Let $A_1, A_2,..., A_m$ be a collection of real $n \times n$ matrices.
What are some conditions (necessary or sufficient) on $A_1,...,A_m$ for any product $A_{i_1} A_{i_2} \cdots A_{i_k}$ to have non-negative trace?
One sufficient condition is if there is some matrix $P$ so that $PA_iP^{-1}$ has positive entries for each $i$.
However, this condition is not necessary: Let $A_1 = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ and $A_2 = \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix}.$ Then products of $A_1$ and $A_2$ are of the form $\begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix}$ for $a \in \mathbb{Z}$, and thus have trace 2. But there is no matrix $P$ so that $P A_1 P^{-1}$ and $PA_2 P^{-1}$ have positive entries.
This question is inspired by the paper Proof of a conjecture on immanants of the Jacobi-Trudi matrix, where Greene shows that irreducible characters evaluated on certain elements of $\mathbb{C}[\mathfrak{S}_n]$ are non-negative by showing that the corresponding elements of the irreducible representation can be written as matrices with positive entries.
Edit: The case with only one matrix isn't obvious to me. If $\mathrm{Tr}(A^k) \ge 0$ for all $k$, then \begin{equation} p_k(\lambda) = \lambda_1^k + \lambda_2^k +...+\lambda_n^k \ge 0, \end{equation} where $\lambda_i$ are the eigenvalues of $A$ and $p_k(\lambda)$ is the $k^{th}$ power-sum symmetric function evaluated at $\lambda_1,\lambda_2,...,\lambda_n$. It might be possible to use Newton's identities to simplify this, but I'm not sure how.