a (left) group action $phi$ of $G$ on $X$ is a function:
$$Gtimes X to X:(g,x) to phi (g,x)$$
In order to show the theorm we should prove identity and compatibility. I just don't understand how.
group-theory group-actions
group-theory group-actions
asked Dec 22 '18 at 15:20
TTaJTa4TTaJTa4
1666
asked Dec 22 '18 at 15:20
TTaJTa4TTaJTa4
1666
edited Dec 22 '18 at 15:44
TTaJTa4
asked Dec 22 '18 at 15:20
TTaJTa4TTaJTa4
1666
asked Dec 22 '18 at 15:20
TTaJTa4TTaJTa4
1666
asked Dec 22 '18 at 15:20
TTaJTa4TTaJTa4
1666
1666
2
$begingroup$
$PA$ is the definition of the action, not a matrix product.
$endgroup$
– Randall
Dec 22 '18 at 15:22
$begingroup$
What do you mean with the "theorem behind the $curvearrowright$ operator"?
$endgroup$
– Hagen von Eitzen
Dec 22 '18 at 15:22
$begingroup$
At present, this question is unclear, so, regrettably, I've voted to close it.
$endgroup$
– Shaun
Dec 22 '18 at 15:25
$begingroup$
Please check out my edit
$endgroup$
– TTaJTa4
Dec 22 '18 at 15:27
1
$begingroup$
You mean you want to prove that the conjugation formula you gave acutally defines an action of the genral linear group on the groups of square matrices, is that it?
$endgroup$
– A.Rod
Dec 22 '18 at 15:30
|
show 4 more comments
2
$begingroup$
$PA$ is the definition of the action, not a matrix product.
$endgroup$
– Randall
Dec 22 '18 at 15:22
$begingroup$
What do you mean with the "theorem behind the $curvearrowright$ operator"?
$endgroup$
– Hagen von Eitzen
Dec 22 '18 at 15:22
$begingroup$
At present, this question is unclear, so, regrettably, I've voted to close it.
$endgroup$
– Shaun
Dec 22 '18 at 15:25
$begingroup$
Please check out my edit
$endgroup$
– TTaJTa4
Dec 22 '18 at 15:27
1
$begingroup$
You mean you want to prove that the conjugation formula you gave acutally defines an action of the genral linear group on the groups of square matrices, is that it?
$endgroup$
– A.Rod
Dec 22 '18 at 15:30
2
2
$begingroup$
$PA$ is the definition of the action, not a matrix product.
$endgroup$
– Randall
Dec 22 '18 at 15:22
$begingroup$
$PA$ is the definition of the action, not a matrix product.
$endgroup$
– Randall
Dec 22 '18 at 15:22
$begingroup$
What do you mean with the "theorem behind the $curvearrowright$ operator"?
$endgroup$
– Hagen von Eitzen
Dec 22 '18 at 15:22
$begingroup$
What do you mean with the "theorem behind the $curvearrowright$ operator"?
$endgroup$
– Hagen von Eitzen
Dec 22 '18 at 15:22
$begingroup$
At present, this question is unclear, so, regrettably, I've voted to close it.
$endgroup$
– Shaun
Dec 22 '18 at 15:25
$begingroup$
At present, this question is unclear, so, regrettably, I've voted to close it.
$endgroup$
– Shaun
Dec 22 '18 at 15:25
$begingroup$
Please check out my edit
$endgroup$
– TTaJTa4
Dec 22 '18 at 15:27
$begingroup$
Please check out my edit
$endgroup$
– TTaJTa4
Dec 22 '18 at 15:27
1
1
$begingroup$
You mean you want to prove that the conjugation formula you gave acutally defines an action of the genral linear group on the groups of square matrices, is that it?
$endgroup$
– A.Rod
Dec 22 '18 at 15:30
$begingroup$
You mean you want to prove that the conjugation formula you gave acutally defines an action of the genral linear group on the groups of square matrices, is that it?
$endgroup$
– A.Rod
Dec 22 '18 at 15:30
|
show 4 more comments
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2
$begingroup$
$PA$ is the definition of the action, not a matrix product.
$endgroup$
– Randall
Dec 22 '18 at 15:22
$begingroup$
What do you mean with the "theorem behind the $curvearrowright$ operator"?
$endgroup$
– Hagen von Eitzen
Dec 22 '18 at 15:22
$begingroup$
At present, this question is unclear, so, regrettably, I've voted to close it.
$endgroup$
– Shaun
Dec 22 '18 at 15:25
$begingroup$
Please check out my edit
$endgroup$
– TTaJTa4
Dec 22 '18 at 15:27
1
$begingroup$
You mean you want to prove that the conjugation formula you gave acutally defines an action of the genral linear group on the groups of square matrices, is that it?
$endgroup$
– A.Rod
Dec 22 '18 at 15:30