Suppose V is a real vector space Define a new space V2 such

Suppose V is a real vector space. Define a new space V^2 such that V^2:= (v_1, v_2): v_1, v_2 elementof V}, and for all (v_1, v_2), (w_1, w_2) elementof V^2 and lambda elementof R, addition and scalar multiplication in V^2 are defined by (v_1, v_2) (w_1, w_2):= (v_1, w_1, v_2 w_2) and lambda (v_1, v_2) (lambda v_1, lambda v_2). It can be shown (but you do not need to prove this that V^2 is a vector space. (i) If 0 elementof V denotes the zero vector in V, what is the zero vector in V^2? Justify your answer. (ii) Let B:= {z_1, z_2} be a basis for V. Show that S:= {(z_1, 0), (z_2, 0), (0, z_1), (0, z_2)} is a basis for V^2. (iii) What is the dimension of V^2?

Solution

(i). If 0 V denotes the 0 vector in V, then the zero vector in V² is (v₁,v₂) such that v₁,v₂ V and v₁,v₂ are both 0 vectors i.e. the zero vector in V² is (0,0).

(ii). If B = {z₁ ,z₂} is a basis for V , then every vector v V is a linear combination of z₁ and z₂. Let v = az₁+bz₂ and v₂ = cz₁+dz₂ be 2 arbitrary vectors in V, where a,b,c,d are arbitrary scalars.. Then (v₁,v₂) is an arbitrary element of V² . Further, (v₁,v₂) = (az₁+bz₂, cz₁+dz₂)= a(z₁,0)+b(z₂,0) +c(0,z₁) +d(0,z₂). This means that (v₁,v₂), which is an arbitrary element of V², is a linear combination of (z₁,0), (z₂,0), (0,z₁) and (0,z₂). Now, since B = {z₁ ,z₂} is a basis for V, hence z₁ ,z₂ are linearly independent so that (z₁,0), (z₂,0), (0,z₁) and (0,z₂) are also linearly independent. Hence, {(z₁,0), (z₂,0), (0,z₁), (0,z₂)} is a basis for V².

(iii). The dimension of V² being equal to the number of elements in its basis, is 4.