The generators of $SO(n)$ are pure imaginary antisymmetric $n \times n$ matrices.
How can this fact be used to show that the dimension of $SO(n)$ is $\frac{n(n-1)}{2}$?
I know that an antisymmetric matrix has $\frac{n(n-1)}{2}$ degrees of freedom, but I can't take this idea any further in the demonstration of the proof.
Thoughts?
The group SO(n) consists of orthogonal matrices with unit determinant. Hence for $A\in SO(n)$, $A^{T}A=AA^{T}=1$, $\det(A)=1$. These matrices perform rotations in an n-dimensional space. To find the number of independent generators of the group, consider the group's fundamental representation in a real, n dimensional, vector space. Any rotation can be parametrised from the magnitude of its angle and the axis of rotation, which in turn, is parametrised via a pair of axis, which form a 2 dimensional surface. The number of independent surfaces is $\begin{pmatrix} n\\2 \end{pmatrix}$, hence the number of independent generators.