Why does an $n \times n$ rotation matrix have $\frac{1}{2}n(n-1)$ undetermined parameters?

Question

Consider an orthogonal transformation between Cartesian coordinate systems in $n$-dimensional space. The $n \times n$ rotation matrix $$R = \left(a_{ij}\right)$$ has $n^2$ entries. These are not independent; they are related by the orthogonality conditions $$a_{ik}a_{jk} = \delta_{ij},$$ which are $\frac{1}{2}\!\!\left(n^2 + n\right)$ independent equations in the $a_{ij}$. Thus, $$n^2 - \frac{1}{2}\!\!\left(n^2 + n\right) = \frac{1}{2}n\left(n-1 \right)$$ of the $a_{ij}$ are left undetermined.

Why is the last step justified? If there are $n$ independent equations, is it always possible to solve them for $n$ unknowns (even if the equations involve the sums of quadratic terms, as they do here)? If so, why? Under what general conditions is it possible to solve a system of $n$ independent equations for $n$ unknowns?

Answer 1

If you have a system of $m$ equations with $M$ unknowns then it is usually possible to solve it locally, meaning that the solution set around a particular point is parametrized by $M-m$ parameters. Precise conditions for "usually" are given by the implicit function theorem, namely functions involved in the equations should be continuously differentiable, and the Jacobian matrix of the system has to be non-singular at the point around which the solution set is parametrized, see http://en.wikipedia.org/wiki/Implicit_function_theorem.

The functions in the orthogonality conditions are quadratic polynomials, so clearly continuously differentiable, but technically one has to check that the Jacobian is non-singular at every point (it is) before doing the subtraction. It is somewhat confusing because the "points" here will be elements of the rotation group, i.e. they are themselves matrices.

Answer 2

If you want something a bit more concrete, pick one row at a time. The first row is any norm 1 vector, so that is $n-1$ parameters. The second row is any norm 1 vector perpendicular to the first row, so that is $n-2$ parameters. The $k$ row is any norm 1 vector perpendicular to the first $k-1$ columns, so that is $n-k$ parameters. You end up with $(n-1) + (n-2) + \cdots + 2 + 1 + 0 = \frac12 n(n-1)$ parameters.

Answer 3

A general rotation can be written in terms of a "rotor" in a clifford algebra. These rotors take the form

$$q = q_0 + \sum_{i<j} q_{ij} e_i \wedge e_j, \quad q_0^2 + \sum_{i<j} q_{ij}^2 = 1$$

so that the rotation, written in terms of the geometric product, is, for any vector $a$,

$$a' = q a q^{-1}$$

The astute reader should recognize this is a generalization of quaternions. The numbers $q_{ij}$ number $n(n-1)/2$ in total. This is the number of degrees of freedom that characterizes the rotor $q$, for $q_0$ is determined by the normalization condition.

Interpreting the degrees of freedom of a rotation--which applies equally to a matrix that describes the rotation--is quite easy to do from this geometric perspective: the wedge product $e_i \wedge e_j$ represents a planar subspace. Each of the numbers $q_{ij}$ is a coefficient of a planar subspace, representing how much of the rotation involves rotation in that specified plane.

Thus, $n(n-1)/2$ comes from the fact that there are $n(n-1)/2$ linearly independent planes in an $n$-dimensional space.