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?
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