Is there any symmetric real matrix $A$ such that $A^{\otimes n}$ is copositive for all positive integers $n$, but such that A is neither positive semidefinite nor has just non-negative entries?

Comments:

1) $A$ matrix "$A$" is copositive if, for any vector $x$ with non-negative entries, $x^T·A·x\geq0$.

2) For matrices of dimension $2$, the solution to this question is "no". This can be seen by taking Hadamard products of $A$ with itself and the matrix $A$' that results when we permute $A$'s rows and columns [i.e., $A'(2,2)=A(1,1), A'(1,1)=A(2,2), A'(1,2)=A'(2,1)=A(1,2)]$.

3) The sets of positive semidefinite matrices and matrices with non-negative entries are both closed under tensors and copositive. They are "maximal" in the sense that, for any matrix $B$ that is not positive semidefinite (non-negative), there exists a positive semidefinite (non-negative) matrix $A$ such that $A\otimes B$ is not copositive.

4) I encountered this problem while I was working on quantum decoherence functionals. The original question was how to find a matrix (or decoherence functional) $A$ that is not positive semidefinite and such that $x^T·A^{\otimes n}·x\geq0$ for all vectors $x\in\{0,1\}^m$. Non-negative matrices didn't work, since, due to certain linear constraints satisfied by decoherence functionals, all decoherence functionals with non-negative entries must be positive semidefinite. I relaxed this question to vectors with non-negative entries, in the hope that someone has considered this problem in the past.

copositivemean? $\endgroup$ – Igor Rivin Nov 7 '14 at 20:52