Suppose that v1 vm is a linearly independent list in a vect

Suppose that v_1, ..., v_m is a linearly independent list in a vector space V and omega elementof V. Prove that dim span V_1 + omega, ..., v_m + omega greaterthanorequalto m = 1.

Solution

Let us assume that dim span v₁+w,v₂+w,…,v_m+w < m-1.This means that the set S = {v₁+w, v₂+w,…,v_m-1+w } is linearly dependent so that, there exist scalars a₁,a₂,…,a_m-1, not all zero, such that a₁(v₁+w) +a₂(v₂+w) +…+ a_m-1(v_m-1+w) = 0.Then a₁ v₁+ a₂ v₂+ …+a _m-1v_m-1 + (a₁+a₂+…+a_m-1 )w = 0….(1). Since a₁,a₂,…,a_m are not all zero, and since the set {v₁,v₂,…v_m} is linearly independent, hence the set{v₁,v₂,…v_m-1} is also linearly independent. Therefore, a₁v₁+ a₂ v₂+ …+a_m-1 v_m-1 0. Then a₁+a₂+…+a_m-1 0 (otherwise,(a₁+a₂+…+a_m-1 )w = 0w = 0 and then a₁v₁+ a₂ v₂+ …+a_m-1 v_m-1 = 0 which implies that the set { v₁,v₂,…v_m-1 } is linearly dependent, a contradiction). On dividing both the sides of equation number (1) by a₁+a₂+…+a_m-1, we get w = -[a₁/( a₁+a₂+…+a_m-1) v1+ a₂/( a₁+a₂+…+a_m-1) v₂ +…+ a_m/( a₁+a₂+…+a_m-1) v_m-1]. This means that w is in span { v₁,v₂,…v_m-1 }. However, this is not necessarily correct as w can be in span { v₁,v₂,…v_m }. Therefore, this is a contradiction and hence dim span v₁+w,v₂+w,…,v_m+w m-1.