TITLE: Walks of Even Length on a Bipartite Graph
QUESTION [5 upvotes]: 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?

REPLY [2 votes]: Yes, because $M^2$ is block diagonal, so
$$
M^{2k}=(M^2)^k=d^{-2k}\pmatrix{
A^TA&0\\
0&AA^T
}^k
=
d^{-2k}\pmatrix{
(A^TA)^k&0\\
0&(AA^T)^k
}.
$$
To put it in a more combinatorial way, each even-length walk on $G$ consists of $2k$ steps, so it can be split into $k$ bisteps.  A bistep travels from one partite set of $G$ to the other and then back again, so ends up remaining within the same partite set.  The matrix $A^T A$ tells you how many ways to take a bistep from one vertex to another, within one of the partite sets, and $A A^T$ does the same for the other partite set.