Let V be a vector space and suppose that U and W are subspac

Let V be a vector space and suppose that U and W are subspaces of V such that V = U W. Suppose that u_1, ..., u_m is a basis of U and that w_1, ..., w_n is a basis of W. Prove that u_1, ..., u_m, w_1, ..., w_n is a basis of V.

Solution

Let X+Y U W , where X and Y be arbitrary vectors in U and W respectively. Then X = a₁u₁+a₂u₂ +…+a_mu_m and Y = b₁w₁+b₂w₂ +…+b_nw_n , where a,a,…,a, b,b,…, b are arbitrary scalars. Further, X+Y = a₁u₁+a₂u₂ +…+a_mu_m + b₁w₁+b₂w₂ +…+b_nw_n so that X+Y span{ u₁,u₂,…,u_m ,w₁, w₂,…w_n}.Hence U W span{ u₁,u₂,…,u_m ,w₁, w₂,…w_n. Also, an arbitrary vector in span{ u₁,u₂,…,u_m ,w₁, w₂,…w_n} is of the form a₁u₁+a₂u₂ +…+a_mu_m + b₁w₁+b₂w₂ +…+b_nw_n = X +Y, where X = a₁u₁+a₂u₂ +…+a_mu_m and Y = b₁w₁+b₂w₂ +…+b_nw_n. Thus, span{ u₁,u₂,…,u_m ,w₁, w₂,…w_n} U W =V. Hence V= U W= span{ u₁,u₂,…,u_m ,w₁, w₂,…w_n}. Further, since u₁,u₂,…,u_m form a basis for U, these vectors are linearly independent. Similarly, w,w,..w are limnearly independent. Now, since, U W = {0}, hence the vectors u₁,u₂,…,u_m ,w₁, w₂,…w_n are linearly independent. Therefore,{ u₁,u₂,…,u_m ,w₁, w₂,…w_n}is a basis for U W.