Question

Let V = R2 with the following operations: (zı, yı) + (2 2,32) = (x1 +T2-1, yı +B2) (addition) c(x1, y) = (czi-e+ 1, cy) where c E R (scalar multiplication). Then V is a vector space with these operations (you can take this as given). (a) (2) Let (-2,4) and (2,3) belong to V and let c -2 R. Find ca + y using the operations defined on V. (b) (2) What is the zero vector in V? Justify. (c) (2) If 코= (-3,2), what is the additive inverse ofz? Justify (d) (2) For an arbitrary vector = (zi,yi), what is the additive inverse of z? Justify. (e) (3) Prove property 7. for vector spaces: If c E R and (,v), (x2,y2) E V, then c(z+め= cz +cji (Hint. work out both sides of the equation separately to show they are equal.)

Answer 1

V = R^2 (x_1, y_1) + (x_2, y_2) = (x_1 + x_2 - 1, y_1 + y_2) & (x_1, y_1) = ((x_1 - c + 1, cy_1), c elementof R (a) x = (-2, 4) & y = (2, 3) c = -2 c x = -2 (-2, 4) = (-2 (-2) - (-2) + 1, -2 (4)) = (4 + 2 + 1, -8) = (7, -8) cx + y = (7, -8) + (2, 3) = (7 + 2 - 1, -8 + 3) c x + = (8, -5) (b) (x_1, y_1) + (a, b) = (x_1, y_1) Implies (x_1 + a - 1, y_1 + b) = (x_1, y_1) Implies x_1 + a - 1 = x_1 & y_1 + b = y_1 Implies a - 1 = 0 & b = 0 Implies a = 1 & b = 0 Implies 0 = (1, 0) is zero vector in V since (x_1, y_1 + (1, 0) = (x_1 + 1 - 1, y_1 + 0) = (x_1, y_1) (c) x = (-3, 2) , y = (y_1, y_2) be additive inverse of x then x + y = (1, 0) = 0 Implies (-3, 2) + (y_1, y_2) = (1, 0) (-3 + y_1 - 1, 2