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Section12.1Matrix Groups

¶

SubsectionSome Facts from Linear Algebra

¶

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 $\alpha \in {\mathbb R}\text{,}$

$$\begin{align*} T({\mathbf x}+{\mathbf y}) & = T({\mathbf x}) + T({\mathbf y})\\ T(\alpha {\mathbf y}) & = \alpha T({\mathbf y}). \end{align*}$$

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

$$\begin{equation*} A = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{pmatrix} \end{equation*}$$

maps the vectors to ${\mathbb R}^m$ linearly by matrix multiplication. Observe that if $\alpha$ is a real number,

$$\begin{equation*} A({\mathbf x} + {\mathbf y} ) = A {\mathbf x }+ A {\mathbf y} \qquad \text{and} \qquad \alpha A {\mathbf x} = A ( \alpha {\mathbf x}), \end{equation*}$$

where

$$\begin{equation*} {\mathbf x} = \begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix}. \end{equation*}$$

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

$$\begin{align*} {\mathbf e}_1 & = (1, 0, \ldots, 0)^{\rm t}\\ {\mathbf e}_2 & = (0, 1, \ldots, 0)^{\rm t}\\ & \vdots & \\ {\mathbf e}_n & = (0, 0, \ldots, 1)^{\rm t}. \end{align*}$$

We can write any vector ${\mathbf x} = (x_1, \ldots, x_n)^{\rm t}$ as

$$\begin{equation*} x_1 {\mathbf e}_1 + x_2 {\mathbf e}_2 + \cdots + x_n {\mathbf e}_n. \end{equation*}$$

Consequently, if

$$\begin{align*} T({\mathbf e}_1) & = (a_{11}, a_{21}, \ldots, a_{m1})^{\rm t},\\ T({\mathbf e}_2) & = (a_{12}, a_{22}, \ldots, a_{m2})^{\rm t},\\ & \vdots & \\ T({\mathbf e}_n) & = (a_{1n}, a_{2n}, \ldots, a_{mn})^{\rm t}, \end{align*}$$

then

$$\begin{align*} T({\mathbf x} ) & = T(x_1 {\mathbf e}_1 + x_2 {\mathbf e}_2 + \cdots + x_n {\mathbf e}_n)\\ & = x_1 T({\mathbf e}_1) + x_2 T({\mathbf e}_2) + \cdots + x_n T({\mathbf e}_n)\\ & = \left( \sum_{k=1}^{n} a_{1k} x_k, \ldots, \sum_{k=1}^{n} a_{mk} x_k \right)^{\rm t}\\ & = A {\mathbf x}. \end{align*}$$

Example12.1

If we let $T : {\mathbb R}^2 \rightarrow {\mathbb R}^2$ be the map given by

$$\begin{equation*} T(x_1, x_2) = (2 x_1 + 5 x_2, - 4 x_1 + 3 x_2), \end{equation*}$$

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

$$\begin{equation*} A = \begin{pmatrix} 2 & 5 \\ -4 & 3 \end{pmatrix}. \end{equation*}$$

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

$$\begin{equation*} I = \begin{pmatrix} 1 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \end{pmatrix} \end{equation*}$$

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 .

Example12.2

If $A$ is the matrix

$$\begin{equation*} \begin{pmatrix} 2 & 1 \\ 5 & 3 \end{pmatrix}, \end{equation*}$$

then the inverse of $A$ is

$$\begin{equation*} A^{-1} = \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix}. \end{equation*}$$

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.

SubsectionThe General and Special Linear Groups

¶

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 $SL_n({\mathbb R})\text{.}$

Example12.3

Given a $2 \times 2$ matrix

$$\begin{equation*} A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}, \end{equation*}$$

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

$$\begin{equation*} A^{-1} = \frac{1}{ad-bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}. \end{equation*}$$

If $A$ is in $SL_2({\mathbb R})\text{,}$ then

$$\begin{equation*} A^{-1} = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}. \end{equation*}$$

Geometrically, $SL_2({\mathbb R})$ is the group that preserves the areas of parallelograms. Let

$$\begin{equation*} A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \end{equation*}$$

be in $SL_2({\mathbb R})\text{.}$ In Figure 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.

