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

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

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

which directly yields four algebraic equations:

$\displaystyle a^2 + b^2$ $\displaystyle = 1$ (4.12)
$\displaystyle a c + b d$ $\displaystyle = 0$ (4.13)
$\displaystyle c a + d b$ $\displaystyle = 0$ (4.14)
$\displaystyle c^2 + d^2$ $\displaystyle = 1 .$ (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.

Steven M LaValle 2012-04-20