What is the probability a random orthonormal matrix will be a rotation matrix?

Question

Construct an orthonormal matrix in $\Bbb{R}^n$ in the following manner:

Choose a point uniformly on the unit sphere in $n$ dimensional space (can be done by sampling from an $n$ dimensional isotropic Gaussian and then normalizing the vector).
Take the plane orthogonal to this vector. This intersects the sphere in a lower dim sphere.
Choose a point uniformly at random on this lower dim sphere.
Repeat until you get to 1-d.

Now we have a set of orthonormal vectors (the position vectors of the points we chose). They could form a rotation matrix or another kind of matrix. What is the probability we'll get a rotation matrix? And what are the other kinds of matrices that this algorithm can give us?

Per the question linked above, we know for two dimensions, this probability is 50%.

Isn't it obviously 50%? You can just flip the sign of (say) the last column to convert between the two kinds of matrices (with determinant 1 or -1). And this choice of sign is what you do randomly in the last step. — Hans Lundmark, Jun 04 '23 at 07:31
Ah, nice. So an orthonormal matrix with determinant $1$ is always a rotation matrix? — Rohit Pandey, Jun 04 '23 at 16:46
Yes, and with determinant -1 it's a rotation composed with a reflection. — Hans Lundmark, Jun 04 '23 at 17:37
I see. Thanks. This is very useful info but not obvious to me. — Rohit Pandey, Jun 04 '23 at 17:41
It's nothing particularly deep; you can find it explained in many basic linear algebra textbooks. — Hans Lundmark, Jun 04 '23 at 17:45

score 1 · Answer 1 · answered Jun 04 '23 at 17:54

$\dfrac{1}{2}$.

Take the standard basis $\{e_1,e_2,\dots,e_n\}$ of $\mathbb{R}^n$.

Rotate the entire basis in any way you want, to send $e_1$ to the first point you chose. This is possible (for large $n$) because the $n-1$-dimensional sphere is connected.
Keeping $e_1$ fixed, rotate the entire basis any way you want, to send $e_2$ to the second point you chose. This is possible (for large $n$) because the $n-2$-dimensional sphere is connected.
Etc.
In this way, you will be able to rotate the basis vectors $e_1,e_2,\dots,e_{n-1}$ to the first $n-1$ points you chose, because the $k$-dimensional sphere is connected for $k \ge 1$.
There will be two antipodal points which are orthogonal to each of the first $n-1$ points you chose. In the last step, you chose one or the other of these points, each with probability $\frac{1}{2}$. Either you chose the one which is the image of $e_n$ under all the rotations I described above (in which case your $n$ vectors form a rotation matrix), or you chose the other one (in which case they don't).

If your $n$ vectors do not form a rotation matrix, they will form a reflection matrix.

What is the probability a random orthonormal matrix will be a rotation matrix?

1 Answers1