Dimension of $SO_n(\mathbb{R})$ 
Is there a simple proof that the dimension of $SO_n(\mathbb{R})$,
  a.k.a the group of rotations in $n$-dimensional space is $(n-1)n/2$?

It would be great to see some proofs based only on the algebraic definition:
$$R \mid \left\{ R^T=R^{-1} \land \det(R)=1 \right\}$$
or alternatively proofs invoking geometrical arguments (though I'd like to stay away from proofs using Lie Algebra methods).
Any takers?

The orthogonality condition is equivalent to the columns of $R$ being an orthonormal basis. You can pick 


*

*first column $v_1$ from $S^{n-1}$, with $(n-1)$ dimensions to choose from.

*second column $v_2$ from $S^{n-1}\cap \{v_1\}^\perp = S^{n-2}$, with $(n-2)$ dimensions.

*third column $v_3$ from $S^{n-1}\cap \{v_1,v_2\}^\perp = S^{n-3}$, with $(n-3)$ dimensions.


and so on, until you have $S^0$ for the last column (two vectors, of which only one gives positive determinant). The total is
$$0+1+2+\dots+(n-1) = \frac{n(n-1)}{2}$$
