Normalizing a normally distributed vector to unit length If I have a random vector $\mathbf{y}$ generated from multivariate gaussian distribution $\mathcal{N}(\mathbf{0}, \mathbf{C})$, then I normalized it to unit length, which is,
$$\mathbf{y} \sim \mathcal{N}(\mathbf{0}, \mathbf{C}),$$
$$ \mathbf{z} = \frac{\mathbf{y}}{||\mathbf{y}||}.$$ 
Is it possible to determine the type of the distribution of $\mathbf{z}$? 
If $\mathbf{y} \sim \mathcal{N}(\mathbf{0},\mathbf{I})$, I think $\mathbf{z}$ should satisfy a multivariate uniform distribution with each of its components is identically and independently distributed from $\mathcal{U}(-1, 1)$. Am I correct? and how to make a proof?
 A: No, if $y\sim N(0,I)$, then $z$ is uniformly distributed on the unit sphere. So the components can not be independent.
A: The random vector $\mathbf{Y}$ taking values in $\mathbb{R}^d$ is radially symmetric if $\forall A \in     \mathbb{R}^{d\times d} $ orthogonal we have that $\mathbf{Y}$ and $A \mathbf{Y}$ have the same distribution.
Now we show the following: If $\mathbf{Y}$ is radially symmetric, then $\mathbf{Y}/||\mathbf{Y}|| \sim Uniform(S^{d-1})$ where $S^{d-1}$ denotes the unit sphere in $\mathbb{R}^d$.
Proof: Let $A$ be orthogonal, then
$A \frac{\mathbf{Y}}{||\mathbf{Y}||} = \frac{A\mathbf{Y}}{||\mathbf{Y}||} = \text{Proj}(A\mathbf{Y})$
where $\text{Proj}(\mathbf{Y}) =  \frac{\mathbf{Y}}{||\mathbf{Y}||}$ is the projection on the unit sphere. Since $\mathbf{Y}$ is radially symmetric, we have that $\text{Proj}(A\mathbf{Y})$  and $\text{Proj}(\mathbf{Y})$ follow the same distribution. Thus, the random vector $\frac{\mathbf{Y}}{||\mathbf{Y}||}$ takes all its values over $S^{d-1}$ and is radially symmetric. This is exactly the definition of a uniform distribution over $S^{d-1}$
Remark: One can show that the only radially symmetric random vector that has independent components is the multidimensional standard normal.
