Hw5 Problem 1

### Problem 1

Determine if the following sets of invertible $n\times n$ matrices with real entries defines a subgroup of
$GL(n , \mathbb{R})$

A: $\{A \in GL(n, \mathbb{R}) \ | \ \det A = \pm 1\}$
B: For $n=2$, the set of upper triangular invertible matrices. (If $A = [a_{ij}]$, $a_{ij} = 0$ for $i>j$)

Solution

A: $\{A \in GL(n, \mathbb{R}) \ | \ \det A = \pm 1\}$

Let $S=\{A \in GL(n, \mathbb{R}) \ | \ \det A = \pm 1\}$
Clearly, $S \subseteq GL(n, \mathbb{R})$. Thus, it is sufficient to show that $S$ is closed under matrix multiplication, contains $I_n$, the identity of $GL(n, \mathbb{R})$, and has inverses for all of its elements.
Closed

Let $A,B \in S$
Then by properties of determinants $\det (AB) = \det(A) \cdot \det(B)$
if $\det(A), \det(B) = 1$, then $\det(AB) = 1 \cdot 1 = 1$
if $\det(A), \det(B) = -1$, then $\det(AB) = -1 \cdot -1 = 1$
if $\det(A) = 1, \det(B) = -1$, then $\det(AB) = 1 \cdot -1 = -1$
and if $det(A) = -1, \det(B) = 1$, then $\det(AB) = -1 \cdot 1 = -1$
so, $S$ is closed under multiplication

Identity

$\det(I_{n}) = 1$ so the identity element is in $S$

Inverse

Let $A \in S$, then $\det(I_{n}) = \det(AA^{-1}) = \det(A) \cdot \det(A^{-1})$
If $\det(A) = 1$, then $1 = 1\det(A^{-1})$ then $det(A^{-1}) = 1$
If $\det(A) = -1$, then $1 = -1\det(A^{-1})$, then $\det(A^{-1}) = -1$
so, the inverse element is also in $S$, and therefore $S$ is a subgroup of $GL(n , \mathbb{R})$.

B: For $n=2$, the set of upper triangular invertible matrices. (If $A = [a_{ij}]$, $a_{ij} = 0$ for $i>j$)
For notational convenience, let $S$ denote the set of upper triangular invertible matrices.

Closed
Let $\begin{bmatrix} a_{11} & a_{12} \\ 0 & a_{22} \end{bmatrix}, \begin{bmatrix} b_{11} & b_{12} \\ 0 & b_{22} \end{bmatrix} \in S$.
Then $\begin{bmatrix} a_{11} & a_{12} \\ 0 & a_{22} \end{bmatrix}$ $\cdot$ $\begin{bmatrix} b_{11} & b_{12} \\ 0 & b_{22} \end{bmatrix}$ $=$ $\begin{bmatrix} a_{11}b_{11} & a_{11}b_{12}+a_{12}b_{22} \\ 0 & a_{22}b_{22} \end{bmatrix}$ which is upper triangular, by definition of upper triangular.
Furthermore, $\begin{bmatrix} a_{11}b_{11} & a_{11}b_{12}+a_{12}b_{22} \\ 0 & a_{22}b_{22} \end{bmatrix}$ is invertible, since the product of any two invertible matrices is invertible.
Thus, $\begin{bmatrix} a_{11}b_{11} & a_{11}b_{12}+a_{12}b_{22} \\ 0 & a_{22}b_{22} \end{bmatrix} \in S$.
Therefore, $S$ is closed under matrix multiplication.

Identity
Consider the identity matrix in $GL(2, \mathbb{R})$ is $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
This matrix is invertible and upper triangular, and therefore in $S$.
Hence, $S$ contains the identity element under the operation of matrix multiplication.

Inverse

By definition of invertible matrix, every element of $S$ must possess an inverse in $GL(2, \mathbb{R})$.
Observe that the matrix $\frac{1 }{(a_{11}a_{22})}$ $\cdot$ $\begin{bmatrix} a_{22} & -a_{12} \\ 0 & a_{11} \end{bmatrix}$ serves as the inverse of some matrix $\begin{bmatrix} a_{11} & a_{12} \\ 0 & a_{22} \end{bmatrix}$ in $S$, and is upper triangular.
Furthermore, since $\frac{1 }{(a_{11}a_{22})}$ $\cdot$ $\begin{bmatrix} a_{22} & -a_{12} \\ 0 & a_{11} \end{bmatrix}$ is the inverse of $\begin{bmatrix} a_{11} & a_{12} \\ 0 & a_{22} \end{bmatrix}$ in $GL(2, \mathbb{R})$, it is clear that $\frac{1 }{(a_{11}a_{22})}$ $\cdot$ $\begin{bmatrix} a_{22} & -a_{12} \\ 0 & a_{11} \end{bmatrix}$ is invertible, by definition of matrix inverse.
Therefore, all elements of $S$ have inverses in $S$.

All conditions hold, so $S$ is a subgroup of $GL(2, \mathbb{R})$.