Figure12.4$SL_2(\mathbb R)$ acting on the unit square

SubsectionThe Orthogonal Group $O(n)$

¶

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 $GL_n( {\mathbb R})\text{.}$

Example12.5

The following matrices are orthogonal:

$$\begin{equation*} \begin{pmatrix} 3/5 & -4/5 \\ 4/5 & 3/5 \end{pmatrix}, \quad \begin{pmatrix} 1/2 & -\sqrt{3}/2 \\ \sqrt{3}/2 & 1/2 \end{pmatrix}, \quad \begin{pmatrix} -1/\sqrt{2} & 0 & 1/ \sqrt{2} \\ 1/\sqrt{6} & -2/\sqrt{6} & 1/\sqrt{6} \\ 1/ \sqrt{3} & 1/ \sqrt{3} & 1/ \sqrt{3} \end{pmatrix}. \end{equation*}$$

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

$$\begin{equation*} \langle {\mathbf x}, {\mathbf y} \rangle = {\mathbf x}^{\rm t} {\mathbf y} = (x_1, x_2, \ldots, x_n) \begin{pmatrix} y_1 \\ y_2 \\ \vdots \\ y_n \end{pmatrix} = x_1 y_1 + \cdots + x_n y_n. \end{equation*}$$

We define the length of a vector ${\mathbf x}=(x_1, \ldots, x_n)^{\rm t}$ to be

$$\begin{equation*} \| {\mathbf x} \| = \sqrt{\langle {\mathbf x}, {\mathbf x} \rangle} = \sqrt{x_1^2 + \cdots + x_n^2}. \end{equation*}$$

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.

Proposition12.6

Let ${\mathbf x}\text{,}$ ${\mathbf y}\text{,}$ and ${\mathbf w}$ be vectors in ${\mathbb R}^n$ and $\alpha \in {\mathbb R}\text{.}$ Then

1. $\langle {\mathbf x}, {\mathbf y} \rangle = \langle {\mathbf y}, {\mathbf x} \rangle\text{.}$

2. $\langle {\mathbf x}, {\mathbf y} + {\mathbf w} \rangle = \langle {\mathbf x}, {\mathbf y} \rangle + \langle {\mathbf x}, {\mathbf w} \rangle\text{.}$

3. $\langle \alpha {\mathbf x}, {\mathbf y} \rangle = \langle {\mathbf x}, \alpha {\mathbf y} \rangle = \alpha \langle {\mathbf x}, {\mathbf y} \rangle\text{.}$

4. $\langle {\mathbf x}, {\mathbf x} \rangle \geq 0$ with equality exactly when ${\mathbf x} = 0\text{.}$

5. If $\langle {\mathbf x}, {\mathbf y} \rangle = 0$ for all ${\mathbf x}$ in ${\mathbb R}^n\text{,}$ then ${\mathbf y} = 0\text{.}$

Example12.7

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

$$\begin{equation*} A= \begin{pmatrix} 3/5 & -4/5 \\ 4/5 & 3/5 \end{pmatrix} \end{equation*}$$

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

$$\begin{equation*} {\mathbf a}_j = \begin{pmatrix} a_{1j} \\ a_{2j} \\ \vdots \\ a_{nj} \end{pmatrix} \end{equation*}$$

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

$$\begin{equation*} \delta_{rs} = \left\{ \begin{array}{cc} 1 & r = s \\ 0 & r \neq s \end{array} \right. \end{equation*}$$

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.

Theorem12.8

Let $A$ be an $n \times n$ matrix. The following statements are equivalent.

1. The columns of the matrix $A$ form an orthonormal set.

2. $A^{-1} = A^{\rm t}\text{.}$

3. For vectors ${\mathbf x}$ and ${\mathbf y}\text{,}$ $\langle A{\mathbf x}, A {\mathbf y} \rangle = \langle {\mathbf x}, {\mathbf y} \rangle\text{.}$

4. For vectors ${\mathbf x}$ and ${\mathbf y}\text{,}$ $\| A{\mathbf x}- A{\mathbf y} \| = \| {\mathbf x}- {\mathbf y} \|\text{.}$

5. For any vector ${\mathbf x}\text{,}$ $\| A{\mathbf x} \| = \| {\mathbf x}\|\text{.}$

Proof

We have already shown (1) and (2) to be equivalent.

