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Before we study matrix groups, we must recall some basic facts from linear algebra. One of the most fundamental ideas of linear algebra is that of a linear transformation. A or $T : {\mathbb R}^n \rightarrow {\mathbb R}^m$ is a map that preserves vector addition and scalar multiplication; that is, for vectors ${\mathbf x}$ and ${\mathbf y}$ in ${\mathbb R}^n$ and a scalar $\alpha \in {\mathbb R}\text{,}$
An $m \times n$ matrix with entries in ${\mathbb R}$ represents a linear transformation from ${\mathbb R}^n$ to ${\mathbb R}^m\text{.}$ If we write vectors ${\mathbf x} = (x_1, \ldots, x_n)^{\rm t}$ and ${\mathbf y} = (y_1, \ldots, y_n)^{\rm t}$ in ${\mathbb R}^n$ as column matrices, then an $m \times n$ matrix
maps the vectors to ${\mathbb R}^m$ linearly by matrix multiplication. Observe that if $\alpha$ is a real number,
where
We will often abbreviate the matrix $A$ by writing $(a_{ij})\text{.}$
Conversely, if $T : {\mathbb R}^n \rightarrow {\mathbb R}^m$ is a linear map, we can associate a matrix $A$ with $T$ by considering what $T$ does to the vectors
We can write any vector ${\mathbf x} = (x_1, \ldots, x_n)^{\rm t}$ as
Consequently, if
then
If we let $T : {\mathbb R}^2 \rightarrow {\mathbb R}^2$ be the map given by
the axioms that $T$ must satisfy to be a linear transformation are easily verified. The column vectors $T {\mathbf e}_1 = (2, -4)^{\rm t}$ and $T {\mathbf e}_2 = (5,3)^{\rm t}$ tell us that $T$ is given by the matrix
Since we are interested in groups of matrices, we need to know which matrices have multiplicative inverses. Recall that an $n \times n$ matrix $A$ is exactly when there exists another matrix $A^{-1}$ such that $A A^{-1} = A^{-1} A = I\text{,}$ where
is the $n \times n$ identity matrix. From linear algebra we know that $A$ is invertible if and only if the determinant of $A$ is nonzero. Sometimes an invertible matrix is said to be .
If $A$ is the matrix
then the inverse of $A$ is
We are guaranteed that $A^{-1}$ exists, since $\det(A) = 2 \cdot 3 - 5 \cdot 1 = 1$ is nonzero.
Some other facts about determinants will also prove useful in the course of this chapter. Let $A$ and $B$ be $n \times n$ matrices. From linear algebra we have the following properties of determinants.
The determinant is a homomorphism into the multiplicative group of real numbers; that is, $\det( A B) = (\det A )(\det B)\text{.}$
If $A$ is an invertible matrix, then $\det(A^{-1}) = 1 / \det A\text{.}$
If we define the transpose of a matrix $A = (a_{ij})$ to be $A^{\rm t} = (a_{ji})\text{,}$ then $\det(A^{\rm t}) = \det A\text{.}$
Let $T$ be the linear transformation associated with an $n \times n$ matrix $A\text{.}$ Then $T$ multiplies volumes by a factor of $|\det A|\text{.}$ In the case of ${\mathbb R}^2\text{,}$ this means that $T$ multiplies areas by $|\det A|\text{.}$
Linear maps, matrices, and determinants are covered in any elementary linear algebra text; however, if you have not had a course in linear algebra, it is a straightforward process to verify these properties directly for $2 \times 2$ matrices, the case with which we are most concerned.
