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