Here, we call a random vector $X\in\mathbb{R^n}$ follows a spherical distribution (or rotationally invariance), if $X$ satisfies that for any orthogonal matrix $Q$, $QX\stackrel{d}=X$. Is it true that $\|X\|_2$ is independent of $\frac{X}{\|X\|_2}$?
I know that in a traditional way, if $X$ has a density function $f(x^Tx)$ with $f(\cdot)\geq 0$, then by polar coordinate transformation, i.e. $x\to(r,\theta)$, we can prove that $r$ is independent of $\theta$, which implies the independence of $\|X\|_2$ and $\frac{X}{\|X\|_2}$.
But in a general case, a spherical distribution $X$ does not necessarily have density, and I have no idea in this case.
Here is a proof
For convenience define two new random variables $R = \|X\|_2$ and $U = \frac XR$. I wish to show that the distribution functions factor. That is
$$F_{(R,U)}(r,u) = F_R(r)F_U(u)$$
Now pick arbitrary $u$ and $v$ on the unit sphere. We can then pick an orthogonal matrix $Q$ such that $Qu = v$. Then we have the following
$$\frac {F_{(R,U)}(r,u)}{F_R(r)} = \frac{F_X(ru)}{F_R(r)} = \frac{F_X(Qru)}{F_R(r)} =\frac{F_X(rv)}{F_R(r)} =\frac{F_{(R,U)}(r,v)}{F_R(r)}$$
This means that we can pick some function $g$ such that
$$ \frac{F_{(R,U)}(r,u)}{F_R(r)} = g(u) $$
From here we simply note that
$$F_U(u) = \lim_{r \to \infty} \frac{F_{(R,U)}(r,u)}{F_R(r)} = g(u)$$
Hence we have the desired factorisation, and $U$ and $R$ are independent