Matrices such that their sum equals their product

Question

Let $N > 1$ and consider square matrices of size $N$. Let $J$ be the matrix full of $1$'s. Suppose you have $n \geq 1$ commuting matrices $A_i$ over some finite field and such that $\sum_i A_i = \prod_i A_i = J$. Is there anything interesting that we can say about these matrices? Thanks!

Answer 1

1. If $n=2$, then it is easy. Assume that $char(K)\not= 2$. We obtain $A_2=J-A_1$ and $(A_1-1/2J)^2=(N/4-1)J$. It remains to solve an equation in the form $U^2=\alpha J$ where $\alpha\in\mathbb{R}$. Note that $J$ is diagonalizable over $K$ with $spectrum(J)=\{N,0,\cdots,0\}$ except if $char(K)$ divides $N$.
2. $n>2$. The $(A_i)_i$ are simultaneously triangularizable over $L$, an algebraic closure of $K$. Let $spectrum(A_i)=(\lambda_{i,j})_j$. Then, there are orderings of the previous spectra s. t. $\sum_i\lambda_{i,1}=\Pi_i\lambda_{i,1}=N$ and, for every $j>1$, $\sum_i\lambda_{i,j}=\Pi_i\lambda_{i,j}=0$.