How to prove that singular vectors have uniform distribution on the sphere? Problem
I have a matrix
$$X = diag(x_1, x_2, x_3), \quad
x_i \sim N \left( 0, 1 \right) \quad
i = \overline{1, 3}$$
and $x_i$ are independent.
SVD is used to find singular vectors:
$$X = U \cdot \Sigma \cdot V^T,$$
where $V$ and $U$ are orthogonal matrices and $\Sigma$ is diagonal matrix,
such that
$$\Sigma = diag\left(\sigma_1, \sigma_2, \sigma_3\right), \,
\sigma_1 \geq \sigma_2 \geq \sigma_3.$$
I need to find distribution of matrix $V \cdot U^T$.
But first I need to find distribution of left and right singular vectors.
In question
Singular vector of random Gaussian matrix
on this forum that singular vectors have uniform distribution
on the sphere of radius $1$. 
But I don't understand how to prove it.
My solution (probably wrong)
I tried to show that distribution of $X$
is invariant with respect to rotations.
To do this we need an orthogonal matrix $A \, : \, A^T = A^{-1}$
and form new matrix
$$X' = A \cdot X,$$
so that distribution of $X$ and $X'$ would be the same.
Though, I've found out that they don't have the same distribution.
$$x_{ij}' =
\sum \limits_{s = 1}^s a_{is} \cdot x_{sj} = a_{ij} \cdot x_{jj}$$
as the matrix $X$ is diagonal.
Then the mathematical expectation
$$\mathbb{E} x_{ij}' =
a_{ij} \cdot \mathbb{E} x_{jj} =
a_{ij} \cdot \mathbb{E} x_j =0$$
as $\mathbb{E} x_i = 0 \, i = \overline{1, 3}$.
But the variance of $x_{ij}'$ doesn't equal 1:
$$Dx_{ij}' =
a_{ij}^2 \cdot Dx_{jj} =
a_{ij}^2 \cdot Dx_j =
a_{ij}^2 \leq 1.$$ 
It can be equal to $1$, but not always.
Questions
What is the right way to prove the statement,
that singular vectors have uniform distribution
on the sphere of radius $1$?
Will the distribution of singular vectors change if diagonal matrix elements $x_i$ has normal distribution with expectation $a$ and variance $\sigma^2 \, : \,
x_i \sim N \left( a, \sigma^2 \right) $?
Am I right that then sphere would have radius $a$?
What if matrix $X$ isn't diagonal? 
 A: First, let's write down the probability density function (PDF) of the matrix $X$; this is just the joint density of $x_1, x_2, x_3$. Since those are independent normals, we have
$$
\prod_{i=1}^3 \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2} x_i^2} \, dx_i = \frac{1}{(\sqrt{2\pi})^3} e^{-\frac{1}{2} (x_1^2+x_2^2 + x_3^2)} \, dx_1 dx_2 dx_3 = \frac{1}{(\sqrt{2\pi})^3} e^{-\frac{1}{2} \operatorname{tr}(X X^T)} \, dX.
$$
Now, the singular value decomposition $X = U \Sigma V^T$ is actually a change of variables (up to ordering the singular values and signs of rows/columns of $U, V^T$, but let's gloss over that) $X \to (U, \Sigma, V^T)$. In other words, you can go from the probability density of the variable $X$ to the probability density of the three variables $U, \Sigma, V^T$. To do so, substitute $X = U \Sigma V^T$ into the PDF of $X$ to obtain
$$
\frac{1}{(\sqrt{2\pi})^3} e^{-\frac{1}{2} \operatorname{tr}(X X^T)} \, dX = \frac{1}{(\sqrt{2\pi})^3} e^{-\frac{1}{2} \operatorname{tr}(\Sigma^T \Sigma)} \, dX
$$
using the cyclic property of $\operatorname{tr}$ and the fact that $U, V$ belong to the orthogonal group.
Recall that the differentials $dX$ and $dU \, d\Sigma \, dV$ are related by the determinant of the Jacobian matrix for the change of variables; according to this, we have 
$$
dX = \prod_{1 \leq i < j \leq 3} (\sigma_i^2 - \sigma_j^2) \, d\Sigma \, dU \, dV,
$$
so substituting gives
$$
\frac{1}{(\sqrt{2\pi})^3} e^{-\frac{1}{2} \operatorname{tr}(X X^T)} \, dX = \frac{1}{(\sqrt{2\pi})^3} e^{-\frac{1}{2} \operatorname{tr}(\Sigma^T \Sigma)} \prod_{1 \leq i < j \leq 3} (\sigma_i^2 - \sigma_j^2) \, d\Sigma \, dU \, dV.
$$
Now, note that the last term is the product of three PDFs (up to normalization): one for $\Sigma$ (namely $e^{-\frac{1}{2} \operatorname{tr}(\Sigma^T \Sigma)} \prod_{1 \leq i < j \leq 3} (\sigma_i^2 - \sigma_j^2) \, d\Sigma$), one for $U$ (namely $1 \, dU$), and one for $V$ (namely $1 \, dV$). Hence, those random variables (matrices) are independent, and moreover, $U$ and $V$ are distributed uniformly over the orthogonal group. Since $U$ and $V$ are independent and uniformly distributed over the orthogonal group, so is $V U^T$.
