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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!

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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.

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  • $\begingroup$ OK, thanks for mentioning MANOVA (never heard of that!). Any explicity reference of the how this ratio makes its way into MANOVA ? About low ranksity, more generally, it's not hard to prove that $1 / rank(\Sigma) \le \sigma_1 / \|\Sigma\|_F \le 1$. For example, see math.stackexchange.com/a/3638563/168758. $\endgroup$
    – dohmatob
    Apr 22, 2020 at 17:40
  • $\begingroup$ In 1-way MANOVA one wants to test equality of means when the response is multivariate. The statistics are functions of the eigenvalues of $E^{-1}H$ where $H = n\sum_i(Y_{i.} - Y_{..})(Y_{i.} - Y_{..})^T$. Now, if the centroids of the groups are nearly aligned, a test statistic such as the Roy's maximum root will have better power than alternative statistics, such as Pillai's or Lawley-Hotelling. Too long to explain in detail in a comment. $\endgroup$
    – F. Tusell
    Apr 22, 2020 at 18:01
  • $\begingroup$ OK, great. Thanks! BTW, the comment could be made part of the answer. $\endgroup$
    – dohmatob
    Apr 22, 2020 at 18:19

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