 **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](http://www.wolframalpha.com/input/?i=eigenvalues%20of%20%7B%7B000.96523,002.6398,000.012905,000.059013%7D,%7B002.6398,%20010.053,%20003.0808,%20000.029887%7D,%7B000.012905,%20003.0808,%20026.252,%20003.2929%7D,%7B000.059013,%20000.029887,%20003.2929,%20000.52308%7D%7D), however $M^{1.5}$ has [negative entries](http://www.wolframalpha.com/input/?i=%7B%7B000.96523,002.6398,000.012905,000.059013%7D,%7B002.6398,%20010.053,%20003.0808,%20000.029887%7D,%7B000.012905,%20003.0808,%20026.252,%20003.2929%7D,%7B000.059013,%20000.029887,%20003.2929,%20000.52308%7D%7D%5E%281.5%29). This example is [due to Koenraad Audenaert](http://personal.rhul.ac.uk/usah/080/QITNotes_files/pospos.pdf). 


A [nicer looking matrix of the same kind](http://www.wolframalpha.com/input/?i=diagonalize%20%7B%7B2.,%206.,%200.1,%200.1%7D,%20%7B6.,%2030.,%206.,%200.1%7D,%20%7B0.1,%206.,%2030.,%206.%7D,%20%7B0.1,%200.1,%206.,%202%7D%7D%5E(1.5)) 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



Positivity of power of positive PSD matrices
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.


