Another exercise on vector space

[trixitium]

Homework Statement

Determine if the following set is a vector space under the the given operations.

The set V of all pairs of real numbers of the form (1,x) with the operations:

(1, y) + (1, y') = (1, y + y')

k(1, y) = (1, ky)

Homework Equations

The Attempt at a Solution

Axiom 4: There is an object 0 in V, called a zero vector for V, such that 0 + u = u + 0 = u.

u = (1, y)

(1, y) + (1, 0) = (1, y+ 0) = (1, u) = u

May I consider (1,0) as the zero vector? I have a doubt if the zero vector 0 has to be (0,0) always or the zero can be defined as any vector (in this case (1,0) ) that remains the vector untouched under addition.

Axim 5: for each u in V, there is an object -u in V, called a negative of u, such that u + (-u) = (-u) + u = 0

u = (1,x)
-u = (-1)(1,x) = (1, -x) by definition of scalar multiplication given for the set V. (Is this correct?)

then u + (-u) = (1,x) + (-1)(1,x) = (1,x) + (1,-x) = (1,x+(-x)) = (1,0) = 0

in this case (1,0) is the zero vector.

 

[Simon Bridge]

Homework Statement

Determine if the following set is a vector space under the the given operations.

The set V of all pairs of real numbers of the form (1,x) with the operations:

(1, y) + (1, y') = (1, y + y')

k(1, y) = (1, ky)

Homework Equations

The Attempt at a Solution

Axiom 4: There is an object 0 in V, called a zero vector for V, such that 0 + u = u + 0 = u.

u = (1, y)

(1, y) + (1, 0) = (1, y+ 0) = (1, u) = u

May I consider (1,0) as the zero vector? I have a doubt if the zero vector 0 has to be (0,0) always or the zero can be defined as any vector (in this case (1,0) ) that remains the vector untouched under addition.

In this case (0,0) cannot be the zero vector, since (0,0) is not in V.

Axiom 5: for each u in V, there is an object -u in V, called a negative of u, such that u + (-u) = (-u) + u = 0

u = (1,x)
-u = (-1)(1,x) = (1, -x) by definition of scalar multiplication given for the set V. (Is this correct?)

yep - defn was k(1,y)=(1,ky) so put k=-1 ... but just like "0" does not need to be (0,0), so -u need not be literally "-1.u". It's a notation. However:

then u + (-u) = (1,x) + (-1)(1,x) = (1,x) + (1,-x) = (1,x+(-x)) = (1,0) = 0

in this case (1,0) is the zero vector.

... well done. You have discovered that the negative vector [-u] is the same as -1.u

You'd normally want to say:

if v=(1,y) is the negative of u=(1,x), then u+v=0

observe: u+v=0 iff y=-x, therefore: -u=(1,-x)=-1.(1,x)=-1.u