Will augmenting a linearly independent set with vectors from the standard basis result in a new basis

Given a linearly independent set S={v1, v2, ... , vn} in R^m where m>n, can a set D, which contains only vectors that are elements of the standard basis for R^m, be defined such that S∪D forms a basis for R^m?

This isn't an exam question or anything; it's just a question I've been wondering about for a little while and can't quite figure out. Thanks in advance!

Yes it can. If S = {v_1, v_2, ..., v_n} is linearly independent then it forms a basis for some n-dimensional subspace of R^m. Now, consider what would happen if no set D, containing only elements for the standard basis of R^m, could be defined such that S∪D forms a basis for R^m.

This would imply that no combination of (m - n) vectors from the standard basis can be defined to all be linearly independent from S = {v_1, v_2, ..., v_n}. The equivalent statement is that there exists a collection of at least n + 1 vectors from the standard basis each of them not linearly independent from S = {v_1, v_2, ..., v_n}. This is a contradiction because these n + 1 vectors from the standard basis would span the same space as S, but S can only span an n-dimensional subspace of R^m while the n+1 collection spans an n+1 dimensional subspace.

This means out initial assumption was incorrect and the set D has to exist. Hope that helped :)