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. 



No arbitrary product of matrices has eigenvalue 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 $0,0,1,1$.


