The Group of Invertible n x n Matrices

# The Group of Invertible n x n Matrices

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

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

We will now look at the group of invertible $n \times n$ matrices $M_{nn}^{-1}$ under matrix multiplication $*$. Let $A, B, C \in M_{nn}^{-1}$.

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

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

The identity for $*$ 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.:

(1)
\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 $I_n \in M_{nn}^{-1}$.

For each element $A \in M_{nn}^{-1}$ we denote the inverse under $*$ to be matrix inverse of $A$ which we denote $A^{-1} \in M_{nn}^{-1}$ such that $A * A^{-1} = I_n$ and $A^{-1} * A = I_n$.

Hence, $(M_{nn}^{-1}, *)$ is a group.