Let $n\in \mathbb{N}$, $c > 1$ and $T = T (n) = \lceil cn \rceil$. Let $p \in (0, 2]$. Givendiagonal $T \times T$ matrix $\Lambda$ with diagonal entries $\lambda_j =j^p$, $j = 1, \ldots, T$, let $\gamma_n$ be the minimum of $n {Tr} ((Q^T\Lambda Q)^2) - {Tr} (Q^T \Lambda Q)^2$ over all $T \times n$ matrices$Q$ with $Q^T Q = I$. I'm interested in whether $\gamma_n \rightarrow \infty$,and at what rate. I suspect $\gamma_n \asymp n^{2 p + 2}$ since this is trueif we take the minimum over $Q$ with standard basis vectors as columns, butI'm not sure how to prove this. The stationarity equation from Lagrangemultipliers is$$(I - QQ^T) \Lambda Q \left( Q^T \Lambda Q - \frac{{Tr}(Q^T \Lambda Q)}{n} I \right) = 0.$$If $Q^T \Lambda Q - \frac{{Tr} (Q^T\Lambda Q)}{n} I$ were invertible, we could deduce $(I - QQ^T) \Lambda Q = 0$,which is equivalent to $\Lambda : Q \rightarrow Q$, which isequivalent to saying that the columns of $Q$ are eigenvectors of $\Lambda$.But I'm not sure whether $Q^T \Lambda Q - \frac{{Tr} (Q^T \Lambda Q)}{n}I$ is invertible at every minimizer. It isn't for arbitrary $\Lambda$, but maybeit is for this specific $\Lambda$?
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