Dimension of the group O(n,R) - How to calc?

[JuanC97]

Hi, I want to find the number of parameters needed to define an orthogonal transformation in Rⁿ.
As I suppose, this equals the dimension of the orthogonal group O(n,R) - but, correct me if I'm wrong.

I haven't been able to figure out how to do this yet. If it helps, I know that an orthogonal matix should have n(n+1)/2 "free components".
That said, I'd appreciate any hint from this point.

 

[micromass]

An orthogonal tranformation $A$ satisfies $AA^T = I$. Those are initially $n^2$ constraints, but some coincide since the matrix $AA^T$ always is symmetric even if the matrix is orthogonal. So the amount of constraints consist of an upper diagonal matrix, which has $\frac{n(n+1)}{2}$ entries.

The amount of freedom is then $n^2 - \frac{n(n+1)}{2} = \frac{n(n-1)}{2}$, which is the dimension.

 

[JuanC97]

An orthogonal tranformation $A$ satisfies $AA^T = I$. Those are initially $n^2$ constraints, but some coincide since the matrix $AA^T$ always is symmetric even if the matrix is orthogonal. So the amount of constraints consist of an upper diagonal matrix, which has $\frac{n(n+1)}{2}$ entries.

The amount of freedom is then $n^2 - \frac{n(n+1)}{2} = \frac{n(n-1)}{2}$, which is the dimension.

Thanks, very concise.

 

[mathwonk]

you can also use induction and geometry. such a matrix consists of n orthogonal unit length columns. so just ask how many ways there are to choose these. If n = 1, only two choices exist of unit vectors, so the dimension is zero. then to go from n to n+1, we begin by choosing a unit vector in n+1 space, i.e. a point on the n dimensional sphere in n+1 space. so the inductive step giving the answer fore n+1 space, just adds n to the answer for n space, so just add n to n(n-1)/2 and you get (n+1)n/2.