Question

If G is a simple graph, the vector space of g is the vector space over the field Z2 of integers modulo 2, whose elements are subsets of E (G). The sum is E+F of two subsets E and F is the set odd edges in E or F but not both, and scalar multiplication is defined by1×E=E and 0xE=∅. Show that this defines a vector space over Z2 and find the basis of it.

Answer 1

well, we see that addition is commutative, and that both addition and scalar multiplication are clearly closed. show that scalar multiplication over addition distributes, i.e. k(E + F) = kE + kF. well, this is easy -- 0 = 0(E + F) = 0E + 0F = ∅. similarly, E + F = 1(E + F) = 1E + 1F = E + F. we get from this a free additive identity ∅ and a free scalar multiplicative identity 1. since 0 + 0 = 0, 0 + 1 = 1, and 1 + 1 = 0 are an exhaustive list of addition in Z/2Z, distribution of scalar sums over scalar multiplication is easy to demonstrate. show associativity, which is probably the hardest (and yet still overwhelmingly easy) to do

Answer 2

since addition is defined as a symmetric difference (i.e. xor, https://en.wikipedia.org/wiki/Symmetric_difference) we see that the sets of singletons consisting of individual edges (i.e. {E_1}, {E_2}, {E_3}, etc.) are a very nice basis to use here, since we can construct every other set of edges possible using these and symmetric difference

Answer 3

please can you elaborate it further ?

Answer 4

that's the last question i need to answer to finish my problem set in graph theory

Answer 5

it has something to do with linear algebra right ?