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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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I am uncertain of what "free parameters" is exactly referring to here, but I would suggest looking for an argument of the form "To fix we parameter we must pick 2 values (order does not matter) from a sample size of n, without repetition". – Ragib Zaman Sep 10 '11 at 15:35
@Ragib: I think what the question means is that one must specify $n(n-1)/2$ values to uniquely determine an orthogonal matrix. user5198: Consider choosing the columns of the matrix one by one. How many degrees of freedom does the first column have? How about the second? And so on... – Rahul Sep 10 '11 at 15:41
The three-dimensional case is discussed here, for example – t.b. Sep 10 '11 at 15:42
up vote 4 down vote accepted

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! – user5198 Sep 10 '11 at 16:42

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.

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