No arbitrary product of matrices has eigenvalue 1?

Question

Consider the matrix $D$, adjacency matrix of an undirected graph $G$ on $n$ vertices, with the notation that $d_{i,i}=0,\forall i$.

The matrices $A_i$ are constructed from Identity matrices, $I_{n*n}$ with $i^{th}$ column replaced with $i^{th}$ column of $D$.

I want to consider matrix $A=\prod_{s\in S} A_{s}$, where $S$ is a sequence containing all the entries in $\{1,2,\ldots, n \}$ at least twice. For example for $n=4$, $S$ can be (3,1,3,1,4,2,4,1) and in that case $A=A_{3}A_{2}A_{1}A_{3}A_{1}A_{4}A_{2}A_{4}A_{1}$. i.e., the only condition is that in the product every entry $k\leq n$ should occur at least twice.

I want to know that if there is a sequence $S$ such that $A$ has eigenvalue $1$, or that for every sequence $S$, $A$ doesn't have the eigenvalue $1$.

I tried examples by hand and also ran few simulations on matlab, but never found an $A$ with eigenvalue $1$.

I want to know if these kind of matrices have been studied before, or any remarks or hints on how to approach this.

Answer 1

In the case $n=4$ you could have $D = \pmatrix{0 & 0 & 0 & 1\cr 0 & 0 & 1 & 0\cr 0 & 1 & 0 & 0\cr 1 & 0 & 0 & 0}$, in which case $A_1 A_2 A_3 A_4 A_1 A_2 A_3 A_4 = \pmatrix{0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0\cr 0 & 1 & 1 & 0\cr 1 & 0 & 0 & 1\cr}$ has eigenvalues $0$ and $1$, both with multiplicity $2$.

EDIT: In fact, in this case for any permutation of $A_1 \ldots A_4 A_1 \ldots A_4$ you always get the eigenvalues $0,0,1,1$.