Unformatted text preview: be read in one direction only, speciﬁed by an arrow. For instance, consider the graph G = ( V, E ) where V = { 1 , 2 , 3 } and E = { (1 → 2) , (2 → 1) , (2 → 3) } . Then there exists a path from 1 to 3 but no path from 3 to 1. G is still a simple graph, as (1 → 2) ± = (2 → 1) . Its adjacency matrix is then M = 1 1 1 . In other words, we set M ij = 1 if ( i → j ) ∈ E, and M ij = 0 otherwise. Note that in general M is no longer symmetric. Let then G = ( V, E ) be an oriented graph with V = { 1 , . . . , n } . Consider the adjacency matrix M of G . Assume that M n ± = 0 where 0 is here the matrix with all its entries being null. Show that G has then a cycle. Hint: prove that (c) still holds for an oriented graph. (f) Examine the reciprocal. 1...
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 Spring '07
 Peche
 Math, Combinatorics, Graph Theory, Hilbert space, Graph theory objects, Eulerian path, Hamiltonian path

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