Let $A\in\{0,1\}^{n\times n}$ be a random, symmetric matrix, in which each upper triangular entry is sampled iid. from Bernoulli($p$).
Let $\lambda=(\lambda_1, \dots, \lambda_n)$ be the vector of eigen values of $A$; let $\|\lambda\|_k = (\sum_i \lambda_i^k)^{1/k}$ be the $k$ norm of $\lambda$; and for $S\subseteq\{1,\dots,n\}^2$ let $A_S$ be the sub-matrix of $A$ that is 1 only in the entries $(i,j)$ only if $A_{i,j}=1$ and $(i,j)\in S$. Let $\lambda_S$ be the eigenvalues of the sub-matrix.
Now $E\|\lambda\|_3 \approx np$ and $E\|\lambda\|_2 \approx n\sqrt{p}$, so for $p=o(1)$ we expect $\|\lambda\|_3 << \|\lambda\|_2$. I would like to show that this continues to hold for sub-matrices of $A$. In particular I'm interested in showing:
$$ \frac{\|\lambda\|_3}{\|\lambda_S\|_3} \left(\frac{\|\lambda_S\|_2}{\|\lambda\|_2}\right)^{1+o(1)} + \frac{\|\lambda_S\|_2^{2/3+o(1)}}{\|\lambda_S\|_3} \ge 1. $$
Which would naturally follow if $\|\lambda_S\|_3/\|\lambda_S\|_2$ is not suddently too large.
My first idea was that the worst case would be if $A_S$ was close to the first principal component of $A$. In that case $\lambda_S = (\lambda_1, 0, \dots)$, and so $\|\lambda_S\|_3=\|\lambda_S\|_2$, which is clearly not good.
However, perhaps it is not usually possible for a sub-matrix to be close to the principal component.
Is there a way I might try to bound the eigenvalue norm-ratio of a sub-matrix in the above sense?