Matrix groups

The first step is to consider the set of transformations as a group, in addition to a topological space.^4.8 We now derive several important groups from sets of matrices, ultimately leading to $SO(n)$ , the group of $n \times n$ rotation matrices, which is very important for motion planning. The set of all nonsingular $n \times n$ real-valued matrices is called the general linear group, denoted by $GL(n)$ , with respect to matrix multiplication. Each matrix $A \in GL(n)$ has an inverse $A^{-1} \in GL(n)$ , which when multiplied yields the identity matrix, $A A^{-1} = I$ . The matrices must be nonsingular for the same reason that 0 was removed from ${\mathbb{Q}}$ . The analog of division by zero for matrix algebra is the inability to invert a singular matrix.

Many interesting groups can be formed from one group, $G_1$ , by removing some elements to obtain a subgroup, $G_2$ . To be a subgroup, $G_2$ must be a subset of $G_1$ and satisfy the group axioms. We will arrive at the set of rotation matrices by constructing subgroups. One important subgroup of $GL(n)$ is the orthogonal group, $O(n)$ , which is the set of all matrices $A \in GL(n)$ for which $A A^T = I$ , in which $A^T$ denotes the matrix transpose of $A$ . These matrices have orthogonal columns (the inner product of any pair is zero) and the determinant is always $1$ or $-1$ . Thus, note that $A A^T$ takes the inner product of every pair of columns. If the columns are different, the result must be 0 ; if they are the same, the result is $1$ because $A A^T = I$ . The special orthogonal group, $SO(n)$ , is the subgroup of $O(n)$ in which every matrix has determinant $1$ . Another name for $SO(n)$ is the group of $n$ -dimensional rotation matrices.

A chain of groups, $SO(n) \leq O(n) \leq GL(n)$ , has been described in which $\leq$ denotes ``a subgroup of.'' Each group can also be considered as a topological space. The set of all $n \times n$ matrices (which is not a group with respect to multiplication) with real-valued entries is homeomorphic to ${\mathbb{R}}^{n^2}$ because $n^2$ entries in the matrix can be independently chosen. For $GL(n)$ , singular matrices are removed, but an $n^2$ -dimensional manifold is nevertheless obtained. For $O(n)$ , the expression $A A^T = I$ corresponds to $n^2$ algebraic equations that have to be satisfied. This should substantially drop the dimension. Note, however, that many of the equations are redundant (pick your favorite value for $n$ , multiply the matrices, and see what happens). There are only ${(^{n}_{2})}$ ways (pairwise combinations) to take the inner product of pairs of columns, and there are $n$ equations that require the magnitude of each column to be $1$ . This yields a total of $n (n+1) / 2$ independent equations. Each independent equation drops the manifold dimension by one, and the resulting dimension of $O(n)$ is $n^2 - n(n+1)/2 = n (n-1)/2$ , which is easily remembered as ${(^{n}_{2})}$ . To obtain $SO(n)$ , the constraint $\det A = 1$ is added, which eliminates exactly half of the elements of $O(n)$ but keeps the dimension the same.

Example 4..12 (Matrix Subgroups)   It is helpful to illustrate the concepts for $n=2$ . The set of all $2 \times 2$ matrices is

$$\left\{ \begin{pmatrix}a & b c & d \end{pmatrix} \;\bigg\vert\; a,b,c,d \in {\mathbb{R}}\right\} ,$$ (4.10)

which is homeomorphic to ${\mathbb{R}}^4$ . The group $GL(2)$ is formed from the set of all nonsingular $2 \times 2$ matrices, which introduces the constraint that $a d - b c \not = 0$ . The set of singular matrices forms a 3D manifold with boundary in ${\mathbb{R}}^4$ , but all other elements of ${\mathbb{R}}^4$ are in $GL(2)$ ; therefore, $GL(2)$ is a 4D manifold.

Next, the constraint $A A^T = I$ is enforced to obtain $O(2)$ . This becomes

$$\begin{pmatrix}a & b c & d \end{pmatrix} \begin{pmatrix}a & c b & d \end{pmatrix} = \begin{pmatrix}1 & 0 0 & 1 \end{pmatrix} ,$$ (4.11)

which directly yields four algebraic equations:

$$a^2 + b^2 = 1$$ (4.12)

$$a c + b d = 0$$ (4.13)

$$c a + d b = 0$$ (4.14)

$$c^2 + d^2 = 1 .$$ (4.15)

Note that (4.14) is redundant. There are two kinds of equations. One equation, given by (4.13), forces the inner product of the columns to be 0 . There is only one because ${(^{n}_{2})} = 1$ for $n=2$ . Two other constraints, (4.12) and (4.15), force the rows to be unit vectors. There are two because $n=2$ . The resulting dimension of the manifold is ${(^{n}_{2})} = 1$ because we started with ${\mathbb{R}}^4$ and lost three dimensions from (4.12), (4.13), and (4.15). What does this manifold look like? Imagine that there are two different two-dimensional unit vectors, $(a,b)$ and $(c,d)$ . Any value can be chosen for $(a,b)$ as long as $a^2 + b^2 = 1$ . This looks like ${\mathbb{S}}^1$ , but the inner product of $(a,b)$ and $(c,d)$ must also be 0 . Therefore, for each value of $(a,b)$ , there are two choices for $c$ and $d$ : 1) $c = b$ and $d = -a$ , or 2) $c = -b$ and $d = a$ . It appears that there are two circles! The manifold is ${\mathbb{S}}^1 \sqcup {\mathbb{S}}^1$ , in which $\sqcup$ denotes the union of disjoint sets. Note that this manifold is not connected because no path exists from one circle to the other.

The final step is to require that $\det A = a d - b c = 1$ , to obtain $SO(2)$ , the set of all 2D rotation matrices. Without this condition, there would be matrices that produce a rotated mirror image of the rigid body. The constraint simply forces the choice for $c$ and $d$ to be $c = -b$ and $a = d$ . This throws away one of the circles from $O(2)$ , to obtain a single circle for $SO(2)$ . We have finally obtained what you already knew: $SO(2)$ is homeomorphic to ${\mathbb{S}}^1$ . The circle can be parameterized using polar coordinates to obtain the standard 2D rotation matrix, (3.31), given in Section 3.2.2.