1. ## Sub-space test

Hi can some one please verify that my work is correct for this, its really important that I get this right.

So we call the set W then

since its not closed under addition so its not a subspace, and I did not bother to check the other axioms but I believe it should be closed under scalar multiplication. Also of course its not empty since (0,0) is in the set

so it basically failed the closure under addition axiom thus making it not a subspace.

Is this solution correct and sufficient

2. ## Re: Sub-space test

you should give an example of two vectors (x,y) that are both in W with (x,y) not in W.

my suggestion is: (x,-x) and (x,x) with x non-zero. but if you indeed show it's not closed under addition, yes, that is sufficient.

3. ## Re: Sub-space test

I did give examples of two vectors (x,y) that are both in W, I think the picture was too small so you might have missed it, i just figured out right now how to make them bigger.
"(x,-x) and (x,x) with x non-zero" thats exactly what I did so yeah I am assuming you missed the picture
so is it good now?

4. ## Re: Sub-space test

sure, that's all you need.