How to prove this problem

1.2.7 ([Ber14, Ex. 1.3.4)). Suppose V is a real vector space. Let W = V x V be the (real) product vector space and define multiplication by complex scalars by the formula (a + bi) (u, v) = (au - bu, bu +av), a, bER, (u, v) EW. Show that W satisfies the axioms for a complex vector space.Denition 1.2.1 (Vector space). A vector space over F is o a set V (whose objects are called vectors), 0 a binary operation + on V called vector addition, and a scalar multiplication: for each c E F and v E V, an element cv E V, called the scalar product of c and 1:), such that the following axioms are satised: (V1) (V2) (V3) (V4) (V5) (V6) (V7) (V8) For all u, v E V, we have u | v = v | u. [commutativity of vector addition) For all u, v, w E V, we have (u + v) + w : u | (v + 10) (associativity of vector addition) There is an element 0 E V such that, for all '0 E V, v + 0 : 0 + v = v. The element 0 is unique and is called the zero vector. For any v E V, there exists an element ~v E V such that v + (av) = 0. The element #v is uniquely determined by v and is called the additive inverse of v. For all a E F and u, v ES V, we have a(u + v) 2 an + av. (distributivity of scalar multiplication over vector addition) For all a, l) E F and v E V, we have (a | tr)v : av + bv. (distributivity of scalar multiplication over eld addition) For all a, l) E F and v E V, we have a(bv) : (ab)v. (compatibility of scalar multiplication with eld multiplication) For all v E V, we have 1v : v, where 1 denotes the multiplicative identity in F. (unity lavJ)