Suppose thatcv = 0 and c (not =) 0. We must show that v = 0. Now there exists an element c^-1
ofK satisfying c^-1 c = 1, since any non-zero element of a field has a multiplicative inverse. It then follows from the vector space axioms and (Let V be a vector space over a field K. Then c0 = 0 and 0v = 0 for all elements c of K and elements v of V . )that
v = 1v = (c^-1c)v = c^-1(cv) = c^-10 = 0, as required.