Matrix Groups

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 \(\alpha \in {\mathbb R}\), \[\begin{aligned}T({\mathbf x}+{\mathbf y}) & = T({\mathbf x}) + T({\mathbf y}) \\ T(\alpha {\mathbf y}) & = \alpha T({\mathbf y})\end{aligned}\]. An \(m \times n\) matrix with entries in \({\mathbb R}\) represents a linear transformation from \({\mathbb R}^n\) to \({\mathbb R}^m\). If we write vectors \({\mathbf x} = (x_1, \ldots, x_n)^\transpose\) and \({\mathbf y} = (y_1, \ldots, y_n)^\transpose\) in \({\mathbb R}^n\) as column matrices, then an \(m \times n\) matrix \[\begin{aligned}\end{aligned}\] maps the vectors to \({\mathbb R}^m\) linearly by matrix multiplication. Observe that if \(\alpha\) is a real number, \[\begin{aligned}\end{aligned}\], where \[\begin{aligned}\end{aligned}\]. We will often abbreviate the matrix \(A\) by writing \((a_{ij})\). \((a_{ij})\) matrix

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 \[\begin{aligned}{\mathbf e}_1 & = (1, 0, \ldots, 0)^\transpose \\ {\mathbf e}_2 & = (0, 1, \ldots, 0)^\transpose \\ & \aatavdots{=} & \\ {\mathbf e}_n & = (0, 0, \ldots, 1)^\transpose\end{aligned}\]. We can write any vector \({\mathbf x} = (x_1, \ldots, x_n)^\transpose\) as \[\begin{aligned}\end{aligned}\]. Consequently, if \[\begin{aligned}T({\mathbf e}_1) & = (a_{11}, a_{21}, \ldots, a_{m1})^\transpose, \\ T({\mathbf e}_2) & = (a_{12}, a_{22}, \ldots, a_{m2})^\transpose, \\ & \aatavdots{=} & \\ T({\mathbf e}_n) & = (a_{1n}, a_{2n}, \ldots, a_{mn})^\transpose\end{aligned}\], then \[\begin{aligned}T({\mathbf x} ) & = T(x_1 {\mathbf e}_1 + x_2 {\mathbf e}_2 + \cdots + x_n {\mathbf e}_n) \\ & = x_1 T({\mathbf e}_1) + x_2 T({\mathbf e}_2) + \cdots + x_n T({\mathbf e}_n) \\ & = \left( \sum_{k=1}^{n} a_{1k} x_k, \ldots, \sum_{k=1}^{n} a_{mk} x_k \right)^\transpose \\ & = A {\mathbf x}\end{aligned}\].

Condensed — the full section is in Judson, Abstract Algebra: Theory and Applications.

The set of all \(n \times n\) invertible matrices forms a group called the general linear group. We will denote this group by \(GL_n({\mathbb R})\). 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 \(\det(B) = 1\). Then \(\det(AB) = \det(A) \det (B) = 1\) and \(\det(A^{-1}) = 1 / \det A = 1\). This subgroup is called the special linear group and is denoted by \(SL_n({\mathbb R})\).

Another subgroup of \(GL_n({\mathbb R})\) is the orthogonal group. A matrix \(A\) is orthogonal if \(A^{-1} = A^\transpose\). The orthogonal group consists of the set of all orthogonal matrices. We write \(O(n)\) for the \(n \times n\) orthogonal group. \(O(n)\) orthogonal group We leave as an exercise the proof that \(O(n)\) is a subgroup of \(GL_n( {\mathbb R})\).

There is a more geometric way of viewing the group \(O(n)\). 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 \({\mathbf x}=(x_1, \ldots, x_n)^\transpose\) and \({\mathbf y}=(y_1, \ldots, y_n)^\transpose\) is \[\begin{aligned}\end{aligned}\]. We define the length of a vector \({\mathbf x}=(x_1, \ldots, x_n)^\transpose\) to be \(\| {\mathbf x} \|\) length of a vector \(\mathbf x\) \[\begin{aligned}\end{aligned}\]. 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 \(\| {\mathbf x}-{\mathbf y} \|\). We leave as an exercise the proof of the following proposition about the properties of Euclidean inner products.

Condensed — the full section is in Judson, Abstract Algebra: Theory and Applications.