Suppose $G \in \mathbb{R}^{m \times n}$ has i.i.d. rows $g_i \sim \mathcal{N}(0, \Sigma)$ for some diagonal matrix $\Sigma = \text{diag}(\lambda_1,\dots,\lambda_n)$ where the diagonal entries satisfy $\lambda_1 \geqslant \dots \geqslant \lambda_n > 0$. Can anything be said about the distribution of the columns of $V$ where $G = U\Lambda V^T$ is the singular value decomposition of $G$? The following question considers the case when the entries of $G$ are i.i.d. $\mathcal{N}(0,1)$ and shows that the singular vectors are uniformly distributed on a sphere of radius $1$:

Singular vector of random Gaussian matrix

The proof capitalizes on the rotational invariance of the standard normal distribution but I don't think the same argument can be made here due to the form of $\Sigma$. Any help or references would be greatly appreciated.

  • $\begingroup$ Did you ever find your answer? $\endgroup$ Aug 3, 2022 at 3:35


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