Question

Let V = R^2 (all ordered pairs of real numbers) with addition defined in the usual way by (x1+y1) + (x2,y2) = (x1+x2, y1+y2) but with scalar multiplication defined by alpha (x,y) = (alpha x, y). Show that V is not a vector space by demonstrating how one of its axioms fail to hold.

Answer 1

alpha x hmmmm been a while is alpha, a constant?

Answer 2

alpha is a scalar

Answer 3

alright so you're going to have to go through each axiom, i know it's going to be one of the multiplication axioms i believe, let me find the list so i can go through them aso

Answer 4

on my exam, I thought A5 didn't hold which was alpha (x+y) = alpha x + alpha y for each scalar alpha alpha and any x and y in V

but it turns out...

its A6 that doesn't hold:
(alpha + beta) x = alpha x + beta x for any scalars alpha and beta and any x in V.

I just don't understand the reason.

Answer 5

so for multiplication you have

\[u=(\alpha x, y)\]?

Answer 6

yes

Answer 7

\[cu+ku=c(x,y)+k(x,y)=(cx,y)+(kx,y)=(cx+kx,2y)\]

Answer 8

is that A5 or A6?

Answer 9

so A5 is c(x,y) = c(x1,x2) + c(y1,y2) = (cx1,x2)+(cy1,y2) = c((x1,y1)+(x2,y2))