ah, nvm, I see it - you can see $p_n$ is orthogonal to any monomial in $\{1, ..., x^{n-1}\}$ by just taking the expectation and noting that the bottom row is now a bunch of moments equal to one of the above rows, so the determinant's zero, and that's enough. sweet trick. thanks!

amazing! this is what I was looking for. is there any easy way to see why these polynomials are orthogonal (e.g. using some rule for the product of two determinants)?

I see! Do you know of anywhere the next few terms might be written out?

yep! it's actually strictly positive

In particular, these conditions seem like they might imply that the tops of the spectra of $A$ and $\Lambda$ are approximately the same, which is the sort of result I'm looking for

@dohmatob Hmm, maybe indirectly! That provides one constraint on the spectrum of $A$, but there are many ways to satisfy that constraint without giving a spectrum of the type I describe. However, using an informal physicist's analysis, I find that there are actually a family of related conditions! I find that $\mathbb{E}\left[\text{trace}(A^k)\right] = \text{trace}(\Lambda^k) + \mathcal{O}(\text{trace}(\Lambda)^k/n)$ when $k \ll n$. These seem like strong enough conditions to say something interesting about the spectrum of $A$, though I'm not sure how best to do so.

@CarloBeenakker Sure! I tend to see this behavior whenever the eigenvalues decay fairly quickly with index. The systematic gap is smaller the faster the decay (which agrees with it being quite large when the decay rate is zero). In general, I find that large, standout eigenvalues are faithfully recovered by this random projection procedure.