Just to reword TPH's idea...
So the basis of vector space V is a subset that can represent any vector in V as a linear combination of vectors from B. In order for a set to be linear independent, no linear combination of vectors in that set can produce 0 (other than the trivial case where all the coefficients are 0).
Thus since C contains ~v, not in B, but in V, we can represent ~v as a linear combination of vectors in the basis, which means we could represent -(~v), and the sum of the linear combinations could now be zero non-trivially.
How bad did I mess up TPH?