Unlocking the walk matrix of a graph

Let G be a graph of order n, with vertex set $$V=\{v_{1},\dots ,v_{n}\}$$ and adjacency matrix A. For non-empty S vertices let $$\mathrm{e}=(x_{1}, \dots , x_{n})^\mathtt{T}$$ the characteristic vector S, that is, $$x_{\ell }=1$$ if $$v_{\ell }\in S$$ }=0$$ otherwise. Then $$n\times n$$ $$\begin{aligned} W^{S}:=\big [{\mathrm{e}}, A{\mathrm{e}}, A^{2}{\mathrm{e}},\dots ,A^{n-1}{\mathrm{e}}\big ] \end{aligned}$$ is walk for S. This term refers to fact in $$W^{S}$$ $$k{\mathrm{th}}$$ entry row corresponding }$$ number walks length $$k-1$$ from some $$\mu _{1},\dots \mu _{s}$$ distinct eigenvalues given $$\mathrm{e}$$ we re-arrange these such way 1 {\mathrm{SD}}(S)\!:\mathrm{e}=\mathrm{e}_{1}+\mathrm{e}_{2}+\dots +\mathrm{e}_{r} certain $$r\le s,$$ where $$\mathrm{e}_{i}$$ are eigenvectors A eigenvalue _{i},$$ all $$1\le i\le r.$$ We refer (1) as spectral decomposition or more properly, its $$\mathrm{e}.$$ show determines vice versa. Explicit algorithms which establish this correspondence. In particular, r appear equal rank $$W^{S}.$$ Various results can derived theorem. has $$\ge n-1.$$ Another application n-1,$$ then another $$G^{*}$$ on V isomorphic only there $$S^{*}\subseteq V$$ so same $$W^{S^{*}}\!,$$ up reordering rows . The assumption two theorems met by almost graphs when $$S=V,$$ it known $${\mathrm{rank}}(W^{V})=n$$ graphs.