Positivity of power of positive PSD matrices

Question

Background: Let $M$ be an $n\times n$ matrix with nonnegative entries. It is immediate that for any integer $k$, $M^k$ has nonnegative entries.

Suppose now that, on top of having nonnegative entries, $M$ is a positive semi-definite matrix (i.e., it is symmetric and has nonnegative eigenvalues). Now one may ask whether for any real $p\ge 1$, $M^p$ has nonnegative entries. Surprisingly, the answer is a no. The matrix $$ M= \begin{bmatrix} 0.96523 & 2.6398 & 0.012905 & 0.059013\\ 2.6398 & 10.053 & 3.0808 & 0.029887\\ 0.012905 & 3.0808 & 26.252 & 3.2929\\ 0.059013 & 0.029887 & 3.2929 & 0.52308 \end{bmatrix} $$ has positive eigenvalues, however $M^{1.5}$ has negative entries. This example is due to Koenraad Audenaert.

A nicer looking matrix of the same kind is $$ M= \begin{bmatrix} 2 &6 &0.1 &0.1\\ 6 &30 &6 &0.1\\ 0.1 &6 &30 &6\\ 0.1 &0.1 &6 &2 \end{bmatrix} $$

Question: Are there such (i.e., $M^p$ has negative values) positive PSD matrices whose eigenbasis is the discrete Fourier transform?

To be more explicit, let us fix the Fourier transform on $\mathbb F_2^n$ as $$H_{x,y} = (-1)^{\langle x,y\rangle}\quad x,y\in\mathbb F_2^n$$

Is there a nonnegative function $f:\mathbb F_2^n\to\mathbb R_+$ such that $Hf\geq 0$ however, $Hf^p$ has negative values for some $p\geq 1$? Here, $f^p$ denotes the coordinate-wise powering of $f$.

Recall that $H/2^{n/2}$ diagonalizes matrices of the form $M(x,y)=f(x+y)$, so the existence of such an $f$ is equivalent to existence of positive PSD matrices diagonalized in the Fourier basis whose $p$th power contains negative entries.

Thank you

Answer 1

There is an $f$ such that $f\ge 0$, $Hf\ge 0$ however $Hf^{p}$ contains negative values for some $p\ge 1$. Take $$ f = \begin{bmatrix} 727 & 200 & 163 & 234 & 429 & 448 & 437 & 6 \end{bmatrix}^T $$ The vector $Hf$ is positive, however $(Hf^{8/7})_{100}\approx -3.35$. Here the subscript $100\in \mathbb F_2^3$ is the 5th index in lexicographic order.