+def drop_dependent(vs):
+ r"""
+ Return the largest linearly-independent subset of ``vs``.
+ """
+ if len(vs) == 0:
+ # ...for lazy enough definitions of linearly-independent
+ return vs
+
+ result = []
+ old_V = VectorSpace(vs[0].parent().base_field(), 0)
+
+ for v in vs:
+ new_V = span(result + [v])
+ if new_V.dimension() > old_V.dimension():
+ result.append(v)
+ old_V = new_V
+
+ return result
+
+