The set of all $n \times n$ invertible matrices forms a group called the . We will denote this group by $GL_n({\mathbb R})\text{.}$ The general linear group has several important subgroups. The multiplicative properties of the determinant imply that the set of matrices with determinant one is a subgroup of the general linear group. Stated another way, suppose that $\det(A) =1$ and $\det(B) = 1\text{.}$ Then $\det(AB) = \det(A) \det (B) = 1$ and $\det(A^{-1}) = 1 / \det A = 1\text{.}$ This subgroup is called the and is denoted by $SL_n({\mathbb R})\text{.}$
Given a $2 \times 2$ matrix
the determinant of $A$ is $ad-bc\text{.}$ The group $GL_2({\mathbb R})$ consists of those matrices in which $ad-bc \neq 0\text{.}$ The inverse of $A$ is
If $A$ is in $SL_2({\mathbb R})\text{,}$ then
Geometrically, $SL_2({\mathbb R})$ is the group that preserves the areas of parallelograms. Let
be in $SL_2({\mathbb R})\text{.}$ In FigureĀ 12.4, the unit square corresponding to the vectors ${\mathbf x} = (1,0)^{\rm t}$ and ${\mathbf y} = (0,1)^{\rm t}$ is taken by $A$ to the parallelogram with sides $(1,0)^{\rm t}$ and $(1, 1)^{\rm t}\text{;}$ that is, $A {\mathbf x} = (1,0)^{\rm t}$ and $A {\mathbf y} = (1, 1)^{\rm t}\text{.}$ Notice that these two parallelograms have the same area.
Another subgroup of $GL_n({\mathbb R})$ is the orthogonal group. A matrix $A$ is if $A^{-1} = A^{\rm t}\text{.}$ The consists of the set of all orthogonal matrices. We write $O(n)$ for the $n \times n$ orthogonal group. We leave as an exercise the proof that $O(n)$ is a subgroup of $GL_n( {\mathbb R})\text{.}$
The following matrices are orthogonal:
There is a more geometric way of viewing the group $O(n)\text{.}$ The orthogonal matrices are exactly those matrices that preserve the length of vectors. We can define the length of a vector using the , or , of two vectors. The Euclidean inner product of two vectors ${\mathbf x}=(x_1, \ldots, x_n)^{\rm t}$ and ${\mathbf y}=(y_1, \ldots, y_n)^{\rm t}$ is
We define the length of a vector ${\mathbf x}=(x_1, \ldots, x_n)^{\rm t}$ to be
Associated with the notion of the length of a vector is the idea of the distance between two vectors. We define the between two vectors ${\mathbf x}$ and ${\mathbf y}$ to be $\| {\mathbf x}-{\mathbf y} \|\text{.}$ We leave as an exercise the proof of the following proposition about the properties of Euclidean inner products.
Let ${\mathbf x}\text{,}$ ${\mathbf y}\text{,}$ and ${\mathbf w}$ be vectors in ${\mathbb R}^n$ and $\alpha \in {\mathbb R}\text{.}$ Then
$\langle {\mathbf x}, {\mathbf y} \rangle = \langle {\mathbf y}, {\mathbf x} \rangle\text{.}$
$\langle {\mathbf x}, {\mathbf y} + {\mathbf w} \rangle = \langle {\mathbf x}, {\mathbf y} \rangle + \langle {\mathbf x}, {\mathbf w} \rangle\text{.}$
$\langle \alpha {\mathbf x}, {\mathbf y} \rangle = \langle {\mathbf x}, \alpha {\mathbf y} \rangle = \alpha \langle {\mathbf x}, {\mathbf y} \rangle\text{.}$
$\langle {\mathbf x}, {\mathbf x} \rangle \geq 0$ with equality exactly when ${\mathbf x} = 0\text{.}$
If $\langle {\mathbf x}, {\mathbf y} \rangle = 0$ for all ${\mathbf x}$ in ${\mathbb R}^n\text{,}$ then ${\mathbf y} = 0\text{.}$
The vector ${\mathbf x} =(3,4)^{\rm t}$ has length $\sqrt{3^2 + 4^2} = 5\text{.}$ We can also see that the orthogonal matrix
preserves the length of this vector. The vector $A{\mathbf x} = (-7/5,24/5)^{\rm t}$ also has length 5.
