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$O(n)$ 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.

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    Do you know how to prove that $O(n)$ is a manifold? One easy way is to use the implicit function theorem and to consider the map $A \mapsto A^{T}A$ from the $n \times n$-matrices to the *symmetric* $n \times n$ matrices and observing that $O(n)$ is the pre-image of the *regular value* given by the identity matrix, then counting dimensions $n(n-1)/2 = n^2 - n(n+1)/2$.2011-05-23
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    A different argument than Theo's would be to observe there's the fibre bundle $$O(n-1) \to O(n) \to S^{n-1}$$. Iterating that bundle and counting the dimensions of the spheres gives you $dim(O(n)) = n-1 + n-2 + \cdots + 3 + 2 + 1 = n(n-1)/2$.2011-05-23
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    A more elementary way of phrasing Ryan's comment might be that the columns of an $O(n)$ matrix form an ON basis in $\mathbf{R}^n$, and you have $n-1$ degrees of freedom when choosing the first vector in such a basis (since the only restriction is that it has to be of length one), then $n-2$ degrees of freedom when choosing the second basis vector (since now there is also the restriction that it has to be orthogonal to the first vector), $n-3$ degrees of freedom for the third vector (since it must have unit length and be orthogonal to the first two), etc.2011-05-23

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