The Set of Units in Mnn

Recall from the Units (Multiplicatively Invertible Elements) in Rings page that if $(R, +, *)$ is a ring then an element $a \in R$ is said to be unit or a multiplicatively invertible element if there exists an element $b \in R$ such that $a * b = 1$ and $b * a = 1$ where we denote $b = a^{-1}$.

Also recall that the set of all $n \times n$ matrices with real coefficients we denote by $M_{nn}$, that is:

(1)

\begin{align} \quad M_{nn} = \left \{ \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{bmatrix} : a_{ij} \in \mathbb{R} \: \mathrm{for} \: i, j \in \{ 1, 2, ..., n \} \right \} \end{align}

We will examine further the multiplicatively invertible elements of the ring $(M_{nn}, +, *)$ where $+$ is defined to be standard matrix addition and $*$ is defined to be standard square matrix multiplication.

Recall from linear algebra that if $A \in M_{nn}$ and $\det A \neq 0$ then there exists a matrix $B \in M_{nn}$ such that $AB = BA = I$ where $I$ is the $n \times n$ identity matrix. We've already seen that for standard matrix multiplication on $M_{nn}$ that the $n \times n$ matrix $I = \begin{bmatrix} 1 & 0 & \cdots & 0\\ 0 & 1 & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & 1 \end{bmatrix}$ is the identity for $*$.

Therefore, if $A$ is an $n \times n$ matrix where $\det A \neq 0$ and if $C_{ij}$ represents the cofactors of $A$ for each $i, j \in \{1, 2, ..., n \}$ and the adjoint of $A$ is denoted as $\mathrm{adj} (A) = \begin{bmatrix} C_{11} & C_{12} & \cdots & C_{1n}\\ C_{21} & C_{22} & \cdots & C_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ C_{n1} & C_{n2} & \cdots & C_{nn} \end{bmatrix}$ and we can determine a well-defined inverse with respect to $*$ given by the formula:

(2)

\begin{align} \quad A^{-1} = \frac{1}{\det A} \mathrm{adj} (A) \end{align}

So $A^{-1}$ for an $n \times n$ matrix $A$ where $\det A \neq 0$ has the property that $AA^{-1} = A^{-1}A = I$. Therefore, the subset of $n \times n$ invertible matrices of $M_{nn}$ is precisely all of the multiplicatively invertible elements in $M_{nn}$.

For the set of $2 \times 2$ matrices we can easily verify this. If $A = \begin{bmatrix} a & b\\ c & d \end{bmatrix}$ then we define $A^{-1} = \begin{bmatrix} \frac{d}{ad - bc} & \frac{-b}{ad - bc} \\ \frac{-c}{ad - bc} & \frac{a}{ad - bc} \end{bmatrix}$ which is well-defined if $\det A = ad - bc \neq 0$. Therefore:

(3)

\begin{align} \quad A * A^{-1} = \begin{bmatrix} a & b\\ c & d \end{bmatrix} \begin{bmatrix} \frac{d}{ad - bc} & \frac{-b}{ad - bc} \\ \frac{-c}{ad - bc} & \frac{a}{ad - bc} \end{bmatrix} = \begin{bmatrix} \frac{ad - bc}{ad - bc} & \frac{-ab + ab}{ad - bc} \\ \frac{cd - cd}{ad - bc} & \frac{-bc + ad}{ad - bc} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0& 1 \end{bmatrix} = I \end{align}

And similarly:

(4)

\begin{align} \quad A^{-1} * A = \begin{bmatrix} \frac{d}{ad - bc} & \frac{-b}{ad - bc} \\ \frac{-c}{ad - bc} & \frac{a}{ad - bc} \end{bmatrix} \begin{bmatrix} a & b\\ c & d \end{bmatrix} = \begin{bmatrix} \frac{ad - bc}{ad - bc} & \frac{bd - bd}{ad - bc} \\ \frac{-ac + ac}{ad - bc} & \frac{-bc + ad}{ad - bc} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0& 1 \end{bmatrix} = I \end{align}

Therefore the multiplicatively invertible elements in $M_{22}$ are the set of $2 \times 2$ matrices whose determinants are nonzero.