Finding the orbits of the orthogonal group $O(n)$ on $\Bbb R^n$

Question

Let $O(n)=\{M\in GL_n(\mathbb{R}):MM^t=M^tM=I\}$ an orthogonal group. I need please an explain why each orbits consists of all vectors with the same length.

I know that an orbit is defined by $$G\triangleright x:=\{g\triangleright x:g\in G\},\qquad x\in X.$$

A student tells me that I should take a vector $x\in\mathbb{R}^n$ and apply the Gram-Schmidt process.

I don't understand this theme and I need an explanation with an example please. My next exercise will be to find orbits of operators whose matrices in the basis $(e_1,\ldots,e_n)$ are diagonal.

Answer 1

O(n) acts on n-dimensional space by rotations. Rotations preserve length. Now remark that given two vectors of the same length they can be rotated onto one another. This Is easy to see by taking the plane through the two vectors. So the problem is now to see that two vectors in the plane can be rotated onto another and this is trivial, just rotate the plane through the angle between the two vectors.

Answer 2

Hint If one writes $M$ in terms of its column vectors, i.e., as $$\newcommand{\mbx}{{\mathbf x}} M = \pmatrix{\mbx_1 & \cdots & \mbx_n},$$ and substitutes this expression in the defining equation $M^T M = I$ of $O(n)$, we see that $M$ is orthogonal iff $(\mbx_1, \ldots, \mbx_n)$ is an orthonormal basis of $\Bbb R^n$, which is exactly the output of the Gram-Schmidt Algorithm.

Additional hint Now, pick any unit vector ${\bf y} \in \Bbb R^n$ and extend it to a basis, and apply Gram-Schmidt: The resulting basis $({\bf y}, \mbx_2, \ldots, \mbx_n)$ is orthonormal, so $$A := \pmatrix{{\bf y} & \mbx_2 & \cdots & \mbx_n}$$ is orthogonal. What is $A {\bf e}_1$? This shows that $O(n)$ acts transitively on the set of vectors with unit length. Can you extend this argument to vectors of arbitrary length?