Matrix Groups

Fields

We haven't really defined Field's yet, but this will be a lot like that definition.

Suppose $F$ is a Field like $Q$ , $R$ , $C$ , or $\frac{Z}{p Z}$ (where $p$ is prime). Then:

Mat (m, n; F) = {m \times n matrices with entries in F}

where $m$ is the number of rows, and $n$ is the number of columns. We have matrix addition (by just adding corresponding entries), have the zero matrix $0$ as the identity, and the inverse is just the matrix with negative entries. This satisfies the Group definition, so then this is a valid group. Notably, this is an infinite Abelian Group.

If you drop it down to square matrices:

Mat (n, n; F)

where we consider matrix multiplication, then:

It's still associative (annoying but possible to verify)
The identity matrix $I$ is the identity
However, not all matrices have an inverse!
So this is almost a group, but not quite. You can extend the idea to throw away the duds that don't have inverses, using the following notation:

(Mat (n, n; F))^{\times} := {invertible n \times n matrices with entries in F}

We call this the general linear group of degree $n$ . Instead of using the above we use:

G L (n, F)

You can drop the $F$ if the field is known from context.

Some considerations:

It's not an Abelian Group (for $n \geq 2$ ) since matrix multiplication isn't commutative (ex: $(A + B)^{2} = A^{2} + B^{2} + A B + B A \neq A^{2} + B^{2} + 2 A B$ )
It's infinite if $F$ is infinite.
Finite if $F$ is finite.

You may recall that the determinant is a function:

det : Mat (n, n; F) \to F

Where if the determinant is 0 then there's no inverse, while if it isn't then there is an inverse:

det (M) \neq 0 ⟺ M invertible

So you could rewrite:

G L (n, F) := {M \in Mat (n, n; F) | det (M) \neq 0}