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Let $A$ be an $N\times N$ Hermitian matrix. For $p+q$ even, consider the following inequality: $$\frac{1}{N}\sum_{i=1}^N (A^p)_{ii} (A^q)_{ii} \geq \Big(\frac{1}{N}\sum_{i=1}^N (A^p)_{ii} \Big) \Big(\frac{1}{N}\sum_{i=1}^N (A^q)_{ii} \Big) = \frac{1}{N^2} \mathrm{tr}(A^p) \mathrm{tr}(A^q).$$ A probabilistic interpretation of this inequality is that the diagonal elements of $A^p$ are positively correlated with the diagonal elements of $A^q$. Therefore, it is clearly true in the case of $p=q$.

I checked it numerically on random Hermitian matrices (both uniform eigenvalues and uniform entries) with $p=1$ and $q=3$, and found this inequality mostly holds, but not always. The ratio of counterexamples seems to decay exponentially with $N$: 1.9%, 0.7%, 0.24%, 0.09%, 0.003%, 0.001% for $N=3$ to $8$. The ratio of counterexamples also decays fast when $p$ and $q$ both get large.

My questions are: is there a proof why this inequality is almost true? What are the necessary and sufficient conditions for it to hold?

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  • $\begingroup$ Did you consider the rank 1 case? I would start by looking at that first. Just a comment. $\endgroup$
    – Malkoun
    Commented Jun 12 at 20:03
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    $\begingroup$ @Malkoun That's a good point. In this case, A can be scaled to a projector. Whenever A is a projector, we have $A^p=A$ for any p, and this inequality always holds. $\endgroup$
    – WunderNatur
    Commented Jun 13 at 1:30
  • $\begingroup$ Let $f_i$ denote the unit eigenvectors for $A$ for $\lambda_i$. Then $e_k\otimes e_k=\sum_{i,j} \overline{f_i(k)f_j(k)}(f_i\otimes f_j)$. Thus the LHS is $N^{-1}\sum_{i,j,k} \lambda_i^p\lambda_j^q |f_i(k)|^2|f_j(k)|^2$ while the RHS is $N^{-2}\sum_{i,j,k,l}\lambda_i^p\lambda_j^q|f_i(k)|^2|f_j(l)|^2$. If $f_i$ (or the unitary element $[f_i(k)]_{i,k}$) are random, then by Levy's lemma, $|f_i(k)|$ are close to $N^{-1/2}$ and LHS $\approx$ RHS. For the $i=j$ summand, LHS $\geq$ RHS by Cauchy--Schwarz. This perhaps hints why LHS is likely larger than RHS. $\endgroup$
    – Narutaka OZAWA
    Commented Jun 17 at 0:57
  • $\begingroup$ @NarutakaOZAWA This looks interesting, but it might be helpful if your notations ($e_k$, $f_i(k)$, overline, etc.) could be explained a little more. $\endgroup$
    – WunderNatur
    Commented Jun 18 at 2:33
  • $\begingroup$ $e_k$ is for the standard basis and $f_i(k)$ is the value of $f_i$ at $k$. $\{ f_i\otimes f_j : i,j\}$ is an orthonormal basis for $\ell_2^N \otimes \ell_2^N$. $\endgroup$
    – Narutaka OZAWA
    Commented Jun 18 at 3:08

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