# Proving a Vector Space

• Oct 29th 2007, 06:25 PM
Proving a Vector Space
I just can't completely answer this question, does it want me to prove it using all 10 axioms? Or is there another way

http://i89.photobucket.com/albums/k2...atrixHelp2.jpg

Thanks
• Oct 29th 2007, 08:52 PM
ThePerfectHacker
Here is an easier way to do this without using all 10 properties of a vector space.

Theorem: Let $V$ be a vector space over a field (you probably learned it over $\mathbb{R}$). Let $W$ be a subset of $V$ so that $\bold{a}+\bold{b} \in W$ for all $\bold{a},\bold{b} \in W$ and $c\bold{a} \in W$ for all $\bold{a}\in W$ and $c\in \mathbb{R}$. Then $W$ is a vector space over the field ( $\mathbb{R}$).

So you only need to check closure of vector addition and scalar multiplication.
• Oct 29th 2007, 09:00 PM
I'm not sure if I'm correct but, the closure under addition wouldn't be true because the final matrix has to end up as

[a+c 2]
[ 2 b+d]
???

I'm almost positive axiom 6 for scalar multiplication holds, but I'm not sure about axiom 1, if you could tell me whether my above theory is correct or not that would be great. I am not very good at proving axioms at least 1 and 6, because there is no left and right hand side of the equation to compare, so I get really lost.
• Oct 30th 2007, 01:32 AM
kalagota
Quote:

I'm not sure if I'm correct but, the closure under addition wouldn't be true because the final matrix has to end up as

[a+c 2]
[ 2 b+d]
???

I'm almost positive axiom 6 for scalar multiplication holds, but I'm not sure about axiom 1, if you could tell me whether my above theory is correct or not that would be great. I am not very good at proving axioms at least 1 and 6, because there is no left and right hand side of the equation to compare, so I get really lost.

no, that is not the way the definition of addition gives you.. the definition says that you will just add the entries (1,1) and (2,2), while entries (1,2) and (2,1) are fixed to 1.
also, the same thing with closure of multiplication..