Square Matrices with +1 or -1 Determinant under Multiplication forms Group

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $n \in \Z_{> 0}$ be a strictly positive integer.

Let $S$ be the set of square matrices of order $n$ of real numbers whose determinant is either $1$ or $-1$.

Let $\struct {S, \times}$ denote the algebraic structure formed by $S$ whose operation is (conventional) matrix multiplication.


Then $\struct {S, \times}$ is a group.


Proof

Taking the group axioms in turn:

$G \, 0$: Closure

Let $\mathbf A, \mathbf B \in S$.

By definition of matrix product, $\mathbf {A B}$ is a square matrix of order $n$.


From Determinant of Matrix Product:

$\det \mathbf A \det \mathbf B = \map \det {\mathbf {A B} }$

Hence the determinant of $\mathbf {A B}$ is either $1$ or $-1$.

Thus $\mathbf {A B} \in S$ and so $\struct {S, \times}$ is closed.

$\Box$


$G \, 1$: Associativity

We have that Matrix Multiplication is Associative.

Thus $\times$ is associative on $\struct {S, \times}$.

$\Box$


$G \, 2$: Identity

From Unit Matrix is Unity of Ring of Square Matrices, the unit matrix $\mathbf I$ serves as the identity element of $\struct {S, \times}$.

$\Box$


$G \, 3$: Inverses

Because the determinants of the elements of $S$ are $1$ or $-1$, they are by definition invertible.

We have that $\mathbf I$ is the identity element of $\struct {\R, \circ}$.


From the definition of invertible matrix, the inverse of any invertible matrix $\mathbf A$ is $\mathbf A^{-1}$.

$\Box$


All the group axioms are thus seen to be fulfilled, and so $\struct {S, \times}$ is a group.

$\blacksquare$


Sources