2.3 Matrix groups

Linear transformations

Let ${\mathbb{R}}^n$ be the space of $n\times 1$ real column vectors

$$\mathbf{x}=\left(\begin{array}{c}{x}_1\\ x_2\\ ⋮\\ x_n\end{array}\right).$$

A mapping $A:{\mathbb{R}}^n\to {\mathbb{R}}^n$ is said to be linear if

$$A(a\mathbf{x}+b\mathbf{y})=aA(\mathbf{x})+bA(\mathbf{y})$$

for all vectors x, $\mathbf{y}\in {\mathbb{R}}^n$ and all real numbers $a,b\in \mathbb{R}$ . Writing

$$\mathbf{x}=\sum _{i=1}^n{x}_i{\mathbf{e}}_i\text{ where }{\mathbf{e}}_1=\left(\begin{array}{c}1\\ 0\\ ⋮\\ 0\end{array}\right),{\mathbf{e}}_2=\left(\begin{array}{c}0\\ 1\\ ⋮\\ 0\end{array}\right),\dots ,{\mathbf{e}}_n=\left(\begin{array}{c}0\\ 0\\ ⋮\\ 1\end{array}\right),$$

we have

$$A(\mathbf{x})=\sum _{i=1}^n{x}_{i}A({\mathbf{e}}_i).$$

If we set

$$A({\mathbf{e}}_i)=\sum _{j=1}^n{a}_{ji}{\mathbf{e}}_j$$

the components $x_i$ of the vector x transform according to the formula

$$\mathbf{x}\mapsto {\mathbf{x}}^{\prime }=A(\mathbf{x})\text{ where }{x}_i^{\prime }=\sum _{j=1}^n{a}_{ij}{x}_j(i=1,\dots ,n).$$

It is common to write this mapping in the form

$${\mathbf{x}}^{\prime }=\mathsf{Ax}$$

where $\mathsf{A}=[a_{ij}]$ is the $n\times n$ array

$$\mathsf{A}=\left(\begin{array}{ccccc}{a}_{11}& a_{12}& a_{13}& \dots & a_{1n}\\ a_{21}& a_{22}& a_{23}& \dots & a_{2n}\\ \dots & \dots & \dots & \dots & \dots \\ a_{n1}& a_{n2}& a_{n3}& \dots & a_{nn}\end{array}\right).$$

A is called the matrix of the linear mapping A, and $a_{ij}$ are its components. The matrix AB of the product transformation AB is then given by the matrix multiplication rule,

$$(\mathsf{AB}{)}_{ij}=\sum _{k=1}^n{a}_{ik}{b}_{kj}.$$

Exercise

Prove this formula.

Linear maps on ${\mathbb{R}}^n$ and $n\times n$ matrices as essentially identical concepts, the latter being little more than a notational device for the former. Be warned, however, when we come to general vector spaces in Chapter 3 such an identification cannot be made in a natural way. In later chapters we will often adopt a different notation for matrix components in order to take account of this difficulty, but for the time being it is possible to use standard matrix notation as we are only concerned with the particular vector space ${\mathbb{R}}^n$ for the rest of this chapter.

A linear transformation A is a one-to-one linear mapping from ${\mathbb{R}}^n$ onto itself. Such a map is invertible, and its matrix A has non-zero determinant, det $\mathsf{A}\ne 0$ . Such a matrix is said to be non-singular and have an inverse matrix ${\mathsf{A}}^{-1}$ given by

$$({\mathsf{A}}^{-1}{)}_{ij}=\frac{A_{ji}}{\det \mathsf{A}},$$

where $A_{ji}$ is the $(j,i)$ cofactor ofthe matrix A, defined as the determinant ofthe submatrix of A formed by removing its jth row and ith column and multiplied by the factor $(-1{)}^{i+j}$ The inverse of a matrix acts as both right and left inverse:

$$\mathsf{A}{\mathsf{A}}^{-1}={\mathsf{A}}^{-1}\mathsf{A}=\mathsf{I},\text{\hspace{1em}(2.2)}$$

where I is the $n\times n$ unit matrix

$$I=\left(\begin{array}{cccc}1& 0& \dots & 0\\ 0& 1& \dots & 0\\ ⋮& & \ddots & \\ 0& 0& \dots & 1\end{array}\right).$$

