# Positivity of power of positive PSD matrices

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

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.