A set S of vectors is given. Find a subset of S with the same span as S that is as small as possible. No idea how to do this. If you do your the best! Thanks Solution The smallest subset with the same span is { (1,-1,2) , (2,-3,0) }. To see this we take any representable vector in the span of the given set. a*(1,-1,2)+b*(2,-3,0)+c*(0,0,0). We may write the same arbitrary vector using linear combinations of { (1,-1,2) , (2,-3,0) }, ie. a*(1,-1,2)+b*(2,-3,0) = a*(1,-1,2)+b*(2,-3,0)+c*(0,0,0). Therefore the span is the same as any vector generated by the set is generated by the subset. We may show that the subset is the smallest such subset if it is linearly independant, ie. there is no constant c for which c*(1,-1,2) = (2,-3,0). If such a c were to exist then from the first component we have that c*1 = 2 therefore c=2, but in the third component we have c*2 = 0 there c=0, which shows no such c can satisfy both. Therefore the subset is as small as is possible whilst still having the same span as the given set. .
A set S of vectors is given. Find a subset of S with the same span as S that is as small as possible. No idea how to do this. If you do your the best! Thanks Solution The smallest subset with the same span is { (1,-1,2) , (2,-3,0) }. To see this we take any representable vector in the span of the given set. a*(1,-1,2)+b*(2,-3,0)+c*(0,0,0). We may write the same arbitrary vector using linear combinations of { (1,-1,2) , (2,-3,0) }, ie. a*(1,-1,2)+b*(2,-3,0) = a*(1,-1,2)+b*(2,-3,0)+c*(0,0,0). Therefore the span is the same as any vector generated by the set is generated by the subset. We may show that the subset is the smallest such subset if it is linearly independant, ie. there is no constant c for which c*(1,-1,2) = (2,-3,0). If such a c were to exist then from the first component we have that c*1 = 2 therefore c=2, but in the third component we have c*2 = 0 there c=0, which shows no such c can satisfy both. Therefore the subset is as small as is possible whilst still having the same span as the given set. .