The Set Of All 2x2 Invertible Matrices

Part 1-Define our group:

GL2($\mathbb{R}$) is the set of all 2x2 Invertible Matrices

Part 2-Show our group is a group:

We say GL2($\mathbb{R}$) is a group if…
1) matrix multiplication is associative on the set
2) matrix multiplication has an identity that can be found in the set
3) every element in the set has an inverse for matrix multiplication.

1) We have proved previously that function composition is associative, and since matrix multiplication is function composition on matrices we can say that matrix multiplication is associative.

2) We know that the 2X2 matrix identity is

$$\begin{bmatrix}1 & 0 \\[0.3em]0 & 1 \end{bmatrix}$$

Let's just check to be sure this is the identity and that it is in the set.

$$\begin{bmatrix}1 & 0 \\[0.3em]0 & 1 \end{bmatrix}\begin{bmatrix} a & b \\[0.3em] c & d \end{bmatrix}=\begin{bmatrix} a & b \\[0.3em] c & d \end{bmatrix}$$ for $$\begin{bmatrix} a & b \\[0.3em] c & d \end{bmatrix}$\in$GL2($\mathbb{R}$) arbitrary.

Similiarly,

$$\begin{bmatrix}a & b \\[0.3em]c & d \end{bmatrix}\begin{bmatrix} 1 & 0 \\[0.3em] 0 & 1 \end{bmatrix}=\begin{bmatrix} a & b \\[0.3em] c & d \end{bmatrix}$$ for $$\begin{bmatrix} a & b \\[0.3em] c & d \end{bmatrix}$\in$ GL2($\mathbb{R}$) arbitrary.

If $det$\begin{bmatrix}1 & 0 \\[0.3em]0 & 1 \end{bmatrix}$\neq0$
$det$\begin{bmatrix}1 & 0 \\[0.3em]0 & 1 \end{bmatrix}$=2$ therefore it is in invertible and is and element of GL2($\mathbb{R}$)

3) Let $A=\begin{bmatrix}a & b \$0.3em] c & d \end{bmatrix}\inGL2(\mathbb{R}) arbitrary. The inverse of A=\frac{1}{det(A)}\begin{bmatrix}d & -b \\[0.3em] -c & a \end{bmatrix}=\frac{1}{ad-bc}\begin{bmatrix}d & -b \\[0.3em] -c & a \end{bmatrix} A-1=\begin{bmatrix}\frac{d}{ad-bc} & \frac{-b}{ad-bc} \\[0.3em] \frac{-c}{ad-bc} & \frac{a}{ad-bc} \end{bmatrix}. As we stated above det(A)\neq 0, therefore this is true. Therefore GL2(\mathbb{R}) is indeed a group. Part 3-Show interesting subgroups of our group: Let G=GL2(\mathbb{R}) H= {A | det (A)=1} 1. Let A\in H and B\in H be arbitrary. Therefore det (A) =1 and det (B)=1. Therefore det (A)\cdot det (B)= 1\cdot1 = 1, so det (AB)=1. This demonstrates closure as needed. 2. We know I = \[\begin{bmatrix}1 & 0 \\[0.3em]0 & 1 \end{bmatrix}$$ $\in G$

Since the determinate is determined by the entries ad-bc, we can see that the determinate of the identity is $(1\cdot1) -(0\cdot0) = 1$. Since det(I)=1 I$\in$H.

3. Since the det $A$-1= $\frac{1}{det(A)}$ and the det (A) = 1 (as demonstrated in part 2 earlier), then the det $A$-1= $\frac{1}{1}$ = 1 therefore A-1 $\in H$

Therefore H is a subgroup of G.

Now $H=${$A | AA$T$=I$}
Let $A\in H$ and $B \in H$ be arbitrary.
1. $AB(AB)$T$=ABB$T$A$T=$AIA$T$=I.$ Therefore $AB\in H$ and so $H$ is closed under matrix multiplication.
2. We know$I=\begin{bmatrix} 1 & 0 \$0.3em] 0 & 1 \end{bmatrix}$ \in G.$
$I$T$=I,$ therefore $II$T$=II=I$. Therefore $I\in H$.
3. Let $M\in H$ be arbitrary. Therefore $MM$T$=1.$ By definition of Inverse we can say that $M$T$=M$-1. Therefore $M$-1$\in H$.
Therefore $H$ is a subgroup of $G$.

page revision: 26, last edited: 02 Nov 2010 22:26