Independence of Eigenvalues of Wishart

Question

This question regards a previous post, but it is not immediately obvious the two are related, so I ask it anyways: are the eigenvalues of a Wishart matrix $\mathbf{S}$ $=$ $\frac{1}{N}\frac{\mathbf{X'}\mathbf{1}\mathbf{1}'\mathbf{X}}{N}$ independent in the finite size case (i.e., $N$ $\approx$ 3-6, quite small)?

I do realize that the covariance between elements could be computed from the joint PDF, but this seems exceedingly laborious; a reference would be time better spent, or yet, a simple argument.

The implication is that the distribution of the sum of the eigenvalues could be computed from the characteristic function method if that were true.

Answer 1

Perhaps too late for the original questioner's purposes, but in case anyone else has the same question, a couple of relevant results are:

Let $nS\sim W(n,\Sigma)$, and let $\lambda_1,\dots,\lambda_p$ the population eigenvalues.

• In the case $\lambda_1>\dots>\lambda_p$, the ordered eigenvalues $l_1>\dots>l_p$ of $S$ satisfy $\text{Var}(l_i)=2\lambda_i^2n^{-1}+O(n^{-2})$ and $\text{Cov}(l_i,l_j) = 2\frac{\lambda_i\lambda_j}{(\lambda_i-\lambda_j)^2} n^{-2} + O(n^{-3})$.
• In the case $\lambda_1=\dots=\lambda_p=1$, the unordered eigenvalues $l_1,\dots,l_p$ of $S$ satisfy $\text{Var}(l_i)=\frac{p+1}{n}$ and $\text{Cov}(l_i,l_j) = -\frac{1}{n}$.

So the sample eigenvalues are not independent, but their covariance decreases asymptotically as $n\rightarrow\infty$.

Source: Khatri and Srivastava, An introduction to multivariate statistics (1979), Section 9.4.