What is degrees of freedom of a real orthogonal marix?

Question

How can I find the degrees of freedom of a $n \times n$ real orthogonal matrix?

I have tried to proceed by principle of induction but I fail.Please tell me the right way to proceed.

Thank you in advance.

Answer 1

For an $n \times n$ matrix with column vectors $v_1 \dots v_n$, we say that it is orthogonal if, $$ \langle v_i, v_j \rangle = 0,\ \ \ \forall (i,j)\ \Big{|}\ i\neq j $$

Recalling that the inner product is commutative, the above gives us this many unique constraint equations: $$\sum_{k=1}^{n-1} k = \frac{n(n-1)}{2}$$

The matrix has $n^2$ elements, so in the orthogonal case, $$ \text{dof}_\text{orthogonal} = n^2 - \frac{n(n-1)}{2}$$

If we also require that the matrix is orthonormal, then there are $n$ more constraints of the form, $$ \langle v_i, v_i \rangle = 1,\ \ \ \forall i $$

and so in the orthonormal case, \begin{align} \text{dof}_\text{orthonormal} &= n^2 - \frac{n(n-1)}{2} - n\\ &= \frac{n(n-1)}{2} \end{align}

This proves that $2 \times 2$ orthonormal matrices have 1 degree of freedom and that $3 \times 3$ orthonormal matrices have 3 degrees of freedom, which agrees with our intuition about rotations.