Question

3. Suppose S is a linearly independent generating set for a vector space V . Show that S is an efficient generating set, i.e., any proper subset of S is not a generating set.

Answer 1

10 omwo- then P.. one extra but least element in s (3) beivens § is a lineavily Independent gene realing set fore a vector space Vio To show: S is efficient set že s is the minimal - generating set of v o Proof: Since s is a lineverly Independent generating set for a vector space v t = S is a basis fore V . xloo let Pes such that P generates V Kloco since PES = 151> IPI -iş must exist at least one extora ele xloco since P generates 5) Every member of v lice ter ve V) can be wrilten as a linear combination of vectors of P But as s is also a generating set fore V for VEV o can be written as linear combination of vectors of But SzP = Some vectors in s are linearly dependent (as P generates u Cassumption) => S is a lineavily dependent set. +6 which is a contradiction as s is linearly Indebendent (Geiven) Hence. out assumption is wrong that pes generates v Hence S is an minimal efficient se generating est as. Pęs is arbitary Hence Proved