Let $k$ be any field and $R\subseteq M_s(k)$ be a subring of $s\times s$ matrices over $k$. Identify $k$ with the scalar matrices, so that $k\subseteq R$. Let $A\in M_n(R)$ be an $n\times n$ matrix.

The Question: Is it possible to construct a matrix $A=[a_{ij}]$, as above, which satisfies Properties P1-P3 below?

Property 1: For every $m=1,2,\ldots$ all the elements on the diagonal of $A^m$ belong to $k$.

Property 2: There exists a cycle $(i\to j\to \ldots\to z\to i)$ such that $a_{ij}a_{jk}\ldots a_{zi}+a_{iz}\ldots a_{kj}a_{ji}$ is not in $k$.

Property 3: Let $L\subseteq R$ denote the set of products of the form $a=a_{ij}a_{jk}\ldots a_{zi}$. For every nonzero vector $v\in k^s$ the set $\{av\mid a\in L\}$ spans all of $k^s$.

Some observations:

1. Negation of P2 implies P1.

2. If $R\subseteq M_s(k)$ is isomorphic to the ring of Quaternions and $A$ is a Hermitian matrix, that is $a_{ij}=\overline{a_{ji}}$, then $A$ does not satisfy P2, hence it satisfies P1. It is easy to find such a matrix which satisfies also P3. This can be done for several other rings with sufficiently good conjugation.

3. If $A\in M_{n}(R_1)$ and $B\in M_{n}(R_2)$ satisfy P1, and additionally, the diagonals of $A^m$ and $B^m$ coincide for every $m$, then $C=A\oplus B \in M_n(R_1\oplus R_2)$ defined entrywise as: $$ c_{ij}= \begin{bmatrix} a_{ij} & \vert & 0 \\ \hline 0 & \vert & b_{ij} \end{bmatrix} $$ also satisfies P1. Thus it is easy to construct a matrix which satisfies P1 and P2.

4. Conjugation by a diagonal matrix or a permutation matrix does not affect P1-P3.


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