Calcuation $Stab_G$ in $mathbb{R}^3$
$begingroup$
Consider action of $X=GL_3(mathbb{R})$ on $mathbb{R}^3$ by $Acdot v = Av$.
I'm trying to deteiminate $Stab_X(H)$ were $H$ is some one-dimensional linear space of $mathbb{R}^3$.
How should I approach this question?
abstract-algebra group-theory
asked Jan 20 at 1:06
abuka123abuka123
344
$endgroup$
add a comment |
$begingroup$
Consider action of $X=GL_3(mathbb{R})$ on $mathbb{R}^3$ by $Acdot v = Av$.
I'm trying to deteiminate $Stab_X(H)$ were $H$ is some one-dimensional linear space of $mathbb{R}^3$.
How should I approach this question?
abstract-algebra group-theory
asked Jan 20 at 1:06
abuka123abuka123
344
$endgroup$
add a comment |
$begingroup$
Consider action of $X=GL_3(mathbb{R})$ on $mathbb{R}^3$ by $Acdot v = Av$.
I'm trying to deteiminate $Stab_X(H)$ were $H$ is some one-dimensional linear space of $mathbb{R}^3$.
How should I approach this question?
abstract-algebra group-theory
asked Jan 20 at 1:06
abuka123abuka123
344
$endgroup$
Consider action of $X=GL_3(mathbb{R})$ on $mathbb{R}^3$ by $Acdot v = Av$.
I'm trying to deteiminate $Stab_X(H)$ were $H$ is some one-dimensional linear space of $mathbb{R}^3$.
How should I approach this question?
abstract-algebra group-theory
abstract-algebra group-theory
asked Jan 20 at 1:06
abuka123abuka123
344
asked Jan 20 at 1:06
abuka123abuka123
344
asked Jan 20 at 1:06
abuka123abuka123
344
asked Jan 20 at 1:06
abuka123abuka123
344
asked Jan 20 at 1:06
abuka123abuka123
344
344
add a comment |
add a comment |
1 Answer
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$begingroup$
First observe that the stabilizer of a line $H$ in $mathbb{R}^3$ is the same as the stabilizer of a non-zero vector $vin H$, i.e. Stab$_X(H)=$Stab$_X(v)$. This is true since the action is linear.
Now for $v=(x,y,z)$, direct computation shows that $$H=Bigg{begin{bmatrix} a_1&a_2&a_3\ b_1&b_2&b_3\ c_1&c_2&c_3end{bmatrix}in GL_3quadbig|quad a_1x+a_2y+a_3z=x, b_1x+b_2y+b_3c=y, c_1x+c_2y+c_3z=zBigg}$$
I don't know if that is the expression you want.
answered Jan 20 at 1:20
LeventLevent
2,729925
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First observe that the stabilizer of a line $H$ in $mathbb{R}^3$ is the same as the stabilizer of a non-zero vector $vin H$, i.e. Stab$_X(H)=$Stab$_X(v)$. This is true since the action is linear.
Now for $v=(x,y,z)$, direct computation shows that $$H=Bigg{begin{bmatrix} a_1&a_2&a_3\ b_1&b_2&b_3\ c_1&c_2&c_3end{bmatrix}in GL_3quadbig|quad a_1x+a_2y+a_3z=x, b_1x+b_2y+b_3c=y, c_1x+c_2y+c_3z=zBigg}$$
I don't know if that is the expression you want.
answered Jan 20 at 1:20
LeventLevent
2,729925
$endgroup$
add a comment |
$begingroup$
First observe that the stabilizer of a line $H$ in $mathbb{R}^3$ is the same as the stabilizer of a non-zero vector $vin H$, i.e. Stab$_X(H)=$Stab$_X(v)$. This is true since the action is linear.
Now for $v=(x,y,z)$, direct computation shows that $$H=Bigg{begin{bmatrix} a_1&a_2&a_3\ b_1&b_2&b_3\ c_1&c_2&c_3end{bmatrix}in GL_3quadbig|quad a_1x+a_2y+a_3z=x, b_1x+b_2y+b_3c=y, c_1x+c_2y+c_3z=zBigg}$$
I don't know if that is the expression you want.
answered Jan 20 at 1:20
LeventLevent
2,729925
$endgroup$
add a comment |
$begingroup$
First observe that the stabilizer of a line $H$ in $mathbb{R}^3$ is the same as the stabilizer of a non-zero vector $vin H$, i.e. Stab$_X(H)=$Stab$_X(v)$. This is true since the action is linear.
Now for $v=(x,y,z)$, direct computation shows that $$H=Bigg{begin{bmatrix} a_1&a_2&a_3\ b_1&b_2&b_3\ c_1&c_2&c_3end{bmatrix}in GL_3quadbig|quad a_1x+a_2y+a_3z=x, b_1x+b_2y+b_3c=y, c_1x+c_2y+c_3z=zBigg}$$
I don't know if that is the expression you want.
answered Jan 20 at 1:20
LeventLevent
2,729925
$endgroup$
First observe that the stabilizer of a line $H$ in $mathbb{R}^3$ is the same as the stabilizer of a non-zero vector $vin H$, i.e. Stab$_X(H)=$Stab$_X(v)$. This is true since the action is linear.
Now for $v=(x,y,z)$, direct computation shows that $$H=Bigg{begin{bmatrix} a_1&a_2&a_3\ b_1&b_2&b_3\ c_1&c_2&c_3end{bmatrix}in GL_3quadbig|quad a_1x+a_2y+a_3z=x, b_1x+b_2y+b_3c=y, c_1x+c_2y+c_3z=zBigg}$$
I don't know if that is the expression you want.
answered Jan 20 at 1:20
LeventLevent
2,729925
answered Jan 20 at 1:20
LeventLevent
2,729925
answered Jan 20 at 1:20
LeventLevent
2,729925
answered Jan 20 at 1:20
LeventLevent
2,729925
2,729925
add a comment |
add a comment |
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