I was reading this proof, and I do not know what does E stands for, could you help me please?
Theorem: Raising an adjacency matrix A of simple graph G to the n-th power gives the number of n-length walks between two vertices vi, vj of G in the resulting matrix.
Proof by induction: Let P(n) be the predicate that the theorem is true for n. We let F(n)ij be the number of n-length walks between vertex vi and vj. P(n) is then the predicate that F(n)ij=Anij. We proceed by induction on n.
Base case: P(1)
Case 1: F(1)ij=A(1)ij=1 if {vi,vj}∈E, so there is is a walk of length 1 between vi, vj.
Case 2: F(1)ij=A(1)ij=0 if {vi,vj}∉E, so there can't be any walk of length 1 between vi and vj.
In both cases F(1)ij=A(1)ij holds, hence P(1) is true.
Inductive step: P(n+1) For purpose of induction, we assume P(n) is true, that is F(n)ij=A(n)ij holds for n.
We can express a walk of length n+1 from vi to vj as a n-length walk from vi to vk and a walk of length 1 from vk to vj.
That means, the number of n+1-length walks from vi to vj is the sum over all walks from vi to vk times the number of ways to walk in one step from vk to vj. Thus:
F(n+1)ij=∑k=1|V|AkjF(n)ik=∑k=1|V|AkjA(n)ik
$E$ is simply the edge set of the graph $G$ here. Saying that $\{v_i,v_j\}\in E$ is saying that pair is an edge.