Is $GL_{n}^{+}(\mathbb R)$ a subgroup of $GL_{n}(\mathbb R)$?

Question

Is $GL_{n}^{+}(\mathbb R)$ a subgroup of $GL_{n}(\mathbb R)$?

Let $A=\begin{bmatrix}3 & 4\\5 & 7\end{bmatrix}$ and $B=\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}$

$\det(A) = 1$ and $\det(B)=1$.

$AB=\begin{bmatrix}3 & 4\\5 & 7\end{bmatrix}$$\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix} = \begin{bmatrix}3 & 7\\5 & 12\end{bmatrix}$.

$BA=\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}$$\begin{bmatrix}3 & 4\\5 & 7\end{bmatrix} = \begin{bmatrix}8 & 5\\5 & 1\end{bmatrix}$.

Since $AB \ne BA$, $GL_{n}^{+}(\mathbb R)$ is not a subgroup of $GL_{n}(\mathbb R)$.

Answer 1

HINT:

To check whether $GL_n^{+}(\mathbb{R})$ is a subgroup of $GL_n(\mathbb{R})$, prove the following:

1. $I \in GL_n^{+}(\mathbb{R})$

2. $A,B \in GL_n^{+}(\mathbb{R}) \implies AB^{-1} \in GL_n^{+}(\mathbb{R})$