The General Linear Group, GLn(R)

The General Linear Group, GLn(R)

Recall from the Groups page that a group a set $G$ with a binary operation $\cdot: G \times G \to G$ where:

  • 1) For all $a, b, c \in G$ we have that $(a \cdot b) \cdot c = a \cdot(b \cdot c)$ (Associativity of $\cdot$).
  • 2) There exists an element $e \in G$ such that $a \cdot e = a$ and $e \cdot a = a$ (The existence of an identity for $\cdot$).
  • 3) For all $a \in G$ there exists a $a^{-1} \in G$ such that $a \cdot a^{-1} = e$ and $a^{-1} \cdot a = e$ (The existence of inverses for each element in $G$).

We will now look at the group of invertible $n \times n$ matrices with real entries under matrix multiplication $\cdot$ which is often called the **General Linear Group $GL_n(\mathbb{R})$. Let $A, B, C \in GL_n(\mathbb{R})$.

Consider the product $A \cdot B$. Since $A$ and $B$ are invertible $n \times n$ matrices, we know from linear algebra that $A \cdot B$ will be an invertible $n \times n$ matrix (whose inverse is $(A \cdot B)^{-1} = B^{-1} \cdot A^{-1}$) and so $(A \cdot B) \in GL_n(\mathbb{R})$ so $GL_n(\mathbb{R})$ is closed under $\cdot$

We also already know that matrix multiplication is associative from linear algebra, and so $A \cdot (B \cdot C) = (A \cdot B) \cdot C$.

The identity for $\cdot$ is the $n \times n$ identity matrix $I_{n}$ whose main diagonal entries are all $1$s and every other entry is a $0$, i.e.:

\begin{align} \quad I_n = \begin{bmatrix} 1 & 0 & \cdots & 0\\ 0 & 1 & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & 1 \end{bmatrix}_{n\times n} \end{align}

We know from linear algebra that for any $n \times n$ square diagonal matrix $A$ that $\det A = \prod_{i=1}^{n} a_{ii}$. So $\det I_{n} = \prod_{i=1}^{n} a_{ii} = \prod_{i=1}^{n} 1 = 1 \neq 0$ and so indeed $I_n$ is invertible, so $I_n \in GL_n(\mathbb{R})$.

For each element $A \in GL_n(\mathbb{R})$ we denote the inverse under $\cdot$ to be matrix inverse of $A$ which we denote $A^{-1} \in GL_n(\mathbb{R})$ such that $A \cdot A^{-1} = I_n$ and $A^{-1} \cdot A = I_n$.

Hence, $(GL_n(\mathbb{R}), \cdot)$ is a group.

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