$(2) \Rightarrow (3)\text{.}$

$$\begin{align*} \langle A{\mathbf x}, A{\mathbf y} \rangle & = (A {\mathbf x})^{\rm t} A {\mathbf y}\\ & = {\mathbf x}^{\rm t} A^{\rm t} A {\mathbf y}\\ & = {\mathbf x}^{\rm t} {\mathbf y}\\ & = \langle {\mathbf x}, {\mathbf y} \rangle. \end{align*}$$

$(3) \Rightarrow (2)\text{.}$ Since

$$\begin{align*} \langle {\mathbf x}, {\mathbf x} \rangle & = \langle A{\mathbf x}, A{\mathbf x} \rangle\\ & = {\mathbf x}^{\rm t} A^{\rm t} A {\mathbf x}\\ & = \langle {\mathbf x}, A^{\rm t} A{\mathbf x} \rangle, \end{align*}$$

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

$$\begin{align*} \| A{\mathbf x} - A{\mathbf y} \|^2 & = \| A({\mathbf x} - {\mathbf y}) \|^2\\ & = \langle A({\mathbf x} - {\mathbf y}), A({\mathbf x} - {\mathbf y}) \rangle\\ & = \langle {\mathbf x} - {\mathbf y}, {\mathbf x} - {\mathbf y} \rangle\\ & = \| {\mathbf x} - {\mathbf y} \|^2. \end{align*}$$

$(4) \Rightarrow (5)\text{.}$ If $A$ is distance-preserving, then $A$ is length-preserving. Letting ${\mathbf y} = 0\text{,}$ we have

$$\begin{equation*} \| A{\mathbf x}\| = \| A{\mathbf x}- A{\mathbf y} \| = \| {\mathbf x}- {\mathbf y} \| = \| {\mathbf x} \|. \end{equation*}$$

$(5) \Rightarrow (3)\text{.}$ We use the following identity to show that length-preserving implies inner product-preserving:

$$\begin{equation*} \langle {\mathbf x}, {\mathbf y} \rangle = \frac{1}{2} \left[ \|{\mathbf x} +{\mathbf y}\|^2 - \|{\mathbf x}\|^2 - \|{\mathbf y}\|^2 \right]. \end{equation*}$$

Observe that

$$\begin{align*} \langle A {\mathbf x}, A {\mathbf y} \rangle & = \frac{1}{2} \left[ \|A {\mathbf x} + A {\mathbf y} \|^2 - \|A {\mathbf x} \|^2 - \|A {\mathbf y} \|^2 \right]\\ & = \frac{1}{2} \left[ \|A ( {\mathbf x} + {\mathbf y} ) \|^2 - \|A {\mathbf x} \|^2 - \|A {\mathbf y} \|^2 \right]\\ & = \frac{1}{2} \left[ \|{\mathbf x} + {\mathbf y}\|^2 - \|{\mathbf x}\|^2 - \|{\mathbf y}\|^2 \right]\\ & = \langle {\mathbf x}, {\mathbf y} \rangle. \end{align*}$$

Figure12.9$O(2)$ acting on $\mathbb R^2$

Example12.10

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

$$\begin{equation*} A = \begin{pmatrix} a & -b \\ b & a \end{pmatrix} = \begin{pmatrix} \cos \theta & - \sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}, \end{equation*}$$

where $0 \leq \theta \lt 2 \pi\text{.}$ A matrix $T$ in $O(2)$ either reflects or rotates a vector in ${\mathbb R}^2$ (Figure 12.9). A reflection about the horizontal axis is given by the matrix

$$\begin{equation*} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, \end{equation*}$$

whereas a rotation by an angle $\theta$ in a counterclockwise direction must come from a matrix of the form

$$\begin{equation*} \begin{pmatrix} \cos \theta & \sin \theta \\ \sin \theta & -\cos \theta \end{pmatrix}. \end{equation*}$$

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

$$\begin{equation*} (A, {\mathbf x}) (B, {\mathbf y}) = (AB, A {\mathbf y} +{\mathbf x}). \end{equation*}$$

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 Exercise 12.3.6, you are asked to check that $E(n)$ is indeed a group under this operation.

Figure12.11Translations in $\mathbb R^2$