Given a random walk on a simple $d$-regular bipartite graph $G$. The adjacency matrix $A'$ of $G$ may be split into blocks $ A'=\pmatrix{ 0 &A^T\\ A&0 }, $
The propagation operator $M=A'/d$ is used for random walks on such a graph. It describes where you can get (and the probality) in one step, starting from a given vertex.
Now $M^2$ describes where you get with two steps and has the following form $ M^2=d^{-2}\pmatrix{ A^TA&0\\ 0&AA^T }. $
I think I may interpret $A^TA$ as the adjacency matrix of a $d^2$-regular, not necessarily simple graph, $\phantom{(EDIT2:)}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!$ since it will have $d$ self-loops for each vertex, graph, so my question is:
May I use $\phantom{(EDIT:)}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!$ one of the blocks of $M^2$ as propagation operator, when I restrict a given problem to walks of even length?
It feels to me that I'm loosing some information because the graph that $\phantom{(EDIT:)}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!$ one of the blocks of $M^2$ acts on has only half of the vertices of $G$. Is this related to a certain kind of symmetry adapted basis?