Unlocking the walk matrix of a graph

Let $G$ be a graph with vertex set $V=\{v_{1},\dots,v_{n}\}$ and adjacency matrix $A.$ For a subset $S$ of $V$ let $\e=(x_{1},\,\dots,\,x_{n})^{\tt T}$ be the characteristic vector of $S,$ that is, $x_{\ell}=1$ if $v_{\ell}\in S$ and $x_{\ell}=0$ otherwise. Then the $n\times n$ matrix $W^{S}:=\big[{\rm e},\,A{\rm e},\,A^{2}{\rm e},\dots,A^{n-1}{\rm e}\big]$ is the {\it walk matrix} of $G$ for $S.$ This name relates to the fact that in $W^{S}$ the $k^{\rm th}$ entry in the row corresponding to $v_{\ell}$ is the number of walks of length $k-1$ from $v_{\ell}$ to some vertex in $S$. Since $A$ is symmetric the characteristic vector of $S$ can be written uniquely as a sum of eigenvectors of $A.$ In particular, we may enumerate the distinct eigenvalues $μ_{1},\dots, μ_{s}$ of $A$ so that \begin{eqnarray}\label{SSA}{\rm SD}(S)\!:\,\e&=&\e_{1}+\e_{2}+\dots+\e_{r}\, \end{eqnarray} where $r\leq s$ and $\e_{i}$ is an eigenvector of $A$ of $μ_{i}$ for all $1\leq i\leq r. We refer to (\ref{SSA}) as the {\it spectral decomposition} of $S,$or more properly, of its characteristic vector. The key result of this paper is that the walk matrix$W^{S}$determines the spectral decomposition of$S$and {\it vice versa.} This holds for any non-empty set$S$ of vertices of the graph and explicit algorithms which establish this correspondence are given. In particular, we show that the number of distinct eigenvectors that appear in \,(\ref{SSA})\, is equal to the rank of $W^{S}.$Several theorems can be derived from this result. We show that$W^{S}$determines the adjacency matrix of$G$if$W^{S}$has rank$\geq n-1$. This theorem is best possible as there are examples of pairs of graphs with the same walk matrix of rank$n-2$ but with different adjacency matrices.

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