Many groups have matrices as their elements. The operation is usually either matrix addition or matrix multiplication.

* Example.* Let G denote the set of all matrices with real entries. (Remember that " " means the matrices have 2 *rows* and 3
*columns*.) Here are some elements of G:

Show that G is a group under matrix addition.

If you add two matrices with real entries, you obtain another matrix with real entries:

That is, addition yields a binary operation on the set.

You should know from linear algebra that matrix addition is associative.

The identity element is the zero matrix:

The inverse of a matrix under this operation is the matrix obtained by negating the entries of the original matrix:

Notice that I *don't* get a group if I try to apply matrix
addition to the set of *all* matrices with real entries. This
does not define a binary operation on the set, because matrices of
different dimensions can't be added.

In general, the set of matrices with real entries --- or entries in , , , or for form a group under matrix addition.

As a special case, the matrices with real entries forms a group under matrix addition. This group is denoted . As you might guess, denotes the group of matrices with rational entries (and so on).

* Example.* Let G be the group of matrices with entries in under matrix addition.

(a) What is the order of G?

(b) Find the inverse of in G.

(a) A matrix has entries. Each entry can be any one of the 3 elements of . Therefore, there are elements.

(b)

Hence, the inverse is .

* Example.* Let

In words, G is the set of matrices with real entries having zeros in the first column.

Show that G is a group under matrix addition.

First,

That is, if you add two elements of G, you get another element of G. Hence, matrix addition gives a binary operation on the set G.

From linear algebra, you know that matrix addition is associative.

The zero matrix is the identity under matrix addition; it's an element of G, since its first column is all-zero.

Finally, the additive inverse of an element is , which is also an element of G. Thus, every element of G has an inverse.

All the axioms for a group have been verified, so G is a group under matrix addition.

* Example.* Consider the set of matrices

(Notice that x must be *nonnegative*). Is G a group under
matrix multiplication?

First, suppose that , . Then

Now , so . Therefore, matrix multiplication gives a binary operation on G.

I'll take for granted the fact that matrix multiplication is associative.

The identity for multiplication is , and this is an element of G.

However, not all elements of G have inverses. To give a specific counterexample, suppose that for

Then

Hence, and . This contradicts . Hence, the element of G does not have an inverse.

Therefore, G is *not* a group under matrix multiplication.

* Example.* denotes the
set of invertible matrices with real entries, the
* general linear group*. Show that is a group under matrix multiplication.

First, if , I know from linear algebra that and . Then

Hence, so . This proves that is closed under matrix multiplication.

I will take it as known from linear algebra that matrix multiplication is associative.

The identity matrix is the matrix

It is the identity for matrix multiplication: for all .

Finally, since *is* the set of
*invertible* matrices, every element of has an inverse under matrix multiplication.

* Example.* denotes
the set of invertible matrices with entries in . The operation is matrix multiplication --- but note
that all the arithmetic is performed in .

For example,

The proof that is a group under
matrix multiplication follows the proof in the last example. (In
fact, the same thing works with any * commutative
ring* in place of or ; commutative
rings will be discussed later.)

(a) What is the order of ?

(b) Find the inverse of .

(a) Notice that

Therefore, has order 3 in .

(b) Recall the formula for the inverse of a matrix:

The formula works in this situation, but you have to interpret the
fraction as a *multiplicative inverse*:

Thus,

On the other hand, the matrix is not an element of . It has determinant , so it's not invertible.

* Example.* Show that the following set is a
subgroup of :

Suppose . Then

Hence, .

Since , the identity matrix is in .

Finally, if , then implies that

But , so , and hence .

Therefore, is a subgroup of .

Copyright 2018 by Bruce Ikenaga