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Let $A $ denotes the set of all lines through the origin.Let $G=GL_{n}(\mathbb{R})$ be the general linear group of $ 2 \times 2 $ matrices and $H$ be the subgroup of $G$ containing all lower triangular matrices. Then find a line $l$ in $A$ whose stabilizer in $G$ is exactly H , i.e; $ stab_{G}(a)=H$. $$$$ I have tried in this way, If a group $G$ acts on a set $X$ then for $ x \in X$ , $stab_{G}(x)=\{g \in G : gx=x \}$. Replacing $ X$ by $A \ and \ x \ by \ l$ , we get $ stab_{G}(l)=\{g \in G: gl=l \}$. In order to show that $ Stab_{G}(l)=H$, we have to show that for any $ s \in Stab_{G}(l) \implies s \in H$. Right here i can't proceed further. please help me from here, thanks.

mmath
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  • You need to pick $l$ in such a way that $gl=l$ if and only if $g$ is lower triangular. (You really need to show the "if" and the "only if".) Well, let $v=(x,y)$ be any non-zero vector along $l$. Let $g = \left( \begin{array}{cc} a & 0 \ b & c \end{array} \right) $ (with $ac \neq 0$) be an invertible lower triangular matrix. How does $g$ act on $l$? – Kenny Wong Mar 04 '17 at 12:09
  • the operation in $g l $ is matrix multiplication – mmath Mar 04 '17 at 12:19
  • Yes. And what do you get when you multiply the matrices that I wrote? – Kenny Wong Mar 04 '17 at 12:22
  • But we have to take a general $ 2 \times 2 $ matrix from $ G $ instead of a lower triangular matrix, is not it? – mmath Mar 04 '17 at 12:27
  • Yes, but focus on the lower triangular one for now. Can you think of a $v$ such that $gv $ is parallel to $v$? – Kenny Wong Mar 04 '17 at 12:29
  • Yes, consider $ v=(0,1)^{T}$, then $ gv =\frac{1}{c} v$ – mmath Mar 04 '17 at 12:36
  • Excellent! (Actually I think $gv = cv$, but that doesn't matter.) Now, can you show that the ONLY matrices $g$ such that $gv$ is a multiple of $v$ are of lower triangular form? – Kenny Wong Mar 04 '17 at 12:39
  • ok, i understood now, that matrices with lower triangular form only satisfies the above property, lot of thanks – mmath Mar 04 '17 at 12:45
  • Great, glad it helped! – Kenny Wong Mar 04 '17 at 12:46

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