12.1: Matrix Groups
Some 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 linear transformation or linear map \(T : {\mathbb R}^n \rightarrow {\mathbb R}^m\) 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 \({\mathbf x} = (x_1, \ldots, x_n)^{t}\) and \({\mathbf y} = (y_1, \ldots, y_n)^{t}\) 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 \({\mathbf x} = (x_1, \ldots, x_n)^{t}\) 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.
Solution
The column vectors \(T {\mathbf e}_1 = (2, -4)^{t}\) and \(T {\mathbf e}_2 = (5,3)^{t}\) 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 invertible 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 nonsingular .
If \(A\) is the matrix
then the inverse of \(A\) is
Solution
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 \(A^{t} = (a_{ji})\text{,}\) then \(\det(A^{t}) = \det A\text{.}\)
- 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 General and Special Linear Groups
The set of all \(n \times n\) invertible matrices forms a group called the general linear group . 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 special linear group 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
Solution
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 Figure \(12.4\), the unit square corresponding to the vectors \({\mathbf x} = (1,0)^{t}\) and \({\mathbf y} = (0,1)^{t}\) is taken by \(A\) to the parallelogram with sides \((1,0)^{t}\) and \((1, 1)^{t}\text{;}\) that is, \(A {\mathbf x} = (1,0)^{t}\) and \(A {\mathbf y} = (1, 1)^{t}\text{.}\) Notice that these two parallelograms have the same area.
\(Figure \text { } 12.4.\) \(SL_2(\mathbb R)\) acting on the unit square
The Orthogonal Group \(O(n)\)
Another subgroup of \(GL_n({\mathbb R})\) is the orthogonal group. A matrix \(A\) is orthogonal if \(A^{-1} = A^{t}\text{.}\) The orthogonal group 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:
Solution
\[ \begin{pmatrix} 3/5 & -4/5 \\ 4/5 & 3/5 \end{pmatrix}, \quad \begin{pmatrix} 1/2 & -\sqrt{3}/2 \\ \sqrt{3}/2 & 1/2 \end{pmatrix}, \quad \begin{pmatrix} -1/\sqrt{2} & 0 & 1/ \sqrt{2} \\ 1/\sqrt{6} & -2/\sqrt{6} & 1/\sqrt{6} \\ 1/ \sqrt{3} & 1/ \sqrt{3} & 1/ \sqrt{3} \end{pmatrix}\text{.} \nonumber \]
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 Euclidean inner product , or dot product , of two vectors. The Euclidean inner product of two vectors \({\mathbf x}=(x_1, \ldots, x_n)^{t}\) and \({\mathbf y}=(y_1, \ldots, y_n)^{t}\) is
We define the length of a vector \({\mathbf x}=(x_1, \ldots, x_n)^{t}\) to be
Associated with the notion of the length of a vector is the idea of the distance between two vectors. We define the distance 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 \({\mathbf x} =(3,4)^{t}\) 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 \(A{\mathbf x} = (-7/5,24/5)^{t}\) also has length 5.
Since \(\det(A A^{t}) = \det(I) = 1\) and \(\det(A) = \det( A^{t} )\text{,}\) 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 \(AA^{t} = I\text{,}\) \(\langle {\mathbf a}_r, {\mathbf a}_s \rangle = \delta_{rs}\text{,}\) 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 orthonormal set . Conversely, given an \(n \times n\) matrix \(A\) whose columns form an orthonormal set, it follows that \(A^{-1} = A^{t}\text{.}\)
We say that a matrix \(A\) is distance-preserving , length-preserving , or inner product-preserving when \(\| A{\mathbf x}- A{\mathbf y} \| =\| {\mathbf x}- {\mathbf y} \|\text{,}\) \(\| A{\mathbf x} \| =\| {\mathbf x} \|\text{,}\) or \(\langle A{\mathbf x}, A{\mathbf y} \rangle = \langle {\mathbf x},{\mathbf y} \rangle\text{,}\) respectively. The following theorem, which characterizes the orthogonal group, says that these notions are the same.
