# Thread: Proving a vector space

1. ## Proving a vector space

I've attached an image of my problem set. I'm confused about what they mean to define the operation of addition or scalar multiplication, do they mean which of the eight axioms? Thank you.

2. Originally Posted by pakman
I've attached an image of my problem set. I'm confused about what they mean to define the operation of addition or scalar multiplication, do they mean which of the eight axioms? Thank you.
You need to show that all of the vectorspace axioms hold.

Identity There exists a Vector $\vec{0}$ such that

$\vec{0}+\vec u =\vec u$ for all vectors u

in this vector space 1 is the additive identity.

let u be any vector in the vector space then

$1+u=1 \cdot u=u \iff 1+u=u$

So this axiom holds.

Try to verify all of the other Axioms.

3. Originally Posted by TheEmptySet
You need to show that all of the vectorspace axioms hold.

Identity There exists a Vector $\vec{0}$ such that

$\vec{0}+\vec u =\vec u$ for all vectors u

in this vector space 1 is the additive identity.

let u be any vector in the vector space then

$1+u=1 \cdot u=u \iff 1+u=u$

So this axiom holds.

Try to verify all of the other Axioms.
So for the first thing about alpha*x = x^alpha, what operation of scalar multiplication is that? I don't see that in any of the axioms?

4. Originally Posted by pakman
So for the first thing about alpha*x = x^alpha, what operation of scalar multiplication is that? I don't see that in any of the axioms?
let k be a scalar

Distibutive Axiom:

$k(\vec u + \vec v)=k\vec u + k\vec v$

now useing the def on operations we need to show that this holds so

$k(\vec u + \vec v)=\underbrace{k(u\cdot v)}_{\mbox{Vector add}}= \underbrace{(uv)^k}_{\mbox{ Scalar mult}}=u^kv^k=k\vec u + k\vec v$

The the distributive law of scalar mult holds.