Statistical intepretation of ratio of spectral norm of covariance matrix to its Frobenius norm In statistics, given a covariance matrix $\Sigma$ with singular values $\sigma_1 \ge \sigma_2 \ge \ldots \ge \sigma_p$, is the ratio of its spectral norm to the Frobenius norm, i.e the ratio $\dfrac{\sigma_1}{\sqrt{\sum_{i=1}^p \sigma_j^2}}$ of any interest ? That is, does this ratio appear naturally in the analysis of certain procedures, algorithms, convergence limits, etc. ?
Thanks in advance!
 A: It could be taken as an indicator of how close the matrix is to a rank 1 matrix. In MANOVA you have cases where this matters.
A: Here's one statistical interpretation of Frobenius norm of $\Sigma$.
Lets say that there's a dataset with $b$ rows $X$ whose empirical covariance matches $\Sigma$. We have
$$\|\Sigma\|^2_F=\|\frac{1}{b}X^TX\|_F^2=\frac{1}{b^2}\|XX^T\|_F^2$$
You can rewrite the last term as
$$\frac{1}{b^2}\sum_{ij} \langle x_i, x_j\rangle^2$$
In other words, $\|\Sigma\|^2_F$ is the average dot product between two pairs of examples in any representative dataset.
Meanwhile, spectral norm of $\Sigma$ is the largest variance among all directions, a kind of normalizing factor.
If we use trace instead of spectral norm, it gives us the quantity which occurs in linear estimation literature:
$$R=\frac{(\text{Tr}\Sigma)^2}{\text{Tr}\Sigma^2}$$
It's kind of a measure of degrees of freedom, giving the size of the largest batch such that samples are mostly pairwise uncorrelated.
When observations come from distributions with high value of $R$ the corresponding estimation problem is easy, for details, see definition 3 and Theorem 2 of Uniform Convergence of Interpolators: Gaussian Width, Norm Bounds, and Benign Overfitting
