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Prove that the number of free parameters in an $n\times n$ orthogonal transformation matrix is equal to $\frac{n(n-1)}{2}$. For example parametrization of $2 \times 2$ orthogonal matrix requires only one parameter, ie $\theta$.

And the parametric form is

$ M_2 = \pm\left(\begin{array}{cc} \cos\theta & \sin\theta\\ -\sin\theta & \cos\theta \end{array}\right) .$

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    The three-dimensional case is discussed [here](http://math.stackexchange.com/questions/28189/freedoms-of-real-orthogonal-matrices), for example2011-09-10

2 Answers 2

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Use the dictionary "orthogonal transformation $\leftrightarrow$ orthonormal basis". For the first basis vector, choose any point on the unit sphere ($n-1$ free parameters). For the second basis vector, choose any point on the unit sphere which is orthogonal to the first basis vector ($n-2$ free parameters). Etc

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    Thank you for the easy to understand proof that can be visualised!2011-09-10
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There are $n^2$ free parameters in an $n\times n$ matrix. Orthogonality imposes $n(n+1)/2$ constraints, namely $n$ normalization constraints on the columns and $\binom n2=n(n-1)/2$ orthogonality constraints, one for each pair of columns. That leaves $n^2-n-n(n-1)/2=n(n-1)/2$ degrees of freedom.