Let A be an m times n matrix and let v1 vk be a basis for Nu

Let A be an m times n matrix, and let {v_1, ....v_k} be a basis for Nul A. If C is an invertible n times n matrix, let w_j = C^-1v_j for j = 1, .., k. (a) Show that the vectors w_1, ... w_k are linearly independent (b) Show that w_1 ..., w_k span Nul AC. (c) Hence show that AC has the same rank as A.

Solution

(a) Let a₁w₁+ a₂ w₂+…+a w_k = 0.Then a₁C^-1 v₁ +a₂ C^-1 v₂ +…+a_k C^-1 v_k=0 or, C^-1(a₁v₁+a₂v₂+…+a_kv_k)= 0. Now, since C is an invertible matrix, hence, C^-1 0. Therefore, a₁v₁+a₂v₂+…+a_kv_k = 0. Since, v₁,v₂,…, v_k are linearly independent, hence a₁ = a₂ = …= a_k = 0. This implies that w₁,w₂,…, w_k are linearly independent.

(b) Let w span {w₁,w₂,…, w_k }. Then w = C^-1 v, where v span {v₁,v₂,…, v_k}. Now, ACw = AC(a₁w₁+ a₂w₂ +… +a_k w_k) where a₁ , a₂,…,a_k are arbitrary real numbers.Then ACw =AC (a₁C^-1v₁+a₂C^-1 v₂ +…+a_k C^-1v_k)= AC C ^-1(a₁v₁+a₂v₂+…+a_kv_k) = Av . However, Av = 0 as v span { v₁,v₂,…, v_k } = Nul(A). Therefore, ACw =0. This implies that w Nul(AC). Hence span { w₁,w₂,…, w_k } Nul(AC). Further, let w be an arbitrary vector in Nul(AC). Then (AC)w = 0 or, A(Cw) = 0. This means that Cw Nul (A) so that Cw = a₁v₁+a₂v₂+…+a_kv_k, where a₁ , a₂,…,a_k are arbitrary real numbers. Then, on multiplying both the sides to the left by C^-1 , we get C^-1 (Cw) = C^-1(a₁v₁+a₂v₂+…+a_kv_k) or, (C^-1C )w =a₁C^-1 v₁ +a₂ C^-1 v₂ +…+a_k C^-1 v_k, or, w = a₁C^-1 v₁ +a₂ C^-1v₂+…+a_k C^-1v_k i.e. w span { w₁,w₂,…, w_k }. Thus, Nul(AC) span { w₁,w₂,…, w_k }. Thus, Nul(AC) = span { w₁,w₂,…, w_k } i.e. w₁,w₂,…, w_k span Nul(AC).

(c) By virtue of part(b) above, nullity (AC) = k . Hence, as per the rank-nullity theorem, rank(AC) = n-k. Further nullity (A) is also k so that rank (A) = n-k. Hence rank(AC) = rank(A).