The components of the unit matrix are frequently written as the Kronecker delta

$${\delta }_{ij}=\{\begin{array}{ll}1& \text{ if }i=j,\\ 0& \text{ if }i\ne j.\end{array}\text{\hspace{1em}(2.3)}$$

The inverse of AB is given by the matrix identity

$$(\mathsf{AB}{)}^{-1}={\mathsf{B}}^{-1}{\mathsf{A}}^{-1}.\text{\hspace{1em}(2.4)}$$

Matrix groups

The set of all $n\times n$ non-singular real matrices is a group, denoted $GL(n,\mathbb{R})$ . The key to this result is the product law of determinants

$$\det (\mathsf{AB})=\det (\mathsf{A})\det (\mathsf{B}).\text{\hspace{1em}(2.5)}$$

Closure: this follows from the fact that det $\mathsf{A}\ne 0$ and det $\mathtt{B}\ne 0$ implies that $\begin{array}{r}\det (\mathsf{AB})=\end{array}$ det A det $\mathtt{B}\ne 0$

Associative law: $(\mathsf{AB})\mathsf{C}=\mathsf{A}(\mathsf{BC})$ is true of all matrices, singular or not.

Identity: the $n\times n$ unit matrix I is an identity element since $\mathsf{IA}=\mathsf{Al}=\mathsf{A}$ for all $n\times n$ matrices A.

Inverse: from Eq. (2.2) ${\mathsf{A}}^{-1}$ clearly acts as an inverse element to A. Equation (2.5) ensures that ${\mathsf{A}}^{-1}$ is non-singular and also belongs to $GL(n,\mathbb{R})$ , since

$$\det {\mathsf{A}}^{-1}=\frac{1}{\det \mathsf{A}}.$$

A similar discussion shows that the set of $n\times n$ non-singular matrices with complex components, denoted $GL(n,\mathbb{C})$ , also forms a group. Except for the case $n=1$ , these groups are non-abelian since matrices do not in general commute, $\mathsf{AB}\ne \mathsf{BA}$ . The groups $GL(n,\mathbb{R})$ and $GL(n,\mathbb{C})$ are called the general linear groups of order n. Subgroups of these groups, whose elements are matrices with the law of composition being matrix multiplication, are generically called matrix groups [5].

In the following examples the associative law may be assumed, since the law of compo sition is matrix multiplication. Frequent use will be made of the concept of the transpose ${\mathsf{A}}^T$ of a matrix A, defined as the matrix formed by reflecting A about its diagonal,

$$({\mathsf{A}}^T{)}_{ij}=a_{ji}\text{ where }\mathsf{A}=[a_{ij}].$$

The following identities can be found in many standard references such as Hildebrand [6], and should be known to the reader:

$$(\mathsf{AB}{)}^T={\mathsf{B}}^T{\mathsf{A}}^T,\text{\hspace{1em}(2.6)}$$

$$\det {\mathbf{A}}^T=\det \mathbf{A},\text{\hspace{1em}(2.7)}$$

and if A is non-singular then the inverse of its transpose is the transpose of its inverse,

$$({\mathsf{A}}^{-1}{)}^T=({\mathsf{A}}^T{)}^{-1}.\text{\hspace{1em}(2.8)}$$

Example 2.10

The special linear group or unimodular group of degree n, denoted $SL(n,\mathbb{R})$ , is defined as the set of $n\times n$ unimodular matrices, real matrices having deter minant 1. Closure with respect to matrix multiplication follows from Eq. (2.5),

$$\det \mathsf{A}=\det \mathsf{B}=1⟹\det (\mathsf{AB})=\det \mathsf{A}\det \mathsf{B}=1.$$

The identity $\mathsf{I}\in SL(n,\mathbb{R})$ since det $\mathsf{I}=1$ , and closure with respect to inverses follows from

$$\det \mathsf{A}=1⟹\det {\mathsf{A}}^{-1}=\frac{1}{\det \mathsf{A}}=1.$$

Example 2.11

A matrix A is called orthogonal if its inverse is equal to its transpose,

$$\mathsf{A}{\mathsf{A}}^T={\mathsf{A}}^T\mathsf{A}=\mathsf{I}.\text{\hspace{1em}(2.9)}$$

The set of real orthogonal $n\times n$ matrices, denoted $O(n)$ , forms a group known as the orthogonal group of order n:

