I know if G is bipartite, all cycles in the graph will have even length. Hence the characteristic vectors of all the cycles in G will all have an even number of non-zero entries, which means such a basis exists. How do I prove the converse? Ie, 'if there is a basis of the cycle space of G such that each vector has an even number of non-zero entries, then G is bipartite'?
Prove a graph G is bipartite iff its cycle space has a basis s.t. each vector in the basis has an even number of non-zero entries?
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$\begingroup$
combinatorics
graph-theory
vector-spaces
bipartite-graphs
1 Answers
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We can proceed:
Assume there is a basis of the cycle space of G such that each vector has an even number of non-zero entries.
Assume G is not bipartite, and thus G has an odd cycle, with characteristic vector v and an odd number of non-zero entries.
By definition of a basis, v is a linear combination of the basis vectors.
Find a contradiction by counting non-zero entries mod 2 in this linear combination.