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Let $W = aI_{n\times n} + bJ_{n\times n}$, where $I$ is an identity matix, $J$ is the matrix of all ones, $a,b\in\mathbb{R}$ and a+b>0. Also, let $A = \mathbf{P} - \mathbf{p}\mathbf{p}^{T}$, where $\mathbf{p} = (p_{1},\ldots,p_{n})^{T}$ and $\mathbf{P} = \rm{diag}(\mathbf{p})$ with $\sum_{i=1}^{n}p_i = 1$ and $0<p_i<1$. Is it possible to show that $trace((AW)^K)$ for any $K>1$ is symmteric function of $\{p_1,\ldots,p_k\}$?

I tried for $K=2,3$ and its true but in general not able to show.

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  • $\begingroup$ you write $AV$ but probably you mean $AW$ (because $V$ is undefined) $\endgroup$ Commented Feb 2, 2018 at 20:51

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To check the symmetry of the trace under a given permutation of the $p_i$'s, take the corresponding permutation matrix $S$ and insert in the trace noting that $S^TS=I$ and $SWS^T=W$:

$${\rm tr}\,(AW)^K={\rm tr}\,(AS^T SWS^TS)^K={\rm tr}\,(\tilde{A}W)^K$$

with $\tilde{A}=SAS^T$ the matrix $A$ after the desired permutation of the $p_i$'s.

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    $\begingroup$ Very nice argument! $\endgroup$
    – Igor Rivin
    Commented Feb 2, 2018 at 23:12

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