Dimension of the manifold O(n) $O(n)$ is the manifold of the orthogonal $n \times n$ matrices. 
How do I prove that its dimension is $\displaystyle \frac{n(n-1)}{2}$?
Edit: Thanks to all your answers. I appreciate your help.
 A: I think the fastest way might be to consider it's Lie Algebra $\mathcal{o}(n)$.  Which is the set of skew symmetric matrices.  Their dimensions will be the same since $\mathcal{o}(n)$ is the tangent space at any point on the manifold $O(n)$.  You get the result you want by counting the dimensions and taking into account the skew symmetry. 
A: One way is to use singular value decomposition.  Let $D(n)$ be the space of $n\times n$ diagonal matrices, and let $F\colon O(n) \times D(n) \times O(n) \to \mathbb{R}^{n\times n}$ be the function $F(O_1,D,O_2) = O_1DO_2$.  Then $F$ is a local homeomorphism at each point $(O_1,D,O_2)$ for which the diagonal entries of $D$ are all different, so it follows that
$$
\dim\;O(n) \,+\, \dim\;D(n) \,+\, \dim\;O(n) \;=\; \dim\;\mathbb{R}^{n\times n}
$$
Since $\dim\;D(n) = n$ and $\dim\;\mathbb{R}^{n\times n} = n^2$, you can solve to get $\dim\;O(n) = (n^2 - n)/2$.
If you don't like singular value decomposition, it works just as well to use diagonalization of symmetric matrices.  Let $S(n)$ be the space of all $n\times n$ symmetric matrices, and define $G\colon O(n) \times D(n) \to S(n)$ by $G(O,D) = O^T D O$.  Again, $G$ is a local homeomorphism whenever the entries of $D$ are all different, so
$$
\dim\;O(n) \,+\, \dim\;D(n) \;=\; \dim\;S(n)
$$
Since $\dim\;S(n) = (n^2 + n)/2$ and $\dim\;D(n) = n$, we conclude that $\dim\;O(n) = (n^2-n)/2$.
Edit: Of course, the solution using symmetric matrices is very similar to the one that Theo gives above in the comments. The difference is that making the local product structure explicit avoids the use of the Implicit Function Theorem.
