How many orthogonal basis does a set of vectors have?

Question

I wonder how many orthogonal bases does a set of vectors have in a space?

Answer 1

Excluding the trivial case of set with just the zero vector, orthogonal bases are always infinitely many since we can scale the basis vectors as we want preserving orthogonality.

Otherwise for orhonormal bases we have

• $2$ choices for $1$ dimensional subspaces
• infinitely many choices for $n-$dimensional subspaces with $n\ge 2$

Answer 2

You have $n(n-1)/2$ degrees of freedom in choosing an orthogonal basis.

The first vector in the basis is a unit vector in $\mathbb{R}^n$. It sits on the unit sphere, which is $n-1$-dimensional, and there are $n-1$ degrees of freedom to choose the first vector.

The rest of the basis vectors lie in a subspace orthogonal to the first, which is an $n-1$-dimensional space. So, by induction, we have $n-2$ degrees of freedom to choose the second basis vector, and so on. The total number of degrees of freedom is $$n-1 + (n-2) + (n-3) + ...+2+1+0 = n(n-1)/2$$

If we put the vectors of an orthogonal basis as columns in an $n\times n$ matrix, we get an 'orthogonal' matrix. These matrices form the set $O(n)$, and all these matrices satisfy the equation $A^T A= I$, the identity matrix.

$O(n)$ is an $n(n-1)/2$-dimensional submanifold of the $n^2$-dimensional set of $n\times n$ matrices.