Group Action on GL(2,R) by Conjugation: Orbit and Isotropy Group of a Matrix

[Oxymoron]

Question

Let G=GL(2,\mathbb{R}) be the group of invertible 2\times 2 martrices with real entries. Consider the action of G on itself by conjugation. For the element

A= \left(\begin{array}{cc}<br /> 2 &amp; 1 \\ <br /> 0 &amp; 3<br /> \end{array}\right)

of G, describe (i) the orbit and (ii) the isotropy group of A

Sorry, I have no working out because I am completely stumped. Can anyone give me some helpful hints or pointers. Thanks

 

[Oxymoron]

The orbit is going to be of the form

O_A = \{gA\,|\, g\in GL(2,\mathbb{R})\}

Here g can be found by the following way

gA = Ag where g \in GL(2,\mathbb{R})

\left(\begin{array}{cc}a &amp; b \\ c &amp; d\end{array}\right)\left(\begin{array}{cc}2 &amp; 1 \\ 0 &amp; 3\end{array}\right) = \left(\begin{array}{cc}2 &amp; 1 \\ 0 &amp; 3\end{array}\right)\left(\begin{array}{cc}a &amp; b \\ c &amp; d\end{array}\right)

\left(\begin{array}{cc}2a &amp; a+3b \\ 2c &amp; c+3d\end{array}\right) = \left(\begin{array}{cc}2a+c &amp; 2b+d \\ 3c &amp; 3d\end{array}\right)

Which implies that 2c = 3c = 0 \Leftrightarrow c = 0. Hence

\left(\begin{array}{cc}2a &amp; a+3b \\ 0 &amp; 3d\end{array}\right) = \left(\begin{array}{cc}2a &amp; 2b+d \\ 0 &amp; 3d\end{array}\right)

And we can write

g = \left(\begin{array}{cc}a &amp; d-a \\ 0 &amp; d\end{array}\right)

Therefore the orbit can be described as

O_A = \{gA\,|\, g\in GL(2,\mathbb{R})\}

So

O_A = \left(\begin{array}{cc}2a &amp; -2a + 3d \\ 0 &amp; 3d\end{array}\right)\quad \forall a,b \in \mathbb{R}

how does this look?

 

[Oxymoron]

The isotropy group is the subgroup of GL(2,\mathbb{R}) consisting of the elements that do not move A. That is

G_A = \{gA = A\,|\,g\in GL(2,\mathbb{R})\}

Therefore we have

\left(\begin{array}{cc}2a &amp; -2a+3d \\ 0 &amp; 3d\end{array}\right) = \left(\begin{array}{cc}2 &amp; 1 \\ 0 &amp; 3\end{array}\right)

So a = 1, \, d = 1. Therefore

G_A = \left(\begin{array}{cc}1 &amp; 0 \\ 0 &amp; 1\end{array}\right) = e

the isotropy subgroup consists of the identity element.

 

[Oxymoron]

Anyone know if I have done this correctly? Anyone?