In a paper I'm reading I saw the argument below:
Let $G\sim\mathcal{N}(\mathbf{0},\mathbf{I})$ and $A$ is a square matrix with SVD $A=U\Sigma V^T$ where $\Sigma=diag(\sigma_1,\dots,\sigma_n)$. Then $||AG||_2^2$ has the same distribution with $\sum \sigma_j^2G_j^2$ by rotational invariance of $G$.
Now, by rotation invariance, we can already say that $V^TG\sim\mathcal{N}(\mathbf{0},\mathbf{I})$ as $V$ is orthogonal. But when I consider the whole product $U\Sigma V^TG$, I can have $U\Sigma V^TG\sim \mathcal{N}(\mathbf{0},U\Sigma V^T\mathbf{I}(U\Sigma V^T)^T)=\mathcal{N}(\mathbf{0},U\Sigma^2 U^T)$. (saying that $\Sigma^2=diag(\sigma_1^2,\dots,\sigma_n^2)$). And I cannot derive the mentioned equivalence of distributions from there. I'd be glad for any help.
I guess I figured it out, by just using $||Uv||_2^2=||v||_2^2$ and $V^TG\sim \mathcal{N}(\mathbf{0},\mathbf{I}) \sim G$ we can handle it. $||AG||_2^2=||U\Sigma V^T G||_2^2=||\Sigma V^T G||_2^2 \sim ||\Sigma G||_2^2 =\sum \sigma_j^2G_j^2$.