# Elem lin algebra (Vector space question)

1. Mar 1, 2012

### Miike012

Let V be the set of all ordered pairs of real numbers, and consider the following addition and scalar multiplication operations on u = (u1,u2) and v = (1,v2)

u + v = (u1+v1 ,u2+v2)

ku = ( 0 , ku2)

The book says that the axiom -u + u = 0 holds true for the given addition and scalar mult.

Which it obviously does not by the given scalar mult...
Hence: u + (-u) = (u1,u2) + (0,-u2) = (u1,0) ≠ 0.

Am I right or wrong?

2. Mar 1, 2012

### HallsofIvy

Staff Emeritus
What you have shown is that -v is NOT equal to (-1)v. "-v" is defined as "the additive inverse" of v and since the "addition" here is the usual "coordinate wise" addition, if v= <x, y> then -v= <-x, -y> while (-1)v= <0, -y>. Of course, since you can prove -v= (-1)v from the basic properties of a vector space, this says that this is not a vector space.

3. Mar 1, 2012

### Dick

-u usually means the additive inverse of (u1,u2). That would be (-u1,-u2). That's actually different from the 'scalar multiple' you've defined. (-1)*u=(0,-u2).