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

• Oxymoron
In summary, the orbit of element A in the group G is described as O_A = \{gA\,|\, g\in GL(2,\mathbb{R})\}where g can be found by the following way gA = Ag where g \in GL(2,\mathbb{R})
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

$$A= \left(\begin{array}{cc} 2 & 1 \\ 0 & 3 \end{array}\right)$$

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

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 $$g \in GL(2,\mathbb{R})$$

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

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

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

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

And we can write

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

Therefore the orbit can be described as

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

So

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

how does this look?

Last edited:
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

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

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

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

the isotropy subgroup consists of the identity element.

Last edited:
Anyone know if I have done this correctly? Anyone?

## What is group action on a set?

Group action on a set is a mathematical concept that describes the interaction between elements of a group and elements of a set. It involves a group acting on a set by transforming its elements in a consistent and structured way.

## What is the purpose of group action on a set?

The purpose of group action on a set is to study the symmetries and transformations of a set by using the structure of a group. It allows for a deeper understanding of the properties and characteristics of both the group and the set.

## What are the key components of a group action on a set?

The key components of a group action on a set are the group, the set, and the action itself. The group is a collection of elements that satisfies certain algebraic properties, the set is a collection of elements that are acted upon, and the action is the transformation of elements in the set by elements of the group.

## How is group action on a set related to other mathematical concepts?

Group action on a set is closely related to other mathematical concepts such as symmetry, group theory, and symmetry groups. It is also used in various areas of mathematics, including geometry, topology, and abstract algebra.

## What are some real-world applications of group action on a set?

Group action on a set has many real-world applications, including in physics, chemistry, computer science, and cryptography. It is used to study the symmetries of physical systems, model chemical reactions, and design efficient algorithms for data processing and encryption.

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