Exercise:
Let v_...v_nin V be s.t. Spv_...v_nV and let u_...u_min V be linear indepent. Then mleq n.

Solution:
Proof. The proof will have at most n steps. In the first step we'll replace u_ with one of the vectors v_...v_n: bf step : Consider the list of n+ vectors u_v_...v_n. By Lemma C this list is linear depent. By Lemma A one of the vectors in this list is in the span of the vectors appearing before it in the list and also if we drop that vector from the list the span doesn't change. By Lemma B that unnecessary vector that can be dropped is not u_ because u_ is lineary indepent since u_...u_m are. We drop that vector from the list u_v_...v_n. We now have a new list of n vectors in which the sublist consisting of the first vector is linearly indepent. And this new list of n vectors still spans V. bf step leq j: From the previous step number j- we have a list of length n that spans V and s.t. the first j- vectors are u_...u_j- and the other n-j- vectors are taken from v_...v_n. This list can be written as u_...u_j-w_...w_n-j-. We consider now the list u_...u_j-u_jw_...w_n-j- which has n+ vectors. By Lemma C this list is linearly depent. By Lemma A again we can drop one of the vectors in the list without changing the span and s.t. this vector is in the span of the vectors standing vefore it in the list. By Lemma B that vector that can be dropped cannot be one of u_...u_j. It must be one of the vectors w_...w_n-j- because u_...u_j are linearly indepent. Now ase by contradiction that m n. Then we can perform the stepts j...n and after step jn we obtain a list that looks like u_...u_n and Spu_...u_nV in other words if m n then after step #n all the vectors from the original list v_...v_n are replaced by u_...u_n. But since m n we still can use the vectors u_n+...u_m as follows.Add u_n+ to the list: we get u_...u_nu_n+. Since Spu_...u_nV by Lemma C we deduce that u_n+u_...u_n is linearly depent contradiction lightning This proves that mleq n as claimed by Lemma D.