could you guys plese solve question 35 and 36 Suppose that S

could you guys plese solve question 35 and 36!

Suppose that S = {v_1 vector, v_2 vector, v_3 vector, v_4 vector} subsetoforequalto R^m is linearly independent. a. Prove that S^/= {v_1 vector, v_3 vector, v_4 vector} is also linearly independent. b. Prove that S^//= {v_2 vector, v_4 vector} is also linearly independent. We will generalize the previous Exercise: Prove that if S = {v_1 vector, v_2 vector, ..., v_n vector} subsetoforequalto R^m is linearly independent, then any subset of S is still linearly independent. What is the contrapositive of this statement?

Solution

35. Let the set {v₁,v₂,v₃,v₄} be linearly independent.

(a). Let us assume that the set {v₁,v₃,v₄} is linearly dependent. Then there exist scalars a₁,a₃,a₄, not all 0 such that a₁v₁+a₃ v₃+a₄ v₄ = 0. Then, a₁v₁+ 0v₂+ a₃ v₃+a₄ v₄ = 0+0 = 0. This means that the set {v₁,v₂,v₃,v₄} is linearly dependent, which is a contradiction. Hence the set {v₁,v₃,v₄} is linearly independent.

(b) Let us assume that the set {v₂,v₄} is linearly dependent. Then there exist scalars a₂,a₄, not both 0 such that a₂v₂+a₄v₄ = 0. Then, 0v₁+ a₂v₂+0v₃+a₄ v₄ = 0+0+0 = 0. This means that the set {v₁,v₂,v₃,v₄} is linearly dependent, which is a contradiction. Hence the set {v₂,v₄} is linearly independent.

36. Let the set S = {v₁,v₂,…v_n} be linearly independent. Let us also assume that a subset of S ,say, S’= { v_i ,v_j ,…,v_k} 1 i,j,k n is linearly dependent. Then there exist scalars a_i,a_j,…,a_k, not all 0 such that a_iv_i+a_jv_j+ …+a_k v_k = 0. Then, 0v₁+ 0v₂+…+ a_iv_i+0v _i+1+ …+a_j v_j +0v_j+1+…+a_k v_k + 0v_k+1+…+0v_n= 0. This means that the set {v₁,v₂,…v_n} is linearly dependent, which is a contradiction. Hence the set { v_i ,v_j ,…,v_k} 1 i,j,k n is linearly independent.