Closure: if A and B are orthogonal matrices, $\mathsf{A}{\mathsf{A}}^T=\mathsf{B}{\mathsf{B}}^T=\mathsf{I},$ , then so is their product AB,

$$(\mathsf{AB})(\mathsf{AB}{)}^T={\mathsf{ABB}}^T{\mathsf{A}}^T={\mathsf{AIA}}^T={\mathsf{AA}}^T=\mathsf{I}.$$

Identity: the unit matrix I is clearly orthogonal since ${|}^T|={|}^2=|$

Inverse: if A is an orthogonal matrix then ${\mathsf{A}}^{-1}$ is also orthogonal for, using Eq. (2.8) and Eq. (2.4),

$${\mathsf{A}}^{-1}({\mathsf{A}}^{-1}{)}^T={\mathsf{A}}^{-1}({\mathsf{A}}^T{)}^{-1}=({\mathsf{A}}^T\mathsf{A}{)}^{-1}={\mathsf{I}}^{-1}=\mathsf{I}.$$

The determinant of an orthogonal matrix is always $\pm 1$ since

$$\mathsf{A}{\mathsf{A}}^T=\mathsf{I}⟹\det \mathsf{A}\det {\mathsf{A}}^T=\det (\mathsf{A}{\mathsf{A}}^T)=\det \mathsf{I}=1.$$

Hence $(\det \mathsf{A}{)}^2=1$ by Eq. (2.7) and the result $\det \mathsf{A}=\pm 1$ follows at once. The orthogonal matrices with determinant 1 are called proper orthogonal matrices, while those with determinant −1 are called improper. The proper orthogonal matrices, denoted $SO(n)$ form a group themselves called the proper orthogonal group of order n. This group is often known as the rotation group in n dimensions – see Section 2.7. It is clearly a subgroup of the special linear group $SL(n,\mathbb{R})$.

Example 2.12

Let $p$ and $q$ be non-negative integers such that $p+q=n$ , and define ${\mathsf{G}}_p$ to be the $n\times n$ matrix whose components ${\mathsf{G}}_p=[g_{ij}]$ are defined by

$$g_{ij}=\{\begin{array}{ll}1& \text{ if }i=j\le p,\\ -1& \text{ if }i=j>p,\\ 0& \text{ if }i\ne j.\end{array}$$

We use $O(p,q)$ to denote the set of matrices A such that

$${\mathsf{A}}^T{\mathsf{G}}_p\mathsf{A}={\mathsf{G}}_p.\text{\hspace{1em}(2.10)}$$

It follows from this equation that any matrix belonging to $O(p,q)$ is non-singular, for on taking determinants,

$$\det {\mathbf{A}}^T\det {\mathbf{G}}_p\det \mathbf{A}=\det {\mathbf{G}}_p.$$

Since det ${\mathsf{G}}_p=\pm 1\ne 0$ we have det ${\mathsf{A}}^T$ det $\mathsf{A}=(\det \mathsf{A}{)}^2=1$ , and consequently

$$\det \mathsf{A}=\pm 1.$$

The group properties of $O(p,q)$ follow:

Closure: if A and B both satisfy Eq. (2.10), then so does their product AB, for

$$(\mathsf{AB}{)}^T{\mathsf{G}}_p(\mathsf{AB})={\mathsf{B}}^T{\mathsf{A}}^T{\mathsf{G}}_p\mathsf{AB}={\mathsf{B}}^T{\mathsf{G}}_p\mathsf{B}={\mathsf{G}}_p.$$

Identity: the unit matrix $\mathsf{A}=\mathsf{I}$ clearly satisfies Eq. (2.10).

Inverse: if Eq. (2.10) is multiplied on the right by ${\mathsf{A}}^{-1}$ and on the left by $({\mathsf{A}}^{-1}{)}^T$ , we have from Eq. (2.8)

$${\mathbf{G}}_p=({\mathsf{A}}^{-1}{)}^T{\mathbf{G}}_p{\mathsf{A}}^{-1}.$$

Hence ${\mathsf{A}}^{-1}$ satisfies Eq. (2.10) and belongs to $O(p,q)$

The group $O(p,q)$ is known as the pseudo-orthogonal group of type $(p,q)$ . The case $q=0,p=n$ reduces to the orthogonal group $O(n)$ . As for the orthogonal group, those elements of $O(p,q)$ having determinant 1 form a subgroup denoted $SO(p,q)$

