# Let V be a vector space over `CC`. Let W = V x V = {(v1, v2) | v1, v2 `in` V}. Determine whether or not W is a vector space over ` ``CC` with addition defined by (u1, u2) + (v1, v2) = (u1+v1,...

Let V be a vector space over `CC`. Let W = V x V = {(v1, v2) | v1, v2 `in` V}. Determine whether or not W is a vector space over ` ``CC` with addition defined by (u1, u2) + (v1, v2) = (u1+v1, u2+v2) `AA` (u1, u2), (v1, v2) `in` W and multiplication defined by (a + bi).(v1, v2) = (a.v1 – b.v2, b.v1+a.v2) for all a+bi `in` `CC` and (v1, v2) `in` W. Here `i^2 = -1` and a, b `in` `RR` . If so, prove that all the conditions in the definition of a vector space are satisfied. If not, show by counter-example that at least one condition does not hold.

### 1 Answer | Add Yours

The only condition that isn't obvious is the associativity of multiplication, i.e. whether `(z_1*z_2)*w =z_1*(z_2*w)` for any complex numbers `z_1, z_2` and any vector `w in W.`

Let's multiply numbers first and then the resulting number by a vector:

`((a_1+b_1i)*(a_2+b_2i))*(v_1,v_2) = ((a_1a_2-b_1b_2)+(a_1b_2+a_2b_1)i)*(v_1,v_2) = `

`= ((a_1a_2-b_1b_2)v_1-(a_1b_2+a_2b_1)v_2, (a_1b_2+a_2b_1)v_1+(a_1a_2-b_1b_2)v_2).`

Now multiply vector by the second number and then the resulting vector by the first number:

`(a_1+b_1i)*` `((a_2+b_2 i)` `*(v_1, v_2)) =`

`=(a_1+b_1i)*(a_2v_1-b_2v_2,b_2v_1+a_2v_2) =`

`= (a_1(a_2v_1-b_2v_2)-b_1(b_2v_1+a_2v_2), b_1(a_2v_1-b_2v_2)+a_1(b_2v_1+a_2v_2)) =`

`= ((a_1a_2-b_1b_2)v_1-(a_1b_2+b_1a_2)v_2, (b_1a_2+a_1b_2)v_1+(a_1a_2-b_1b_2)v_2).`

As we see, the results are the same. So **yes**, W is a complex vector space.