Theorem \(12.8\)
Let \(A\) be an \(n \times n\) matrix. The following statements are equivalent.
- The columns of the matrix \(A\) form an orthonormal set.
- \(A^{-1} = A^{t}\text{.}\)
- 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 \({\mathbf x}\text{,}\) \(\| A{\mathbf x} \| = \| {\mathbf x}\|\text{.}\)
- 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})^{t} A {\mathbf y}\\ & = {\mathbf x}^{t} A^{t} A {\mathbf y}\\ & = {\mathbf x}^{t} {\mathbf y}\\ & = \langle {\mathbf x}, {\mathbf y} \rangle\text{.} \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}^{t} A^{t} A {\mathbf x}\\ & = \langle {\mathbf x}, A^{t} A{\mathbf x} \rangle\text{,} \end{align*}we know that \(\langle {\mathbf x}, (A^{t} A - I){\mathbf x} \rangle = 0\) for all \({\mathbf x}\text{.}\) Therefore, \(A^{t} A -I = 0\) or \(A^{-1} = A^{t}\text{.}\)
\((3) \Rightarrow (4)\text{.}\) 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\text{.} \end{align*}\((4) \Rightarrow (5)\text{.}\) If \(A\) is distance-preserving, then \(A\) is length-preserving. Letting \({\mathbf y} = 0\text{,}\) we have
\[ \| A{\mathbf x}\| = \| A{\mathbf x}- A{\mathbf y} \| = \| {\mathbf x}- {\mathbf y} \| = \| {\mathbf x} \|\text{.} \nonumber \]\((5) \Rightarrow (3)\text{.}\) We use the following identity to show that length-preserving implies inner product-preserving:
\[ \langle {\mathbf x}, {\mathbf y} \rangle = \frac{1}{2} \left[ \|{\mathbf x} +{\mathbf y}\|^2 - \|{\mathbf x}\|^2 - \|{\mathbf y}\|^2 \right]\text{.} \nonumber \]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\text{.} \end{align*}
\(Figure \text { } 12.9.\) \(O(2)\) acting on \(\mathbb R^2\)
Let us examine the orthogonal group on \({\mathbb R}^2\) a bit more closely. An element \(A \in O(2)\) is determined by its action on \({\mathbf e}_1 = (1, 0)^{t}\) and \({\mathbf e}_2 = (0, 1)^{t}\text{.}\) If \(A{\mathbf e}_1 = (a,b)^{t}\text{,}\) then \(a^2 + b^2 = 1\text{,}\) since the length of a vector must be preserved when it is multiplied by \(A\text{.}\) Since multiplication of an element of \(O(2)\) preserves length and orthogonality, \(A{\mathbf e}_2 = \pm(-b, a)^{t}\text{.}\) If we choose \(A{\mathbf e}_2 = (-b, a)^{t}\text{,}\) then
where \(0 \leq \theta \lt 2 \pi\text{.}\) The matrix \(A\) rotates a vector in \(\mathbb R^2\) counterclockwise about the origin by an angle of \(\theta\) (Figure 12.9).
If we choose \(A{\mathbf e}_2 = (b, -a)^{t}\text{,}\) then we obtain the matrix
Here, \(\det B = -1\) and
A reflection about the horizontal axis is given by the matrix
and \(B = AC\) (see Figure 12.9). Thus, a reflection about a line \(\ell\) is simply a reflection about the horizontal axis followed by a rotation.
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 special orthogonal group , \(SO(n)\text{,}\) is just the intersection of \(O(n)\) and \(SL_n({\mathbb R})\text{;}\) that is, those elements in \(O(n)\) with determinant one. The Euclidean group , \(E(n)\text{,}\) 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 Exercise 12.4.6, you are asked to check that \(E(n)\) is indeed a group under this operation.
\(Figure \text { } 12.11.\) Translations in \(\mathbb R^2\)