Example 2.13

Let J be the $2n\times 2n$ matrix

$$J=\left(\begin{array}{cc}0& I\\ -I& 0\end{array}\right),$$

where 0 is the $n\times n$ zero matrix and I is the $n\times n$ unit matrix. A $2n\times 2n$ matrix A is said to be symplectic if it satisfies the equation

$${\mathsf{A}}^T\mathsf{JA}=\mathsf{J}.\text{\hspace{1em}(2.11)}$$

The argument needed to show that these matrices form a group is essentially identical to that just given for $O(p,q)$ . Again, since det $\mathsf{J}=1$ , it follows immediately from Eq. (2.11) that det $\mathsf{A}=\pm 1$ , and A is non-singular. The group is denoted $Sp(2n)$ , called the symplectic group of order $2n$

Exercise

Show that the symplectic matrices of order 2 are precisely the unimodular matrices of order 2. Hence in the $2\times 2$ case ($n=1$ in the notation $Sp(2n)$ above), all symplectic matrices have determinant 1. It turns out that symplectic matrices of any order have determinant 1, but the proof of this is more complicated.

Example 2.14

The general complex linear group, $GL(n,\mathbb{C})$ , is defined exactly as for the reals. It is the set ofnon-singular complex $n\times n$ matrices, where the law ofcomposition is matrix product using multiplication of complex numbers. We define special subgroups of this group the same way as for the reals:

$SL(n,\mathbb{C})$ is the complex unimodular group of degree n, consisting of complex $n\times n$ matrices having determinant 1;

$O(n,\mathbb{C})$ is the complex orthogonal group of degree n, whose elements are complex $n\times n$ matrices A satisfying ${\mathsf{A}}^T\mathsf{A}=\mathsf{I};$

$SO(n,\mathbb{C})$ is the complex proper orthogonal group, which is the intersection of the above two groups.

There is no complex equivalent ofthe pseudo-orthogonal groups since these are all isomor phic to $O(n,\mathbb{C})$ – see Problem 2.7.

Example 2.15

The adjoint of a complex matrix A is defined as its complex conjugate transpose ${\mathsf{A}}^{†}=\overline{{\mathsf{A}}^T}$ , whose components are $a_{ij}^{†}=\overline{a_{ji}}$ where a bar over a complex number refers to its complex conjugate. An $n\times n$ complex matrix U is called unitary if

$${\mathsf{UU}}^{†}=\mathsf{I}.\text{\hspace{1em}(2.12)}$$

It follows immediately that

$$\det \mathsf{U}\overline{\det \mathsf{U}}=|\det \mathsf{U}{|}^2=1,$$

and there exists a real number $\varphi$ with $0\le \varphi <2\pi$ such that det $\mathsf{U}={\mathsf{e}}^{i\varphi }$ . Hence all unitary matrices are non-singular and the group properties are straightforward to verify. The group of all $n\times n$ unitary matrices is called the unitary group of order n, denoted $U(n)$ . The subgroup ofunitary matrices having det $U=1$ is called the special unitary group oforder n, denoted $SU(n)$ .

Problems

Problem 2.6

Show that the following sets of matrices form groups with respect to addition of matrices, but that none ofthem is a group with respect to matrix multiplication: (i) real antisymmetric $n\times n$ matrices $({\mathsf{A}}^T=-\mathsf{A})$ , (ii) real $n\times n$ matrices having vanishing trace $(\mathrm{tr}\mathsf{A}={\sum }_{i=1}^n{a}_{ii}=0)$ (iii) complex hermitian $n\times n$ matrices $({\mathsf{H}}^{†}=\mathsf{H})$

Problem 2.7

Find a diagonal complex matrix S such that

$$\mathsf{I}={\mathsf{S}}^T{\mathsf{G}}_p\mathsf{S}$$

where ${\mathsf{G}}_p$ is defined in Example 2.12. Show that:

(a) Every complex matrix A satisfying Eq. (2.10) can be written in the form

$$\mathsf{A}={\mathsf{SBS}}^{-1}$$

where B is a complex orthogonal matrix (i.e. a member o $O(n,\mathbb{C}))$

(b) The complex versions of the pseudo-orthogonal groups, $O(p,q,\mathbb{C})$ , are all isomorphic to each other if they have the same dimension,

$$O(p,q,\mathbb{C})\cong O(n,\mathbb{C})\text{ where }n=p+q.$$

Problem 2.8

Show that every element of $SU(2)$ has the form

$$\mathsf{U}=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\text{ where }a=\overline{d}\text{ and }b=-\overline{c}.$$