Question

Given two sets S and T, the direct product of S and T is the set of ordered pairs S × T = {(s, t)|s ∈ S, t ∈ T}.Let V, W be two vector spaces over F.

(a) Prove that V × W is a vector space over F under componentwise addition and scalar multiplication (i.e. if (v1, w1),(v2, w2) ∈ V × W, then (v1, w1) + (v2, w2) = (v1+w1, v2+w2) and a(v, w) = (av, aw) for any (v, w) ∈ V ×W, a ∈ F).

(b) If dim V = n and dim W = m, prove that dim V × W = n + m by constructing a basis.

Answer 1

(a) Let V and W be 2 vector spaces over a field F and let (v₁, w₁),(v₂, w₂) be 2 arbitrary elements of VxW and also, let a be an arbitrary scalar. Then (v₁, w₁)+(v₂, w₂) = (v₁+v₂, w₁+w₂). Further,since V and Ware vector spaces, hence v₁+v₂ ∈ V and w₁+w₂ ∈ W so that(v₁+v₂, w₁+w₂) ∈ VxW. Thus, VxW is closed under addition. Also, since V and Ware vector spaces, hence av₁∈ V and aw₁ ∈ W so that (av₁,aw₁)= a(v₁, w₁) ∈ VxW. Thus, VxW is closed under scalar multiplication. Further, the zero vector (0,0) is apparently in VxW. Therefore VxW is a vector space.

(b).Let V and W be 2 finite vector spaces of dimensions m and n respectively over a field F and let {v₁,v₂,…,v_m} and {w₁,w₂,…,w_m} be bases for V and W respectively. Then any vector in V is a linear combination of v₁,v₂,…,v_m and any vector in W is a linear combination of w₁,w₂,…,w_m. Thus, any vector in VxW is of the form (v,w) where v ∈ V, w ∈ W. Now, for each choice of v, there are n choices of w so that the dimension of VxW is mn.