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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.

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  • $\begingroup$ Recall that if $A$ is a complex matrix, then $\overline{A}$ is the matrix obtained from $A$ by replacing each entry with its complex conjugate. i.e. $\overline{A}=(A^T)^*$. If $A_1,\dots,A_r$ are complex matrices, then every product $A_{i_1}\otimes\overline{A_{i_1}},\dots,A_{i_n}\otimes\overline{A_{i_n}}$ has real non-negative trace. Furthermore, in quantum information theory, the completely positive superoperators can be put into a linear one-to-one correspondence with the sums of the form $A_1\otimes\overline{A_1}+\dots+A_r\otimes\overline{A_r}$. $\endgroup$
    – Joseph Van Name
    Commented Dec 16, 2022 at 22:30
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    $\begingroup$ One sufficient condition is that there is some real matrix $P$ so that $PA_iP^{-1}$ is upper triangular with non-negative diagonal entries for each $i$. $\endgroup$
    – Aaron Meyerowitz
    Commented Dec 17, 2022 at 0:21

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