Since $\det(A A^{\rm t}) = \det(I) = 1$ and $\det(A) = \det( A^{\rm t} )\text{,}$ the determinant of any orthogonal matrix is either 1 or $-1\text{.}$ Consider the column vectors
of the orthogonal matrix $A= (a_{ij})\text{.}$ Since $AA^{\rm t} = I\text{,}$ $\langle {\mathbf a}_r, {\mathbf a}_s \rangle = \delta_{rs}\text{,}$ where
is the Kronecker delta. Accordingly, column vectors of an orthogonal matrix all have length 1; and the Euclidean inner product of distinct column vectors is zero. Any set of vectors satisfying these properties is called an . Conversely, given an $n \times n$ matrix $A$ whose columns form an orthonormal set, it follows that $A^{-1} = A^{\rm t}\text{.}$
We say that a matrix $A$ is , , or when $\| T{\mathbf x}- T{\mathbf y} \| =\| {\mathbf x}- {\mathbf y} \|\text{,}$ $\| T{\mathbf x} \| =\| {\mathbf x} \|\text{,}$ or $\langle T{\mathbf x}, T{\mathbf y} \rangle = \langle {\mathbf x},{\mathbf y} \rangle\text{,}$ respectively. The following theorem, which characterizes the orthogonal group, says that these notions are the same.
Let $A$ be an $n \times n$ matrix. The following statements are equivalent.
The columns of the matrix $A$ form an orthonormal set.
$A^{-1} = A^{\rm t}\text{.}$
For vectors ${\mathbf x}$ and ${\mathbf y}\text{,}$ $\langle A{\mathbf x}, A {\mathbf y} \rangle = \langle {\mathbf x}, {\mathbf y} \rangle\text{.}$
For vectors ${\mathbf x}$ and ${\mathbf y}\text{,}$ $\| A{\mathbf x}- A{\mathbf y} \| = \| {\mathbf x}- {\mathbf y} \|\text{.}$
For any vector ${\mathbf x}\text{,}$ $\| A{\mathbf x} \| = \| {\mathbf x}\|\text{.}$
We have already shown (1) and (2) to be equivalent.
$(2) \Rightarrow (3)\text{.}$
$(3) \Rightarrow (2)\text{.}$ Since
we know that $\langle {\mathbf x}, (A^{\rm t} A - I){\mathbf x} \rangle = 0$ for all ${\mathbf x}\text{.}$ Therefore, $A^{\rm t} A -I = 0$ or $A^{-1} = A^{\rm t}\text{.}$
$(3) \Rightarrow (4)\text{.}$ If $A$ is inner product-preserving, then $A$ is distance-preserving, since
$(4) \Rightarrow (5)\text{.}$ If $A$ is distance-preserving, then $A$ is length-preserving. Letting ${\mathbf y} = 0\text{,}$ we have
$(5) \Rightarrow (3)\text{.}$ We use the following identity to show that length-preserving implies inner product-preserving:
Observe that
Let us examine the orthogonal group on ${\mathbb R}^2$ a bit more closely. An element $T \in O(2)$ is determined by its action on ${\mathbf e}_1 = (1, 0)^{\rm t}$ and ${\mathbf e}_2 = (0, 1)^{\rm t}\text{.}$ If $T({\mathbf e}_1) = (a,b)^{\rm t}\text{,}$ then $a^2 + b^2 = 1$ and $T({\mathbf e}_2) = (-b, a)^{\rm t}\text{.}$ Hence, $T$ can be represented by
where $0 \leq \theta \lt 2 \pi\text{.}$ A matrix $T$ in $O(2)$ either reflects or rotates a vector in ${\mathbb R}^2$ (FigureĀ 12.9). A reflection about the horizontal axis is given by the matrix
whereas a rotation by an angle $\theta$ in a counterclockwise direction must come from a matrix of the form
A reflection about a line $\ell$ is simply a reflection about the horizontal axis followed by a rotation. If $\det A =-1\text{,}$ then $A$ gives a reflection.
Two of the other matrix or matrix-related groups that we will consider are the special orthogonal group and the group of Euclidean motions. The , $SO(n)\text{,}$ is just the intersection of $O(n)$ and $SL_n({\mathbb R})\text{;}$ that is, those elements in $O(n)$ with determinant one. The , $E(n)\text{,}$ can be written as ordered pairs $(A, {\mathbf x})\text{,}$ where $A$ is in $O(n)$ and ${\mathbf x}$ is in ${\mathbb R}^n\text{.}$ We define multiplication by
The identity of the group is $(I,{\mathbf 0})\text{;}$ the inverse of $(A, {\mathbf x})$ is $(A^{-1}, -A^{-1} {\mathbf x})\text{.}$ In ExerciseĀ 12.3.6, you are asked to check that $E(n)$ is indeed a group under this operation.
